Artin's axioms

Source proofs by the Stacks Project authors, as distributed in the AI Integrated Stacks Project. Source copyright: Copyright (C) 2005 -- 2025 Johan de Jong. This modified course edition is published by the Open Math Courses project, KokunoYumeto/open-math-courses. Writing, adaptation and integration: GPT-6.1 Sol (OpenAI), Codex, Ultra, October 2026. Permission is granted to copy, distribute and modify this modified chapter under the GNU Free Documentation License, Version 1.2 or any later version, with no Invariant Sections, Front-Cover Texts or Back-Cover Texts. Eligible independently written additions retain their CC0 1.0 dedication. Self-checked by the writing AI. The complete licence accompanies this edition.

An algebraic stack has smooth coordinates, whereas a moduli problem usually arrives as a rule assigning a groupoid to each scheme. Artin's criterion constructs coordinates from that rule. The construction has three stages: obtain a versal deformation over a complete local ring, approximate it by a family of finite type, and enlarge its versal locus to an open set. A second argument explains why a flat presentation, even one with inseparable fibres, can be replaced by a smooth presentation.

Read Algebraic stacks first. We use the quotient and bootstrap theorems of The bootstrap theorem, Schlessinger's existence theorem for versal formal objects, the deformation theory of square-zero extensions and the naive cotangent complex, and the commutative algebra of completion and henselization. Precise imported statements appear at the end. All fibre products of groupoids below are 2-fibre products.

1. Infinitesimal objects, arrows and patching

Except in Section 6, let \(S\) be locally Noetherian and let \(\mathcal X\) be a category fibred in groupoids on schemes over \(S\). The subscript “fppf” on this category specifies its underlying site; it does not by itself assert descent.

A finite type field over \(S\) means a field \(k\) with a morphism \(\operatorname{Spec}k\to S\) of finite type. This is more restrictive than being a finitely generated extension of the residue field at an arbitrary point of \(S\). A point of a scheme is a finite type point if its inclusion as the spectrum of its residue field is of finite type. Such a point is closed in some open neighbourhood.

Fix \(x_0\in\mathcal X(k)\). Its predeformation groupoid \(\mathcal F_{k,x_0}(A)\), for an Artinian local \(S\)-algebra \(A\) with specified residue field \(k\), consists of pairs

An arrow respects \(\iota\). Pullback along a ring map gives a covariant functor on Artinian rings. The fibre at \(k\) is equivalent to the terminal groupoid: the marking removes the automorphisms of the chosen special object. Automorphisms of its deformations can remain.

1.1. The Rim–Schlessinger condition

For a diagram of Artinian local \(S\)-algebras

whose spectra are of finite type over \(S\), condition (RS) requires the restriction functor

to be an equivalence. The right side remembers the identification on \(A\). Applying (1.2) to marked objects proves that \(\mathcal F_{k,x_0}\) is a deformation category whenever \(\mathcal X\) has (RS).

Condition (RS\(^*\)) requires the same equivalence for arbitrary \(S\)-algebras, with \(A_2\to A\) surjective and its kernel square zero. Filtering a nilpotent ideal into square-zero quotients shows that it implies (RS). The quantifier “arbitrary” is essential in Section 5, where products of modules need not be finite. These definitions are [Stacks, Tags 06L9 and 0CXN].

Lemma 1.1 (the groupoid operations preserve patching). If \(\mathcal X,\mathcal Y,\mathcal Z\) satisfy (RS), then \(\mathcal X\times_{\mathcal Y}\mathcal Z\) does too. The assertion also holds for (RS\(^*\)).

Proof. A patched object consists of an \(\mathcal X\)-object and a \(\mathcal Z\)-object on each side, and an arrow between their images in \(\mathcal Y\). Patch the two objects by essential surjectivity for \(\mathcal X\) and \(\mathcal Z\). Full faithfulness for \(\mathcal Y\) supplies the unique arrow between their images which restricts to the specified arrows. For morphisms, patch the two component arrows by full faithfulness and test their compatibility after restriction; faithfulness for \(\mathcal Y\) proves the compatibility upstairs. This proves both essential surjectivity and full faithfulness. \(\square\)

Lemma 1.2 (algebraic stacks have (RS)). Every algebraic stack satisfies (RS).

Proof. First, a diagram (1.1) is a pushout of affine schemes, and maps from its pushout to an algebraic space patch uniquely. One can check the latter assertion on étale scheme coordinates: after a finite étale cover of the Artinian local pushout, lift the common residue-field point to an étale chart; formal étaleness lifts that choice through the nilpotent thickenings. The two maps then land in a common affine chart, where the assertion is exactly the fibre product identity for rings. Unique lifts make the resulting maps compatible on overlaps, and the algebraic-space sheaf condition descends them. This also proves uniqueness.

For objects of an algebraic stack, use a smooth scheme atlas. After a finite étale cover of \(\operatorname{Spec}P\), the object over \(A_1\) lifts to the atlas: choose a finite separable residue-field point of its smooth atlas fibre and lift through the Artinian thickening. Its restriction to \(A\), together with the given identification, lifts across \(A_2\to A\) by smoothness. The two atlas maps agree on \(A\), and their closed points lie in a common affine atlas chart. The affine pushout patches them to a map from \(\operatorname{Spec}P\).

This constructs the desired object étale locally. The Isom spaces of two such objects are algebraic spaces, so the first paragraph proves that (1.2) is fully faithful. Consequently the local objects have unique compatible descent identifications, including the cocycle equality, and descend to the required object. \(\square\)

The stronger assertion for arbitrary affine pushouts along a thickening is [Stacks, Tag 07WM]. Its general flat-space patching input is not needed for the proof just given.

1.2. Two linear invariants

Write \(k[\epsilon]=k[\epsilon]/(\epsilon^2)\). For a deformation category, define

Patching the split extensions \(k\oplus M\) makes these \(k\)-vector spaces. Addition uses the map \(k[M\oplus N]\to k[M]\) induced by addition of vectors, and scalar multiplication uses multiplication on the square-zero ideal. For infinitesimal automorphisms, addition agrees with composition; terms involving two elements of that ideal vanish.

More generally, under (RS\(^*\)) there are \(A\)-linear functors \(T_x(M)\) and \(\operatorname{Inf}_x(M)\) for \(x\in\mathcal X(A)\). The first is the set of marked lifts to \(A[M]=A\oplus M\), and the second is the automorphism group of the trivial marked lift. If \(A'\twoheadrightarrow A\) has square-zero kernel \(I\), then, when lifts exist, their isomorphism classes form a torsor under \(T_x(I)\). The automorphisms of every marked lift are canonically \(\operatorname{Inf}_x(I)\).

To see the torsor assertion, use

Applying (RS\(^*\)) produces a difference between two lifts. On three lifts, the identity \((a_3-a_1)=(a_2-a_1)+(a_3-a_2)\) proves the addition law. Fixing one lift converts differences into the claimed simply transitive action. All constructions commute with maps of square-zero extensions. This is the mechanism behind [Stacks, Tag 07Y6].

Lemma 1.3 (field change and fibre products). Under (RS), a finite extension \(l/k\) gives

For \(w=(x,z,\alpha)\) in \(\mathcal W=\mathcal X\times_{\mathcal Y}\mathcal Z\) there is an exact sequence

Proof. For an Artinian \(l\)-algebra \(B\), apply (RS) to \(B\times_l k\). It identifies marked \(k\)-deformations over that ring with marked \(l\)-deformations over \(B\). The field-change result for linear deformation functors then gives (1.5), including inseparable finite extensions.

In (1.6), the second map is the difference between the two induced infinitesimal automorphisms in \(\mathcal Y\). The boundary twists the trivial pair of deformations by an infinitesimal automorphism of its identifying arrow. Its kernel consists precisely of twists obtainable from component automorphisms. A pair of component deformations comes from \(\mathcal W\) exactly when their \(\mathcal Y\)-images are isomorphic as marked deformations; choices of such an isomorphism differ by the boundary. These observations verify exactness at each term. The constructions by split extensions make all the maps linear. The same proof gives (1.6) with \(T_x(M)\) and \(\operatorname{Inf}_x(M)\) under (RS\(^*\)). \(\square\)

References: [Stacks, Tags 07WW, 07WY and 07YT]. For a naive obstruction complex \(E_x\) as in Section 5.3, the analogous field-change maps on \(H^i(E_x\otimes^{\mathbf L}k)\) are isomorphisms for \(i=0,1\): they are dual to the identifications of \(T\) and \(\operatorname{Inf}\). The indices here are \(0,1\), not \(-1,0\).

Lemma 1.4 (finiteness for an algebraic stack). If an algebraic stack has a smooth atlas locally of finite type over \(S\), the two spaces (1.3) are finite dimensional.

Proof. After a finite separable field extension, lift the object to the atlas. Its tangent space is a quotient of the atlas tangent space, since smoothness lifts deformations. On an affine atlas chart \(\operatorname{Spec}B\) over \(\operatorname{Spec}\Lambda\), that space is \(\operatorname{Hom}_B(\Omega_{B/\Lambda},k)\), which is finite dimensional because \(B\) is of finite type. The infinitesimal automorphisms inject into the tangent space at the identity point of the relation space; an étale scheme chart of that space is again locally of finite type. Apply the same argument and descend finiteness by (1.5). \(\square\)

This proof uses the finite module of differentials, so it also covers inseparable residue-field extensions. The reference is [Stacks, Tag 07X1].

2. Limits, complete local rings and approximation

2.1. Limit preservation includes arrows

The category \(\mathcal X\) is limit preserving if, for every directed inverse system of affine \(S\)-schemes with limit \(T=\lim T_i\),

is an equivalence. An object must descend to some stage. An arrow between descended objects must descend after increasing that stage, and two arrows equal on \(T\) must become equal at a later stage. Requiring only descent of objects is weaker. For a morphism \(\mathcal X\to\mathcal Y\), “limit preserving on objects” refers to descent of a lift of an already descended \(\mathcal Y\)-object, including its specified identifying arrow. See [Stacks, Tags 07XK and 06CT].

Finite presentation of algebras, modules, étale charts and their finite diagrams gives the following useful test. An algebraic space \(Z\to B\) is locally of finite presentation exactly when its functor on affine \(B\)-schemes preserves directed limits. One checks this on étale affine charts; finite presentation descends the chart and its equality data, and the étale sheaf condition glues the descended morphisms. The converse on an affine chart is the ring criterion for finite presentation, followed by étale descent.

Lemma 2.1 (testing on finite presentation bases). Suppose \(\mathcal Y\) is limit preserving on objects and both categories are Zariski stacks. To establish representability of \(f:\mathcal X\to\mathcal Y\) by algebraic spaces, it suffices to establish it over affine tests \(V\) locally of finite presentation over \(S\), with a uniform allowable size bound. Once representability is known, any property stable under base change and fppf local on the target can be tested on those same \(V\).

Proof. An affine \(V\) mapping into \(\operatorname{Spec}\Lambda\subset S\) is a limit of affine schemes of finite presentation over \(\Lambda\). The object testing \(f\) descends to a stage \(V_i\); its entire fibre category is therefore represented by the base change of the representing space over \(V_i\). For general \(V\), do this on affine opens. On overlaps the representing spaces have canonical compatible isomorphisms, since they represent the same category. Zariski descent glues them. The size bound permits the coproduct of their charts in the chosen site.

For the property assertion, the same calculation identifies the general fibre morphism with a base change of a finite presentation test. Base change and locality prove the claim. \(\square\)

References: [Stacks, Tags 07WI and 06CT]. Fibre products preserve limit preservation: descend both component objects, then their identifying arrow, and finally the finitely many equalities required for a morphism.

2.2. Formal effectiveness is a groupoid assertion

A formal object over a Noetherian complete local \(S\)-algebra \(R\), with residue field of finite type over \(S\), is a compatible system \(\xi_n\in\mathcal X(R/\mathfrak m^n)\), including its transition isomorphisms. It is effective if it is the completion of an object of \(\mathcal X(R)\). The stronger assertion used in axiom [4] is the equivalence

The limit on the right is a groupoid of compatible systems. The equivalence must include all compatible arrows, not only the existence of algebraizations. These equivalences also respect the local base morphisms in the category of formal objects. See [Stacks, Tag 07X3].

Lemma 2.2 (formal objects of algebraic stacks). Every algebraic stack satisfies (2.2).

Proof. A compatible system of maps to a scheme has its closed point in one affine open. Ring maps into the complete ring \(R=\lim R/\mathfrak m^n\) then give the unique algebraized map. For an algebraic space, choose an étale chart at the formal closed point. After a finite étale extension \(R'/R\), formal étaleness gives compatible lifts to that chart. The scheme assertion algebraizes them. Their identifying arrows algebraize by the same argument on the chart relation, and faithfulness verifies the cocycle. Étale descent gives the unique map over \(R\).

For a stack, two objects have an algebraic-space Isom sheaf. The preceding paragraph proves full faithfulness in (2.2). To algebraize a formal object, choose a smooth atlas and a finite separable residue-field point of its first fibre. The corresponding finite étale extension \(R'\) of the complete local ring lifts this point. Smoothness successively lifts it at every order. The resulting maps to the scheme atlas algebraize. Over \(R'\otimes_RR'\), a finite product of complete local rings, full faithfulness algebraizes the formal descent arrows. It also proves their cocycle on the triple product and identifies the descended object's completion with the specified formal object. Effective étale descent finishes the proof. \(\square\)

Equivalences commute with 2-fibre products. Thus (2.2), just like (RS), passes to a fibre product of three categories satisfying it.

2.3. The approximation input and its use

A Noetherian ring is a G-ring if every localization has geometrically regular formal fibres. Appendix B proves the following approximation theorem, including the desingularization and G-ring permanence inputs, with both residue-field cases retained.

Artin approximation. A regular map of Noetherian rings is a filtered colimit of smooth algebras (Popescu's theorem). If \(B\) is a henselian Noetherian local G-ring, a finite polynomial system over \(B\) with a solution in \(\widehat B\) has, for every \(N\), a solution in \(B\) congruent to it modulo \(\mathfrak m^N\). For a local G-ring which is not henselian, the solution lies in an étale neighbourhood inducing the same residue field. These are [Stacks, Tags 07GC, 07QY and 07QZ].

The object version needed here also preserves associated graded rings.

Lemma 2.3 (approximating a family). Let \(\mathcal X\) be limit preserving on objects. Let \(x_R\in\mathcal X(R)\), where \(R\) is as in (2.2), and let \(s\) be the image of its closed point in \(S\). If \(\mathcal O_{S,s}\) is a G-ring, then, for every \(N\), there are a finite type \(S\)-algebra \(A\), a maximal ideal \(\mathfrak n\), and \(x_A\in\mathcal X(A)\), together with

Proof. Choose an affine open $\operatorname{Spec}\Lambda\subset S$ containing $s$. The map from the local scheme $\operatorname{Spec}R$ factors through it: its inverse image is an open containing the closed point of a local spectrum, hence the entire spectrum. The finite-type hypothesis on the residue field says that $k$ is a finitely generated $\Lambda$-algebra on this chart. Since $S$ is locally Noetherian, $\Lambda$ is Noetherian.

Write $R$ as the filtered colimit of its finite-type $\Lambda$-subalgebras. Limit preservation on objects gives a finite-type algebra $C$, a map $C\to R$, and an object $x_C$, with a chosen isomorphism $x_C|_R\simeq x$. No descent assertion about all arrows of $\mathcal X$ is being assumed. Choose a finite presentation

and write $\bar a_i\in R$ for the images of its generators.

Choose finitely many generators of $k$ as a $\Lambda$-algebra and lift them to $R$. Add lifts of a finite generating set of $\mathfrak m_R$. These choices define a polynomial ring $T=\Lambda[z_1,\ldots,z_e]$ and a map $T\to R$ whose reduction to $k$ is surjective and whose image contains generators of $\mathfrak m_R$. Thus $\mathfrak n=\ker(T\to k)$ is maximal. Elements outside $\mathfrak n$ map to units of $R$, so the map extends to the local ring $Q=T_{\mathfrak n}$ and then continuously to

Here is a direct proof of its surjectivity. The chosen elements generate $\mathfrak m_R^j/\mathfrak m_R^{j+1}$ by their degree-$j$ monomials with coefficients in $k$. Those coefficients lift through $Q\to k$. Starting with any element of $R$, lift its residue and then its successive errors by such monomials. At the $j$th correction the added element lies in $\mathfrak m_Q^j$. The partial lifts are Cauchy in $Q$, hence define an element of $P$, and their images converge to the prescribed element of the complete ring $R$. This proves surjectivity; it also proves the usual surjective-cotangent-space criterion in this instance.

The complete ring $P$ is Noetherian. Choose generators $b_1,\ldots,b_r$ of the kernel, and lifts $a_i\in P$ of the $\bar a_i$. Choose coefficients $c_{ji}$ such that

The syzygy module of $(b_1,\ldots,b_r)$ is finite. Choose generators $k_\ell=(k_{\ell1},\ldots,k_{\ell r})$, giving an exact presentation

The map $K$ sends the $\ell$th basis vector to $k_\ell$; in particular $\sum_i k_{\ell i}b_i=0$.

Let $c$ be one Artin–Rees constant for both displayed maps relative to $\mathfrak m_P$, and choose $M\geq\max(N,c+1)$. The common constant exists by Artin–Rees for finite modules, and increasing a constant preserves its defining inclusions. The finite-complex perturbation and canonical graded-quotient statements are the written Lesson 7, Appendix A, Lemmas A.1–A.2. We use their full statements here and do not reproduce their proof.

Put $\mathfrak p=\mathfrak n\cap\Lambda$. The hypothesis says that $\Lambda_{\mathfrak p}$ is a G-ring. The permanence theorem proved below makes its essentially finite-type algebra $Q$ a G-ring. Apply the pointed étale-neighbourhood approximation theorem of Section 5 simultaneously to the following finite polynomial system:

The tuple $(a_i,b_i,c_{ji},k_{\ell i})$ is an exact formal solution in $P$. We obtain an étale $Q$-algebra $B$, a maximal ideal $\mathfrak q$ above $\mathfrak n$ with residue field exactly $k$, and a solution $(a'_i,b'_i,c'_{ji},k'_{\ell i})$ whose entries differ from the old ones by $\mathfrak m_P^M$. The induced identification $\widehat{B_{\mathfrak q}}=P$ preserves the map from $Q$, its residue field and every finite-order quotient. There is no need to inject the entire possibly disconnected algebra $B$ into $P$.

Because $b_i$ lies in $\mathfrak m_P$ and $M\geq1$, each $b'_i$ lies in $\mathfrak q$. Set

The exact polynomial relations define $C\to A^*$ by $y_i\mapsto a'_i$, so they give $x^*=x_C|_{A^*}$. Completion of a quotient of a Noetherian local ring gives

This uses exactness of Noetherian completion for finite modules, supplied by the written completion prerequisite. As $b'_i-b_i\in\mathfrak m_P^M$, the ideals generated by the two lists become identical after adding $\mathfrak m_P^M$. Consequently

Localization at the selected maximal ideal does not change these quotients: every element outside that ideal is already a unit modulo any of its powers. The two maps from $C$ to the displayed common quotient agree, because $a'_i-a_i\in\mathfrak m_P^M$. Restricting the chosen $x_C|_R\simeq x$ therefore gives the required marked comparison of objects at order $M$, hence at order $N$.

The matrix $K'$ and row $(b'_i)$ still form a complex, since all the syzygy equations were preserved exactly. Their entries differ from those of the original exact presentation by $\mathfrak m_P^M\subset\mathfrak m_P^{c+1}$. Lesson 7, Lemma A.2 now identifies the initial ideals inside $\operatorname{gr}_{\mathfrak m_P}P$. Its canonical identity on that graded algebra yields

This is an isomorphism of graded algebras. The quotient filtration is the maximal-ideal filtration, so completion identifies its right side with $\operatorname{gr}_{\mathfrak m^*}A^*$. This explains why the entire relation complex was approximated: approximating only the ideal generators would not provide the reverse initial-ideal containment.

It remains to obtain a finite-type algebra over the original affine chart. The algebra $A^*$ is of finite presentation over $Q=T_{\mathfrak n}$. Its finitely many generators, relations and denominators descend to a finite-type $T$-algebra $A_0$ with

Products of denominators direct this system. Limit preservation on objects descends $x^*$ to an object $x_A$ at some stage $A=(A_0)_f$, with a specified isomorphism after base change to $A^*$. This algebra is finite type over $\Lambda$ and hence over $S$. Let $\mathfrak m_A$ be the inverse image of $\mathfrak m^*$. The image of $A\to k$ contains the image of $T$, which already surjects onto $k$ by the original residue-field generators. Thus $A/\mathfrak m_A=k$: the selected prime is maximal and has exactly the prescribed residue field.

Since $A^*$ is a localization of $A$ away from elements outside the selected point,

Localization at a maximal ideal preserves every quotient by its powers and therefore its associated graded algebra. All the ring comparisons constructed above consequently descend to $A$. Restricting the specified isomorphism $x_A|_{A^*}\simeq x^*$ to their common finite-order quotient supplies the required marking on objects; no finite-stage descent of an arbitrary arrow was needed. This proves all assertions. $\square$

The construction follows the Stacks Project proof of family approximation, Tag 07XB. The proof above spells out the complete-local presentation and the final finite-stage point and object comparisons. Its Popescu, G-ring and étale-completion inputs are proved in this reader or bound to the exact written commutative-algebra providers. The separate finite-complex comparison is bound to the actual written Lesson 7 appendix.

Reference: [Stacks, Tag 07XB]. Appendix A proves the finite-complex comparison as commutative algebra before applying it to algebraicity.

3. From versality to a smooth chart

A formal object \(\xi\) over \(R\) is versal if its functor of marked pullbacks to Artinian local rings is smooth: a pullback over \(B\) can be lifted across every small extension \(B'\twoheadrightarrow B\), together with the specified identification with a deformation over \(B'\). A small extension has one-dimensional residue-field kernel.

For \(x\in\mathcal X(U)\), with \(U\) locally of finite type over \(S\), say that \(x\) is versal at \(u\) when

is smooth. Maps from Artinian local rings with that closed point factor uniquely through the completion of \(\mathcal O_{U,u}\); thus (3.1) is exactly versality of the completed object.

The Schlessinger theorem used as a prerequisite says that (RS) and finite-dimensional tangent space give a versal formal object with Noetherian complete local parameter ring [Stacks, Tag 06IW]. It concerns groupoids and retains their markings; it does not require absence of automorphisms.

Lemma 3.1 (algebraizing a versal chart). If \(\xi\) is effective and versal, \(\mathcal X\) is limit preserving on objects, and \(\mathcal O_{S,s}\) is a G-ring at its image point, then \(\xi\) is isomorphic to the completion of a finite type family \(x_A\) at a point with the same residue field.

Proof. Apply Lemma 2.3 with \(N=2\). Versality successively lifts the identifying map at order two to a compatible map from \(\xi\) to the completion of \(x_A\). Its ring map \(R\to\widehat{A_{\mathfrak n}}\) is surjective: it is surjective on the residue field and cotangent space, and complete Nakayama gives surjectivity at each order, then on the inverse limit. The two rings have equal dimensions of all graded pieces by (2.3). The surjections on their Artinian quotients therefore have equal finite lengths and are isomorphisms. Their inverse limit is an isomorphism, including the family and its transition identifications. \(\square\)

Reference: [Stacks, Tag 07XH]. Approximation alone would not preserve versality; the graded-ring comparison supplies the missing step.

Lemma 3.2 (versal everywhere implies smooth). Suppose \(\mathcal X\) has representable diagonal, satisfies (RS), and is limit preserving. If \(U\) is locally of finite type over \(S\) and \(x\in\mathcal X(U)\) is versal at every finite type point, then \(U\to\mathcal X\) is smooth.

Proof. Its base change by an affine \(V\to\mathcal X\) locally of finite presentation over \(S\) is an algebraic space \(Z=U\times_{\mathcal X}V\). The finite presentation limit test and the fibre product argument in Section 2 show that \(Z\) is locally of finite presentation over \(S\). Choose an étale scheme chart \(W\to Z\).

At a finite type point \(w\), the induced field-valued point of \(U\) has finite residue extension of the image point. Lemma 1.3 and smoothness under field change for deformation categories therefore preserve versality. The deformation category of \(Z\) is the fibre product of those of \(U\) and \(V\) over that of \(\mathcal X\). Smoothness of (3.1) passes to this fibre product and then through \(W\to Z\). The Artinian lifting test for a morphism of schemes locally of finite type over a locally Noetherian scheme [Stacks, Tag 02HX] makes \(W\to V\) smooth at \(w\).

The smooth locus is open. Every nonempty closed subset of a scheme contains a point closed in an affine open, hence a finite type point. Thus this locus is all of \(W\). Smoothness descends through the étale chart to \(Z\to V\). Lemma 2.1 extends the conclusion to arbitrary tests \(V\). \(\square\)

The same argument at a single finite type point proves the version used later: for finite type \(U,V\), versality at \(u\) makes \(U\times_{\mathcal X}V\to V\) smooth at each finite type point above \(u\). The representable diagonal is locally of finite type because its Isom functors preserve limits. These statements explain [Stacks, Tags 07XD and 07XP].

Openness of versality asks that a versal finite type point of every finite type family have an open neighbourhood on whose finite type points the family is versal. For \(f:\mathcal X\to\mathcal Y\), it means openness for every base change by a finite type scheme over \(\mathcal Y\). In particular, openness for a diagonal is a separate condition.

4. The axioms and the two Artin criteria

Here is the numbering of [Stacks, Tag 07XJ].

Fix small models of the field groupoids when recording the bound in [−1]; the construction chooses representatives of their isomorphism classes. For a functor in sets, [0] says étale sheaf, [3] concerns only tangents, [4] is bijectivity of completion, and [5] asks openness for the functor itself. This is [Stacks, Tag 07XZ].

Theorem 4.1 (Artin with representable diagonal). Suppose \(\mathcal O_{S,s}\) is a G-ring for every finite type point \(s\) of \(S\). Suppose \(\mathcal X\) has representable diagonal, satisfies [−1]–[3], every formal object is effective, and openness of versality holds for \(\mathcal X\). Then \(\mathcal X\) is an algebraic stack. If it is a functor in sets, it is an algebraic space.

Proof. For every finite type field object \(x_0\), Schlessinger's theorem gives a versal formal object. Effectiveness and Lemma 3.1 give a finite type scheme family through \(x_0\), at a point with exactly its residue field. Shrink that scheme by openness of versality. Lemma 3.2 makes the resulting morphism to \(\mathcal X\) smooth. Axiom [−1] allows their disjoint union \(U\).

For an affine finite presentation test \(V\to\mathcal X\), the image of \(U\times_{\mathcal X}V\to V\) is open. It contains every finite type point \(v\): the chart chosen for the object over \(\kappa(v)\), and its specified special point, give a point of this fibre product above \(v\). The complement must consequently be empty. Lemma 2.1 proves surjectivity on every test.

The relation \(R=U\times_{\mathcal X}U\) is an algebraic space, with smooth projections. The quotient stack of this smooth groupoid is algebraic by Lesson 5. It agrees with \(\mathcal X\) as an étale stack: smooth surjections have local sections for the étale topology, so every \(\mathcal X\)-object has local coordinates in \(U\); its arrows and their composition are exactly \(R\). Étale descent identifies the two stacks, proving also the fppf stack condition.

For a functor in sets the quotient has no automorphisms. The smooth relation is an equivalence relation. The flat bootstrap of Lesson 2 identifies its sheaf quotient as an algebraic space, and the same étale local-coordinate argument identifies that space with the original functor. \(\square\)

This proves [Stacks, Tags 07Y1 and 07Y4], including the functor criterion's assumption that its diagonal is representable by algebraic spaces.

4.1. Recovering the diagonal

The second diagonal

is the identity section of the inertia category. Its representability says that the condition \(\alpha=\operatorname{id}\), for a family of automorphisms, is represented by an algebraic space over its parameter scheme. For two isomorphisms \(\alpha,\beta:x\to y\), equality is the condition \(\alpha^{-1}\beta=\operatorname{id}_x\). Consequently (4.1) is representable exactly when every Isom functor has representable diagonal. This is the direct equality test in [Stacks, Tag 07WG].

Theorem 4.2 (Artin's stack criterion). Let \(S\) be locally Noetherian, with \(\mathcal O_{S,s}\) a G-ring at every finite type point. Let \(\mathcal X\) satisfy [−1]–[5]. If its second diagonal (4.1) is representable by algebraic spaces, then \(\mathcal X\) is an algebraic stack.

Proof. Take an affine \(V\) locally of finite presentation over \(S\), and two objects \(x_1,x_2\in\mathcal X(V)\). Their Isom category is a setoid: an object is an isomorphism between the two fixed objects, and it has no nontrivial automorphisms after the two identifications are fixed. Thus its isomorphism-class functor \(I\) is an étale sheaf, by [0]. Its diagonal is representable by the equality test above.

We verify the remaining functor axioms for \(I\). The bound [−1] follows from the bound on arrows in \(\mathcal X\). Limits and (RS) follow by forming the fibre product of \(\mathcal X\to\mathcal X\times_S\mathcal X\) with \(V\), using Lemma 1.1 and the full arrow version of limit preservation. The six-term sequence (1.6) shows that its tangent space is an extension of subquotients of the tangents and infinitesimal automorphisms of \(\mathcal X\) and \(V\); all are finite dimensional. Lemma 2.2 gives effectiveness for the scheme \(V\), and axiom [4] for \(\mathcal X\), together with full faithfulness, algebraizes the identifying arrows. Completion therefore commutes with this fibre product. Finally [5] for \(\Delta_{\mathcal X}\) is exactly openness of versality for \(I\).

The base \(V\) is locally Noetherian. Its local rings at finite type points are essentially of finite type over the corresponding local rings of \(S\), so they are G-rings [Stacks, Tag 07PV]. The functor case of Theorem 4.1 now makes \(I\) an algebraic space over \(V\). Lemma 2.1 extends this assertion from finite presentation tests to all tests; its size bound is the field-arrow bound just used, together with the size of the finite type charts constructed in Theorem 4.1. Thus \(\Delta_{\mathcal X}\) is representable.

Apply the stack case of Theorem 4.1. Its hypotheses are now established, and its smooth atlas proves the conclusion. \(\square\)

Reference: [Stacks, Tag 07Y5]. Neither representability of the first diagonal nor algebraicity is assumed in this argument. The G-ring condition is used at the approximation stage, before constructing the atlas.

5. Two ways to prove openness

5.1. A failed lift can be placed in a family

We first isolate the argument which lets infinitesimal information detect an open locus. Suppose \(\mathcal X\) has (RS\(^*\)), representable diagonal, and limit preservation. Work with \(x\in\mathcal X(A)\), where \(U=\operatorname{Spec}A\) is of finite type over \(\operatorname{Spec}\Lambda\subset S\).

Lemma 5.1 (a witness to nonversality). If \(x\) is not versal at a finite type point \(u\), there is a square-zero extension \(C\twoheadrightarrow A\) with kernel \(\kappa(u)\), and a marked lift \(y\in\mathcal X(C)\), which admits no retraction to the family \(x\) on any open neighbourhood of \(u\).

Here a retraction means a morphism \(r:\operatorname{Spec}C\to U\) restricting to the identity on \(U\), and an isomorphism \(y\simeq r^*x\) restricting to the marking \(y|_U\simeq x\).

Proof. Failure of versality is witnessed by a small extension \(B'\twoheadrightarrow B\), a map \(A\to B\) with closed point \(u\), and a lift of the resulting object over \(B'\), for which the specified lifting problem has no solution. Take \(C=A\times_B B'\). Its kernel is the one-dimensional \(\kappa(u)\)-module of the small extension. Condition (RS\(^*\)) patches \(x\) and the object over \(B'\) to \(y\). A retraction near \(u\) pulls back to \(\operatorname{Spec}B'\), whose sole point lies over \(u\). It would supply exactly the forbidden lift, with its required identification. \(\square\)

Lemma 5.2 (retraction near a versal point). Let \(u_0\) be a versal finite type point. Every marked lift of \(x\) over a square-zero extension \(D\twoheadrightarrow A\) which is of finite type over \(\Lambda\) admits a retraction after shrinking around \(u_0\). The same assertion holds for a possibly infinite extension whenever its marked lift descends to such a finite type extension.

Proof. For the descended family \(x_D\), form \(Z=U\times_{\mathcal X}\operatorname{Spec}D\). The marking gives a section \(i:U\to Z\) over the thickening \(U\subset\operatorname{Spec}D\). The single-point version of Lemma 3.2 shows that \(Z\to\operatorname{Spec}D\) is smooth at \(i(u_0)\). Shrink \(U\) so it is smooth along the whole section. The infinitesimal lifting property of this smooth algebraic space lifts \(i\) across the affine square-zero thickening. The lifted map to \(Z\) supplies both a retraction to \(U\) and the desired isomorphism of objects. Pullback gives the final assertion. \(\square\)

These lemmas also show that versality is preserved under generalization among finite type points. Indeed, descend the witness of Lemma 5.1 to a finite type \(\Lambda\)-subalgebra of \(C\) containing lifts of generators of \(A\). Its map to \(A\) is surjective and its kernel square zero. If \(u\) specialized to a versal \(u_0\), Lemma 5.2 would retract it on a neighbourhood of \(u_0\), hence on a neighbourhood of \(u\), a contradiction.

The following modest Noetherian topological facts will be used. Every subset has the same closure as its points maximal under generalization, and an infinite dense subset of a Noetherian scheme has a countable dense subset [Stacks, Tags 0G2R and 0G2F]. Consequently, if the nonversal finite type points accumulate at a versal \(u_0\), one can choose a countable sequence \(u_i\) of nonversal points, with no specializations between them, whose closure contains \(u_0\). No individual \(u_i\) specializes to \(u_0\).

This gives the strong-effectiveness criterion of [Stacks, Tag 0CXU]. If every compatible system over surjective ring towers \(R_n\), with square-zero \(\ker(R_m\to R_n)\) for \(m\geq n\), is effective over \(\lim R_n\), then openness of versality holds. To prove it, take the witnesses \(C_i\to A\) just constructed and their finite fibre products

Their kernels over \(A\) are direct sums of \(\kappa(u_i)\), with square-zero multiplication. Condition (RS\(^*\)) patches their marked objects compatibly. Strong effectiveness gives an object over \(\lim R_n\). Limit preservation descends that object to a finite type subalgebra surjecting onto \(A\). Lemma 5.2 gives a retraction near \(u_0\). Pulling it to any \(C_i\) whose point lies in that neighbourhood contradicts Lemma 5.1. Thus the accumulating sequence cannot exist. This proves the criterion, with its stronger tower hypothesis explicitly distinguished from axiom [4].

5.2. Product-compatible obstruction modules

An obstruction theory consists of \(A\)-linear functors \(\mathcal O_x:\operatorname{Mod}_A\to\operatorname{Mod}_A\), functorial in the object, base ring and module, and elements

for each square-zero extension. They must commute with maps of deformation situations, and must satisfy

Functoriality includes the identity and composition laws for maps \((A,x,M)\to(B,x|_B,N)\), with an \(A\)-linear map \(M\to N\). An obstruction module without the equivalence (5.3) is insufficient for the theorem below. See [Stacks, Tag 07YG].

Theorem 5.3 (the product criterion). Suppose \(\mathcal X\) has representable diagonal, (RS\(^*\)), and limit preservation. Suppose it has an obstruction theory such that for every object \(x\) and every countable collection of modules \(M_i\),

Then \(\mathcal X\) satisfies openness of versality.

Proof. If openness fails at \(u_0\), choose the sequence and marked witnesses \(y_i/C_i\) of Section 5.1. Form the ring

It surjects onto \(A\) with square-zero kernel \(M=\prod_i\kappa(u_i)\). The obstruction \(o_x(D)\) maps to \(o_x(C_i)=0\) for every \(i\). Injectivity in (5.4) makes it zero, so choose a marked lift \(y/D\).

For each \(i\), its restriction and \(y_i\) differ by \(t_i\in T_x(\kappa(u_i))\), using the torsor of Section 1.2. The first condition of (5.4) supplies \(t\in T_x(M)\) with all these components. Replace \(y\) by \(t\cdot y\). Its restrictions are now isomorphic to the specified \(y_i\) as marked lifts. In particular the isomorphisms respect \(x\), which is necessary for the next step.

Limit preservation descends \(y\) to a finite type \(\Lambda\)-subalgebra \(D_0\subset D\). Increase \(D_0\) to include lifts of generators of \(A\). Then \(D_0\twoheadrightarrow A\) is a finite type square-zero extension, and its family restricts to \(x\) with the inherited marking. Lemma 5.2 retracts it near \(u_0\). Pull this retraction through \(D_0\to D\to C_i\). Since \(u_0\) lies in the closure of the \(u_i\), some \(u_i\) lies in that neighbourhood. The retraction of the corresponding marked \(y_i\) contradicts its defining property.

This proves the theorem directly; no effectiveness assertion for objects over infinite towers is assumed. \(\square\)

This is [Stacks, Tag 0CYF]. The proof only needs compatibility of obstruction classes for maps with fixed quotient \(A\), but (5.2) has been stated with the usual full functoriality.

5.3. Naive cotangent obstruction theories

Use cohomological indexing. For \(A=\Lambda[z_1,\ldots,z_r]/J\), the naive cotangent complex is

A square-zero surjection \(A'\to A\) with kernel \(I\) gives \(\mathrm{NL}_{A/A'}\simeq I[1]\). The successive maps of naive cotangent complexes have zero composition, but in general do not form a distinguished transitivity triangle.

A naive obstruction theory assigns \(E_x\in D^-(A)\) and \(\xi_x:E_x\to\mathrm{NL}_{A/\Lambda}\), together with

Its required compatibilities are as follows. Base change \(A\to B\) gives maps \(E_x\to E_{x|_B}\) in \(D(A)\), compatible with \(\xi\), identities and compositions. The identifications (5.6) respect the module and base-change maps of Section 1.2. The class of \(E_x\to\mathrm{NL}_{A/\Lambda}\to\Omega_{A/\Lambda}\) is the canonical deformation obtained by pulling \(x\) along \(a\mapsto(a,da)\) into \(A[\Omega_{A/\Lambda}]\). Finally, \(x\) lifts across \(A'\to A\) exactly when

is zero in \(D(A)\). All these requirements, not just a complex of finite modules, enter the definition [Stacks, Tag 07YP].

Lemma 5.4 (the fibre criterion). Assume (RS\(^*\)) and the tangent and obstruction properties just specified for \(x/A\). At a finite type point \(u\) with residue field \(k\), consider

These conditions imply versality. If \(u\) is closed in \(U\), versality implies them.

Proof. For a bounded-above complex \(E\), \[ \operatorname{Ext}^i_A(E,k) \simeq\operatorname{Hom}_k(H^{-i}(E\otimes_A^{\mathbf L}k),k). \tag{5.9} \] Resolve by free modules and split the resulting complex of vector spaces to obtain this identity.

The second condition of (5.8), by (5.9) and the canonical-element compatibility, says that \(\operatorname{Der}_\Lambda(A,k)\to T_x(k)\) is surjective. For a small extension \(B'\to B\) with residue field \(k\), the obstruction to lifting \(A\to B\) is in \(\operatorname{Ext}^1_A(\mathrm{NL}_{A/\Lambda},k)\). The first condition makes its map to \(\operatorname{Ext}^1_A(E_x,k)\) injective. If the pulled-back object lifts to \(B'\), (RS\(^*\)) gives a lift of \(x\) to \(A\times_BB'\), so (5.7) kills that image. The ring obstruction is therefore zero. Choose a lift of the ring map. Its object may differ from the prescribed marked object by a tangent class, but the surjectivity on tangents adjusts the lift by a derivation to remove the difference. This solves the lifting problem, including its identifying arrow, and proves sufficiency.

For necessity, tangent surjectivity follows immediately from versality. Suppose the first map in (5.8) were not onto. Choose a functional on \(H^{-1}(\mathrm{NL}\otimes^{\mathbf L}k)\) which kills the image and is nonzero. Extend it to \(J/J^2\otimes_Ak\), and denote the resulting functional by \(\lambda\). With \(P=\Lambda[z_1,\ldots,z_r]\), set

Since \(u\) is closed, the nonzero image is all of \(k\); the kernel \(I\) of \(A'\to A\) is \(k\) and has square zero. The differential of the discarded relation lies in the kernel of \(J/J^2\otimes k\to k^{\oplus r}\). Equivalently, the image under this differential of \(\ker\lambda\) equals the image of all \(J/J^2\otimes k\): subtract a multiple of an element in the differential kernel on which \(\lambda=1\). Thus \(\Omega_{A'/\Lambda}\otimes k\to\Omega_{A/\Lambda}\otimes k\) is an isomorphism. The extension is essential: a section would add the nonzero derivation to \(I=k\), contrary to that isomorphism of differential fibres.

By (5.9), the choice of \(\lambda\) makes \(E_x\to I[1]\) zero. Therefore \(x\) lifts to \(A'\). Artin–Rees gives \(n\) with \((\mathfrak m')^n\cap I=0\). The extension \[ B'=A'/(\mathfrak m')^n\longrightarrow B=A/\mathfrak m^n \] still has kernel \(k\), and remains essential. Versality lifts the quotient map from the completed local ring of \(A\) to \(B\) to a map into \(B'\), with the lifted object. Because the maximal ideal of \(B'\) has \(n\)-th power zero, that map factors through \(B\) and is a section, a contradiction. The first map of (5.8) must be surjective. \(\square\)

This supplies the essential-extension argument behind [Stacks, Tags 07YJ and 07YN], without assuming a transitivity triangle for (5.5).

Theorem 5.5 (naive obstruction theory gives openness). Suppose \(\mathcal X\) has (RS\(^*\)) and a naive obstruction theory. If every \(E_x\) over a finite type base has finitely generated cohomology modules, then \(\mathcal X\) satisfies openness of versality.

Proof. Shrink a finite type family so its versal point \(u_0\) is closed and the base is affine. Set \(C=\operatorname{Cone}(\xi_x)\). It is bounded above with finite cohomology, since \(A\) is Noetherian and the naive cotangent complex has finite modules. By Lemma 5.4 and the cohomology exact sequence of the cone, \(H^{-1}(C\otimes^{\mathbf L}k(u_0))=0\).

Represent \(C\) by a bounded-above complex of finite free modules locally near \(u_0\). Over the residue field it is exact in degree \(-1\). Choose bases for the image of the incoming differential and a complement mapping isomorphically onto the image of the outgoing differential. Invert the finitely many corresponding matrix minors. Cancelling these invertible blocks removes the whole degree \(-1\) term. The remaining complex consequently has \(H^{-1}(C\otimes^{\mathbf L}M)=0\) for every module on that open neighbourhood.

In particular the fibre cone has zero \(H^{-1}\) at every point there. Its cohomology sequence gives both conditions (5.8), and their sufficient direction proves versality at every finite type point in the neighbourhood. \(\square\)

Reference: [Stacks, Tag 07YU]. This criterion needs the functoriality and canonical tangent class in (5.6)–(5.7). It does not replace them by the bare assertion that obstructions happen to lie in a finite module.

6. Flat groupoids over an arbitrary base

In this section \(S\) is any scheme. No Noetherian or G-ring hypothesis is imposed. The flat bootstrap for sheaves from Lesson 2 will be used, while the stack version will be proved.

6.1. Finite sources and restriction of scalars

Lemma 6.1 (maps from a finite locally free space). If \(Z\to B\) is finite locally free and \(X\to Z\) is an algebraic space, the functor \[ \operatorname{Res}_{Z/B}(X)(T)= \operatorname{Mor}_{Z}(Z_T,X) \tag{6.1} \] is an algebraic space. Restriction of scalars carries a surjective étale morphism to a surjective étale morphism.

Proof. We explain the étale assertion first. For \(W\to Z\) étale, its section functor over \(B\) is étale and representable. This can be checked over an affine \(B\). If \(W\to Z\) is separated, a section is an open-and-closed subscheme of \(W_T\), finite locally free over \(T\), mapping isomorphically to \(Z_T\).

The finite-part construction of Lesson 2 applies to the separated, flat, locally finitely presented, locally quasi-finite map \(W\to B\), without the extra requirement of containing an identity section. Over the affine base, separated locally quasi-finite recognition first makes \(W\) a scheme. The construction's proof is unchanged: over a henselian local base, isolate a selected finite union of fibre components; scheme Zariski main and idempotent lifting extend it to a finite open-and-closed part; finite presentation descends it to an étale neighbourhood. Equality of two such parts is an open-and-closed condition, computed by the ranks of their two finite locally free differences. The étale quotient of these charts represents the finite-part functor.

Within that functor, the condition of mapping isomorphically to \(Z_T\) is open. The map from a selected finite part to \(Z_T\) is finite étale. Its rank is one exactly on the isomorphism locus; the complement of that locus is closed in \(Z_T\), and its finite image in \(T\) is closed. Thus the section functor is an open subspace of the finite-part space and is étale over \(B\).

For a general \(W\), cover it by a disjoint union \(W'\) of affine schemes over the affine base. Its map to \(Z\), and its map to \(W\), are separated. Sections of \(W'\to Z\) therefore form an algebraic space by the preceding argument. They cover the section functor of \(W\to Z\) étale locally: over a strictly henselian local base, the finite scheme \(Z\) is a product of henselian local factors with separably closed residue fields. A surjective étale cover has a section on each factor. Their finite union of choices descends to an étale neighbourhood of the base. The same separated calculation describes every fibre of the map of section functors. The étale bootstrap of Lesson 2 now proves the general assertion.

For an étale map \(X'\to X\), the fibre of its restriction of scalars at a map \(Z_T\to X\) is precisely the section functor of \(Z_T\times_XX'\to Z_T\). This proves representability and étaleness, and the preceding local section argument proves surjectivity when the original map is surjective.

Now trivialize the finite locally free algebra of \(Z\) on affine base opens, and take an étale cover \(X'\to X\) with \(X'\) a disjoint union of affine schemes. For a finite union of these affines, maps from \(Z_T\) are described by algebra maps into the finite free algebra of \(Z_T\). Choose its basis; coordinates for the images of generators and polynomial relations describe an affine scheme. The requirement that a map be over \(Z\) imposes the corresponding equations. Infinite presentations cause no difficulty for representability. Every map from a quasi-compact \(Z_T\) uses only finitely many of the affine components. The finite-component functors are open subfunctors, since \(Z_T\to T\) is finite and the excluded image has closed image in \(T\). They glue to represent restriction of scalars for \(X'\).

The étale assertion already proved gives a representable surjective étale map from that space to (6.1). The sheaf condition follows from descent of morphisms. The étale bootstrap proves (6.1). \(\square\)

This proves the restriction-of-scalars input [Stacks, Tag 05YF], including the étale assertion [Stacks, Tag 05YD]. The finite-part argument also explains why separatedness was needed only in an intermediate chart. No separatedness of \(X\) is assumed.

6.2. The finite Hilbert stack used here

Let \(\mathcal H_d\), for \(d\geq1\), classify finite locally free schemes \(Z\to T\) of degree \(d\). Let \(\mathcal H_d(X)\) also remember a map \(Z\to X\). These objects are finite maps, not necessarily closed subschemes.

Lemma 6.2. Both \(\mathcal H_d\) and \(\mathcal H_d(X)\), for an algebraic space \(X\), are algebraic stacks.

Proof. Put an arbitrary commutative unital algebra structure on the free module of rank \(d\). Its multiplication coefficients \(c_{ij}^k\) and unit coefficients satisfy finitely many equations expressing commutativity, associativity and the two unit identities. They define an affine scheme \(\operatorname{Alg}_d\) over \(\mathbf Z\). Change of basis gives an action of the smooth group \(\operatorname{GL}_d\).

A finite locally free algebra has a Zariski locally chosen basis. The basis changes and all their compatibility equations identify its stack with \([\operatorname{Alg}_d/\operatorname{GL}_d]\), after base change to \(S\). This quotient is algebraic by the smooth quotient theorem of Lesson 5. Passing between the algebra groupoid and the finite-scheme groupoid uses \(\varphi\mapsto\operatorname{Spec}(\varphi^{-1})\) on isomorphisms; this inversion makes the functor covariant.

For a fixed finite scheme \(Z/T\), the fibre of \(\mathcal H_d(X)\to\mathcal H_d\) is the space of maps \(Z\to X\). Lemma 6.1 represents it. Pull back a smooth atlas of \(\mathcal H_d\); its fibre is an algebraic space, and its smooth scheme atlas supplies a smooth atlas of \(\mathcal H_d(X)\). The diagonal is represented by the same finite-source equality construction. \(\square\)

References: [Stacks, Tags 05YQ and 05YS]. Finite locally free algebras and their isomorphisms descend because their modules, multiplication and unit descend.

Suppose \(q:U\to\mathcal Y\) is representable, surjective, flat and locally of finite presentation, with \(U\) an algebraic space and \(\mathcal Y\) an fppf stack. Define \(\mathcal H_d(U/\mathcal Y)\) to have objects

Lemma 6.3. The diagonal of \(\mathcal Y\) is representable, and every \(\mathcal H_d(U/\mathcal Y)\) is an algebraic stack.

Proof. For two objects over a scheme \(T\), choose an fppf scheme cover \(T'\to T\) lifting both to \(U\). Their Isom sheaf over \(T'\) is the pullback of the algebraic space \(U\times_{\mathcal Y}U\) along their two coordinate maps. Its map to the original Isom sheaf is representable, flat, locally of finite presentation and surjective, being the base change of \(T'\to T\). The flat sheaf bootstrap makes the original Isom sheaf algebraic. Thus the diagonal is representable.

The map \[ \mathcal H_d(U/\mathcal Y)\longrightarrow \mathcal H_d(U)\times\mathcal Y \tag{6.3} \] is representable: over a fixed \((Z/T,f,y)\), it is \(\operatorname{Res}_{Z/T}\operatorname{Isom}_{\mathcal Y}(y|_Z,q(f))\), an algebraic space by Lemma 6.1.

Choose a smooth scheme atlas \(P\to\mathcal H_d(U)\), and set \(W=P\times_{\mathcal H_d(U)}\mathcal H_d(U/\mathcal Y)\). This is a stack in setoids. An automorphism over the fixed \(Z\) and \(f\) restricts to the identity on \(y|_Z\); the finite locally free map \(Z\to T\) is an fppf cover because \(d>0\), so descent of arrows makes it the identity on \(y\).

By (6.3), \(V=W\times_{\mathcal Y}U\) is an algebraic space over \(P\times_SU\). Its map to the sheaf associated to \(W\) is representable, flat, locally of finite presentation and surjective. The flat sheaf bootstrap makes \(W\) an algebraic space. Finally \(W\to\mathcal H_d(U/\mathcal Y)\) is representable smooth and surjective, as the base change of \(P\). The smooth stack recognition of Lesson 5 proves algebraicity. \(\square\)

This proves the inputs [Stacks, Tags 05XW, 05YH and 06CI] without presupposing the flat stack theorem.

6.3. Why complete intersections give smooth coordinates

In (6.2), the pair \((f,\alpha)\) is a map \(Z\to U_y=U\times_{\mathcal Y}T\). Let \(\mathcal H_{d,\mathrm{lci}}(U/\mathcal Y)\) be the subcategory where this map is unramified and a local complete intersection.

This is an open substack. To check it over a family, use étale scheme charts. Unramifiedness is the vanishing condition for the finite module of relative differentials, and the complete-intersection locus in a flat finitely presented family is open and commutes with base change. The scheme versions follow from finite presentations and the regular-sequence criterion; their algebraic-space formulations are [Stacks, Tags 05X8 and 06CE]. Since \(Z\to T\) is finite, the image of the bad locus is closed in \(T\). Its complement represents exactly the indicated subfunctor. Thus the inclusion is a representable open immersion, including arbitrary base changes.

Lemma 6.4 (lifting the finite slice). Let \(T\subset T'\) be an affine square-zero thickening. Let \(X'\to T'\) be flat and locally of finite presentation, put \(X=X'\times_{T'}T\), and let \(Z\to X\) be unramified lci with \(Z\to T\) finite locally free of degree \(d\). Then it extends to \(Z'\to X'\), with \(Z'\to T'\) finite locally free of the same degree and \(Z=Z'\times_{T'}T\).

Proof. If \(I\) is the ideal of \(T\) in \(T'\), flatness and the regular-immersion conormal calculation give

The last module is finite locally free because \(Z\to X\) is unramified lci. The scheme conormal statements, transported by étale charts, are [Stacks, Tags 06CB and 06CC]. Since \(Z\) is affine, (6.4) splits.

Construct the universal first-order neighbourhood of \(Z\) over \(X'\). Étale locally an unramified map is a closed immersion into an étale neighbourhood, and this neighbourhood is cut out by the square of its ideal. Uniqueness of infinitesimal lifts between the étale neighbourhoods glues these constructions. Its square-zero ideal on \(Z\) is \(\mathcal C_{Z/X'}\). Quotient this ideal by the chosen summand \(\mathcal C_{Z/X}\). The resulting neighbourhood \(Z'\) has ideal \(I\otimes\mathcal O_Z\) and a map to \(X'\).

The natural multiplication map identifies its ideal with the image of \(I\), so the square-zero flatness criterion gives flatness of \(Z'\to T'\) and the stated base-change identity. A nilpotent thickening of an affine scheme is affine. Lift finitely many module generators for the finite algebra of \(Z/T\); they generate the algebra of \(Z'/T'\), since the cokernel equals \(I\) times itself and \(I^2=0\). Thus \(Z'\to T'\) is finite. Locally lift a basis of its reduction of rank \(d\). The resulting map from the free module of rank \(d\) is surjective by the same argument. Flatness makes its kernel remain exact after reduction modulo \(I\), so that kernel is equal to \(I\) times itself and is zero. The algebra is therefore finite locally free of rank \(d\). The open lci condition above ensures that its map to \(X'\) still belongs to the indicated substack. \(\square\)

This is the independently established lifting argument of [Stacks, Tag 06D8]. It also proves that \(\mathcal H_{d,\mathrm{lci}}(U/\mathcal Y)\to\mathcal Y\) is formally smooth on objects: apply the lemma to \(U_{y'}\to T'\) for a lift \(y'\) of the base object.

These morphisms are limit preserving on objects. A finite locally free algebra, its map into the fixed locally finitely presented algebraic space \(U_y\), and its finitely many compatibility data descend along affine limits. The finite source uses only finitely many charts; the coefficient construction in Lemma 6.1 then involves finite presentations. The open unramified lci condition descends after increasing the stage. This is the finite presentation argument of [Stacks, Tag 06CH].

Their union over \(d\geq1\) is surjective on field objects. Indeed \(U_y\) is a nonempty algebraic space locally of finite presentation over a field \(k\). Take an affine étale chart and a closed point in its Cohen–Macaulay locus, which is nonempty on a finite type scheme. A system of parameters in the Cohen–Macaulay local ring is a regular sequence. After shrinking the chart, its zero scheme is supported only at that point and is finite over \(k\). Call it \(Z\). The regular immersion followed by the étale map is unramified lci; \(Z\) is a nonzero finite \(k\)-algebra, so it is finite locally free of some degree \(d\). It gives an object (6.2). The scheme inputs are [Stacks, Tags 045U and 0570]; they impose no perfection assumption on \(k\).

Theorem 6.5 (flat stack bootstrap). If \(q:U\to\mathcal Y\) is a representable surjective flat locally finitely presented morphism from an algebraic space to an fppf stack, then \(\mathcal Y\) is an algebraic stack.

Proof. Lemma 6.3 gives its representable diagonal and the algebraic stacks \(\mathcal H_d(U/\mathcal Y)\). Their lci opens are algebraic too. Choose smooth scheme atlases for these opens, and take their union \(P\).

The map \(P\to\mathcal Y\) is representable because \(\mathcal Y\) has representable diagonal. It is limit preserving on objects and formally smooth on objects: both properties compose, using the lifted identifying arrows in their definitions, and the atlases are smooth. For a representable morphism these two properties mean locally finite presentation and formal smoothness respectively. Test the definitions on \(P\times_{\mathcal Y}T\); its objects are actual morphisms to that algebraic space, so the limit and infinitesimal criteria apply. The infinitesimal criterion for a finitely presented morphism makes \(P\to\mathcal Y\) smooth.

The finite slices above cover every field object, allowing field extensions to lift to \(P\). This is surjectivity for a representable morphism. Thus \(P\to\mathcal Y\) is a smooth surjective scheme atlas. Lesson 5's recognition theorem proves the conclusion. \(\square\)

This proves [Stacks, Tag 06DC]. The uses of limits, formal smoothness and field-surjectivity are the precise notions of [Stacks, Tags 06CT, 06CZ and 06D4].

Theorem 6.6 (flat groupoids are algebraic). For any groupoid \(R\rightrightarrows U\) in algebraic spaces over \(S\), if both projections are flat and locally of finite presentation, then \([U/R]\) is an algebraic stack.

Proof. Lesson 5's diagonal calculation represents the Isom sheaves of the quotient: they are locally the pullbacks of the endpoint map \(R\to U\times_SU\), and the flat sheaf bootstrap descends them. In particular \(U\to[U/R]\) is representable.

On an fppf local chart \(a:T\to U\) of a quotient object, its pullback is \(R\times_{s,U,a}T\to T\). This is flat and locally of finite presentation by base change, and is surjective because the identity gives a section. These properties descend fppf locally. Theorem 6.5 therefore applies to \(U\to[U/R]\). \(\square\)

Reference: [Stacks, Tags 06FG and 06FI]. This completes the flat-groupoid theorem used as a stated forward reference in Quotient stacks and Deligne–Mumford stacks, including examples such as \(B\mu_p\).

For orientation, call a map of stacks algebraic when all its scheme-base fibres are algebraic stacks [Stacks, Tag 05XX]. If the target is algebraic, an algebraic map has algebraic source: pull back an atlas and then take an atlas of that fibre. A map from an algebraic stack to a stack with representable diagonal is algebraic for the same reason, using a source atlas. Finally an algebraic map with target having representable diagonal gives representable source diagonal: over each target Isom space, the source Isom fibre is the Isom space of the corresponding algebraic fibre stack. These atlas and Isom calculations prove the three permanence statements without changing the meaning of “representable”.

7. A complete example and a failure of effectiveness

7.1. Line bundles on a proper flat curve

Let \(C\to S\) be proper, flat and of finite presentation, with fibres of dimension at most one. Let \(S\) be locally Noetherian with the G-ring condition of Theorem 4.2. Define \(\mathcal{Pic}_{C/S}(T)\) to be the groupoid of invertible sheaves on \(C_T\), with all their isomorphisms. This is the Picard stack; we do not divide out line bundles pulled back from \(T\).

We will use two precise scheme results. For a proper flat finitely presented scheme and a finitely presented sheaf flat over the base, its derived direct image is perfect and commutes with arbitrary base change [Stacks, Tag 0B91]. For a proper scheme over a complete Noetherian ring, completion gives an equivalence between coherent sheaves and compatible systems of coherent sheaves on its infinitesimal neighbourhoods [Stacks, Tag 088C; EGA III, Theorem 5.1.5]. The latter includes morphisms.

For a line bundle \(N\) on \(C_A\), the first result and the dimension bound give

with \(P^i\) finite projective \(A\)-modules. To justify the two-term range, its derived fibres have cohomology only in degrees \(0,1\), by coherent cohomology on proper schemes of dimension at most one. A bounded finite-projective representative can then be shortened: cancel the split surjections at its highest excessive degrees and the split injections at its lowest excessive degrees. Fibrewise exactness and Nakayama make these cancellations valid on the base. On an affine base they give (7.1); the construction also shows the asserted Tor amplitude. Thus (7.1) computes base change to every \(A\)-algebra and tensoring with every \(A\)-module.

[−1]. A finite affine cover of \(C_k\), finite presentations of line bundles on that cover, and matrices describing their transitions give a uniform cardinal bound for their isomorphism classes and arrows as \(k\) ranges over finite type fields over \(S\). The cardinal can be enlarged once to contain the affine base presentations and the field models. This supplies the site-size condition.

[0]. Invertible sheaves and their isomorphisms have effective fpqc descent, by descent of quasi-coherent modules followed by the local rank-one condition. In particular this is an étale stack.

[1]. A line bundle on \(C_A\), with \(A=\operatorname{colim}A_i\), is of finite presentation. Finite affine covers, its local presentations and the finitely many transition maps descend to a stage. Their inverse identities and the rank-one condition hold after increasing the stage. The same finite-presentation argument descends isomorphisms and tests equality of isomorphisms. Thus the full groupoids preserve limits.

[2]. Flatness of \(C/S\) makes the structure rings of its affine charts preserve the fibre product (1.1) under base change. Finite projective modules patch over that fibre product, as proved in Exercise 8.2 below. Apply this to the locally free rank-one modules on the charts. Full faithfulness patches their transition maps and checks the cocycle, so the patched sheaf is invertible. This proves (RS), and the same argument with arbitrary square-zero extensions proves (RS\(^*\)).

[3]. The transition-function calculation gives

Indeed, write a first-order change of transition units as \(g_{ij}(1+\epsilon a_{ij})\). The cocycle equation is the additive Čech cocycle equation for \(a_{ij}\); changing frames adds a coboundary. An automorphism reducing to the identity is \(1+\epsilon a\) with a global section \(a\). The spaces are finite dimensional by proper coherent cohomology.

[4]. For a complete Noetherian local ring \(R\), apply Grothendieck existence to \(C_R\) and a compatible system \(L_n\). It gives a coherent \(L\), including all compatible morphisms. It is invertible near the closed fibre: completing the local stalk along the base ideal gives a free module of rank one, and the local adic criterion, or faithful flatness of completion after localization at the point, gives local freeness there. The non-invertible locus is closed. Its image in \(\operatorname{Spec}R\) is closed by properness and misses the closed point. A nonempty closed subset of a local spectrum contains that point, so the locus is empty. Thus \(L\) is invertible everywhere. Full faithfulness algebraizes isomorphisms and their inverses, proving (2.2).

[5] for the stack. For \(L\in\mathcal{Pic}_{C/S}(A)\) and an \(A\)-module \(M\), the same Čech calculation gives \[ T_L(M)=H^1(C_A,p^*M),\qquad \operatorname{Inf}_L(M)=H^0(C_A,p^*M). \tag{7.3} \] The obstruction to lifting \(L\) across \(A'\to A\) lies in \(H^2(C_A,p^*I)\): lift local frames and transition units; their failure to satisfy the cocycle is a Čech 2-cocycle, and changing the transitions changes it by a coboundary. Its vanishing is precisely the ability to glue a lift. Formula (7.1) for \(\mathcal O_C\) shows that this \(H^2\) is zero for every \(I\). Thus zero obstruction modules give a functorial obstruction theory.

Furthermore \(H^1(C_A,p^*M)=\operatorname{coker}(P^0\otimes_AM\to P^1\otimes_AM)\). Finite projective modules commute with products under tensoring, and products of module sequences are exact. Hence \(T_L\) commutes with countable products, and the zero obstruction functor has the required injectivity. Theorem 5.3 proves openness.

[5] for the diagonal, and the second diagonal. For two line bundles \(L_1,L_2\), put \(N=L_1^\vee\otimes L_2\). Their Hom functor is the affine scheme of vectors in \(P^0\) annihilated by the differential in (7.1). The universal vector gives a section of \(N\). It is an isomorphism exactly where its zero locus in the proper curve has empty image in the parameter scheme. That condition is open, so \(\operatorname{Isom}(L_1,L_2)\) is a scheme locally of finite presentation over the base.

Its diagonal is representable, hence the stack's second diagonal is representable. Openness of versality for these Isom schemes follows from the ordinary Artinian smoothness criterion and the open smooth locus, using étale scheme charts if necessary. This checks the other part of [5].

All the axioms, including their arrow conditions, have now been verified. Theorem 4.2 proves that \(\mathcal{Pic}_{C/S}\) is an algebraic stack. The proof did not replace this stack by its sheaf of isomorphism classes.

7.2. The affine line loses a formal automorphism

Let \(R\) be a complete discrete valuation ring with uniformizer \(t\). On \(\mathbf A^1_R\), the trivial line bundle has automorphism group \(R[x]^\times=R^\times\): \(R\) is a domain, so a unit polynomial has degree zero.

Over \(R/t^n\), however, \(1+tx\) is a unit, with inverse

The units and their inverses form compatible systems. They give an automorphism of the formal trivial line bundle in \(\lim_n\mathcal{Pic}_{\mathbf A^1/R}(R/t^n)\). The corresponding element \(1+tx\) in \(\lim_n(R/t^n)[x]\) is not the image of a unit of \(R[x]\), since that image would be constant. Thus completion is not full on automorphisms, even though the formal object under discussion is already the completion of a trivial line bundle. Axiom [4] fails. This is exactly the groupoid distinction needed in (2.2).

8. Exercises and complete solutions

Exercise 8.1 (medium: limits of algebraic stacks). Show that an algebraic stack locally of finite presentation over \(S\) is limit preserving. Do not assume quasi-compactness or quasi-separatedness.

Solution. Choose a smooth atlas \(U\to\mathcal X\) with \(U\) locally of finite presentation over \(S\). Then \(R=U\times_{\mathcal X}U\) is smooth over \(U\), hence locally of finite presentation over \(S\). The morphism \(R\to U\times_SU\) is locally of finite presentation by the finite-presentation permanence rule on charts. Descent along the atlas therefore shows that the diagonal of \(\mathcal X\) is locally of finite presentation.

For descended objects over \(T_i\), their Isom space is locally of finite presentation over \(T_i\); its affine limit criterion gives descent and eventual equality of all arrows. This proves full faithfulness in (2.1).

For essential surjectivity, pull the atlas back to an object over an affine \(T\). Choose an étale scheme chart of that smooth algebraic-space cover and finitely many affine opens whose images cover \(T\). Their disjoint union \(V\to T\) is a smooth surjection of finite presentation, with a map to \(U\) and descent arrows in \(R\). The affine schemes \(V\), \(V\times_TV\) and its triple product are of finite presentation over \(T\). Their finite equations descend to a sufficiently late \(T_i\), as do their maps to the locally finitely presented spaces \(U\) and \(R\). The finitely many identity and cocycle equalities descend after a further increase. Smoothness and surjectivity of the covering descend too. Effective fppf descent in the already algebraic stack produces the object over that \(T_i\). Its pullback is the original object. Only a finite portion of the atlas was used for this affine test, so no global compactness assumption was introduced.

Exercise 8.2 (medium: module patching). Verify (RS) for finite locally free modules, including the arrows.

Solution. For (1.1), the ring \(P\) is local, with residue field that of \(A_1\). A patched pair of finite free modules must have equal rank, since their restrictions to \(A\) are isomorphic. Choose bases. The gluing isomorphism is an invertible matrix \(g\) over \(A\). Lift its entries to \(A_2\); its determinant lifts a unit because \(\ker(A_2\to A)\) is nilpotent, so the lifted matrix is invertible. Change the second basis by that matrix. The gluing is now the identity, and the pair is the restriction of \(P^r\).

An arrow between two such pairs is a pair of matrices over \(A_1,A_2\) which agree over \(A\). Entrywise it is a unique matrix over \(P\). If the arrow is an isomorphism, its compatible inverse matrices patch as well. This proves full faithfulness and essential surjectivity. Markings at the residue field are preserved by doing the same construction with their specified identifications.

The general finite-projective version, used for (RS\(^*\)) in Section 7, is obtained by adding complements. Write the module over \(A_1\) as the image of an idempotent in \(A_1^n\). Lift its restriction to an idempotent in \(A_2^n\). Its image is isomorphic to the given second module with its specified reduction: maps between finite projectives lift across the nilpotent ideal, and maps inverse modulo that ideal are inverse after multiplying by an invertible correction. Add the two complementary summands. The enlarged pair is free on both sides, and its gluing matrix lifts as above. Its two projection idempotents, now equal over \(A\), patch to an idempotent over \(P\). Its image is the desired finite projective module. Equivalently that image is the fibre product of the original two modules. Maps patch entrywise in the free ambient modules and respect the idempotents. This proves the finite-projective Milnor patching lemma for arbitrary \(A_1\) and a square-zero surjection \(A_2\to A\), including all arrows.

Exercise 8.3 (medium: vector-bundle tangents). For a vector bundle \(E\) on a proper curve \(C/k\), prove \[ T_E\mathcal{Bun}_C=\operatorname{Ext}^1_C(E,E) =H^1(C,\mathcal E nd(E)). \tag{8.1} \]

Solution. On a finite affine trivializing cover, write its matrices as \(g_{ij}\). A first-order change has the form \(g_{ij}+\epsilon h_{ij}\). Linearizing the cocycle and the frame changes identifies its class with an additive Čech 1-cocycle in \(\mathcal E nd(E)\), modulo coboundaries. Since \(C\) is separated, affine intersections compute the cohomology of this quasi-coherent sheaf. This gives the last term of (8.1), and addition is exactly the split-extension addition of Section 1.

There is also an intrinsic description. A first-order bundle \(E'\) gives the extension \[ 0\to E\xrightarrow{\epsilon}E'\to E\to0 \] as an \(\mathcal O_C\)-module; flatness over \(k[\epsilon]\) identifies the kernel with \(E\). Conversely, an extension gives the middle term an \(\epsilon\)-action by the composite of its quotient to \(E\) and its inclusion of \(E\). Locally the extension splits because \(E\) is locally free, and the resulting module is a free deformation over \(\mathcal O_C[\epsilon]\). Thus extensions and marked first-order bundles are equivalent on isomorphism classes. Their Baer addition matches the cocycle addition. Finally \(\mathcal H om(E,-)\) is exact locally, so global derived Hom from \(E\) is cohomology of \(E^\vee\otimes-\). This identifies \(\operatorname{Ext}^1(E,E)\) with \(H^1(\mathcal E nd(E))\). Infinitesimal automorphisms, for comparison, are \(H^0(\mathcal E nd(E))\).

Exercise 8.4 (hard: the formal Picard groupoid). For a smooth proper curve \(C/k\), verify [3] and [4] for its line-bundle stack.

Solution. Formula (7.2) gives the tangent and infinitesimal automorphism spaces at any line bundle after every finite field extension. Proper coherent cohomology makes \(H^0\) and \(H^1\) finite dimensional, proving [3]. No assumption that automorphisms vanish is made.

Given a complete Noetherian local \(k\)-algebra \(R\), Grothendieck existence on the proper scheme \(C_R\) algebraizes every compatible system of line bundles to a coherent sheaf. The completed local modules along the closed fibre are free of rank one, so the sheaf is invertible near that fibre. Its closed non-invertible locus has proper closed image in \(\operatorname{Spec}R\); such an image cannot be nonempty while missing the closed point. The sheaf is therefore a line bundle on all of \(C_R\). Full faithfulness of existence algebraizes every compatible isomorphism. Applying it also to the inverse and using faithfulness proves that the algebraized map is an isomorphism. Consequently completion is an equivalence of groupoids, not merely a surjection on objects. This proves [4], and explains precisely where properness excludes the phenomenon in (7.4).

Appendix A. Artin–Rees perturbation and graded quotients

The approximation in Lemma 2.3 preserves an entire finite relation complex. We prove the two perturbation facts that make its associated graded conclusion valid, with their exact congruence order.

A.1. The perturbation theorem

Let \(A\) be Noetherian and let \(I\subset\operatorname{Jac}(A)\). For a map \(q\colon M\to N\) of finite modules, say that \(c\geq0\) is an Artin–Rees constant for \(q\) when \[ q(M)\cap I^nN\subset q(I^{n-c}M)\qquad(n\geq c). \] Artin–Rees applied to the finite submodule \(q(M)\subset N\) gives such a constant. We may enlarge a constant, so two given maps admit one common constant.

Lemma A.1. Suppose \[ L\xrightarrow f M\xrightarrow g N, \qquad L\xrightarrow{f'}M\xrightarrow{g'}N \] are complexes of finite \(A\)-modules, the first is exact at \(M\), and \(c\) is an Artin–Rees constant for both \(f\) and \(g\). Assume \[ (f-f')(L)\subset I^{c+1}M, \qquad(g-g')(M)\subset I^{c+1}N. \tag{AR.1} \] Then the second complex is exact at \(M\), and the same \(c\) is an Artin–Rees constant for \(g'\).

Proof. Fix \(n\geq c\) and \(a\in M\) such that \(g'(a)\in I^nN\). We show that we can subtract elements of \(f'(L)\) until the remaining element lies in \(I^{n-c}M\). Such subtractions do not change \(g'(a)\), because the perturbed sequence is a complex.

Suppose at an intermediate step that \(a\in I^rM\), with \(0\leq r<n-c\). Equation (AR.1) gives \[ g(a)=g'(a)+(g-g')(a) \in I^nN+I^{r+c+1}N=I^{r+c+1}N. \] Artin–Rees for \(g\) supplies \(a_1\in I^{r+1}M\) with \(g(a_1)=g(a)\). Exactness of the first sequence supplies \(b\in L\) with \(a-a_1=f(b)\). If \(r\geq c\), Artin–Rees for \(f\) permits choosing \(b\in I^{r-c}L\), since \(f(b)\in I^rM\). If \(r<c\), keep any such \(b\).

In either case \[ a=f'(b)+a_2, \qquad a_2=a_1+(f-f')(b)\in I^{r+1}M. \] Indeed, for \(r\geq c\) the error is in \(I^{c+1}I^{r-c}M=I^{r+1}M\); for \(r<c\) it is in \(I^{c+1}M\subset I^{r+1}M\). Replace \(a\) by \(a_2\). The order rises by one while \(g'(a)\) remains fixed. After finitely many steps we obtain \[ (g')^{-1}(I^nN)\subset f'(L)+I^{n-c}M. \tag{AR.2} \] Applying \(g'\) proves \(g'(M)\cap I^nN\subset g'(I^{n-c}M)\), including the case \(n=c\), when no adjustment was needed.

If \(g'(a)=0\), (AR.2) holds for every \(n\geq c\). Its class in the finite module \(M/f'(L)\) lies in every power of \(I\). The Krull-intersection theorem, with \(I\subset\operatorname{Jac}(A)\), makes that class zero. Hence \(\ker g'\subset\operatorname{im}f'\). The converse follows from \(g'f'=0\). No completeness of \(A\), freeness of the modules, or exactness at an endpoint has been used. \(\square\)

A.2. Equality of the initial submodules

Lemma A.2. Under the preceding hypotheses put \(Q=N/g(M)\) and \(Q'=N/g'(M)\). The identity on \(\operatorname{gr}_I N\) induces a canonical isomorphism \[ \operatorname{gr}_I Q\simeq\operatorname{gr}_I Q' \] of graded \(\operatorname{gr}_I A\)-modules. If \(N=A\) and the two images are ideals, this is an isomorphism of graded algebras.

Proof. In degree \(n\geq0\) the quotient filtration gives \[ (\operatorname{gr}_I Q)_n =I^nN/\bigl(I^{n+1}N+(g(M)\cap I^nN)\bigr). \tag{AR.3} \] We compare the two denominator submodules inside \(I^nN\). If \(n\leq c\), (AR.1) says that \(g(a)\) and \(g'(a)\) differ by an element of \(I^{n+1}N\). For \(g(a)\in I^nN\), its perturbed image is therefore also in \(I^nN\). This gives one containment of denominators.

If \(n>c\), write an element of \(g(M)\cap I^nN\) as \(g(a)\) with \(a\in I^{n-c}M\), using the constant for \(g\). Then \[ g(a)-g'(a)\in I^{c+1}I^{n-c}N=I^{n+1}N. \] Again \(g'(a)\in I^nN\), giving the containment. The perturbation lemma supplies the same constant for \(g'\), so exchanging \(g\) and \(g'\) proves the reverse containment in every degree. The quotients in (AR.3) are thus identical subquotients of \(\operatorname{gr}_I N\). Their identifications respect its graded module action. For ideals in \(A\), they respect multiplication as well, proving the algebra assertion. \(\square\)

A.3. Flat base change of the constant

Lemma A.3. Let \(A\to B\) be flat, with both rings Noetherian. An Artin–Rees constant \(c\) for \(q\colon M\to N\) remains a constant for \(q_B\) relative to \(IB\).

Proof. For \(n\geq c\), the constant condition is equivalent to \[ q^{-1}(I^nN)\subset\ker q+I^{n-c}M. \] The left side is the kernel of \(M\to N/I^nN\). Flat tensor product preserves that kernel and the kernel of \(q\), and identifies the quotients and powers with their base changes. Tensor the displayed inclusion with \(B\), then apply \(q_B\); the desired containment follows. \(\square\)

A.4. Use in family approximation

In Lemma 2.3 the relation presentation is a complex of finite free modules over the complete local presentation ring \(P\): \[ P^{\oplus t}\xrightarrow f P^{\oplus r} \xrightarrow{(b_1,\ldots,b_r)}P\longrightarrow R\longrightarrow0. \] The columns of \(f\) generate the syzygies of the ideal generators. Choose a common constant \(c\) for its two displayed maps, relative to the maximal ideal of \(P\). Polynomial approximation must preserve the equations saying that the new generators and new syzygies still form a complex, and must approximate all these coefficients to order at least \(c+1\). Its order must also be at least the requested \(N\).

The perturbation lemma gives exactness at \(P^{\oplus r}\). The graded quotient lemma identifies the two associated graded quotient algebras, rather than merely their Hilbert functions. The quotient filtration is the maximal-ideal filtration of the local quotient, so this is precisely the graded comparison asserted in (2.3). Completion of the approximating Noetherian local ring preserves its residue quotients and associated graded algebra; thus the comparison returns to that local ring. For a finite type ring and its localization at a maximal ideal, every quotient by a power of that maximal ideal is unchanged by localization, since each element outside the ideal is already a unit in the quotient. The comparison consequently survives the finite-stage descent of the approximating object.

This argument explains why approximating only the ideal generators is insufficient: the exact perturbed syzygy equations supply (AR.2), and (AR.2) supplies the Artin–Rees constant for the perturbed map needed for the reverse graded containment.

The mathematical source is the Stacks Project authors, More on Algebra, labels lemma-approximate-complex and lemma-approximate-complex-graded (Tags 07VE–07VF), and lemma-works-flat-extension, at AI Integrated Stacks Project revision 565b10e987aba5969b21145a0833f42d69f96790. That section credits Conrad–de Jong for part of the material. This independently expressed proof is eligible CC0 programme writing; the consulted human source retains its attribution and GFDL terms.

Prerequisites and proof scope

All four acceptance results—Artin's space and stack criteria, openness from naive obstruction theories, and algebraicity of flat groupoid quotients—have been proved above. The product-compatible obstruction criterion and the stronger-effectiveness criterion have also been proved. The imported inputs are:

• Schlessinger's existence theorem for a versal formal object in a predeformation category satisfying (S1), (S2), and finite-dimensional tangent space [Stacks, Tag 06IW], and its field-change, linearity and small-extension tests. These are the assigned formal-deformation prerequisites. The cotangent obstruction to lifting a ring map, the regular-immersion conormal calculation and square-zero flatness criterion are the assigned deformation prerequisites; the space versions are obtained on étale charts [Stacks, Tags 06CB, 06CC and 06BH].
• Appendix B supplies Popescu's desingularization, both polynomial-approximation forms and G-ring permanence [Stacks, Tags 07GC, 07QY, 07QZ and 07PV]. Section 2.3 and Appendix B §B.6.1 give the full marked-family construction. Appendix A proves the finite-complex Artin–Rees perturbation statements [Stacks, Tags 07VE and 07VF]. Lower predecessor references retain the exact written, partial or unbound states recorded in Appendix B.
• Appendix B includes the ordinary-scheme Artinian lifting-at-a-point chain used by the smoothness test [Stacks, Tag 02HX]. The remaining scheme prerequisites include étale lifting over henselian local rings, finite presentation descent along affine limits, scheme Zariski main and idempotent lifting from the earlier lessons. The openness criteria for unramified and lci families with flat finitely presented source and target, and the Cohen–Macaulay slicing inputs, have their precise space or scheme formulations in [Stacks, Tags 05X8, 06CE, 045U and 0570].
• The Noetherian topological statements in Section 5.1 [Stacks, Tags 0G2F and 0G2R]; coherent cohomology finiteness and the dimension bound; perfect direct image with arbitrary base change [Stacks, Tag 0B91]; and Grothendieck existence for proper schemes [Stacks, Tag 088C; the proper-support extension is Tag 088E]. Existence is used with all its morphisms. Its conversion from coherent sheaves to invertible sheaves was proved in Section 7.

The general pushout theorem for algebraic stacks along arbitrary affine thickenings [Stacks, Tag 07WM] is mentioned as a stronger statement. The Artinian assertion needed here was proved in Lemma 1.2. The general theorem's flat-space patching input is not developed in this lesson.

References

• [Stacks] The Stacks project, Artin's Axioms: Tags 07T0, 07T2, 07WM, 06L9, 07WT, 07WW, 07WY, 07X3, 07XA, 07XK, 07XD, 07XP, 07XJ, 07XZ, 07Y0, 07Y1, 07Y3, 07Y4, 07Y5, 0CXN, 0CXR, 07Y6, 07YF, 07YG, 0CYF, 07YJ, 07YP, 07YT and 07YU. Read the chapter in AI Integrated Stacks Project, Artin's Axioms.
• [Stacks] Criteria for Representability: Tags 05XF, 05XH, 06CT, 06CZ, 06D4, 05XX, 06DB, 06DC, 06FG and 06FI, and the finite Hilbert and restriction-of-scalars constructions used in Section 6. See AI Integrated Stacks Project, Criteria for Representability.
• [Stacks] Smoothing Ring Maps, Tags 07GC, 07QY and 07QZ; More on Algebra, Tags 07PV, 07VE and 07VF; Formal Deformation Theory, Tag 06IW; and the scheme-theory inputs listed above. See AI Integrated Stacks Project, Smoothing Ring Maps. AI Integrated Stacks Project retains the upstream tags; its additions and corrections are not reviewed by the Stacks project's maintainers.
• [EGA] A. Grothendieck, with J. Dieudonné, Éléments de géométrie algébrique III, Theorem 5.1.5, for the existence theorem invoked through its proper-scheme form in [Stacks, Tag 088C].

Appendix B. Artin approximation and general Néron desingularization

Section and statement numbers within this appendix are local to the appendix; an explicit course or provider title qualifies every outside reference.

This foundation retains the full assertion needed in Lesson 7: a regular map between arbitrary Noetherian rings is a filtered colimit of smooth algebras; consequently every formal solution over a henselian Noetherian local G-ring can be approximated to every prescribed order. The positive-characteristic argument below includes inseparable residue fields. The version over a local G-ring without henselianity produces a pointed étale neighborhood with the same residue field.

The geometric idea is to replace a finite algebraic problem by a smooth one and then use smooth coordinates to impose an exact solution with a prescribed finite-order residue. The proof develops singularity ideals, lifting and desingularization before separating the two residue-field cases. Supporting commutative-algebra constructions and their precise prerequisites accompany that argument.

B.1. Sources, authorship and proof status

The incorporated proofs adapt the Stacks Project Authors' Smoothing Ring Maps chapter, the marked family-approximation proof from Artin's Axioms, and the specified predecessor proofs from More on Algebra and Commutative Algebra, as distributed by AI Integrated Stacks Project. Popescu proved the main theorem; the Stacks exposition follows Richard Swan's treatment, which credits André's notes and Ogoma's arguments. This appendix reuses the Stacks exposition; it does not draw directly on the separately cited papers.

Human source credit: the Stacks Project Authors, with the source copyright notice Copyright (C) 2005–2025 Johan de Jong. AI source credit: the credited contributors to the pinned AI Integrated Stacks Project edition, to the extent their contributions occur in the retained material. Adaptation, dependency comparison and the explicit supplementary verifications here: OpenAI Codex, GPT-6.1 Sol, Ultra, 5 October 2026. These credits do not assert that an AI independently reviewed the unchanged human proof.

The incorporated and adapted human-source component is distributed under GNU FDL version 1.2 or later, with no Invariant Sections, no Front-Cover Texts and no Back-Cover Texts. The complete licence text accompanies this draft below. Independently written programme material retains the user's CC0 1.0 dedication; no AI rights holder is invented. The combined derivative preserves the human-source terms.

B.2. The finite factorization criterion

All rings and maps in this chapter are commutative and unital. Smooth means finitely presented and formally smooth; the equivalent standard Jacobian presentations and flat-fibre characterization are predecessor results identified in §8. A regular map $R\to\Lambda$ is flat and has geometrically regular Noetherian fibres. A Noetherian algebra over a field is geometrically regular if it remains regular after every finite purely inseparable extension of that field. This does not mean that the residue fields at all its primes are separable over the base field.

Lemma 2.1. A ring map $R\to\Lambda$ is a filtered colimit of smooth $R$-algebras if and only if every $R$-algebra map $A\to\Lambda$ from a finitely presented $R$-algebra factors as $A\to B\to\Lambda$, with $B$ smooth over $R$.

Proof. A finitely presented algebra has finitely many generators and relations. A map from it into a filtered colimit descends to a stage: choose a common stage containing the finitely many images, and then a later common stage where all relations vanish. This proves one implication.

For the converse take the category of smooth $R$-algebras equipped with an $R$-algebra map to $\Lambda$, taking a set of representatives. It is nonempty, since $R$ is smooth over itself. Two objects map to a common object: their tensor product is finitely presented, maps to $\Lambda$, and hence factors through a smooth algebra by the hypothesis. Two parallel morphisms $u,v\colon B\rightrightarrows C$ become equal after a morphism to a smooth algebra: quotient $C$ by the differences $u(b_i)-v(b_i)$ for finitely many algebra generators of $B$; this quotient is finitely presented and maps to $\Lambda$, so apply the hypothesis. The category is therefore filtered.

Its colimit maps to $\Lambda$. Each $\lambda\in\Lambda$ occurs in a stage by applying the hypothesis to $R[T]\to\Lambda$, $T\mapsto\lambda$. If two stage elements have the same image, pass to a common stage $C$, quotient it by their difference, and apply the hypothesis again; they become equal in a later smooth stage. The map on colimits is bijective and is a ring homomorphism, so is an isomorphism. This proves the precise criterion, including its filteredness assertion. ∎

This supplies the categorical verification implicit in the native algebra-lemma-when-colimit. The proof uses no Popescu theorem.

B.3. Popescu's theorem and the proof mechanism

Theorem 3.1 (Popescu). Every regular homomorphism $R\to\Lambda$ of Noetherian rings is a filtered colimit of smooth $R$-algebras.

The complete algebraic constructions establishing the theorem appear in §9, not merely its last paragraph. Their logical order is as follows.

For a finitely presented $R$-algebra $A$, write $H_{A/R}$ for the radical ideal defining its nonsmooth locus, and $\mathfrak h_A=\sqrt{H_{A/R}\Lambda}$ after a specified map $A\to\Lambda$. A resolution at a prime $\mathfrak q\supseteq\mathfrak h_A$ is a factorization $A\to B\to\Lambda$ with $B$ finitely presented, $\mathfrak h_A\subseteq\mathfrak h_B$, and $\mathfrak h_B\nsubseteq\mathfrak q$. Thus a resolution strictly increases the radical ideal. It does not require the target $\Lambda$ to be finitely presented.

The singular-ideal construction expresses its generators through minors of a Jacobian presentation and the relations among the equations. An improved presentation, obtained from the symmetric algebra of the conormal module, admits an algebra retraction and has free differentials wherever the original map is smooth. The comparison lemma then makes suitable powers of an element into strictly standard elements. These constructions are the input to the two principal algebraic steps.

First, flat nilpotent deformations of a filtered colimit of smooth algebras remain such colimits. In the square-zero case, a presentation is lifted and finitely many relations are killed using the equational criterion for flatness; the general nilpotent case follows by successive square-zero quotients. Second, the lifting and desingularization lemmas turn a smooth factorization modulo $\pi^8$ into a factorization whose smooth locus expands, assuming the stated annihilator equalities. The powers $\pi^8,\pi^4,\pi^2$ are retained in the actual constructions. Neither step is a simple-root Hensel argument.

These lemmas reduce the theorem to a geometrically regular Noetherian algebra over a field. To justify the reduction, assume failure and choose, among its failing quotient ideals, an ideal $I\subset R$ maximal under inclusion, such that the regular map $R/I\to\Lambda/I\Lambda$ fails the factorization property. Every quotient by a nonzero ideal now has the property. If the nilradical of this new $R$ were nonzero, it would be nilpotent because $R$ is Noetherian; nilpotent lifting and flatness would give the property for $R$, a contradiction. Thus $R$ is reduced. Its total ring of fractions is a finite product of fields, and the field theorem yields a smooth factorization there. Clearing finitely many denominators produces a strictly standard nonzerodivisor $\pi$ in a finite presentation. The property for $R/\pi^8R$, followed by lifting and desingularization, returns a smooth factorization over $R$. The target is flat over $R$, so a nonzerodivisor of $R$ remains a nonzerodivisor of $\Lambda$, giving the required annihilator equalities. This proves the transfer back from the field case to the original map.

Over a field $k$, choose a prime minimal over $\mathfrak h_A$. In characteristic zero the regular local target has a separable residue field, and the regular-parameter construction resolves it. In characteristic $p>0$, the inseparable proof constructs a finite polynomial subalgebra and a local Artinian approximation which captures the finitely many needed coefficients. The injective map $H_1(L_{K/k})\to\mathfrak m/\mathfrak m^2$, furnished by geometric regularity, gives the finite-dimensional condition needed for this construction even when $K/k$ is inseparable or infinitely generated. Ogoma's annihilator lemma makes the localized parameters usable before localization. The construction with auxiliary variables $t_i$ returns from these Artinian data to the original finitely presented algebra. Both complete proofs and their return maps are retained in §9.

Every resolution strictly increases $\mathfrak h_A$. The ascending-chain condition in the Noetherian target terminates this process at $\mathfrak h_B=\Lambda$. At that point choose $f_i\in H_{B/R}$ and $\lambda_i\in\Lambda$ with $\sum_i f_i\lambda_i=1$. The algebra

maps to $\Lambda$ by $Z_i\mapsto\lambda_i$ and is smooth over $R$. Indeed the opens $D(f_i)$ cover its spectrum, and $C_{f_i}$ is a polynomial algebra in the other $Z_j$ over the smooth algebra $B_{f_i}$. This establishes smoothness on a covering and hence smoothness of $C$. Lemma 2.1 now proves the theorem. The zero target gives the trivial smooth zero-algebra factorization; zero-dimensional local targets are covered explicitly in the supplementary verifications below. ∎

B.4. Approximating a finite polynomial system

Theorem 4.1. Let $(R,\mathfrak m,k)$ be a henselian Noetherian local G-ring and $\widehat R=\varprojlim R/\mathfrak m^q$. Given polynomials $f_1,\ldots,f_s\in R[X_1,\ldots,X_n]$ and $a\in\widehat R^n$ with $f_j(a)=0$, for every integer $N\geq1$ there is $b\in R^n$ with $f_j(b)=0$ and $a_i-b_i\in\mathfrak m^N\widehat R$ for all $i$.

Proof. The G-ring hypothesis at the maximal ideal says precisely that $R\to\widehat R$ is regular; completion is Noetherian and flat by the actual completion provider. Choose $c_i\in R$ with $a_i-c_i\in\mathfrak m^N\widehat R$, and generators $d_1,\ldots,d_M$ of $\mathfrak m^N$. Write $a_i=c_i+\sum_\ell d_\ell a_{i\ell}$. Define the finite polynomial system

It has the formal solution $Y_{i\ell}=a_{i\ell}$. It therefore suffices to prove that every finite polynomial system with a solution in $\widehat R$ has a solution in $R$: any solution $y$ of the new system gives $b_i=c_i+\sum_\ell d_\ell y_{i\ell}$, with the prescribed congruences.

For that existence assertion, let $A\subseteq\widehat R$ be the $R$-subalgebra generated by the finitely many solution coordinates. Since $R$ is Noetherian it is finitely presented over $R$. Popescu's theorem and Lemma 2.1 give $A\to B\to\widehat R$ with $B$ smooth over $R$. Reducing the last map modulo the maximal ideal gives an $R$-algebra map $B\to k$.

The smooth-section lifting lemma supplies an étale $R$-algebra $R'$, an identification $R'/\mathfrak mR'=k$, and an $R$-algebra map $B\to R'$ reducing to that evaluation. Its complete proof is included in §10; it is the finite-projective-conormal and slicing argument, so makes no assumption that $B$ already has a single simple-root presentation. Henselianity, using the proved section criterion AG-CA, Theorem 1.2, gives an $R$-algebra section $R'\to R$. The coordinates' images under $A\to B\to R'\to R$ solve the polynomial system. Apply this assertion to the $g_j$ to obtain the required $b$. ∎

For order $N=0$, a solution given at order one also suffices. The assertion holds for every positive order; no uniform finite upper bound on the requested order is imposed.

B.5. The pointed étale-neighborhood version

Theorem 5.1. Let $(R,\mathfrak m,k)$ be a Noetherian local G-ring, without a henselianity assumption. For the same $f_j$ and formal solution $a$, every $N\geq1$ admits an étale $R$-algebra $R'$, a maximal ideal $\mathfrak m'\subset R'$ over $\mathfrak m$ with specified residue isomorphism $\kappa(\mathfrak m')=k$, and a tuple $b\in(R')^n$ solving $f_j=0$. Under the corresponding local map $R'_{\mathfrak m'}\to\widehat R$, one has $a_i-b_i\in\mathfrak m^N\widehat R$. Equivalently, with the canonical pointed identification $\widehat{R'_{\mathfrak m'}}=\widehat R$, the differences lie in $(\mathfrak m')^N\widehat{R'_{\mathfrak m'}}$.

Proof. Make the same substitution $X_i=c_i+\sum_\ell d_\ell Y_{i\ell}$ as in Theorem 4.1. Factor the finitely presented algebra generated by the new formal coordinates through a smooth $B$. Lift the residue evaluation $B\to k$ to $B\to R'$, where $R\to R'$ is étale and $R'/\mathfrak mR'=k$. Define $b_i=c_i+\sum_\ell d_\ell y_{i\ell}$, where $y_{i\ell}$ are the images of the new coordinates in $R'$. Every $g_j(y)$ vanishes, hence $f_j(b)=0$.

Set $\mathfrak m'=\mathfrak mR'$. The quotient is $k$, so this is maximal with the asserted residue isomorphism. Since $\widehat R$ is complete local it is henselian. Its proved section criterion, applied to $R'\otimes_R\widehat R$ with the selected residue evaluation, gives the local $R$-algebra map $R'_{\mathfrak m'}\to\widehat R$. Each difference is a sum of the $d_\ell\in\mathfrak m^N$ times elements of $\widehat R$, proving the congruence.

For completeness, formal étaleness gives a unique compatible lift of the residue evaluation $R'\to k$ to $R'\to R/\mathfrak m^q$ for each $q$. The selected local branch is thereby identified modulo $\mathfrak m^q$ with $R/\mathfrak m^q$. One way to verify the identification is to pass to the Artinian base $R/\mathfrak m^q$: its selected local étale algebra is finite étale, since its field fibre is finite and the maximal ideal of the base is nilpotent. As a finite flat module over the local Artinian base it is free; its rank is the dimension of its residue algebra, namely one. Its unit generates it by Nakayama, so the structural map is an isomorphism. These identifications respect reduction maps by uniqueness. Taking limits gives $\widehat{R'_{\mathfrak m'}}=\widehat R$ and identifies $\mathfrak m'$ with $\mathfrak m$. This also follows from the same-residue-field pointed neighborhoods used in the written henselization construction AG-CA, §§4 and 6. ∎

Corollary 5.2. If $R$ is Noetherian, $\mathfrak p\subset R$ is a prime, and $R_{\mathfrak p}$ is a G-ring, a formal solution in $\widehat{R_{\mathfrak p}}$ can be approximated to any order in an étale $R$-algebra $R'$ at a prime $\mathfrak p'$ with $\kappa(\mathfrak p')=\kappa(\mathfrak p)$.

Proof. Apply Theorem 5.1 to $R_{\mathfrak p}$. The resulting étale algebra has a finite presentation and its étale equations and selected Jacobian inverses contain only finitely many coefficients from $R_{\mathfrak p}$. Clearing their denominators produces an étale algebra over some $R_s$, $s\notin\mathfrak p$, and hence an étale $R$-algebra. Clear also the finitely many denominators in the solution coordinates and the finitely many relations $f_j(b)=0$ by one further principal localization away from the selected prime. The localized ring at that prime is unchanged, so its completion, residue isomorphism and congruences are unchanged. This is the finite-presentation descent used in the native lemma-approximation-property-variant, whose full proof is retained below. ∎

B.6. Approximation of finite algebraic data

B.6.1. A marked family and its associated graded algebra

Theorem 6.1 (family approximation). Let $S$ be locally Noetherian and let $\mathcal X\to(\mathrm{Sch}/S)_{fppf}$ be a category fibred in groupoids that is limit preserving on objects. Let $x\in\mathcal X(R)$, where $(R,\mathfrak m_R)$ is complete Noetherian local and its residue field $k$ is of finite type over $S$. Write $s$ for the image of its closed point, and assume $\mathcal O_{S,s}$ is a G-ring. For each $N\geq1$ there are a finite-type $S$-algebra $A$, a maximal ideal $\mathfrak m_A$, an object $x_A\in\mathcal X(A)$, and

as $S$-algebras, together with a specified isomorphism of the restricted objects and an isomorphism

of graded $k$-algebras. Neither a perfect residue field nor characteristic zero is assumed.

Proof. Choose an affine open $\operatorname{Spec}\Lambda\subset S$ containing $s$. The map from the local scheme $\operatorname{Spec}R$ factors through it: its inverse image is an open containing the closed point of a local spectrum, hence the entire spectrum. The finite-type hypothesis on the residue field says that $k$ is a finitely generated $\Lambda$-algebra on this chart. Since $S$ is locally Noetherian, $\Lambda$ is Noetherian.

Write $R$ as the filtered colimit of its finite-type $\Lambda$-subalgebras. Limit preservation on objects gives a finite-type algebra $C$, a map $C\to R$, and an object $x_C$, with a chosen isomorphism $x_C|_R\simeq x$. No descent assertion about all arrows of $\mathcal X$ is being assumed. Choose a finite presentation

and write $\bar a_i\in R$ for the images of its generators.

Choose finitely many generators of $k$ as a $\Lambda$-algebra and lift them to $R$. Add lifts of a finite generating set of $\mathfrak m_R$. These choices define a polynomial ring $T=\Lambda[z_1,\ldots,z_e]$ and a map $T\to R$ whose reduction to $k$ is surjective and whose image contains generators of $\mathfrak m_R$. Thus $\mathfrak n=\ker(T\to k)$ is maximal. Elements outside $\mathfrak n$ map to units of $R$, so the map extends to the local ring $Q=T_{\mathfrak n}$ and then continuously to

Here is a direct proof of its surjectivity. The chosen elements generate $\mathfrak m_R^j/\mathfrak m_R^{j+1}$ by their degree-$j$ monomials with coefficients in $k$. Those coefficients lift through $Q\to k$. Starting with any element of $R$, lift its residue and then its successive errors by such monomials. At the $j$th correction the added element lies in $\mathfrak m_Q^j$. The partial lifts are Cauchy in $Q$, hence define an element of $P$, and their images converge to the prescribed element of the complete ring $R$. This proves surjectivity; it also proves the usual surjective-cotangent-space criterion in this instance.

The complete ring $P$ is Noetherian. Choose generators $b_1,\ldots,b_r$ of the kernel, and lifts $a_i\in P$ of the $\bar a_i$. Choose coefficients $c_{ji}$ such that

The syzygy module of $(b_1,\ldots,b_r)$ is finite. Choose generators $k_\ell=(k_{\ell1},\ldots,k_{\ell r})$, giving an exact presentation

The map $K$ sends the $\ell$th basis vector to $k_\ell$; in particular $\sum_i k_{\ell i}b_i=0$.

Let $c$ be one Artin–Rees constant for both displayed maps relative to $\mathfrak m_P$, and choose $M\geq\max(N,c+1)$. The common constant exists by Artin–Rees for finite modules, and increasing a constant preserves its defining inclusions. The finite-complex perturbation and canonical graded-quotient statements are the written Lesson 7, Appendix A, Lemmas A.1–A.2. We use their full statements here and do not reproduce their proof.

Put $\mathfrak p=\mathfrak n\cap\Lambda$. The hypothesis says that $\Lambda_{\mathfrak p}$ is a G-ring. The permanence theorem proved below makes its essentially finite-type algebra $Q$ a G-ring. Apply the pointed étale-neighbourhood approximation theorem of Section 5 simultaneously to the following finite polynomial system:

The tuple $(a_i,b_i,c_{ji},k_{\ell i})$ is an exact formal solution in $P$. We obtain an étale $Q$-algebra $B$, a maximal ideal $\mathfrak q$ above $\mathfrak n$ with residue field exactly $k$, and a solution $(a'_i,b'_i,c'_{ji},k'_{\ell i})$ whose entries differ from the old ones by $\mathfrak m_P^M$. The induced identification $\widehat{B_{\mathfrak q}}=P$ preserves the map from $Q$, its residue field and every finite-order quotient. There is no need to inject the entire possibly disconnected algebra $B$ into $P$.

Because $b_i$ lies in $\mathfrak m_P$ and $M\geq1$, each $b'_i$ lies in $\mathfrak q$. Set

The exact polynomial relations define $C\to A^*$ by $y_i\mapsto a'_i$, so they give $x^*=x_C|_{A^*}$. Completion of a quotient of a Noetherian local ring gives

This uses exactness of Noetherian completion for finite modules, supplied by the written completion prerequisite. As $b'_i-b_i\in\mathfrak m_P^M$, the ideals generated by the two lists become identical after adding $\mathfrak m_P^M$. Consequently

Localization at the selected maximal ideal does not change these quotients: every element outside that ideal is already a unit modulo any of its powers. The two maps from $C$ to the displayed common quotient agree, because $a'_i-a_i\in\mathfrak m_P^M$. Restricting the chosen $x_C|_R\simeq x$ therefore gives the required marked comparison of objects at order $M$, hence at order $N$.

The matrix $K'$ and row $(b'_i)$ still form a complex, since all the syzygy equations were preserved exactly. Their entries differ from those of the original exact presentation by $\mathfrak m_P^M\subset\mathfrak m_P^{c+1}$. Lesson 7, Lemma A.2 now identifies the initial ideals inside $\operatorname{gr}_{\mathfrak m_P}P$. Its canonical identity on that graded algebra yields

This is an isomorphism of graded algebras. The quotient filtration is the maximal-ideal filtration, so completion identifies its right side with $\operatorname{gr}_{\mathfrak m^*}A^*$. This explains why the entire relation complex was approximated: approximating only the ideal generators would not provide the reverse initial-ideal containment.

It remains to obtain a finite-type algebra over the original affine chart. The algebra $A^*$ is of finite presentation over $Q=T_{\mathfrak n}$. Its finitely many generators, relations and denominators descend to a finite-type $T$-algebra $A_0$ with

Products of denominators direct this system. Limit preservation on objects descends $x^*$ to an object $x_A$ at some stage $A=(A_0)_f$, with a specified isomorphism after base change to $A^*$. This algebra is finite type over $\Lambda$ and hence over $S$. Let $\mathfrak m_A$ be the inverse image of $\mathfrak m^*$. The image of $A\to k$ contains the image of $T$, which already surjects onto $k$ by the original residue-field generators. Thus $A/\mathfrak m_A=k$: the selected prime is maximal and has exactly the prescribed residue field.

Since $A^*$ is a localization of $A$ away from elements outside the selected point,

Localization at a maximal ideal preserves every quotient by its powers and therefore its associated graded algebra. All the ring comparisons constructed above consequently descend to $A$. Restricting the specified isomorphism $x_A|_{A^*}\simeq x^*$ to their common finite-order quotient supplies the required marking on objects; no finite-stage descent of an arbitrary arrow was needed. This proves all assertions. $\square$

The construction follows the Stacks Project proof of family approximation, Tag 07XB. The proof above spells out the complete-local presentation and the final finite-stage point and object comparisons. Its Popescu, G-ring and étale-completion inputs are proved in this reader or bound to the exact written commutative-algebra providers. The separate finite-complex comparison is bound to the actual written Lesson 7 appendix.

B.7. Jacobian, parameter and cotangent calculations

The following additions are made in the incorporated proof, at the stated native labels. Each concerns an actual inference used by the theorem.

1. lemma-final-solve: replace the omitted verification by the covering $D(f_i)$ and elimination of $Z_i$ given in §3 above.
2. lemma-product: if $\Lambda_i=\varinjlim B_{i,\alpha}$ with $B_{i,\alpha}$ smooth over $R_i$, use the product of the two filtered index categories. Then $\Lambda_1\times\Lambda_2=\varinjlim(B_{1,\alpha}\times B_{2,\beta})$; each product algebra is smooth over $R_1\times R_2$, since the source and target split into the two open-and-closed idempotent components and smoothness holds on both. All maps are product algebra maps. This proves the product assertion rather than assuming preservation of filteredness.
3. lemma-desingularize-lifting-apply: let $\eta\in\operatorname{Spec}\Lambda$ lie over a point where $\bar C$ is smooth. If $\pi\notin\eta$, the lifting lemma makes $D$ smooth over $R$ there. If $\pi\in\eta$, use the compatible factorization $\bar C/\pi^4\to D/\pi^4\to\Lambda/\pi^4$; the same lifting lemma again makes $D$ smooth at the relevant inverse-image prime. Thus the image of $H_{D/R}$ is not contained in $\eta$. The inclusion $H_{D/R}B\subset H_{B/R}$ implies the same for $H_{B/R}$. On $\operatorname{Spec}(\Lambda/\pi^8)$, the open defined by $H_{\bar C/(R/\pi^8)}$ is therefore contained in the open defined by $H_{B/R}$. Reversing closed complements is precisely the claimed radical ideal inclusion. Away from $\pi$, the construction is smooth everywhere. This supplies the native omitted verification.
4. lemma-ogoma: if $(s^n\pi)^2m=0$, then $\pi^2m$ vanishes after localization, so $m\in K'$. Since $s(K'/K)=0$, one has $s\pi m=0$, and hence $s^n\pi m=0$. The converse is immediate. This proves the assertion for every $n>0$.
5. lemma-enlarge-solution-modulo: choose a power $q=p^r$ at least the nilpotence order. For $u\in\mathfrak m_{\bar\Lambda}$, $(1+u)^{p^r}=1+u^{p^r}=1$. Merely saying that $q$ is divisible by $p$ does not prove the displayed identity. If $\alpha$ is algebraic over $F$, write its irreducible polynomial as $h(T^{p^r})$ with $h'\ne0$; then $\alpha^{p^r}$ is separable over $F$. This proves the field step formerly referenced to an external Fields section. Enlarge $r$ if needed and use the finite-étale lifting of the separable residue extension.
6. lemma-resolve-general: set $e=8c$, and choose

The native choice $n=N+dc$ alone does not imply its later inclusion $\mathfrak p^n\subseteq(\pi_1^e,\ldots,\pi_d^e)$. A regular system of parameters generates $\mathfrak p$ after localization, and every monomial of degree $d(e-1)+1$ has an exponent at least $e$. This proves the inclusion and, because $\mathfrak p\Lambda_{\mathfrak q}=\mathfrak q\Lambda_{\mathfrak q}$, the corresponding target inclusion. Increasing $n$ changes none of the earlier constructions: choose the Artinian approximation at this larger order, keep the same $N,c,e$, and use $\mathfrak q^n\subseteq\mathfrak q^{n-N}H_{A/k}\Lambda_{\mathfrak q}$. The source proof's fifth/sixth-step cross-references are also corrected to express the actual direction: solve modulo $\mathfrak p^n$, then quotient the solution by $J$ to solve modulo $J$. 7. In the induction choosing $\delta_i$, write $v_j=\pi_{t+1}^Nc_j+\pi_{t+1}^Nc'_j\in\Lambda_{\mathfrak q}$. Choose a common $s_0\notin\mathfrak q$ for their finitely many denominators, so $v_j=a_j'/s_0$. The equality $\pi_{t+1}^{2N}=\sum_j a_jv_j$ holds after localization; multiply by one further $u\notin\mathfrak q$ killing the error in $\Lambda$. Choose $s$ divisible by $us_0$, put $\mu_j=(s^{2N}/s_0)a'_j$, and obtain $(s\pi_{t+1})^{2N}=\sum_j a_j\mu_j$ in $\Lambda$. In $\bar\Lambda$, the coefficient $\mu_j$ has image $s^{2N}\pi_{t+1}^Nd_j$, which lies in $D'$ after adjoining the permitted power of $s$. The native text misses the factor $\pi_{t+1}^N$; membership, rather than equality with $s^{2N}d_j$, is what is needed. Multiplying $s$ by a further power preserves the equality after scaling all coefficients, so the enlargement lemma can enforce membership of $s$ itself. 8. At the passage from $R/JR$ to $(R/I)_{\mathfrak r}$, every $t_i$ becomes a unit, because its image $\delta_i\notin\mathfrak q$. Hence $I_{\mathfrak r}=JR_{\mathfrak r}$ and $I\Lambda_{\mathfrak q}=J\Lambda_{\mathfrak q}$. Localizing the solution over $(R/JR)_{\mathfrak p}$ at the remaining elements of $R\setminus\mathfrak r$ therefore gives a solution over $(R/I)_{\mathfrak r}$. The source and target maps are the specified polynomial-algebra maps; their localizations type-check. 9. In the final factorization, $\delta_i\in D'$ has nonzero residue because $D'\to\bar\Lambda$ is a flat local map, hence faithfully flat. It is a unit in the Artinian local ring $D'$. The formula $z_{ij}\mapsto\lambda_{ij}t_i^{2N}/\delta_i^{2N}$ is therefore defined. Its relations follow from $(\pi_i\delta_i)^{2N}=\sum_j a_j\lambda_{ij}$, and evaluation $t_i\mapsto\delta_i$ gives the original coefficients. This establishes both commutative triangles required by the last step. 10. If $d=0$, omit the empty parameter-lifting step (whose stated lemma assumes $r\geq1$). In the separable proof, the localized target is a field with separable extension over $k$, hence a filtered colimit of smooth algebras by the retained field lemma; the height-zero delocalization lemma then gives the resolution. In the inseparable proof, take $c=1$, use the Artinian approximation with no $\pi_i,t_i$, and retain the same localization/height-zero argument. The solution-modulo lemma supplies a factorization through an essentially smooth algebra at order $n\geq1$; its finitely presented source factors through a smooth stage. No positive-dimensional regular-parameter argument is invoked in this case.

11. In more-algebra-lemma-symmetric-algebra-smooth, the two-term presentation complex is $K\to A^{\oplus n}$, tensored with $C$, rather than $K\to M$. It splits because $M$ is projective; its cokernel is $M\otimes_A C$, exactly as the displayed conormal computation says. The derivative corrects this typed-complex typo and the adjacent $R/I$ typo in the projective-lifting lemma.
12. The omitted verification in composition of regular maps is supplied in the retained derivative: localize and quotient at a source prime, then make a finite purely inseparable extension of its residue field. Both resulting intermediate and target rings are Noetherian by the regular-map fibre hypotheses, and the intermediate ring is regular. The base-changed second map has regular fibres. A flat map of Noetherian rings with regular source and regular fibres has regular target, by its exact local regularity predecessor. This verifies every geometric fibre of the composite; composition of flat maps verifies its flatness.

The exponent correction repairs this proof; it is not a counterexample to Popescu's theorem. The native source bodies remain unchanged, and these changes are confined to the incorporated derivative.

B.8. Commutative-algebra prerequisites

We use the following written commutative-algebra proofs at their matched scope. The lesson title and source file identify the provider independently of the course's current numbering. Each provider retains its own prerequisites; the exact file bytes and theorem correspondence are recorded in the conversion manifest.

The following proofs supply the conormal and Koszul transitivity arguments, finite projective and smooth-section lifting, geometric regularity, the finite-over-regular complete-local construction, and G-ring permanence. In particular the first-homology quotient argument is included at its own scope; the AG-CA cotangent-presentation comparison is not substituted for it. Formal smoothness of arbitrary separable field extensions is proved in the included algebraic chain, while the coefficient-ring lesson is used only for the prime-field assertion it actually proves.

Section 6.1 proves the full family-approximation construction used in Lesson 7, Lemma 2.3. Its finite-complex Artin–Rees comparison (Tags 07VE/07VF) is the actual written Lesson 7, Appendix A, Lemmas A.1–A.2; its presentation, completion, point and finite-stage object comparisons are supplied here. The concise prerequisite record below identifies the still unbound lower foundations.

B.9. Desingularization from singularity ideals

We now give the full proof chain from the Stacks source smoothing.tex: singular ideals and improved presentations; flat nilpotent lifting; lifting and desingularization; reduction to fields; localization; separable and inseparable residue fields; Popescu's theorem; both approximation theorems and the prime-localized variant. The independent Néron-DVR interlude is not needed for this chain and has not been imported into this assigned support chapter. Its omission does not narrow the regular Noetherian map theorem. All corrections are identified in §7 and applied in the incorporated derivative.

Singular ideals

Let $R \to A$ be a ring map. The singular ideal of $A$ over $R$ is the radical ideal in $A$ cutting out the singular locus of the morphism $\operatorname{Spec}(A) \to \operatorname{Spec}(R)$. Here is a formal definition.

Definition. The singularity ideal

Let $R \to A$ be a ring map. The singular ideal of $A$ over $R$, denoted $H_{A/R}$ is the unique radical ideal $H_{A/R} \subset A$ with $$V(H_{A/R}) = \{\mathfrak q \in \operatorname{Spec}(A) \mid R \to A \text{ not smooth at }\mathfrak q\}$$

This makes sense because the set of primes where $R \to A$ is smooth is open, see Algebra, Definition Smoothness at a prime ideal. In order to find an explicit set of generators for the singular ideal we first prove the following lemma.

Lemma. A Jacobian presentation near a smooth point

Let $R$ be a ring. Let $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$. Let $\mathfrak q \subset A$ be a prime ideal. Assume $R \to A$ is smooth at $\mathfrak q$. Then there exists an $a \in A$, $a \not \in \mathfrak q$, an integer $c$, $0 \leq c \leq \min(n, m)$, subsets $U \subset \{1, \ldots, n\}$, $V \subset \{1, \ldots, m\}$ of cardinality $c$ such that $$a = a' \det(\partial f_j/\partial x_i)_{j \in V, i \in U}$$ for some $a' \in A$ and $$a f_\ell \in (f_j, j \in V) + (f_1, \ldots, f_m)^2$$ for all $\ell \in \{1, \ldots, m\}$.

Proof. Set $I = (f_1, \ldots, f_m)$ so that the naive cotangent complex of $A$ over $R$ is homotopy equivalent to $I/I^2 \to \bigoplus A\text{d}x_i$, see Algebra, Lemma Kähler differentials, Theorems 3.1–3.3, Proposition 3.4 and Theorem 7.1. We will use the formation of the naive cotangent complex commutes with localization, see Algebra, Section The naive cotangent complex, especially Algebra, Lemma Localization of the naive cotangent complex. By Algebra, Definitions Smooth ring maps and Smoothness at a prime ideal we see that $(I/I^2)_a \to \bigoplus A_a\text{d}x_i$ is a split injection for some $a \in A$, $a \not \in \mathfrak q$. After renumbering $x_1, \ldots, x_n$ and $f_1, \ldots, f_m$ we may assume that $f_1, \ldots, f_c$ form a basis for the vector space $I/I^2 \otimes_A \kappa(\mathfrak q)$ and that $\text{d}x_{c + 1}, \ldots, \text{d}x_n$ map to a basis of $\Omega_{A/R} \otimes_A \kappa(\mathfrak q)$. Hence after replacing $a$ by $aa'$ for some $a' \in A$, $a' \not \in \mathfrak q$ we may assume $f_1, \ldots, f_c$ form a basis for $(I/I^2)_a$ and that $\text{d}x_{c + 1}, \ldots, \text{d}x_n$ map to a basis of $(\Omega_{A/R})_a$. In this situation $a^N$ for some large integer $N$ satisfies the conditions of the lemma (with $U = V = \{1, \ldots, c\}$). $\square$

We will use the notion of a strictly standard element in $A$ over $R$. Our notion is slightly weaker than the one in Swan's paper the original source citation swan. We also define an elementary standard element to be one of the type we found in the lemma above. We compare the different types of elements in Lemma Comparing standard smooth presentations.

Definition. Strict standard smoothness

Let $R \to A$ be a ring map of finite presentation. We say an element $a \in A$ is elementary standard in $A$ over $R$ if there exists a presentation $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ and $0 \leq c \leq \min(n, m)$ such that

$$a = a' \det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$$ for some $a' \in A$ and

$$a f_{c + j} \in (f_1, \ldots, f_c) + (f_1, \ldots, f_m)^2$$ for $j = 1, \ldots, m - c$. We say $a \in A$ is strictly standard in $A$ over $R$ if there exists a presentation $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ and $0 \leq c \leq \min(n, m)$ such that

$$a = \sum\nolimits_{I \subset \{1, \ldots, n\},\ |I| = c} a_I \det(\partial f_j/\partial x_i)_{j = 1, \ldots, c,\ i \in I}$$ for some $a_I \in A$ and

$$a f_{c + j} \in (f_1, \ldots, f_c) + (f_1, \ldots, f_m)^2$$ for $j = 1, \ldots, m - c$.

The following lemma is useful to find implications of (the displayed identity).

Lemma. The Jacobian relation in a strict presentation

Let $R$ be a ring. Let $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ and write $I = (f_1, \ldots, f_m)$. Let $a \in A$. Then (the displayed identity) implies there exists an $A$-linear map $\psi : \bigoplus\nolimits_{i = 1, \ldots, n} A \text{d}x_i \to A^{\oplus c}$ such that the composition $$A^{\oplus c} \xrightarrow{(f_1, \ldots, f_c)} I/I^2 \xrightarrow{f \mapsto \text{d}f} \bigoplus\nolimits_{i = 1, \ldots, n} A \text{d}x_i \xrightarrow{\psi} A^{\oplus c}$$ is multiplication by $a$. Conversely, if such a $\psi$ exists, then $a^c$ satisfies (the displayed identity).

Proof. This is a special case of Algebra, Lemma A left inverse for a matrix. $\square$

Lemma. Controlling a power of the Jacobian determinant (Elkik)

Let $R \to A$ be a ring map of finite presentation. The singular ideal $H_{A/R}$ is the radical of the ideal generated by strictly standard elements in $A$ over $R$ and also the radical of the ideal generated by elementary standard elements in $A$ over $R$.

Proof. Assume $a$ is strictly standard in $A$ over $R$. We claim that $A_a$ is smooth over $R$, which proves that $a \in H_{A/R}$. Namely, let $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ and $c$ be as in Definition Strict standard smoothness. Write $I = (f_1, \ldots, f_m)$ so that the naive cotangent complex of $A$ over $R$ is given by $I/I^2 \to \bigoplus A\text{d}x_i$. Assumption (the displayed identity) implies that $(I/I^2)_a$ is generated by the classes of $f_1, \ldots, f_c$. Assumption (the displayed identity) implies that the differential $(I/I^2)_a \to \bigoplus A_a\text{d}x_i$ has a left inverse, see Lemma The Jacobian relation in a strict presentation. Hence $R \to A_a$ is smooth by definition and Algebra, Lemma Localization of the naive cotangent complex.

Let $H_e, H_s \subset A$ be the radical of the ideal generated by elementary, resp. strictly standard elements of $A$ over $R$. By definition and what we just proved we have $H_e \subset H_s \subset H_{A/R}$. The inclusion $H_{A/R} \subset H_e$ follows from Lemma A Jacobian presentation near a smooth point. $\square$

Example. A singular locus that is not quasi-compact

The set of points where a finitely presented ring map is smooth needn't be a quasi-compact open. For example, let $R = k[x, y_1, y_2, y_3, \ldots]/(xy_i)$ and $A = R/(x)$. Then the smooth locus of $R \to A$ is $\bigcup D(y_i)$ which is not quasi-compact.

Lemma. Base change of a strict standard presentation

Let $R \to A$ be a ring map of finite presentation. Let $R \to R'$ be a ring map. If $a \in A$ is elementary, resp. strictly standard in $A$ over $R$, then $a \otimes 1$ is elementary, resp. strictly standard in $A \otimes_R R'$ over $R'$.

Proof. If $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ is a presentation of $A$ over $R$, then $A \otimes_R R' = R'[x_1, \ldots, x_n]/(f'_1, \ldots, f'_m)$ is a presentation of $A \otimes_R R'$ over $R'$. Here $f'_j$ is the image of $f_j$ in $R'[x_1, \ldots, x_n]$. Hence the result follows from the definitions. $\square$

Lemma. Solving the strict-standard Jacobian relations

Let $R \to A \to \Lambda$ be ring maps with $A$ of finite presentation over $R$. Assume that $H_{A/R} \Lambda = \Lambda$. Then there exists a factorization $A \to B \to \Lambda$ with $B$ smooth over $R$.

Proof. Choose $f_1, \ldots, f_r \in H_{A/R}$ and $\lambda_1, \ldots, \lambda_r \in \Lambda$ such that $\sum f_i\lambda_i = 1$ in $\Lambda$. Set $B = A[x_1, \ldots, x_r]/(f_1x_1 + \ldots + f_rx_r - 1)$ and define $B \to \Lambda$ by mapping $x_i$ to $\lambda_i$. To check that $B$ is smooth over $R$ use that $A_{f_i}$ is smooth over $R$ by definition of $H_{A/R}$ and that $B_{f_i}$ is smooth over $A_{f_i}$. Indeed, $\sum f_ix_i=1$ makes the $D(f_i)$ a cover of $\operatorname{Spec}(B)$. On $D(f_i)$ one eliminates $x_i$, giving a polynomial algebra over $A_{f_i}$. Smoothness on this cover proves the claim. $\square$

Presentations of algebras

Some of the results in this section are due to Elkik. Note that the algebra $C$ in the following lemma is a symmetric algebra over $A$. Moreover, if $R$ is Noetherian, then $C$ is of finite presentation over $R$.

Lemma. Improving a finite presentation

Let $R$ be a ring and let $A$ be a finitely presented $R$-algebra. There exists a finite type $R$-algebra map $A \to C$ which has a retraction with the following two properties

1. for each $a \in A$ such that $R \to A_a$ is a local complete intersection (More on Algebra, Definition Local complete-intersection ring maps) the ring $C_a$ is smooth over $A_a$ and has a presentation $C_a = R[y_1, \ldots, y_m]/J$ such that $J/J^2$ is free over $C_a$, and

2. for each $a \in A$ such that $A_a$ is smooth over $R$ the module $\Omega_{C_a/R}$ is free over $C_a$.

Proof. Choose a presentation $A = R[x_1, \ldots, x_n]/I$ and write $I = (f_1, \ldots, f_m)$. Define the $A$-module $K$ by the short exact sequence $$0 \to K \to A^{\oplus m} \to I/I^2 \to 0$$ where the $j$th basis vector $e_j$ in the middle is mapped to the class of $f_j$ on the right. Set $$C = \text{Sym}^*_A(I/I^2).$$ The retraction is just the projection onto the degree $0$ part of $C$. We have a surjection $R[x_1, \ldots, x_n, y_1, \ldots, y_m] \to C$ which maps $y_j$ to the class of $f_j$ in $I/I^2$. The kernel $J$ of this map is generated by the elements $f_1, \ldots, f_m$ and by elements $\sum h_j y_j$ with $h_j \in R[x_1, \ldots, x_n]$ such that $\sum h_j e_j$ defines an element of $K$. By Algebra, Lemma The transitivity sequence for the naive cotangent complex applied to $R \to A \to C$ and the presentations above and More on Algebra, Lemma The cotangent complex of a symmetric algebra there is an exact sequence

$$I/I^2 \otimes_A C \to J/J^2 \to K \otimes_A C \to 0$$ of $C$-modules. Let $h \in R[x_1, \ldots, x_n]$ be an element with image $a \in A$. We will use as presentations for the localized rings $$A_a = R[x_0, x_1, \ldots, x_n]/I' \quad\text{and}\quad C_a = R[x_0, x_1, \ldots, x_n, y_1, \ldots, y_m]/J'$$ where $I' = (hx_0 - 1, I)$ and $J' = (hx_0 - 1, J)$. Hence $I'/(I')^2 = A_a \oplus (I/I^2)_a$ as $A_a$-modules and $J'/(J')^2 = C_a \oplus (J/J^2)_a$ as $C_a$-modules. Thus we obtain

$$C_a \oplus I/I^2 \otimes_A C_a \to C_a \oplus (J/J^2)_a \to K \otimes_A C_a \to 0$$ as the sequence of Algebra, Lemma The transitivity sequence for the naive cotangent complex corresponding to $R \to A_a \to C_a$ and the presentations above.

Next, assume that $a \in A$ is such that $A_a$ is a local complete intersection over $R$. Then $(I/I^2)_a$ is finite projective over $A_a$, see More on Algebra, Lemma Finite projectivity of a quasi-regular conormal module. Hence we see $K_a \oplus (I/I^2)_a \cong A_a^{\oplus m}$ is free. In particular $K_a$ is finite projective too. By More on Algebra, Lemma Cotangent transitivity with a complete-intersection terminal map the sequence (the displayed identity) is exact on the left. Hence $$J'/(J')^2 \cong C_a \oplus I/I^2 \otimes_A C_a \oplus K \otimes_A C_a \cong C_a^{\oplus m + 1}$$ This proves (1). Finally, suppose that in addition $A_a$ is smooth over $R$. Then the same presentation shows that $\Omega_{C_a/R}$ is the cokernel of the map $$J'/(J')^2 \longrightarrow \bigoplus\nolimits_i C_a\text{d}x_i \oplus \bigoplus\nolimits_j C_a\text{d}y_j$$ The summand $C_a$ of $J'/(J')^2$ in the decomposition above corresponds to $hx_0 - 1$ and hence maps isomorphically to the summand $C_a\text{d}x_0$. The summand $I/I^2 \otimes_A C_a$ of $J'/(J')^2$ maps injectively to $\bigoplus_{i = 1, \ldots, n} C_a\text{d}x_i$ with quotient $\Omega_{A_a/R} \otimes_{A_a} C_a$. The summand $K \otimes_A C_a$ maps injectively to $\bigoplus_{j \geq 1} C_a\text{d}y_j$ with quotient isomorphic to $I/I^2 \otimes_A C_a$. Thus the cokernel of the last displayed map is the module $I/I^2 \otimes_A C_a \oplus \Omega_{A_a/R} \otimes_{A_a} C_a$. Since $(I/I^2)_a \oplus \Omega_{A_a/R}$ is free (from the definition of smooth ring maps) we see that (2) holds. $\square$

The following proposition was proved for smooth ring maps over henselian pairs by Elkik in the original source citation Elkik. For smooth ring maps it can be found in the original source citation Arabia, where it is also proven that ring maps between smooth algebras can be lifted.

Proposition. Lifting a smooth algebra over a quotient

Smooth and syntomic algebras lift along surjections

Let $R \to R_0$ be a surjective ring map with kernel $I$.

1. If $R_0 \to A_0$ is a syntomic ring map, then there exists a syntomic ring map $R \to A$ such that $A/IA \cong A_0$.

2. If $R_0 \to A_0$ is a smooth ring map, then there exists a smooth ring map $R \to A$ such that $A/IA \cong A_0$.

Proof. Assume $R_0 \to A_0$ syntomic, in particular a local complete intersection (More on Algebra, Lemma Syntomic algebras and local complete intersections). Choose a presentation $A_0 = R_0[x_1, \ldots, x_n]/J_0$. Set $C_0 = \text{Sym}^*_{A_0}(J_0/J_0^2)$. Note that $J_0/J_0^2$ is a finite projective $A_0$-module (Algebra, Lemma The conormal module of a syntomic presentation). By Lemma Improving a finite presentation the ring map $A_0 \to C_0$ is smooth and we can find a presentation $C_0 = R_0[y_1, \ldots, y_m]/K_0$ with $K_0/K_0^2$ free over $C_0$. By Algebra, Lemma A presentation realizing a basis of the conormal module we can assume $C_0 = R_0[y_1, \ldots, y_m]/(\overline{f}_1, \ldots, \overline{f}_c)$ where $\overline{f}_1, \ldots, \overline{f}_c$ maps to a basis of $K_0/K_0^2$ over $C_0$. Choose $f_1, \ldots, f_c \in R[y_1, \ldots, y_m]$ lifting $\overline{f}_1, \ldots, \overline{f}_c$ and set $$C = R[y_1, \ldots, y_m]/(f_1, \ldots, f_c)$$ By construction $C_0 = C/IC$. By Algebra, Lemma Localization of a relative complete intersection we can after replacing $C$ by $C_g$ assume that $C$ is a relative global complete intersection over $R$. We conclude that there exists a finite projective $A_0$-module $P_0$ such that $C_0 = \text{Sym}^*_{A_0}(P_0)$ is isomorphic to $C/IC$ for some syntomic $R$-algebra $C$.

Choose an integer $n$ and a direct sum decomposition $A_0^{\oplus n} = P_0 \oplus Q_0$. By More on Algebra, Lemma Lifting a finite projective module we can find an étale ring map $C \to C'$ which induces an isomorphism $C/IC \to C'/IC'$ and a finite projective $C'$-module $Q$ such that $Q/IQ$ is isomorphic to $Q_0 \otimes_{A_0} C/IC$. Then $D = \text{Sym}_{C'}^*(Q)$ is a smooth $C'$-algebra (see More on Algebra, Lemma Smoothness of a symmetric algebra). Picture $$\begin{gathered}\begin{matrix}R & \phantom{X} & C & C' & D \\ R/I & A_0 & C/IC & C'/IC' & D/ID\end{matrix} \\[6pt] \begin{aligned}R & \longrightarrow R/I \\ R & \longrightarrow C \\ C & \longrightarrow C' \\ C & \longrightarrow C/IC \\ C' & \longrightarrow D \\ C' & \longrightarrow C'/IC' \\ D & \longrightarrow D/ID \\ R/I & \longrightarrow A_0 \\ A_0 & \longrightarrow C/IC \\ C/IC & \xrightarrow{\cong} C'/IC' \\ C'/IC' & \longrightarrow D/ID\end{aligned}\end{gathered}$$ Observe that our choice of $Q$ gives $$\begin{aligned} D/ID & = \text{Sym}_{C/IC}^*(Q_0 \otimes_{A_0} C/IC) \\ & = \text{Sym}_{A_0}^*(Q_0) \otimes_{A_0} C/IC \\ & = \text{Sym}_{A_0}^*(Q_0) \otimes_{A_0} \text{Sym}_{A_0}^*(P_0) \\ & = \text{Sym}_{A_0}^*(Q_0 \oplus P_0) \\ & = \text{Sym}_{A_0}^*(A_0^{\oplus n}) \\ & = A_0[x_1, \ldots, x_n] \end{aligned}$$ Choose $f_1, \ldots, f_n \in D$ which map to $x_1, \ldots, x_n$ in $D/ID = A_0[x_1, \ldots, x_n]$. Set $A = D/(f_1, \ldots, f_n)$. Note that $A_0 = A/IA$. We claim that $R \to A$ is syntomic in a neighbourhood of $V(IA)$. If the claim is true, then we can find a $f \in A$ mapping to $1 \in A_0$ such that $A_f$ is syntomic over $R$ and the proof of (1) is finished.

Proof of the claim. Observe that $R \to D$ is syntomic as a composition of the syntomic ring map $R \to C$, the étale ring map $C \to C'$ and the smooth ring map $C' \to D$ (Algebra, Lemmas Composition of syntomic ring maps and Smooth algebras are syntomic). The question is local on $\operatorname{Spec}(D)$, hence we may assume that $D$ is a relative global complete intersection (Algebra, Lemma Local criteria for a syntomic algebra). Say $D = R[y_1, \ldots, y_m]/(g_1, \ldots, g_s)$. Let $f'_1, \ldots, f'_n \in R[y_1, \ldots, y_m]$ be lifts of $f_1, \ldots, f_n$. Then we can apply Algebra, Lemma Localization of a relative complete intersection to get the claim.

Proof of (2). Since a smooth ring map is syntomic, we can find a syntomic ring map $R \to A$ such that $A_0 = A/IA$. By assumption the fibres of $R \to A$ are smooth over primes in $V(I)$ hence $R \to A$ is smooth in an open neighbourhood of $V(IA)$ (Algebra, Lemma Smoothness from flatness and smooth fibres). Thus we can replace $A$ by a localization to obtain the result we want. $\square$

We know that any syntomic ring map $R \to A$ is locally a relative global complete intersection, see Algebra, Lemma Local criteria for a syntomic algebra. The next lemma says that a vector bundle over $\operatorname{Spec}(A)$ is a relative global complete intersection.

Lemma. Complete-intersection presentations of syntomic algebras

Let $R \to A$ be a syntomic ring map. Then there exists a smooth $R$-algebra map $A \to C$ with a retraction such that $C$ is a global relative complete intersection over $R$, i.e., $$C \cong R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$$ flat over $R$ and all fibres of dimension $n - c$.

Proof. Apply Lemma Improving a finite presentation to get $A \to C$. By Algebra, Lemma A presentation realizing a basis of the conormal module we can write $C = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ with $f_i$ mapping to a basis of $J/J^2$. The ring map $R \to C$ is syntomic (hence flat) as it is a composition of a syntomic and a smooth ring map. The dimension of the fibres is $n - c$ by Algebra, Lemma Local criteria for complete intersections (the fibres are local complete intersections, so the lemma applies). $\square$

Lemma. Standard presentations of smooth algebras

Let $R \to A$ be a smooth ring map. Then there exists a smooth $R$-algebra map $A \to B$ with a retraction such that $B$ is standard smooth over $R$, i.e., $$B \cong R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$$ and $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$ is invertible in $B$.

Proof. Apply Lemma Complete-intersection presentations of syntomic algebras to get a smooth $R$-algebra map $A \to C$ with a retraction such that $C = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ is a relative global complete intersection over $R$. As $C$ is smooth over $R$ we have a short exact sequence $$0 \to \bigoplus\nolimits_{j = 1, \ldots, c} C f_j \to \bigoplus\nolimits_{i = 1, \ldots, n} C\text{d}x_i \to \Omega_{C/R} \to 0$$ Since $\Omega_{C/R}$ is a projective $C$-module this sequence is split. Choose a left inverse $t$ to the first map. Say $t(\text{d}x_i) = \sum c_{ij} f_j$ so that $\sum_i \frac{\partial f_j}{\partial x_i} c_{i\ell} = \delta_{j\ell}$ (Kronecker delta). Let $$B' = C[y_1, \ldots, y_c] = R[x_1, \ldots, x_n, y_1, \ldots, y_c]/(f_1, \ldots, f_c)$$ The $R$-algebra map $C \to B'$ has a retraction given by mapping $y_j$ to zero. We claim that the map $$R[z_1, \ldots, z_n] \longrightarrow B',\quad z_i \longmapsto x_i - \sum\nolimits_j c_{ij} y_j$$ is étale at every point in the image of $\operatorname{Spec}(C) \to \operatorname{Spec}(B')$. In $\Omega_{B'/R[z_1, \ldots, z_n]}$ we have $$0 = \text{d}f_j - \sum\nolimits_i \frac{\partial f_j}{\partial x_i} \text{d}z_i \equiv \sum\nolimits_{i, \ell} \frac{\partial f_j}{\partial x_i} c_{i\ell} \text{d}y_\ell \equiv \text{d}y_j \bmod (y_1, \ldots, y_c)\Omega_{B'/R[z_1, \ldots, z_n]}$$ Since $0 = \text{d}z_i = \text{d}x_i$ modulo $\sum B'\text{d}y_j + (y_1, \ldots, y_c)\Omega_{B'/R[z_1, \ldots, z_n]}$ we conclude that $$\Omega_{B'/R[z_1, \ldots, z_n]}/ (y_1, \ldots, y_c)\Omega_{B'/R[z_1, \ldots, z_n]} = 0.$$ As $\Omega_{B'/R[z_1, \ldots, z_n]}$ is a finite $B'$-module by Nakayama's lemma there exists a $g \in 1 + (y_1, \ldots, y_c)$ that $(\Omega_{B'/R[z_1, \ldots, z_n]})_g = 0$. This proves that $R[z_1, \ldots, z_n] \to B'_g$ is unramified, see Algebra, Definition Unramified ring maps. For any ring map $R \to k$ where $k$ is a field we obtain an unramified ring map $k[z_1, \ldots, z_n] \to (B'_g) \otimes_R k$ between smooth $k$-algebras of dimension $n$. It follows that $k[z_1, \ldots, z_n] \to (B'_g) \otimes_R k$ is flat by Algebra, Lemmas Flatness from Cohen–Macaulayness over a regular base and Smoothness after an algebraic closure of the ground field. By the critère de platitude par fibre (Algebra, Lemma The fibrewise criterion for flatness) we conclude that $R[z_1, \ldots, z_n] \to B'_g$ is flat. Finally, Algebra, Lemma Characterizations of étale algebras implies that $R[z_1, \ldots, z_n] \to B'_g$ is étale. Set $B = B'_g$. Note that $C \to B$ is smooth and has a retraction, so also $A \to B$ is smooth and has a retraction. Moreover, $R[z_1, \ldots, z_n] \to B$ is étale. By Algebra, Lemma Étale algebras in standard smooth form we can write $$B = R[z_1, \ldots, z_n, w_1, \ldots, w_m]/(g_1, \ldots, g_m)$$ with $\det(\partial g_j/\partial w_i)$ invertible in $B$. This proves the lemma. $\square$

Lemma. Standard smooth presentations in a filtered colimit

Let $R \to \Lambda$ be a ring map. If $\Lambda$ is a filtered colimit of smooth $R$-algebras, then $\Lambda$ is a filtered colimit of standard smooth $R$-algebras.

Proof. Let $A \to \Lambda$ be an $R$-algebra map with $A$ of finite presentation over $R$. According to Algebra, Lemma Recognizing a filtered colimit of finite presentations we have to factor this map through a standard smooth algebra, and we know we can factor it as $A \to B \to \Lambda$ with $B$ smooth over $R$. Choose an $R$-algebra map $B \to C$ with a retraction $C \to B$ such that $C$ is standard smooth over $R$, see Lemma Standard presentations of smooth algebras. Then the desired factorization is $A \to B \to C \to B \to \Lambda$. $\square$

Lemma. Including prescribed generators in a smooth presentation

Let $R \to A$ be a standard smooth ring map. Let $E \subset A$ be a finite subset of order $|E| = n$. Then there exists a presentation $A = R[x_1, \ldots, x_{n + m}]/(f_1, \ldots, f_c)$ with $c \geq n$, with $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$ invertible in $A$, and such that $E$ is the set of congruence classes of $x_1, \ldots, x_n$.

Proof. Choose a presentation $A = R[y_1, \ldots, y_m]/(g_1, \ldots, g_d)$ such that the image of $\det(\partial g_j/\partial y_i)_{i, j = 1, \ldots, d}$ is invertible in $A$. Choose an enumeration $E = \{a_1, \ldots, a_n\}$ and choose $h_i \in R[y_1, \ldots, y_m]$ whose image in $A$ is $a_i$. Consider the presentation $$A = R[x_1, \ldots, x_n, y_1, \ldots, y_m]/ (x_1 - h_1, \ldots, x_n - h_n, g_1, \ldots, g_d)$$ and set $c = n + d$. $\square$

Lemma. Comparing standard smooth presentations

Let $R \to A$ be a ring map of finite presentation. Let $a \in A$. Consider the following conditions on $a$:

1. $A_a$ is smooth over $R$,

2. $A_a$ is smooth over $R$ and $\Omega_{A_a/R}$ is stably free,

3. $A_a$ is smooth over $R$ and $\Omega_{A_a/R}$ is free,

4. $A_a$ is standard smooth over $R$,

5. $a$ is strictly standard in $A$ over $R$,

6. $a$ is elementary standard in $A$ over $R$.

Then we have

1. $4$ $\Rightarrow$ (3) $\Rightarrow$ (2) $\Rightarrow$ (1),

2. $6$ $\Rightarrow$ (5),

3. $6$ $\Rightarrow$ (4),

4. $5$ $\Rightarrow$ (2),

5. $2$ $\Rightarrow$ the elements $a^e$, $e \geq e_0$ are strictly standard in $A$ over $R$,

6. $4$ $\Rightarrow$ the elements $a^e$, $e \geq e_0$ are elementary standard in $A$ over $R$.

Proof. Part (a) is clear from the definitions and Algebra, Lemma Standard smooth algebras. Part (b) is clear from Definition Strict standard smoothness.

Proof of (c). Choose a presentation $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ such that (the displayed identity) and (the displayed identity) hold. Choose $h \in R[x_1, \ldots, x_n]$ mapping to $a$. Then $$A_a = R[x_0, x_1, \ldots, x_n]/(x_0h - 1, f_1, \ldots, f_m).$$ Write $J = (x_0h - 1, f_1, \ldots, f_m)$. By (the displayed identity) we see that the $A_a$-module $J/J^2$ is generated by $x_0h - 1, f_1, \ldots, f_c$ over $A_a$. Hence, as in the proof of Algebra, Lemma A presentation realizing a basis of the conormal module, we can choose a $g \in 1 + J$ such that $$A_a = R[x_0, \ldots, x_n, x_{n + 1}]/ (x_0h - 1, f_1, \ldots, f_c, gx_{n + 1} - 1).$$ At this point (the displayed identity) implies that $R \to A_a$ is standard smooth (use the coordinates $x_0, x_1, \ldots, x_c, x_{n + 1}$ to take derivatives).

Proof of (d). Choose a presentation $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ such that (the displayed identity) and (the displayed identity) hold. Write $I = (f_1, \ldots, f_m)$. We already know that $A_a$ is smooth over $R$, see Lemma Controlling a power of the Jacobian determinant. By Lemma The Jacobian relation in a strict presentation we see that $(I/I^2)_a$ is free on $f_1, \ldots, f_c$ and maps isomorphically to a direct summand of $\bigoplus A_a \text{d}x_i$. Since $\Omega_{A_a/R} = (\Omega_{A/R})_a$ is the cokernel of the map $(I/I^2)_a \to \bigoplus A_a \text{d}x_i$ we conclude that it is stably free.

Proof of (e). Choose a presentation $A = R[x_1, \ldots, x_n]/I$ with $I$ finitely generated. By assumption we have a short exact sequence $$0 \to (I/I^2)_a \to \bigoplus\nolimits_{i = 1, \ldots, n} A_a\text{d}x_i \to \Omega_{A_a/R} \to 0$$ which is split exact. Hence we see that $(I/I^2)_a \oplus \Omega_{A_a/R}$ is a free $A_a$-module. Since $\Omega_{A_a/R}$ is stably free we see that $(I/I^2)_a$ is stably free as well. Thus replacing the presentation chosen above by $A = R[x_1, \ldots, x_n, x_{n + 1}, \ldots, x_{n + r}]/J$ with $J = (I, x_{n + 1}, \ldots, x_{n + r})$ for some $r$ we get that $(J/J^2)_a$ is (finite) free. Choose $f_1, \ldots, f_c \in J$ which map to a basis of $(J/J^2)_a$. Extend this to a list of generators $f_1, \ldots, f_m \in J$. Consider the presentation $A = R[x_1, \ldots, x_{n + r}]/(f_1, \ldots, f_m)$. Then (the displayed identity) holds for $a^e$ for all sufficiently large $e$ by construction. Moreover, since $(J/J^2)_a \to \bigoplus\nolimits_{i = 1, \ldots, n + r} A_a\text{d}x_i$ is a split injection we can find an $A_a$-linear left inverse. Writing this left inverse in terms of the basis $f_1, \ldots, f_c$ and clearing denominators we find a linear map $\psi_0 : A^{\oplus n + r} \to A^{\oplus c}$ such that $$A^{\oplus c} \xrightarrow{(f_1, \ldots, f_c)} J/J^2 \xrightarrow{f \mapsto \text{d}f} \bigoplus\nolimits_{i = 1, \ldots, n + r} A \text{d}x_i \xrightarrow{\psi_0} A^{\oplus c}$$ is multiplication by $a^{e_0}$ for some $e_0 \geq 1$. By Lemma The Jacobian relation in a strict presentation we see (the displayed identity) holds for all $a^{ce_0}$ and hence for $a^e$ for all $e$ with $e \geq ce_0$.

Proof of (f). Choose a presentation $A_a = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ such that $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$ is invertible in $A_a$. We may assume that for some $m < n$ the classes of the elements $x_1, \ldots, x_m$ correspond to $a_i/1$ where $a_1, \ldots, a_m \in A$ are generators of $A$ over $R$, see Lemma Including prescribed generators in a smooth presentation. After replacing $x_i$ by $a^Nx_i$ for $m < i \leq n$ we may assume the class of $x_i$ is $a_i/1 \in A_a$ for some $a_i \in A$. Consider the ring map $$\Psi : R[x_1, \ldots, x_n] \longrightarrow A,\quad x_i \longmapsto a_i.$$ This is a surjective ring map. By replacing $f_j$ by $a^Nf_j$ we may assume that $f_j \in R[x_1, \ldots, x_n]$ and that $\Psi(f_j) = 0$ (since after all $f_j(a_1/1, \ldots, a_n/1) = 0$ in $A_a$). Let $J = \operatorname{Ker}(\Psi)$. Then $A = R[x_1, \ldots, x_n]/J$ is a presentation and $f_1, \ldots, f_c \in J$ are elements such that $(J/J^2)_a$ is freely generated by $f_1, \ldots, f_c$ and such that $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$ maps to an invertible element of $A_a$. It follows that (the displayed identity) and (the displayed identity) hold for $a^e$ and all large enough $e$ as desired. $\square$

The lifting problem

The goal in this section is to prove (Proposition Lifting an algebraic factorization) that the collection of algebras which are filtered colimits of smooth algebras is closed under infinitesimal flat deformations. The proof is elementary and only uses the results on presentations of smooth algebras from Section The geometric construction.

Lemma. A first lifting step

Let $R \to \Lambda$ be a ring map. Let $I \subset R$ be an ideal. Assume that

1. $I^2 = 0$, and

2. $\Lambda/I\Lambda$ is a filtered colimit of smooth $R/I$-algebras.

Let $\varphi : A \to \Lambda$ be an $R$-algebra map with $A$ of finite presentation over $R$. Then there exists a factorization $$A \to B/J \to \Lambda$$ where $B$ is a smooth $R$-algebra and $J \subset IB$ is a finitely generated ideal.

Proof. Choose a factorization $$A/IA \to \bar B \to \Lambda/I\Lambda$$ with $\bar B$ standard smooth over $R/I$; this is possible by assumption and Lemma Standard smooth presentations in a filtered colimit. Write $$\bar B = A/IA[t_1, \ldots, t_r]/(\bar g_1, \ldots, \bar g_s)$$ and say $\bar B \to \Lambda/I\Lambda$ maps $t_i$ to the class of $\lambda_i$ modulo $I\Lambda$. Choose $g_1, \ldots, g_s \in A[t_1, \ldots, t_r]$ lifting $\bar g_1, \ldots, \bar g_s$. Write $\varphi(g_i)(\lambda_1, \ldots, \lambda_r) = \sum \epsilon_{ij} \mu_{ij}$ for some $\epsilon_{ij} \in I$ and $\mu_{ij} \in \Lambda$. Define $$A' = A[t_1, \ldots, t_r, \delta_{i, j}]/ (g_i - \sum \epsilon_{ij} \delta_{ij})$$ and consider the map $$A' \longrightarrow \Lambda,\quad a \longmapsto \varphi(a),\quad t_i \longmapsto \lambda_i,\quad \delta_{ij} \longmapsto \mu_{ij}$$ We have $$A'/IA' = A/IA[t_1, \ldots, t_r]/(\bar g_1, \ldots, \bar g_s)[\delta_{ij}] \cong \bar B[\delta_{ij}]$$ This is a standard smooth algebra over $R/I$ as $\bar B$ is standard smooth. Choose a presentation $A'/IA' = R/I[x_1, \ldots, x_n]/(\bar f_1, \ldots, \bar f_c)$ with $\det(\partial \bar f_j/\partial x_i)_{i, j = 1, \ldots, c}$ invertible in $A'/IA'$. Choose lifts $f_1, \ldots, f_c \in R[x_1, \ldots, x_n]$ of $\bar f_1, \ldots, \bar f_c$. Then $$B = R[x_1, \ldots, x_n, x_{n + 1}]/ (f_1, \ldots, f_c, x_{n + 1}\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c} - 1)$$ is smooth over $R$. Since smooth ring maps are formally smooth (Algebra, Proposition Formal smoothness of smooth algebras) there exists an $R$-algebra map $B \to A'$ which is an isomorphism modulo $I$. Then $B \to A'$ is surjective by Nakayama's lemma (Algebra, Lemma Nakayama's lemma). Thus $A' = B/J$ with $J \subset IB$ finitely generated (see Algebra, Lemma Finite presentation and finite algebras). $\square$

Lemma. A second lifting step

Let $R \to \Lambda$ be a ring map. Let $I \subset R$ be an ideal. Assume that

1. $I^2 = 0$,

2. $\Lambda/I\Lambda$ is a filtered colimit of smooth $R/I$-algebras, and

3. $R \to \Lambda$ is flat.

Let $\varphi : B \to \Lambda$ be an $R$-algebra map with $B$ smooth over $R$. Let $J \subset IB$ be a finitely generated ideal such that $\varphi(J) = 0$. Then there exists $R$-algebra maps $$B \xrightarrow{\alpha} B' \xrightarrow{\beta} \Lambda$$ such that $B'$ is smooth over $R$, such that $\alpha(J) = 0$ and such that $\beta \circ \alpha = \varphi$.

Proof. If we can prove the lemma in case $J = (h)$, then we can prove the lemma by induction on the number of generators of $J$. Namely, suppose that $J$ can be generated by $n$ elements $h_1, \ldots, h_n$ and the lemma holds for all cases where $J$ is generated by $n - 1$ elements. Then we apply the case $n = 1$ to produce $B \to B' \to \Lambda$ where the first map kills $h_n$. Then we let $J'$ be the ideal of $B'$ generated by the images of $h_1, \ldots, h_{n - 1}$ and we apply the case for $n - 1$ to produce $B' \to B'' \to \Lambda$. It is easy to verify that $B \to B'' \to \Lambda$ does the job.

Assume $J = (h)$ and write $h = \sum \epsilon_i b_i$ for some $\epsilon_i \in I$ and $b_i \in B$. Note that $0 = \varphi(h) = \sum \epsilon_i \varphi(b_i)$. As $\Lambda$ is flat over $R$, the equational criterion for flatness (Algebra, Lemma The equational criterion for flatness) implies that we can find $\lambda_j \in \Lambda$, $j = 1, \ldots, m$ and $a_{ij} \in R$ such that $\varphi(b_i) = \sum_j a_{ij} \lambda_j$ and $\sum_i \epsilon_i a_{ij} = 0$. Set $$C = B[x_1, \ldots, x_m]/(b_i - \sum a_{ij} x_j)$$ with $C \to \Lambda$ given by $\varphi$ and $x_j \mapsto \lambda_j$. Choose a factorization $$C \to B'/J' \to \Lambda$$ as in Lemma A first lifting step. Since $B$ is smooth over $R$ we can lift the map $B \to C \to B'/J'$ to a map $\alpha : B \to B'$. Then $\varphi = \beta \circ \alpha$. To finish the proof we check that $\alpha(h) = 0$. Namely, the fact that $\alpha$ lifts $B \to C \to B'/J'$ implies that $$\alpha(b_i) = \sum a_{ij} \xi_j + \theta_i$$ for some $\xi_j \in B'$ and $\theta_i \in J' \subset IB'$. Hence we see that $$\alpha(h) = \alpha(\sum \epsilon_i b_i) = \sum \epsilon_i a_{ij} \xi_j + \sum \epsilon_i \theta_i = 0$$ because of the relations above and the fact that $I^2 = 0$. $\square$

Proposition. Lifting an algebraic factorization

Ind-smoothness of an algebra is stable under infinitesimal deformations

Let $R \to \Lambda$ be a ring map. Let $I \subset R$ be an ideal. Assume that

1. $I$ is nilpotent,

2. $\Lambda/I\Lambda$ is a filtered colimit of smooth $R/I$-algebras, and

3. $R \to \Lambda$ is flat.

Then $\Lambda$ is a filtered colimit of smooth $R$-algebras.

Proof. Since $I^n = 0$ for some $n$, it follows by induction on $n$ that it suffices to consider the case where $I^2 = 0$. Let $\varphi : A \to \Lambda$ be an $R$-algebra map with $A$ of finite presentation over $R$. We have to find a factorization $A \to B \to \Lambda$ with $B$ smooth over $R$, see Algebra, Lemma Recognizing a filtered colimit of finite presentations. By Lemma A first lifting step we may assume that $A = B/J$ with $B$ smooth over $R$ and $J \subset IB$ a finitely generated ideal. By Lemma A second lifting step we can find a commutative diagram $$\begin{gathered}\begin{matrix}B & \phantom{X} & B' \\ \phantom{X} & \Lambda\end{matrix} \\[6pt] \begin{aligned}B & \xrightarrow{\alpha} B' \\ B & \xrightarrow{\varphi} \Lambda \\ B' & \xrightarrow{\beta} \Lambda\end{aligned}\end{gathered}$$ of $R$-algebras with $B'$ smooth over $R$ such that $\alpha(J) = 0$. Thus $\alpha$ factors as $B \to A \to B'$ and the proof is complete. $\square$

The lifting lemma

Here is a fiendishly clever lemma.

Lemma. The lifting lemma

Let $R$ be a Noetherian ring. Let $\Lambda$ be an $R$-algebra. Let $\pi \in R$ and assume that $\text{Ann}_R(\pi) = \text{Ann}_R(\pi^2)$ and $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^2)$. Suppose we have $R$-algebra maps $R/\pi^2R \to \bar C \to \Lambda/\pi^2\Lambda$ with $\bar C$ of finite presentation. Then there exists an $R$-algebra homomorphism $D \to \Lambda$ and a commutative diagram $$\begin{gathered}\begin{matrix}R/\pi^2R & \bar C & \Lambda/\pi^2\Lambda \\ R/\pi R & D/\pi D & \Lambda/\pi \Lambda\end{matrix} \\[6pt] \begin{aligned}R/\pi^2R & \longrightarrow \bar C \\ R/\pi^2R & \longrightarrow R/\pi R \\ \bar C & \longrightarrow \Lambda/\pi^2\Lambda \\ \bar C & \longrightarrow D/\pi D \\ \Lambda/\pi^2\Lambda & \longrightarrow \Lambda/\pi \Lambda \\ R/\pi R & \longrightarrow D/\pi D \\ D/\pi D & \longrightarrow \Lambda/\pi \Lambda\end{aligned}\end{gathered}$$ with the following properties

1. $D$ is of finite presentation,

2. $R \to D$ is smooth at any prime $\mathfrak q$ with $\pi \not \in \mathfrak q$,

3. $R \to D$ is smooth at any prime $\mathfrak q$ with $\pi \in \mathfrak q$ lying over a prime of $\bar C$ where $R/\pi^2 R \to \bar C$ is smooth, and

4. $\bar C/\pi \bar C \to D/\pi D$ is smooth at any prime lying over a prime of $\bar C$ where $R/\pi^2R \to \bar C$ is smooth.

Proof. We choose a presentation $$\bar C = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$$ We also denote $I = (f_1, \ldots, f_m)$ and $\bar I$ the image of $I$ in $R/\pi^2R[x_1, \ldots, x_n]$. Since $R$ is Noetherian, so is $\bar C$. Hence the smooth locus of $R/\pi^2 R \to \bar C$ is quasi-compact, see Topology, Lemma Noetherian topological spaces (programme binding). Applying Lemma A Jacobian presentation near a smooth point we may choose a finite list of elements $a_1, \ldots, a_r \in R[x_1, \ldots, x_n]$ such that

1. the union of the open subspaces $\operatorname{Spec}(\bar C_{a_k}) \subset \operatorname{Spec}(\bar C)$ cover the smooth locus of $R/\pi^2 R \to \bar C$, and

2. for each $k = 1, \ldots, r$ there exists a finite subset $E_k \subset \{1, \ldots, m\}$ such that $(\bar I/\bar I^2)_{a_k}$ is freely generated by the classes of $f_j$, $j \in E_k$.

Set $I_k = (f_j, j \in E_k) \subset I$ and denote $\bar I_k$ the image of $I_k$ in $R/\pi^2R[x_1, \ldots, x_n]$. By (2) and Nakayama's lemma we see that $(\bar I/\bar I_k)_{a_k}$ is annihilated by $1 + b'_k$ for some $b'_k \in \bar I_{a_k}$. Suppose $b'_k$ is the image of $b_k/(a_k)^N$ for some $b_k \in I$ and some integer $N$. After replacing $a_k$ by $a_k((a_k)^N + b_k)$ we get

1. $(\bar I_k)_{a_k} = (\bar I)_{a_k}$.

Thus, after possibly replacing $a_k$ by a high power, we may write

1. $a_k f_\ell = \sum\nolimits_{j \in E_k} h_{k, \ell}^jf_j + \pi^2 g_{k, \ell}$

for any $\ell \in \{1, \ldots, m\}$ and some $h_{k, \ell}^j, g_{k, \ell} \in R[x_1, \ldots, x_n]$. If $\ell \in E_k$ we choose $h_{k, \ell}^j = a_k\delta_{\ell, j}$ (Kronecker delta) and $g_{k, \ell} = 0$. Set $$D = R[x_1, \ldots, x_n, z_1, \ldots, z_m]/ (f_j - \pi z_j, p_{k, \ell}).$$ Here $j \in \{1, \ldots, m\}$, $k \in \{1, \ldots, r\}$, $\ell \in \{1, \ldots, m\}$, and $$p_{k, \ell} = a_k z_\ell - \sum\nolimits_{j \in E_k} h_{k, \ell}^j z_j

• \pi g_{k, \ell}.$$ Note that for $\ell \in E_k$ we have $p_{k, \ell} = 0$ by our choices above.

The map $R \to D$ is the given one. Say $\bar C \to \Lambda/\pi^2\Lambda$ maps $x_i$ to the class of $\lambda_i$ modulo $\pi^2$. For an element $f \in R[x_1, \ldots, x_n]$ we denote $f(\lambda) \in \Lambda$ the result of substituting $\lambda_i$ for $x_i$. Then we know that $f_j(\lambda) = \pi^2 \mu_j$ for some $\mu_j \in \Lambda$. Define $D \to \Lambda$ by the rules $x_i \mapsto \lambda_i$ and $z_j \mapsto \pi\mu_j$. This is well defined because $$\begin{aligned} p_{k, \ell} & \mapsto a_k(\lambda) \pi \mu_\ell - \sum\nolimits_{j \in E_k} h_{k, \ell}^j(\lambda) \pi \mu_j

• \pi g_{k, \ell}(\lambda) \ & = \pi\left(a_k(\lambda) \mu_\ell - \sum\nolimits_{j \in E_k} h_{k, \ell}^j(\lambda) \mu_j
• g_{k, \ell}(\lambda)\right) \end{aligned}$$ Substituting $x_i = \lambda_i$ in (4) above we see that the expression inside the brackets is annihilated by $\pi^2$, hence it is annihilated by $\pi$ as we have assumed $\text{Ann}\Lambda(\pi) = \text{Ann}\Lambda(\pi^2)$. The map $\bar C \to D/\pi D$ is determined by $x_i \mapsto x_i$ (clearly well defined). Thus we are done if we can prove (b), (c), and (d).

Using (4) we obtain the following key equality $$\begin{aligned} \pi p_{k, \ell} & = \pi a_k z_\ell - \sum\nolimits_{j \in E_k} \pi h_{k, \ell}^jz_j

• \pi^2 g_{k, \ell} \ & =
• a_k (f_\ell - \pi z_\ell) + a_k f_\ell + \sum\nolimits_{j \in E_k} h_{k, \ell}^j (f_j - \pi z_j) - \sum\nolimits_{j \in E_k} h_{k, \ell}^j f_j - \pi^2 g_{k, \ell} \ & = -a_k(f_\ell - \pi z_\ell) + \sum\nolimits_{j \in E_k} h_{k, \ell}^j(f_j - \pi z_j) \end{aligned}$$ The end result is an element of the ideal generated by $f_j - \pi z_j$. In particular, we see that $D[1/\pi]$ is isomorphic to $R[1/\pi][x_1, \ldots, x_n, z_1, \ldots, z_m]/(f_j - \pi z_j)$ which is isomorphic to $R[1/\pi][x_1, \ldots, x_n]$ hence smooth over $R$. This proves (b).

For fixed $k \in \{1, \ldots, r\}$ consider the ring $$D_k = R[x_1, \ldots, x_n, z_1, \ldots, z_m]/ (f_j - \pi z_j, j \in E_k, p_{k, \ell})$$ The number of equations is $m = |E_k| + (m - |E_k|)$ as $p_{k, \ell}$ is zero if $\ell \in E_k$. Also, note that $$\begin{aligned} (D_k/\pi D_k)_{a_k} & = R/\pi R[x_1, \ldots, x_n, 1/a_k, z_1, \ldots, z_m]/ (f_j, j \in E_k, p_{k, \ell}) \\ & = (\bar C/\pi \bar C)_{a_k}[z_1, \ldots, z_m]/ (a_kz_\ell - \sum\nolimits_{j \in E_k} h_{k, \ell}^j z_j) \\ & \cong (\bar C/\pi \bar C)_{a_k}[z_j, j \in E_k] \end{aligned}$$ In particular $(D_k/\pi D_k)_{a_k}$ is smooth over $(\bar C/\pi \bar C)_{a_k}$. By our choice of $a_k$ we have that $(\bar C/\pi \bar C)_{a_k}$ is smooth over $R/\pi R$ of relative dimension $n - |E_k|$, see (2). Hence for a prime $\mathfrak q_k \subset D_k$ containing $\pi$ and lying over $\operatorname{Spec}(\bar C_{a_k})$ the fibre ring of $R \to D_k$ is smooth at $\mathfrak q_k$ of dimension $n$. Thus $R \to D_k$ is syntomic at $\mathfrak q_k$ by our count of the number of equations above, see Algebra, Lemma Localization of a relative complete intersection. Hence $R \to D_k$ is smooth at $\mathfrak q_k$, see Algebra, Lemma Smoothness from flatness and smooth fibres.

To finish the proof, let $\mathfrak q \subset D$ be a prime containing $\pi$ lying over a prime where $R/\pi^2 R \to \bar C$ is smooth. Then $a_k \not \in \mathfrak q$ for some $k$ by (1). We will show that the surjection $D_k \to D$ induces an isomorphism on local rings at $\mathfrak q$. Since we know that the ring maps $\bar C/\pi \bar C \to D_k/\pi D_k$ and $R \to D_k$ are smooth at the corresponding prime $\mathfrak q_k$ by the preceding paragraph this will prove (c) and (d) and thus finish the proof.

First, note that for any $\ell$ the equation $\pi p_{k, \ell} = -a_k(f_\ell - \pi z_\ell) + \sum_{j \in E_k} h_{k, \ell}^j (f_j - \pi z_j)$ proved above shows that $f_\ell - \pi z_\ell$ maps to zero in $(D_k)_{a_k}$ and in particular in $(D_k)_{\mathfrak q_k}$. The relations (4) imply that $a_k f_\ell = \sum_{j \in E_k} h_{k, \ell}^j f_j$ in $I/I^2$. Since $(\bar I_k/\bar I_k^2)_{a_k}$ is free on $f_j$, $j \in E_k$ we see that $$a_{k'} h_{k, \ell}^j - \sum\nolimits_{j' \in E_{k'}} h_{k', \ell}^{j'} h_{k, j'}^j$$ is zero in $\bar C_{a_k}$ for every $k, k', \ell$ and $j \in E_k$. Hence we can find a large integer $N$ such that $$a_k^N\left( a_{k'} h_{k, \ell}^j - \sum\nolimits_{j' \in E_{k'}} h_{k', \ell}^{j'} h_{k, j'}^j \right)$$ is in $I_k + \pi^2R[x_1, \ldots, x_n]$. Computing modulo $\pi$ we have $$\begin{aligned} & a_kp_{k', \ell} - a_{k'}p_{k, \ell} + \sum h_{k', \ell}^{j'} p_{k, j'} \ &

• a_k \sum h_{k', \ell}^{j'} z_{j'}

• a_{k'} \sum h_{k, \ell}^j z_j
• \sum h_{k', \ell}^{j'} a_k z_{j'}

• \sum \sum h_{k', \ell}^{j'} h_{k, j'}^j z_j \ & = \sum \left( a_{k'} h_{k, \ell}^j
• \sum h_{k', \ell}^{j'} h_{k, j'}^j \right) z_j \end{aligned}$$ with Einstein summation convention. Combining with the above we see $a_k^{N + 1} p_{k', \ell}$ is contained in the ideal generated by $I_k$ and $\pi$ in $R[x_1, \ldots, x_n, z_1, \ldots, z_m]$. Thus $p_{k', \ell}$ maps into $\pi (D_k)_{a_k}$. On the other hand, the equation $$\pi p_{k', \ell} = -a_{k'} (f_\ell - \pi z_\ell) + \sum\nolimits_{j' \in E_{k'}} h_{k', \ell}^{j'}(f_{j'} - \pi z_{j'})$$ shows that $\pi p_{k', \ell}$ is zero in $(D_k){a_k}$. Since we have assumed that $\text{Ann}R(\pi) = \text{Ann}R(\pi^2)$ and since $(D_k){\mathfrak q_k}$ is smooth hence flat over $R$ we see that $\text{Ann}{(D_k){\mathfrak q_k}}(\pi) = \text{Ann}{(D_k){\mathfrak q_k}}(\pi^2)$. We conclude that $p_{k', \ell}$ maps to zero as well, hence $D_{\mathfrak q} = (D_k)_{\mathfrak q_k}$ and we win. $\square$

The desingularization lemma

The following construction enlarges the smooth locus.

Lemma. The desingularization lemma

Let $R$ be a Noetherian ring. Let $\Lambda$ be an $R$-algebra. Let $\pi \in R$ and assume that $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^2)$. Let $A \to \Lambda$ be an $R$-algebra map with $A$ of finite presentation. Assume

1. the image of $\pi$ is strictly standard in $A$ over $R$, and

2. there exists a section $\rho : A/\pi^4 A \to R/\pi^4 R$ which is compatible with the map to $\Lambda/\pi^4 \Lambda$.

Then we can find $R$-algebra maps $A \to B \to \Lambda$ with $B$ of finite presentation such that $\mathfrak a B \subset H_{B/R}$ where $\mathfrak a = \text{Ann}_R(\text{Ann}_R(\pi^2)/\text{Ann}_R(\pi))$.

Proof. Choose a presentation $$A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$$ and $0 \leq c \leq \min(n, m)$ such that (the displayed identity) holds for $\pi$ and such that

$$\pi f_{c + j} \in (f_1, \ldots, f_c) + (f_1, \ldots, f_m)^2$$ for $j = 1, \ldots, m - c$. Say $\rho$ maps $x_i$ to the class of $r_i \in R$. Then we can replace $x_i$ by $x_i - r_i$. Hence we may assume $\rho(x_i) = 0$ in $R/\pi^4 R$. This implies that $f_j(0) \in \pi^4R$ and that $A \to \Lambda$ maps $x_i$ to $\pi^4\lambda_i$ for some $\lambda_i \in \Lambda$. Write $$f_j = f_j(0) + \sum\nolimits_{i = 1, \ldots, n} r_{ji} x_i + \text{h.o.t.}$$ This implies that the constant term of $\partial f_j/\partial x_i$ is $r_{ji}$. Apply $\rho$ to (the displayed identity) for $\pi$ and we see that $$\pi = \sum\nolimits_{I \subset \{1, \ldots, n\},\ |I| = c} r_I \det(r_{ji})_{j = 1, \ldots, c,\ i \in I} \bmod \pi^4R$$ for some $r_I \in R$. Thus we have $$u\pi = \sum\nolimits_{I \subset \{1, \ldots, n\},\ |I| = c} r_I \det(r_{ji})_{j = 1, \ldots, c,\ i \in I}$$ for some $u \in 1 + \pi^3R$. By Algebra, Lemma A left inverse for a matrix this implies there exists a $n \times c$ matrix $(s_{ik})$ such that $$u\pi \delta_{jk} = \sum\nolimits_{i = 1, \ldots, n} r_{ji}s_{ik}\quad \text{for all } j, k = 1, \ldots, c$$ (Kronecker delta). We introduce auxiliary variables $v_1, \ldots, v_c, w_1, \ldots, w_n$ and we set $$h_i = x_i - \pi^2 \sum\nolimits_{j = 1, \ldots c} s_{ij} v_j - \pi^3 w_i$$ In the following we will use that $$R[x_1, \ldots, x_n, v_1, \ldots, v_c, w_1, \ldots, w_n]/ (h_1, \ldots, h_n) = R[v_1, \ldots, v_c, w_1, \ldots, w_n]$$ without further mention. In $R[x_1, \ldots, x_n, v_1, \ldots, v_c, w_1, \ldots, w_n]/ (h_1, \ldots, h_n)$ we have $$\begin{aligned} f_j & = f_j(x_1 - h_1, \ldots, x_n - h_n) \\ & = \pi^2 \sum\nolimits_{k = 1}^c \left(\sum\nolimits_{i = 1}^n r_{ji} s_{ik}\right) v_k + \pi^3 \sum\nolimits_{i = 1}^n r_{ji}w_i \bmod \pi^4 \\ & = \pi^3 v_j + \pi^3 \sum\nolimits_{i = 1}^n r_{ji}w_i \bmod \pi^4 \end{aligned}$$ for $1 \leq j \leq c$. Hence we can choose elements $g_j \in R[v_1, \ldots, v_c, w_1, \ldots, w_n]$ such that $g_j = v_j + \sum r_{ji}w_i \bmod \pi$ and such that $f_j = \pi^3 g_j$ in the $R$-algebra $R[x_1, \ldots, x_n, v_1, \ldots, v_c, w_1, \ldots, w_n]/ (h_1, \ldots, h_n)$. We set $$B = R[x_1, \ldots, x_n, v_1, \ldots, v_c, w_1, \ldots, w_n]/ (f_1, \ldots, f_m, h_1, \ldots, h_n, g_1, \ldots, g_c).$$ The map $A \to B$ is clear. We define $B \to \Lambda$ by mapping $x_i \to \pi^4\lambda_i$, $v_i \mapsto 0$, and $w_i \mapsto \pi \lambda_i$. Then it is clear that the elements $f_j$ and $h_i$ are mapped to zero in $\Lambda$. Moreover, it is clear that $g_i$ is mapped to an element $t$ of $\pi\Lambda$ such that $\pi^3t = 0$ (as $f_i = \pi^3 g_i$ modulo the ideal generated by the $h$'s). Hence our assumption that $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^2)$ implies that $t = 0$. Thus we are done if we can prove the statement about smoothness.

Note that $B_\pi \cong A_\pi[v_1, \ldots, v_c]$ because the equations $g_i = 0$ are implied by $f_i = 0$. Hence $B_\pi$ is smooth over $R$ as $A_\pi$ is smooth over $R$ by the assumption that $\pi$ is strictly standard in $A$ over $R$, see Lemma Controlling a power of the Jacobian determinant.

Set $B' = R[v_1, \ldots, v_c, w_1, \ldots, w_n]/(g_1, \ldots, g_c)$. As $g_i = v_i + \sum r_{ji}w_i \bmod \pi$ we see that $B'/\pi B' = R/\pi R[w_1, \ldots, w_n]$. Hence $R \to B'$ is smooth of relative dimension $n$ at every point of $V(\pi)$ by Algebra, Lemmas Localization of a relative complete intersection and Smoothness from flatness and smooth fibres (the first lemma shows it is syntomic at those primes, in particular flat, whereupon the second lemma shows it is smooth).

Let $\mathfrak q \subset B$ be a prime with $\pi \in \mathfrak q$ and for some $r \in \mathfrak a$, $r \not \in \mathfrak q$. Denote $\mathfrak q' = B' \cap \mathfrak q$. We claim the surjection $B' \to B$ induces an isomorphism of local rings $(B')_{\mathfrak q'} \to B_\mathfrak q$. This will conclude the proof of the lemma. Note that $B_\mathfrak q$ is the quotient of $(B')_{\mathfrak q'}$ by the ideal generated by $f_{c + j}$, $j = 1, \ldots, m - c$. We observe two things: first the image of $f_{c + j}$ in $(B')_{\mathfrak q'}$ is divisible by $\pi^2$ and second the image of $\pi f_{c + j}$ in $(B')_{\mathfrak q'}$ can be written as $\sum b_{j_1 j_2} f_{c + j_1}f_{c + j_2}$ by (the displayed identity). Thus we see that the image of each $\pi f_{c + j}$ is contained in the ideal generated by the elements $\pi^2 f_{c + j'}$. Hence $\pi f_{c + j} = 0$ in $(B')_{\mathfrak q'}$ as this is a Noetherian local ring, see Algebra, Lemma Krull's intersection theorem. As $R \to (B')_{\mathfrak q'}$ is flat we see that $$\left(\text{Ann}_R(\pi^2)/\text{Ann}R(\pi)\right) \otimes_R (B'){\mathfrak q'}

\text{Ann}{(B'){\mathfrak q'}}(\pi^2)/\text{Ann}{(B'){\mathfrak q'}}(\pi)$$ Because $r \in \mathfrak a$ is invertible in $(B'){\mathfrak q'}$ we see that this module is zero. Hence we see that the image of $f{c + j}$ is zero in $(B')_{\mathfrak q'}$ as desired. $\square$

Lemma. Desingularization of a strict presentation

Let $R$ be a Noetherian ring. Let $\Lambda$ be an $R$-algebra. Let $\pi \in R$ and assume that $\text{Ann}_R(\pi) = \text{Ann}_R(\pi^2)$ and $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^2)$. Let $A \to \Lambda$ and $D \to \Lambda$ be $R$-algebra maps with $A$ and $D$ of finite presentation. Assume

1. $\pi$ is strictly standard in $A$ over $R$, and

2. there exists an $R$-algebra map $A/\pi^4 A \to D/\pi^4 D$ compatible with the maps to $\Lambda/\pi^4 \Lambda$.

Then we can find an $R$-algebra map $B \to \Lambda$ with $B$ of finite presentation and $R$-algebra maps $A \to B$ and $D \to B$ compatible with the maps to $\Lambda$ such that $H_{D/R}B \subset H_{B/D}$ and $H_{D/R}B \subset H_{B/R}$.

Proof. We apply Lemma The desingularization lemma to $$D \longrightarrow A \otimes_R D \longrightarrow \Lambda$$ and the image of $\pi$ in $D$. By Lemma Base change of a strict standard presentation we see that $\pi$ is strictly standard in $A \otimes_R D$ over $D$. As our section $\rho : (A \otimes_R D)/\pi^4 (A \otimes_R D) \to D/\pi^4 D$ we take the map induced by the map in (2). Thus Lemma The desingularization lemma applies and we obtain a factorization $A \otimes_R D \to B \to \Lambda$ with $B$ of finite presentation and $\mathfrak a B \subset H_{B/D}$ where $$\mathfrak a = \text{Ann}_D(\text{Ann}_D(\pi^2)/\text{Ann}_D(\pi)).$$ For any prime $\mathfrak q$ of $D$ such that $D_\mathfrak q$ is flat over $R$ we have $\text{Ann}_{D_\mathfrak q}(\pi^2)/\text{Ann}_{D_\mathfrak q}(\pi) = 0$ because annihilators of elements commutes with flat base change and we assumed $\text{Ann}_R(\pi) = \text{Ann}_R(\pi^2)$. Because $D$ is Noetherian we see that $\text{Ann}_D(\pi^2)/\text{Ann}_D(\pi)$ is a finite $D$-module, hence formation of its annihilator commutes with localization. Thus we see that $\mathfrak a \not \subset \mathfrak q$. Hence we see that $D \to B$ is smooth at any prime of $B$ lying over $\mathfrak q$. Since any prime of $D$ where $R \to D$ is smooth is one where $D_\mathfrak q$ is flat over $R$ we conclude that $H_{D/R}B \subset H_{B/D}$. The final inclusion $H_{D/R}B \subset H_{B/R}$ follows because compositions of smooth ring maps are smooth (Algebra, Lemma Composition of smooth ring maps). $\square$

Lemma. Combining lifting with desingularization

Let $R$ be a Noetherian ring. Let $\Lambda$ be an $R$-algebra. Let $\pi \in R$ and assume that $\text{Ann}_R(\pi) = \text{Ann}_R(\pi^2)$ and $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^2)$. Let $A \to \Lambda$ be an $R$-algebra map with $A$ of finite presentation and assume $\pi$ is strictly standard in $A$ over $R$. Let $$A/\pi^8A \to \bar C \to \Lambda/\pi^8\Lambda$$ be a factorization with $\bar C$ of finite presentation. Then we can find a factorization $A \to B \to \Lambda$ with $B$ of finite presentation such that $R_\pi \to B_\pi$ is smooth and such that $$H_{\bar C/(R/\pi^8 R)} \cdot \Lambda/\pi^8\Lambda \subset \sqrt{H_{B/R} \Lambda} \bmod \pi^8\Lambda.$$

Proof. Apply Lemma The lifting lemma to get $R \to D \to \Lambda$ with a factorization $\bar C/\pi^4\bar C \to D/\pi^4 D \to \Lambda/\pi^4\Lambda$ such that $R \to D$ is smooth at any prime not containing $\pi$ and at any prime lying over a prime of $\bar C/\pi^4\bar C$ where $R/\pi^8 R \to \bar C$ is smooth. By Lemma Desingularization of a strict presentation we can find a finitely presented $R$-algebra $B$ and factorizations $A \to B \to \Lambda$ and $D \to B \to \Lambda$ such that $H_{D/R}B\subset H_{B/R}$. Away from $\pi$, the lifting lemma makes $D$ smooth over $R$, and the displayed inclusion makes $B$ smooth there too. Let $\eta$ be a prime of $\Lambda$ containing $\pi$ whose inverse image in $\bar C$ is in the smooth locus. The compatible map $\bar C/\pi^4\to D/\pi^4\to\Lambda/\pi^4$ and the lifting lemma show that the inverse image of $\eta$ in $D$ is smooth over $R$. Thus $H_{D/R}\Lambda\nsubseteq\eta$, and the displayed inclusion gives $H_{B/R}\Lambda\nsubseteq\eta$. On $\operatorname{Spec}(\Lambda/\pi^8)$ the smooth open of $\bar C$ is therefore contained in the open defined by $H_{B/R}$. Taking closed complements gives exactly the stated radical ideal inclusion. $\square$

Reduction to the field case

In this section we apply the lemmas in the previous sections to prove that it suffices to prove the main result when the base ring is a field, see Lemma Reducing desingularization to field bases.

Situation. The global desingularization problem

Here $R \to \Lambda$ is a regular ring map of Noetherian rings.

Let $R \to \Lambda$ be as in Situation The global desingularization problem. We say PT holds for $R \to \Lambda$ if $\Lambda$ is a filtered colimit of smooth $R$-algebras.

Lemma. Tensor products and direct sums

Let $R_i \to \Lambda_i$, $i = 1, 2$ be as in Situation The global desingularization problem. If PT holds for $R_i \to \Lambda_i$, $i = 1, 2$, then PT holds for $R_1 \times R_2 \to \Lambda_1 \times \Lambda_2$.

Proof. Write $\Lambda_i=\mathop{\rm colim}_\alpha B_{i,\alpha}$ with smooth $R_i$-algebras. The product of the two filtered index categories is filtered, and its colimit of $B_{1,\alpha}\times B_{2,\beta}$ is $\Lambda_1\times\Lambda_2$: elements and equality of elements are checked componentwise at common stages. Each product algebra is smooth over $R_1\times R_2$, since the idempotent components of both spectra split and the two restrictions are the given smooth maps. This proves the lemma. $\square$

Lemma. Returning from a localized base

Let $R \to A \to \Lambda$ be ring maps with $A$ of finite presentation over $R$. Let $S \subset R$ be a multiplicative set. Let $S^{-1}A \to B' \to S^{-1}\Lambda$ be a factorization with $B'$ smooth over $S^{-1}R$. Then we can find a factorization $A \to B \to \Lambda$ such that some $s \in S$ maps to an elementary standard element (Definition Strict standard smoothness) in $B$ over $R$.

Proof. We first apply Lemma Standard presentations of smooth algebras to $S^{-1}R \to B'$. Thus we may assume $B'$ is standard smooth over $S^{-1}R$. Write $A = R[x_1, \ldots, x_n]/(g_1, \ldots, g_t)$ and say $x_i \mapsto \lambda_i$ in $\Lambda$. We may write $B' = S^{-1}R[x_1, \ldots, x_{n + m}]/(f_1, \ldots, f_c)$ for some $c \geq n$ where $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$ is invertible in $B'$ and such that $A \to B'$ is given by $x_i \mapsto x_i$, see Lemma Including prescribed generators in a smooth presentation. After multiplying $x_i$, $i > n$ by an element of $S$ and correspondingly modifying the equations $f_j$ we may assume $B' \to S^{-1}\Lambda$ maps $x_i$ to $\lambda_i/1$ for some $\lambda_i \in \Lambda$ for $i > n$. Choose a relation $$1 = a_0 \det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c} + \sum\nolimits_{j = 1, \ldots, c} a_jf_j$$ for some $a_j \in S^{-1}R[x_1, \ldots, x_{n + m}]$. Since each element of $S$ is invertible in $B'$ we may (by clearing denominators) assume that $f_j, a_j \in R[x_1, \ldots, x_{n + m}]$ and that $$s_0 = a_0 \det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c} + \sum\nolimits_{j = 1, \ldots, c} a_jf_j$$ for some $s_0 \in S$. Since $g_j$ maps to zero in $S^{-1}R[x_1, \ldots, x_{n + m}]/(f_1, \ldots, f_c)$ we can find elements $s_j \in S$ such that $s_j g_j = 0$ in $R[x_1, \ldots, x_{n + m}]/(f_1, \ldots, f_c)$. Since $f_j$ maps to zero in $S^{-1}\Lambda$ we can find $s'_j \in S$ such that $s'_j f_j(\lambda_1, \ldots, \lambda_{n + m}) = 0$ in $\Lambda$. Consider the ring $$B = R[x_1, \ldots, x_{n + m}]/ (s'_1f_1, \ldots, s'_cf_c, g_1, \ldots, g_t)$$ and the factorization $A \to B \to \Lambda$ with $B \to \Lambda$ given by $x_i \mapsto \lambda_i$. We claim that $s = s_0s_1 \ldots s_ts'_1 \ldots s'_c$ is elementary standard in $B$ over $R$ which finishes the proof. Namely, $s_j g_j \in (f_1, \ldots, f_c)$ and hence $sg_j \in (s'_1f_1, \ldots, s'_cf_c)$. Finally, we have $$a_0\det(\partial s'jf_j/\partial x_i){i, j = 1, \ldots, c} + \sum\nolimits_{j = 1, \ldots, c} (s'_1 \ldots \hat{s'_j} \ldots s'_c) a_j s'_jf_j

s_0s'_1\ldots s'_c$$ which divides $s$ as desired. $\square$

Lemma. Reducing desingularization to field bases

Proving Popescu approximation reduces to algebras over a field

If for every Situation The global desingularization problem where $R$ is a field PT holds, then PT holds in general.

Proof. Assume PT holds for any Situation The global desingularization problem where $R$ is a field. Let $R \to \Lambda$ be as in Situation The global desingularization problem arbitrary. Note that $R/I \to \Lambda/I\Lambda$ is another regular ring map of Noetherian rings, see More on Algebra, Lemma Base change of regular ring maps. Consider the set of ideals $$\mathcal{I} = \{I \subset R \mid R/I \to \Lambda/I\Lambda \text{ does not have PT}\}$$ We have to show that $\mathcal{I}$ is empty. If this set is nonempty, then it contains a maximal element because $R$ is Noetherian. Replacing $R$ by $R/I$ and $\Lambda$ by $\Lambda/I$ we obtain a situation where PT holds for $R/I \to \Lambda/I\Lambda$ for any nonzero ideal of $R$. In particular, we see by applying Proposition Lifting an algebraic factorization that $R$ is a reduced ring.

Let $A \to \Lambda$ be an $R$-algebra homomorphism with $A$ of finite presentation. We have to find a factorization $A \to B \to \Lambda$ with $B$ smooth over $R$, see Algebra, Lemma Recognizing a filtered colimit of finite presentations.

Let $S \subset R$ be the set of nonzerodivisors and consider the total ring of fractions $Q = S^{-1}R$ of $R$. We know that $Q = K_1 \times \ldots \times K_n$ is a product of fields, see Algebra, Lemmas Total rings of fractions without embedded primes and Irreducible components of a Noetherian spectrum. By Lemma Tensor products and direct sums and our assumption PT holds for the ring map $S^{-1}R \to S^{-1}\Lambda$. Hence we can find a factorization $S^{-1}A \to B' \to S^{-1}\Lambda$ with $B'$ smooth over $S^{-1}R$.

We apply Lemma Returning from a localized base and find a factorization $A \to B \to \Lambda$ such that some $\pi \in S$ is elementary standard in $B$ over $R$. After replacing $A$ by $B$ we may assume that $\pi$ is elementary standard, hence strictly standard in $A$. We know that $R/\pi^8R \to \Lambda/\pi^8\Lambda$ satisfies PT. Hence we can find a factorization $R/\pi^8 R \to A/\pi^8A \to \bar C \to \Lambda/\pi^8\Lambda$ with $R/\pi^8 R \to \bar C$ smooth. By Lemma The lifting lemma we can find an $R$-algebra map $D \to \Lambda$ with $D$ smooth over $R$ and a factorization $R/\pi^4 R \to A/\pi^4A \to D/\pi^4D \to \Lambda/\pi^4\Lambda$. By Lemma Desingularization of a strict presentation we can find $A \to B \to \Lambda$ with $B$ smooth over $R$ which finishes the proof. $\square$

Localization and descent of resolutions

Situation. The local desingularization problem

We are given a Noetherian ring $R$ and an $R$-algebra map $A \to \Lambda$ and a prime $\mathfrak q \subset \Lambda$. We assume $A$ is of finite presentation over $R$. In this situation we denote $\mathfrak h_A = \sqrt{H_{A/R} \Lambda}$.

Let $R \to A \to \Lambda \supset \mathfrak q$ be as in Situation The local desingularization problem. We say $R \to A \to \Lambda \supset \mathfrak q$ can be resolved if there exists a factorization $A \to B \to \Lambda$ with $B$ of finite presentation and $\mathfrak h_A \subset \mathfrak h_B \not \subset \mathfrak q$. In this case we will call the factorization $A \to B \to \Lambda$ a resolution of $R \to A \to \Lambda \supset \mathfrak q$.

Lemma. Lifting a local desingularization solution

Let $R \to A \to \Lambda \supset \mathfrak q$ be as in Situation The local desingularization problem. Let $r \geq 1$ and $\pi_1, \ldots, \pi_r \in R$ map to elements of $\mathfrak q$. Assume

1. for $i = 1, \ldots, r$ we have $$\text{Ann}{R/(\pi_1^8, \ldots, \pi{i - 1}^8)R}(\pi_i)

   \text{Ann}{R/(\pi_1^8, \ldots, \pi{i - 1}^8)R}(\pi_i^2)$$ and $$\text{Ann}{\Lambda/(\pi_1^8, \ldots, \pi{i - 1}^8)\Lambda}(\pi_i)

   \text{Ann}{\Lambda/(\pi_1^8, \ldots, \pi{i - 1}^8)\Lambda}(\pi_i^2)$$

2. for $i = 1, \ldots, r$ the element $\pi_i$ maps to a strictly standard element in $A$ over $R$.

Then, if $$R/(\pi_1^8, \ldots, \pi_r^8)R \to A/(\pi_1^8, \ldots, \pi_r^8)A \to \Lambda/(\pi_1^8, \ldots, \pi_r^8)\Lambda \supset \mathfrak q/(\pi_1^8, \ldots, \pi_r^8)\Lambda$$ can be resolved, so can $R \to A \to \Lambda \supset \mathfrak q$.

Proof. We are going to prove this by induction on $r$.

The case $r = 1$. Here the assumption is that there exists a factorization $A/\pi_1^8 \to \bar C \to \Lambda/\pi_1^8$ which resolves the situation modulo $\pi_1^8$. Conditions (1) and (2) are the assumptions needed to apply Lemma Combining lifting with desingularization. Thus we can "lift" the resolution $\bar C$ to a resolution of $R \to A \to \Lambda \supset \mathfrak q$.

The case $r > 1$. In this case we apply the induction hypothesis for $r - 1$ to the situation $R/\pi_1^8 \to A/\pi_1^8 \to \Lambda/\pi_1^8 \supset \mathfrak q/\pi_1^8\Lambda$. Note that property (2) is preserved by Lemma Base change of a strict standard presentation. $\square$

Lemma. Returning from a localized target

Source credit: the original source citation swan (Lemma 12.2) or the original source citation popescu-GND (Lemma 2)

Let $R \to A \to \Lambda \supset \mathfrak q$ be as in Situation The local desingularization problem. Let $\mathfrak p = R \cap \mathfrak q$. Assume that $\mathfrak q$ is minimal over $\mathfrak h_A$ and that $R_\mathfrak p \to A_\mathfrak p \to \Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q$ can be resolved. Then there exists a factorization $A \to C \to \Lambda$ with $C$ of finite presentation such that $H_{C/R} \Lambda \not \subset \mathfrak q$.

Proof. Let $A_\mathfrak p \to C \to \Lambda_\mathfrak q$ be a resolution of $R_\mathfrak p \to A_\mathfrak p \to \Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q$. By our assumption that $\mathfrak q$ is minimal over $\mathfrak h_A$ this means that $H_{C/R_\mathfrak p} \Lambda_\mathfrak q = \Lambda_\mathfrak q$. By Lemma Solving the strict-standard Jacobian relations we may assume that $C$ is smooth over $R_\mathfrak p$. By Lemma Standard presentations of smooth algebras we may assume that $C$ is standard smooth over $R_\mathfrak p$. Write $A = R[x_1, \ldots, x_n]/(g_1, \ldots, g_t)$ and say $A \to \Lambda$ is given by $x_i \mapsto \lambda_i$. Write $C = R_\mathfrak p[x_1, \ldots, x_{n + m}]/(f_1, \ldots, f_c)$ for some $c \geq n$ such that $A \to C$ maps $x_i$ to $x_i$ and such that $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$ is invertible in $C$, see Lemma Including prescribed generators in a smooth presentation. After clearing denominators we may assume $f_1, \ldots, f_c$ are elements of $R[x_1, \ldots, x_{n + m}]$. Of course $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$ is not invertible in $R[x_1, \ldots, x_{n + m}]/(f_1, \ldots, f_c)$ but it becomes invertible after inverting some element $s_0 \in R$, $s_0 \not \in \mathfrak p$. As $g_j$ maps to zero under $R[x_1, \ldots, x_n] \to A \to C$ we can find $s_j \in R$, $s_j \not \in \mathfrak p$ such that $s_j g_j$ is zero in $R[x_1, \ldots, x_{n + m}]/(f_1, \ldots, f_c)$. Write $f_j = F_j(x_1, \ldots, x_{n + m}, 1)$ for some polynomial $F_j \in R[x_1, \ldots, x_n, X_{n + 1}, \ldots, X_{n + m + 1}]$ homogeneous in $X_{n + 1}, \ldots, X_{n + m + 1}$. Pick $\lambda_{n + i} \in \Lambda$, $i = 1, \ldots, m + 1$ with $\lambda_{n + m + 1} \not \in \mathfrak q$ such that $x_{n + i}$ maps to $\lambda_{n + i}/\lambda_{n + m + 1}$ in $\Lambda_\mathfrak q$. Then $$\begin{aligned} F_j(\lambda_1, \ldots, \lambda_{n + m + 1}) & = (\lambda_{n + m + 1})^{\deg(F_j)} F_j(\lambda_1, \ldots, \lambda_n, \frac{\lambda_{n + 1}}{\lambda_{n + m + 1}}, \ldots, \frac{\lambda_{n + m}}{\lambda_{n + m + 1}}, 1) \\ & = (\lambda_{n + m + 1})^{\deg(F_j)} f_j(\lambda_1, \ldots, \lambda_n, \frac{\lambda_{n + 1}}{\lambda_{n + m + 1}}, \ldots, \frac{\lambda_{n + m}}{\lambda_{n + m + 1}}) \\ & = 0 \end{aligned}$$ in $\Lambda_\mathfrak q$. Thus we can find $\lambda_0 \in \Lambda$, $\lambda_0 \not \in \mathfrak q$ such that $\lambda_0 F_j(\lambda_1, \ldots, \lambda_{n + m + 1}) = 0$ in $\Lambda$. Now we set $B$ equal to $$R[x_0, \ldots, x_{n + m + 1}]/ (g_1, \ldots, g_t, x_0F_1(x_1, \ldots, x_{n + m + 1}), \ldots, x_0F_c(x_1, \ldots, x_{n + m + 1}))$$ which we map to $\Lambda$ by mapping $x_i$ to $\lambda_i$. Let $b$ be the image of $x_0 x_{n + m + 1} s_0 s_1 \ldots s_t$ in $B$. Then $B_b$ is isomorphic to $$R_{s_0s_1 \ldots s_t}[x_0, x_1, \ldots, x_{n + m + 1}, 1/x_0x_{n + m + 1}]/ (f_1, \ldots, f_c)$$ which is smooth over $R$ by construction. Since $b$ does not map to an element of $\mathfrak q$, we win. $\square$

Lemma. Returning from a height-zero localization

Let $R \to A \to \Lambda \supset \mathfrak q$ be as in Situation The local desingularization problem. Let $\mathfrak p = R \cap \mathfrak q$. Assume

1. $\mathfrak q$ is minimal over $\mathfrak h_A$,

2. $R_\mathfrak p \to A_\mathfrak p \to \Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q$ can be resolved, and

3. $\dim(\Lambda_\mathfrak q) = 0$.

Then $R \to A \to \Lambda \supset \mathfrak q$ can be resolved.

Proof. By (3) the ring $\Lambda_\mathfrak q$ is Artinian local hence $\mathfrak q\Lambda_\mathfrak q$ is nilpotent. Thus $(\mathfrak h_A)^N \Lambda_\mathfrak q = 0$ for some $N > 0$. Thus there exists a $\lambda \in \Lambda$, $\lambda \not \in \mathfrak q$ such that $\lambda (\mathfrak h_A)^N = 0$ in $\Lambda$. Say $H_{A/R} = (a_1, \ldots, a_r)$ so that $\lambda a_i^N = 0$ in $\Lambda$. By Lemma Returning from a localized target we can find a factorization $A \to C \to \Lambda$ with $C$ of finite presentation such that $\mathfrak h_C \not \subset \mathfrak q$. Write $C = A[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$. Set $$B = A[x_1, \ldots, x_n, y_1, \ldots, y_r, z, t_{ij}]/ (f_j - \sum y_i t_{ij}, zy_i)$$ where $t_{ij}$ is a set of $rm$ variables. Note that there is a map $B \to C[y_i, z]/(y_iz)$ given by setting $t_{ij}$ equal to zero. The map $B \to \Lambda$ is the composition $B \to C[y_i, z]/(y_iz) \to \Lambda$ where $C[y_i, z]/(y_iz) \to \Lambda$ is the given map $C \to \Lambda$, maps $z$ to $\lambda$, and maps $y_i$ to the image of $a_i^N$ in $\Lambda$.

We claim that $B$ is a solution for $R \to A \to \Lambda \supset \mathfrak q$. First note that $B_z$ is isomorphic to $C[z, z^{-1}, t_{ij}]$ as a C-algebra. Choose $c \in H_{C/R}$ whose image in $\Lambda$ is not in $\mathfrak q$. Then $B_{zc}$ is smooth over $R$. On the other hand, $B_{y_\ell} \cong A[x_i, y_i, y_\ell^{-1}, t_{ij}, i \not = \ell]$ which is smooth over $A$. Thus we see that $zc$ and $a_\ell y_\ell$ (compositions of smooth maps are smooth) are all elements of $H_{B/R}$. This proves the lemma. $\square$

Separable residue fields

In this section we explain how to solve a local problem in the case of a separable residue field extension.

Lemma. The annihilator stabilization step (Ogoma)

Let $A$ be a Noetherian ring and let $M$ be a finite $A$-module. Let $S \subset A$ be a multiplicative set. If $\pi \in A$ and $\operatorname{Ker}(\pi : S^{-1}M \to S^{-1}M) = \operatorname{Ker}(\pi^2 : S^{-1}M \to S^{-1}M)$ then there exists an $s \in S$ such that for any $n > 0$ we have $\operatorname{Ker}(s^n\pi : M \to M) = \operatorname{Ker}((s^n\pi)^2 : M \to M)$.

Proof. Let $K = \operatorname{Ker}(\pi : M \to M)$ and $K' = \{m \in M \mid \pi^2 m = 0\text{ in }S^{-1}M\}$ and $Q = K'/K$. Note that $S^{-1}Q = 0$ by assumption. Since $A$ is Noetherian we see that $Q$ is a finite $A$-module. Hence we can find an $s \in S$ such that $s$ annihilates $Q$. If $(s^n\pi)^2m=0$ then $\pi^2m$ vanishes after localization, so $m\in K'$. As $s(K'/K)=0$, we have $s\pi m=0$ and hence $s^n\pi m=0$. The reverse kernel inclusion is immediate, so this $s$ works for every $n>0$. $\square$

Lemma. Parameters adapted to the singularity ideal

Let $\Lambda$ be a Noetherian ring. Let $I \subset \Lambda$ be an ideal. Let $I \subset \mathfrak q$ be a prime. Let $n, e$ be positive integers. Assume that $\mathfrak q^n\Lambda_\mathfrak q \subset I\Lambda_\mathfrak q$ and that $\Lambda_\mathfrak q$ is a regular local ring of dimension $d$. Then there exist $\pi_1, \ldots, \pi_d \in \Lambda$ such that

1. $(\pi_1, \ldots, \pi_d)\Lambda_\mathfrak q = \mathfrak q\Lambda_\mathfrak q$,

2. $\pi_1^n, \ldots, \pi_d^n \in I$, and

3. for $i = 1, \ldots, d$ we have $$\text{Ann}_{\Lambda/(\pi_1^e, \ldots, \pi_{i - 1}^e)\Lambda}(\pi_i) = \text{Ann}_{\Lambda/(\pi_1^e, \ldots, \pi_{i - 1}^e)\Lambda}(\pi_i^2).$$

Proof. Set \(S = \Lambda \setminus \mathfrak q\) so that \(\Lambda_\mathfrak q = S^{-1}\Lambda\). First pick \(\pi_1, \ldots, \pi_d\) with (1) which is possible as \(\Lambda_\mathfrak q\) is regular. By assumption \(\pi_i^n \in I\Lambda_\mathfrak q\). Thus we can find \(s_1, \ldots, s_d \in S\) such that \(s_i\pi_i^n \in I\). Replacing \(\pi_i\) by \(s_i\pi_i\) we get (2). Note that (1) and (2) are preserved by further multiplying by elements of \(S\). Suppose that (3) holds for \(i = 1, \ldots, t\) for some \(t \in \{0, \ldots, d\}\). Note that \(\pi_1, \ldots, \pi_d\) is a regular sequence in \(S^{-1}\Lambda\), see Algebra, Lemma Regular rings are Cohen–Macaulay. In particular \(\pi_1^e, \ldots, \pi_t^e, \pi_{t + 1}\) is a regular sequence in \(S^{-1}\Lambda = \Lambda_\mathfrak q\) by Algebra, Lemma Powers of a regular sequence. Hence we see that

Thus we get (3) for \(i = t + 1\) after replacing \(\pi_{t + 1}\) by \(s\pi_{t + 1}\) for some \(s \in S\) by Lemma The annihilator stabilization step. By induction on \(t\) this produces a sequence satisfying (1), (2), and (3). \(\square\)

Lemma. Desingularization with separable residue fields

Let $k \to A \to \Lambda \supset \mathfrak q$ be as in Situation The local desingularization problem where

1. $k$ is a field,

2. $\Lambda$ is Noetherian,

3. $\mathfrak q$ is minimal over $\mathfrak h_A$,

4. $\Lambda_\mathfrak q$ is a regular local ring, and

5. the field extension $\kappa(\mathfrak q)/k$ is separable.

Then $k \to A \to \Lambda \supset \mathfrak q$ can be resolved.

Proof. Set $d=\dim\Lambda_{\mathfrak q}$. If $d=0$, the localized target is a field separable over $k$, hence a filtered colimit of smooth algebras by Algebra, Lemma Syntomic algebras in a filtered colimit. Factor the finitely presented localized source through one such algebra and apply Lemma Returning from a height-zero localization; this resolves the original situation. We may therefore assume $d>0$. Set $R = k[x_1, \ldots, x_d]$. Choose $n > 0$ such that $\mathfrak q^n\Lambda_\mathfrak q \subset \mathfrak h_A\Lambda_\mathfrak q$ which is possible as $\mathfrak q$ is minimal over $\mathfrak h_A$. Choose generators $a_1, \ldots, a_r$ of $H_{A/k}$. Set $$B = A[x_1, \ldots, x_d, z_{ij}]/(x_i^n - \sum z_{ij}a_j)$$ Each $B_{a_j}$ is smooth over $R$ because it is a polynomial algebra over $A_{a_j}[x_1, \ldots, x_d]$ and $A_{a_j}$ is smooth over $k$. Hence $B_{x_i}$ is smooth over $R$. Let $B \to C$ be the $R$-algebra map constructed in Lemma Improving a finite presentation which comes with a $R$-algebra retraction $C \to B$. In particular a map $C \to \Lambda$ fitting into the diagram below. By construction $C_{x_i}$ is a smooth $R$-algebra with $\Omega_{C_{x_i}/R}$ free. Hence we can find $c > 0$ such that $x_i^c$ is strictly standard in $C/R$, see Lemma Comparing standard smooth presentations. Now choose $\pi_1, \ldots, \pi_d \in \Lambda$ as in Lemma Parameters adapted to the singularity ideal where $n = n$, $e = 8c$, $\mathfrak q = \mathfrak q$ and $I = \mathfrak h_A$. Write $\pi_i^n = \sum \lambda_{ij} a_j$ for some $\lambda_{ij} \in \Lambda$. There is a map $B \to \Lambda$ given by $x_i \mapsto \pi_i$ and $z_{ij} \mapsto \lambda_{ij}$. Set $R = k[x_1, \ldots, x_d]$. Diagram $$\begin{gathered}\begin{matrix}R & B \\ k & A & \Lambda\end{matrix} \\[6pt] \begin{aligned}R & \longrightarrow B \\ B & \longrightarrow \Lambda \\ k & \longrightarrow R \\ k & \longrightarrow A \\ A & \longrightarrow B \\ A & \longrightarrow \Lambda\end{aligned}\end{gathered}$$ Now we apply Lemma Lifting a local desingularization solution to $R \to C \to \Lambda \supset \mathfrak q$ and the sequence of elements $x_1^c, \ldots, x_d^c$ of $R$. Assumption (2) is clear. Assumption (1) holds for $R$ by inspection and for $\Lambda$ by our choice of $\pi_1, \ldots, \pi_d$. (Note that if $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^2)$, then we have $\text{Ann}_\Lambda(\pi) = \text{Ann}_\Lambda(\pi^c)$ for all $c > 0$.) Thus it suffices to resolve $$R/(x_1^e, \ldots, x_d^e) \to C/(x_1^e, \ldots, x_d^e) \to \Lambda/(\pi_1^e, \ldots, \pi_d^e) \supset \mathfrak q/(\pi_1^e, \ldots, \pi_d^e)$$ for $e = 8c$. By Lemma Returning from a height-zero localization it suffices to resolve this after localizing at $\mathfrak q$. But since $x_1, \ldots, x_d$ map to a regular sequence in $\Lambda_\mathfrak q$ we see that $R_\mathfrak p \to \Lambda_\mathfrak q$ is flat, see Algebra, Lemma Flatness over a regular local ring. Hence $$R_\mathfrak p/(x_1^e, \ldots, x_d^e) \to \Lambda_\mathfrak q/(\pi_1^e, \ldots, \pi_d^e)$$ is a flat ring map of Artinian local rings. Moreover, this map induces a separable field extension on residue fields by assumption. Thus this map is a filtered colimit of smooth algebras by Algebra, Lemma Syntomic algebras in a filtered colimit and Proposition Lifting an algebraic factorization. Existence of the desired solution follows from Algebra, Lemma Recognizing a filtered colimit of finite presentations. $\square$

Inseparable residue fields

In this section we explain how to solve a local problem in the case of an inseparable residue field extension.

Lemma. Compatibility of the local approximation data

Let $k$ be a field of characteristic $p > 0$. Let $(\Lambda, \mathfrak m, K)$ be an Artinian local $k$-algebra. Assume that $\dim H_1(L_{K/k}) < \infty$. Then $\Lambda$ is a filtered colimit of Artinian local $k$-algebras $A$ with each map $A \to \Lambda$ flat, with $\mathfrak m_A \Lambda = \mathfrak m$, and with $A$ essentially of finite type over $k$.

Proof. Note that the flatness of $A \to \Lambda$ implies that $A \to \Lambda$ is injective, so the lemma really tells us that $\Lambda$ is a directed union of these types of subrings $A \subset \Lambda$. Let $n$ be the minimal integer such that $\mathfrak m^n = 0$. We will prove this lemma by induction on $n$. The case $n = 1$ is clear as a field extension is a union of finitely generated field extensions.

Pick $\lambda_1, \ldots, \lambda_d \in \mathfrak m$ which generate $\mathfrak m$. As $K$ is formally smooth over $\mathbf{F}_p$ (see Algebra, Lemma Elementary formally smooth extensions) we can find a ring map $\sigma : K \to \Lambda$ which is a section of the quotient map $\Lambda \to K$. In general $\sigma$ is not a $k$-algebra map. Given $\sigma$ we define $$\Psi_\sigma : K[x_1, \ldots, x_d] \longrightarrow \Lambda$$ using $\sigma$ on elements of $K$ and mapping $x_i$ to $\lambda_i$. Claim: there exists a $\sigma : K \to \Lambda$ and a subfield $k \subset F \subset K$ finitely generated over $k$ such that the image of $k$ in $\Lambda$ is contained in $\Psi_\sigma(F[x_1, \ldots, x_d])$.

We will prove the claim by induction on the least integer $n$ such that $\mathfrak m^n = 0$. It is clear for $n = 1$. If $n > 1$ set $I = \mathfrak m^{n - 1}$ and $\Lambda' = \Lambda/I$. By induction we may assume given $\sigma' : K \to \Lambda'$ and $k \subset F' \subset K$ finitely generated such that the image of $k \to \Lambda \to \Lambda'$ is contained in $A' = \Psi_{\sigma'}(F'[x_1, \ldots, x_d])$. Denote $\tau' : k \to A'$ the induced map. Choose a lift $\sigma : K \to \Lambda$ of $\sigma'$ (this is possible by the formal smoothness of $K/\mathbf{F}_p$ we mentioned above). For later reference we note that we can change $\sigma$ to $\sigma + D$ for some derivation $D : K \to I$. Set $A = F'[x_1, \ldots, x_d]/(x_1, \ldots, x_d)^n$. Then $\Psi_\sigma$ induces a ring map $\Psi_\sigma : A \to \Lambda$. The composition with the quotient map $\Lambda \to \Lambda'$ induces a surjective map $A \to A'$ with nilpotent kernel. Choose a lift $\tau : k \to A$ of $\tau'$ (possible as $k/\mathbf{F}_p$ is formally smooth). Thus we obtain two maps $k \to \Lambda$, namely $\Psi_\sigma \circ \tau : k \to \Lambda$ and the given map $i : k \to \Lambda$. These maps agree modulo $I$, whence the difference is a derivation $\theta = i - \Psi_\sigma \circ \tau : k \to I$. Note that if we change $\sigma$ into $\sigma + D$ then we change $\theta$ into $\theta - D|_k$.

Choose a set of elements $\{y_j\}_{j \in J}$ of $k$ whose differentials $\text{d}y_j$ form a basis of $\Omega_{k/\mathbf{F}_p}$. The Jacobi-Zariski sequence for $\mathbf{F}_p \subset k \subset K$ is $$0 \to H_1(L_{K/k}) \to \Omega_{k/\mathbf{F}_p} \otimes K \to \Omega_{K/\mathbf{F}_p} \to \Omega_{K/k} \to 0$$ As $\dim H_1(L_{K/k}) < \infty$ we can find a finite subset $J_0 \subset J$ such that the image of the first map is contained in $\bigoplus_{j \in J_0} K\text{d}y_j$. Hence the elements $\text{d}y_j$, $j \in J \setminus J_0$ map to $K$-linearly independent elements of $\Omega_{K/\mathbf{F}_p}$. Therefore we can choose a $D : K \to I$ such that $\theta - D|_k = \xi \circ \text{d}$ where $\xi$ is a composition $$\Omega_{k/\mathbf{F}_p} = \bigoplus\nolimits_{j \in J} k \text{d}y_j \longrightarrow \bigoplus\nolimits_{j \in J_0} k \text{d}y_j \longrightarrow I$$ Let $f_j = \xi(\text{d}y_j) \in I$ for $j \in J_0$. Change $\sigma$ into $\sigma + D$ as above. Then we see that $\theta(a) = \sum_{j \in J_0} a_j f_j$ for $a \in k$ where $\text{d}a = \sum a_j \text{d}y_j$ in $\Omega_{k/\mathbf{F}_p}$. Note that $I$ is generated by the monomials $\lambda^E = \lambda_1^{e_1} \ldots \lambda_d^{e_d}$ of total degree $|E| = \sum e_i = n - 1$ in $\lambda_1, \ldots, \lambda_d$. Write $f_j = \sum_E c_{j, E} \lambda^E$ with $c_{j, E} \in K$. Replace $F'$ by $F = F'(c_{j, E})$. Then the claim holds.

Choose $\sigma$ and $F$ as in the claim. The kernel of $\Psi_\sigma$ is generated by finitely many polynomials $g_1, \ldots, g_t \in K[x_1, \ldots, x_d]$ and we may assume their coefficients are in $F$ after enlarging $F$ by adjoining finitely many elements. In this case it is clear that the map $A = F[x_1, \ldots, x_d]/(g_1, \ldots, g_t) \to K[x_1, \ldots, x_d]/(g_1, \ldots, g_t) = \Lambda$ is flat. By the claim $A$ is a $k$-subalgebra of $\Lambda$. It is clear that $\Lambda$ is the filtered colimit of these algebras, as $K$ is the filtered union of the subfields $F$. Finally, these algebras are essentially of finite type over $k$ by Algebra, Lemma Finite generation in an Artinian local target. $\square$

Lemma. A factorization modulo a prescribed ideal

Let $k$ be a field of characteristic $p > 0$. Let $\Lambda$ be a Noetherian geometrically regular $k$-algebra. Let $\mathfrak q \subset \Lambda$ be a prime ideal. Let $n \geq 1$ be an integer and let $E \subset \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ be a finite subset. Then we can find $m \geq 0$ and $\varphi : k[y_1, \ldots, y_m] \to \Lambda$ with the following properties

1. setting $\mathfrak p = \varphi^{-1}(\mathfrak q)$ we have $\mathfrak q\Lambda_\mathfrak q = \mathfrak p \Lambda_\mathfrak q$ and $k[y_1, \ldots, y_m]_\mathfrak p \to \Lambda_\mathfrak q$ is flat,

2. there is a factorization by homomorphisms of local Artinian rings $$k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p \to D \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$$ where the first arrow is essentially smooth and the second is flat,

3. $E$ is contained in $D$ modulo $\mathfrak q^n\Lambda_\mathfrak q$.

Proof. Set $\bar \Lambda = \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$. Note that $\dim H_1(L_{\kappa(\mathfrak q)/k}) < \infty$ by More on Algebra, Proposition Characterizations of geometric regularity. Pick $A \subset \bar \Lambda$ containing $E$ such that $A$ is local Artinian, essentially of finite type over $k$, the map $A \to \bar \Lambda$ is flat, and $\mathfrak m_A$ generates the maximal ideal of $\bar \Lambda$, see Lemma Compatibility of the local approximation data. Denote $F = A/\mathfrak m_A$ the residue field so that $k \subset F \subset K$. Pick $\lambda_1, \ldots, \lambda_t \in \Lambda$ which map to elements of $A$ in $\bar \Lambda$ such that moreover the images of $\text{d}\lambda_1, \ldots, \text{d}\lambda_t$ form a basis of $\Omega_{F/k}$. Consider the map $\varphi' : k[y_1, \ldots, y_t] \to \Lambda$ sending $y_j$ to $\lambda_j$. Set $\mathfrak p' = (\varphi')^{-1}(\mathfrak q)$. By More on Algebra, Lemma Geometric regularity over a field the ring map $k[y_1, \ldots, y_t]_{\mathfrak p'} \to \Lambda_\mathfrak q$ is flat and $\Lambda_\mathfrak q/\mathfrak p' \Lambda_\mathfrak q$ is regular. Thus we can choose further elements $\lambda_{t + 1}, \ldots, \lambda_m \in \Lambda$ which map into $A \subset \bar \Lambda$ and which map to a regular system of parameters of $\Lambda_\mathfrak q/\mathfrak p' \Lambda_\mathfrak q$. We obtain $\varphi : k[y_1, \ldots, y_m] \to \Lambda$ having property (1) such that $k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p \to \bar\Lambda$ factors through $A$. Thus $k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p \to A$ is flat by Algebra, Lemma Descent of flatness. By construction the residue field extension $F/\kappa(\mathfrak p)$ is finitely generated and $\Omega_{F/\kappa(\mathfrak p)} = 0$. Hence it is finite separable by More on Algebra, Lemma Cartier's equality for differentials. Thus $k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p \to A$ is finite by Algebra, Lemma Finite generation in an Artinian local target. Finally, we conclude that it is étale by Algebra, Lemma Characterizations of étale algebras. Since an étale ring map is certainly essentially smooth we win. $\square$

Lemma. Enlarging a factorization in positive characteristic

Let $\varphi : k[y_1, \ldots, y_m] \to \Lambda$, $n$, $\mathfrak q$, $\mathfrak p$ and $$k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n \to D \to \Lambda_\mathfrak q/\mathfrak q^n \Lambda_\mathfrak q$$ be as in Lemma A factorization modulo a prescribed ideal. Then for any $\lambda \in \Lambda \setminus \mathfrak q$ there exists an integer $q > 0$ and a factorization $$k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n \to D \to D' \to \Lambda_\mathfrak q/\mathfrak q^n \Lambda_\mathfrak q$$ such that $D \to D'$ is an essentially smooth map of local Artinian rings, the last arrow is flat, and $\lambda^q$ is in $D'$.

Proof. Set $\bar \Lambda = \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$. Let $\bar \lambda$ be the image of $\lambda$ in $\bar \Lambda$. Let $\alpha \in \kappa(\mathfrak q)$ be the image of $\lambda$ in the residue field. Let $k \subset F \subset \kappa(\mathfrak q)$ be the residue field of $D$. If $\alpha$ is in $F$ then we can find an $x \in D$ such that $x \bar\lambda = 1 \bmod \mathfrak q$. Hence $(x\bar\lambda)^q=1$ if $q=p^r$ is at least the nilpotence order of $\mathfrak q\bar\Lambda$: Frobenius gives $(1+u)^{p^r}=1+u^{p^r}=1$. Hence $\bar\lambda^q$ is in $D$. If $\alpha$ is transcendental over $F$, then we can take $D' = (D[\bar \lambda])_\mathfrak m$ equal to the subring generated by $D$ and $\bar \lambda$ localized at $\mathfrak m = D[\bar \lambda] \cap \mathfrak q \bar \Lambda$. This works because $D[\bar \lambda]$ is in fact a polynomial algebra over $D$ in this case. Finally, if $\lambda \bmod \mathfrak q$ is algebraic over $F$, then we can find a $p$-power $q$ such that $\alpha^q$ is separable algebraic over $F$, see the following elementary field argument. The irreducible polynomial of $\alpha$ has the form $h(T^{p^r})$ with $h'\ne0$, after extracting all common powers of $p$ from its exponents; thus $\alpha^{p^r}$ is separable over $F$. Note that $D$ and $\bar\Lambda$ are henselian local rings, see Algebra, Lemma Henselianity in local dimension zero. Let $D \to D'$ be a finite étale extension whose residue field extension is $F(\alpha^q)/F$, see Algebra, Lemma Finite étale algebras over a henselian ring. Since $\bar\Lambda$ is henselian and $F(\alpha^q)$ is contained in its residue field we can find a factorization $D' \to \bar \Lambda$. By the first part of the argument we see that $\bar\lambda^{qq'} \in D'$ for some $q' > 0$. $\square$

Lemma. Desingularization with inseparable residue fields

Let $k \to A \to \Lambda \supset \mathfrak q$ be as in Situation The local desingularization problem where

1. $k$ is a field of characteristic $p > 0$,

2. $\Lambda$ is Noetherian and geometrically regular over $k$,

3. $\mathfrak q$ is minimal over $\mathfrak h_A$.

Then $k \to A \to \Lambda \supset \mathfrak q$ can be resolved.

Proof. If $d=\dim\Lambda_{\mathfrak q}=0$, apply Lemma A factorization modulo a prescribed ideal at order $n=1$, with $E$ the images of algebra generators of $A$. The local target is a field and the polynomial source localizes at a height-zero prime, so its local ring is a field. The resulting $D$ is essentially smooth over that source and contains those generators. Thus $A$ localized at this base factors through a smooth stage; finite-presentation factorization followed by Lemma Returning from a height-zero localization yields the resolution. This is also the construction below with no parameters and no auxiliary $t_i$; the empty lifting step is omitted. Hence assume $d>0$ below.

The lemma is proven by the following steps in the given order. We will justify each of these steps below.

1. Pick an integer $N > 0$ such that $\mathfrak q^N\Lambda_\mathfrak q \subset H_{A/k}\Lambda_\mathfrak q$.

2. Pick generators $a_1, \ldots, a_t \in A$ of the ideal $H_{A/k}$.

Set $d = \dim(\Lambda_\mathfrak q)$.

Set $B = A[x_1, \ldots, x_d, z_{ij}]/(x_i^{2N} - \sum z_{ij}a_j)$.

5. Consider $B$ as a $k[x_1, \ldots, x_d]$-algebra and let $B \to C$ be as in Lemma Improving a finite presentation. We also obtain a section $C \to B$.

Choose $c > 0$ such that each $x_i^c$ is strictly standard in $C$ over $k[x_1, \ldots, x_d]$.

7. Set $e=8c$ and choose $n\geq\max\{N+dc,d(e-1)+1\}$.

8. Let $E \subset \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ be the images of generators of $A$ as a $k$-algebra.

Choose an integer $m$ and a $k$-algebra map $\varphi : k[y_1, \ldots, y_m] \to \Lambda$ and a factorization by local Artinian rings $$k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p
\to D \to
\Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$$ such that the first arrow is essentially smooth, the second is flat, $E$ is contained in $D$, with $\mathfrak p = \varphi^{-1}(\mathfrak q)$ the map $k[y_1, \ldots, y_m]_\mathfrak p \to \Lambda_\mathfrak q$ is flat, and $\mathfrak p \Lambda_\mathfrak q = \mathfrak q \Lambda_\mathfrak q$.

Choose $\pi_1, \ldots, \pi_d \in \mathfrak p$ which map to a regular system of parameters of $k[y_1, \ldots, y_m]_\mathfrak p$.

11. Let $R = k[y_1, \ldots, y_m, t_1, \ldots, t_d]$ and $\gamma_i = \pi_i t_i$.

12. If necessary modify the choice of $\pi_i$ such that for $i = 1, \ldots, d$ we have $$\text{Ann}_{R/(\gamma_1^e, \ldots, \gamma_{i - 1}^e)R}(\gamma_i) = \text{Ann}_{R/(\gamma_1^e, \ldots, \gamma_{i - 1}^e)R}(\gamma_i^2)$$

There exist $\delta_1, \ldots, \delta_d \in \Lambda$, $\delta_i \not \in \mathfrak q$ and a factorization $D \to D' \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ with $D'$ local Artinian, $D \to D'$ essentially smooth, the map $D' \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ flat such that, with $\pi_i' = \delta_i \pi_i$, we have for $i = 1, \ldots, d$

1.  $(\pi_i')^{2N} = \sum a_j\lambda_{ij}$ in $\Lambda$ where $\lambda_{ij} \bmod \mathfrak q^n\Lambda_\mathfrak q$ is an element of $D'$,

2.  $\text{Ann}_{\Lambda/({\pi'}_1^e, \ldots, {\pi'}_{i - 1}^e)}({\pi'}_i) =
    \text{Ann}_{\Lambda/({\pi'}_1^e, \ldots, {\pi'}_{i - 1}^e)}({\pi'}_i^2)$,

3.  $\delta_i \bmod \mathfrak q^n\Lambda_\mathfrak q$ is an element of $D'$.

14. Define $B \to \Lambda$ by sending $x_i$ to $\pi'_i$ and $z_{ij}$ to $\lambda_{ij}$ found above. Define $C \to \Lambda$ by composing the map $B \to \Lambda$ with the retraction $C \to B$.

15. Map $R \to \Lambda$ by $\varphi$ on $k[y_1, \ldots, y_m]$ and by sending $t_i$ to $\delta_i$. Further introduce a map $$k[x_1, \ldots, x_d] \longrightarrow R = k[y_1, \ldots, y_m, t_1, \ldots, t_d]$$ by sending $x_i$ to $\gamma_i = \pi_i t_i$.

16. It suffices to resolve $$R \to C \otimes_{k[x_1, \ldots, x_d]} R \to \Lambda \supset \mathfrak q$$

17. Set $I = (\gamma_1^e, \ldots, \gamma_d^e) \subset R$.

18. It suffices to resolve $$R/I \to C \otimes_{k[x_1, \ldots, x_d]} R/I \to \Lambda/I\Lambda \supset \mathfrak q/I\Lambda$$

19. We denote $\mathfrak r \subset R = k[y_1, \ldots, y_m, t_1, \ldots, t_d]$ the inverse image of $\mathfrak q$.

20. It suffices to resolve $$(R/I)_\mathfrak r \to C \otimes_{k[x_1, \ldots, x_d]} (R/I)_\mathfrak r \to \Lambda_\mathfrak q/I\Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q/I\Lambda_\mathfrak q$$

Set $J = (\pi_1^e, \ldots, \pi_d^e)$ in $k[y_1, \ldots, y_m]$.

It suffices to resolve $$(R/JR)_\mathfrak p \to
C \otimes_{k[x_1, \ldots, x_d]} (R/JR)_\mathfrak p \to
\Lambda_\mathfrak q/J\Lambda_\mathfrak q
\supset
\mathfrak q\Lambda_\mathfrak q/J\Lambda_\mathfrak q$$

23. It suffices to resolve $$(R/\mathfrak p^nR)_\mathfrak p \to C \otimes_{k[x_1, \ldots, x_d]} (R/\mathfrak p^nR)_\mathfrak p \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$$

24. It suffices to resolve $$(R/\mathfrak p^nR)_\mathfrak p \to B \otimes_{k[x_1, \ldots, x_d]} (R/\mathfrak p^nR)_\mathfrak p \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$$

The ring $D'[t_1, \ldots, t_d]$ is given the structure of an $R_\mathfrak p/\mathfrak p^nR_\mathfrak p$-algebra by the given map $k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p
\to D'$ and by sending $t_i$ to $t_i$. It suffices to find a factorization $$B \otimes_{k[x_1, \ldots, x_d]} (R/\mathfrak p^nR)_\mathfrak p
\to D'[t_1, \ldots, t_d] \to
\Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$$ where the second arrow sends $t_i$ to $\delta_i$ and induces the given homomorphism $D' \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$.

26. Such a factorization exists by our choice of $D'$ above.

We now give the justification for each of the steps, except that we skip justifying the steps which just introduce notation.

Ad (the indicated step). This is possible as $\mathfrak q$ is minimal over $\mathfrak h_A = \sqrt{H_{A/k}\Lambda}$.

Ad (the indicated step). Note that $A_{a_j}$ is smooth over $k$. Hence $B_{a_j}$, which is isomorphic to a polynomial algebra over $A_{a_j}[x_1, \ldots, x_d]$, is smooth over $k[x_1, \ldots, x_d]$. Thus $B_{x_i}$ is smooth over $k[x_1, \ldots, x_d]$. By Lemma Improving a finite presentation we see that $C_{x_i}$ is smooth over $k[x_1, \ldots, x_d]$ with finite free module of differentials. Hence some power of $x_i$ is strictly standard in $C$ over $k[x_1, \ldots, x_d]$ by Lemma Comparing standard smooth presentations.

Ad (the indicated step). This follows by applying Lemma A factorization modulo a prescribed ideal.

Ad (the indicated step). Since $k[y_1, \ldots, y_m]_\mathfrak p \to \Lambda_\mathfrak q$ is flat and $\mathfrak p \Lambda_\mathfrak q = \mathfrak q \Lambda_\mathfrak q$ by construction we see that $\dim(k[y_1, \ldots, y_m]_\mathfrak p) = d$ by Algebra, Lemma Dimension of a flat family. Thus we can find $\pi_1, \ldots, \pi_d \in \mathfrak p$ which map to a regular system of parameters in $k[y_1, \ldots, y_m]_\mathfrak p$.

Ad (the indicated step). By Algebra, Lemma Regular rings are Cohen–Macaulay any permutation of the sequence \(\pi_1, \ldots, \pi_d\) is a regular sequence in \(k[y_1, \ldots, y_m]_\mathfrak p\). Hence \(\gamma_1 = \pi_1 t_1, \ldots, \gamma_d = \pi_d t_d\) is a regular sequence in \(R_\mathfrak p = k[y_1, \ldots, y_m]_\mathfrak p[t_1, \ldots, t_d]\), see Algebra, Lemma Regular sequences in a polynomial ring. Let \(S = k[y_1, \ldots, y_m] \setminus \mathfrak p\) so that \(R_\mathfrak p = S^{-1}R\). Note that \(\pi_1, \ldots, \pi_d\) and \(\gamma_1, \ldots, \gamma_d\) remain regular sequences if we multiply our \(\pi_i\) by elements of \(S\). Suppose that

holds for \(i = 1, \ldots, t\) for some \(t \in \{0, \ldots, d\}\). Note that \(\gamma_1^e, \ldots, \gamma_t^e, \gamma_{t + 1}\) is a regular sequence in \(S^{-1}R\) by Algebra, Lemma Powers of a regular sequence. Hence we see that

Thus we get

after replacing \(\pi_{t + 1}\) by \(s\pi_{t + 1}\) for some \(s \in S\) by Lemma The annihilator stabilization step. By induction on \(t\) this produces the desired sequence.

Ad (the indicated step). Let \(S = \Lambda \setminus \mathfrak q\) so that \(\Lambda_\mathfrak q = S^{-1}\Lambda\). Set \(\bar \Lambda = \Lambda_\mathfrak q/\mathfrak q^n \Lambda_\mathfrak q\). Suppose that we have a \(t \in \{0, \ldots, d\}\) and \(\delta_1, \ldots, \delta_t \in S\) and a factorization \(D \to D' \to \bar \Lambda\) as in (the indicated step) such that (a), (b), (c) hold for \(i = 1, \ldots, t\). We have \(\pi_{t + 1}^N \in H_{A/k}\Lambda_\mathfrak q\) as \(\mathfrak q^N \Lambda_\mathfrak q \subset H_{A/k}\Lambda_\mathfrak q\) by (the indicated step). Hence \(\pi_{t + 1}^N \in H_{A/k} \bar\Lambda\). Hence \(\pi_{t + 1}^N \in H_{A/k}D'\) as \(D' \to \bar \Lambda\) is faithfully flat, see Algebra, Lemma Universal injectivity of a faithfully flat ring map. Recall that \(H_{A/k} = (a_1, \ldots, a_t)\). Say \(\pi_{t + 1}^N = \sum a_j d_j\) in \(D'\) and choose \(c_j \in \Lambda_\mathfrak q\) lifting \(d_j \in D'\). Then \(\pi_{t + 1}^N = \sum c_j a_j + \epsilon\) with \(\epsilon \in \mathfrak q^n\Lambda_\mathfrak q \subset \mathfrak q^{n - N}H_{A/k}\Lambda_\mathfrak q\). Write \(\epsilon = \sum a_j c'_j\) for some \(c'_j \in \mathfrak q^{n - N}\Lambda_\mathfrak q\). Hence \(\pi_{t + 1}^{2N} = \sum (\pi_{t + 1}^N c_j + \pi_{t + 1}^N c'_j) a_j\). Note that \(\pi_{t + 1}^Nc'_j\) maps to zero in \(\bar \Lambda\); this trivial but key observation will ensure later that (a) holds. Now we choose \(s \in S\) such that there exist \(\mu_{t + 1j} \in \Lambda\) such that on the one hand \(\pi_{t + 1}^N c_j + \pi_{t + 1}^N c'_j = \mu_{t + 1j}/s^{2N}\) in \(S^{-1}\Lambda\) and on the other \((s \pi_{t + 1})^{2N} = \sum \mu_{t + 1j}a_j\) in \(\Lambda\). To verify this denominator step, write all coefficients \(v_j=\pi_{t+1}^Nc_j+\pi_{t+1}^Nc'_j\) as \(a'_j/s_0\) with a common \(s_0\in S\). The localized equality has an error killed by some \(u\in S\). Choose \(s\) divisible by \(us_0\) and set \(\mu_{t+1,j}=(s^{2N}/s_0)a'_j\). Multiplying the error by \(s^{2N}/s_0\) kills it and gives the asserted equality in \(\Lambda\). We may further replace \(s\) by a power and enlarge \(D'\) such that \(s\) maps to an element of \(D'\). With these choices \(\mu_{t+1,j}\) maps to \(s^{2N}\pi_{t+1}^Nd_j\), an element of \(D'\): both \(\pi_{t+1}\) and \(d_j\) are in \(D'\), and we arranged that the image of \(s\) is in \(D'\). Note that \(\pi_1, \ldots, \pi_d\) are a regular sequence of parameters in \(S^{-1}\Lambda\) by our choice of \(\varphi\). Hence \(\pi_1, \ldots, \pi_d\) forms a regular sequence in \(\Lambda_\mathfrak q\) by Algebra, Lemma Regular rings are Cohen–Macaulay. It follows that \({\pi'}_1^e, \ldots, {\pi'}_t^e, s\pi_{t + 1}\) is a regular sequence in \(S^{-1}\Lambda\) by Algebra, Lemma Powers of a regular sequence. Thus we get

Hence we may apply Lemma The annihilator stabilization step to find an \(s' \in S\) such that

for any \(q > 0\). By Lemma Enlarging a factorization in positive characteristic we can choose \(q\) and enlarge \(D'\) such that \((s')^q\) maps to an element of \(D'\). Setting \(\delta_{t + 1} = (s')^qs\) we conclude that (a), (b), (c) hold for \(i = 1, \ldots, t + 1\). For (a) note that \(\lambda_{t + 1j} = (s')^{2Nq}\mu_{t + 1j}\) works. By induction on \(t\) we win.

Ad (the indicated step). By construction the radical of $H_{(C \otimes_{k[x_1, \ldots, x_d]} R)/R} \Lambda$ contains $\mathfrak h_A$. Namely, the elements $a_j \in H_{A/k}$ map to elements of $H_{B/k[x_1, \ldots, x_d]}$, hence map to elements of $H_{C/k[x_1, \ldots, x_d]}$, hence $a_j \otimes 1$ map to elements of $H_{C \otimes_{k[x_1, \ldots, x_d]} R/R}$. Moreover, if we have a solution $C \otimes_{k[x_1, \ldots, x_d]} R \to T \to \Lambda$ of $$R \to C \otimes_{k[x_1, \ldots, x_d]} R \to \Lambda \supset \mathfrak q$$ then $H_{T/R} \subset H_{T/k}$ as $R$ is smooth over $k$. Hence $T$ will also be a solution for the original situation $k \to A \to \Lambda \supset \mathfrak q$.

Ad (the indicated step). Follows on applying Lemma Lifting a local desingularization solution to $R \to C \otimes_{k[x_1, \ldots, x_d]} R \to \Lambda \supset \mathfrak q$ and the sequence of elements $\gamma_1^c, \ldots, \gamma_d^c$. We note that since $x_i^c$ are strictly standard in $C$ over $k[x_1, \ldots, x_d]$ the elements $\gamma_i^c$ are strictly standard in $C \otimes_{k[x_1, \ldots, x_d]} R$ over $R$ by Lemma Base change of a strict standard presentation. The other assumption of Lemma Lifting a local desingularization solution holds by steps (the indicated step) and (the indicated step).

Ad (the indicated step). Apply Lemma Returning from a height-zero localization to the situation in (the indicated step). In the rest of the arguments the target ring is local Artinian, hence we are looking for a factorization by a smooth algebra $T$ over the source ring.

Ad (the indicated step). Suppose that $C \otimes_{k[x_1, \ldots, x_d]} (R/JR)_\mathfrak p \to T \to \Lambda_\mathfrak q/J\Lambda_\mathfrak q$ is a solution to $$(R/JR)_\mathfrak p \to C \otimes_{k[x_1, \ldots, x_d]} (R/JR)_\mathfrak p \to \Lambda_\mathfrak q/J\Lambda_\mathfrak q \supset \mathfrak q\Lambda_\mathfrak q/J\Lambda_\mathfrak q$$ Each $t_i$ is a unit at $\mathfrak r$, since its image is $\delta_i\notin\mathfrak q$. Hence $I_{\mathfrak r}=JR_{\mathfrak r}$ and $I\Lambda_{\mathfrak q}=J\Lambda_{\mathfrak q}$. Localizing the given solution along $R\setminus\mathfrak r$ therefore gives $C\otimes_{k[x_1,\ldots,x_d]}(R/I)_{\mathfrak r}\to T_{\mathfrak r}\to \Lambda_{\mathfrak q}/I\Lambda_{\mathfrak q}$, with $T_{\mathfrak r}$ smooth over $(R/I)_{\mathfrak r}$. This is the required solution.

Ad (the indicated step). Our choice $n\geq d(e-1)+1$ is large enough so that $\mathfrak p^nk[y_1, \ldots, y_m]_\mathfrak p \subset J_\mathfrak p$ and $\mathfrak q^n \Lambda_\mathfrak q \subset J\Lambda_\mathfrak q$. Hence if we have a solution $C \otimes_{k[x_1, \ldots, x_d]} (R/\mathfrak p^nR)_\mathfrak p \to T \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ of (the indicated step) then we can take $T/JT$ as the solution for (the indicated step). Indeed the regular parameters generate the localized maximal ideal, and each monomial of degree $d(e-1)+1$ has some exponent at least $e$, proving the two asserted ideal inclusions.

Ad (the indicated step). This is true because we have a section $C \to B$ in the category of $R$-algebras.

Ad (the indicated step). This is true because $D'$ is essentially smooth over the local Artinian ring $k[y_1, \ldots, y_m]_\mathfrak p/\mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p$ and $$R_\mathfrak p/\mathfrak p^nR_\mathfrak p = k[y_1, \ldots, y_m]_\mathfrak p/ \mathfrak p^n k[y_1, \ldots, y_m]_\mathfrak p[t_1, \ldots, t_d].$$ Hence $D'[t_1, \ldots, t_d]$ is a filtered colimit of smooth $R_\mathfrak p/\mathfrak p^nR_\mathfrak p$-algebras and $B \otimes_{k[x_1, \ldots, x_d]} (R_\mathfrak p/\mathfrak p^nR_\mathfrak p)$ factors through one of these.

Ad (the indicated step). The final twist of the proof is that we cannot just use the map $B \to D'$ which maps $x_i$ to the image of $\pi_i'$ in $D'$ and $z_{ij}$ to the image of $\lambda_{ij}$ in $D'$ because we need the diagram $$\begin{gathered}\begin{matrix}B & D'[t_1, \ldots, t_d] \\ k[x_1, \ldots, x_d] & R_\mathfrak p/\mathfrak p^nR_\mathfrak p\end{matrix} \\[6pt] \begin{aligned}B & \longrightarrow D'[t_1, \ldots, t_d] \\ k[x_1, \ldots, x_d] & \longrightarrow R_\mathfrak p/\mathfrak p^nR_\mathfrak p \\ k[x_1, \ldots, x_d] & \longrightarrow B \\ R_\mathfrak p/\mathfrak p^nR_\mathfrak p & \longrightarrow D'[t_1, \ldots, t_d]\end{aligned}\end{gathered}$$ to commute and we need the composition $B \to D'[t_1, \ldots, t_d] \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$ to be the map of (the indicated step). This requires us to map $x_i$ to the image of $\pi_i t_i$ in $D'[t_1, \ldots, t_d]$. Hence we map $z_{ij}$ to the image of $\lambda_{ij} t_i^{2N} / \delta_i^{2N}$ in $D'[t_1, \ldots, t_d]$ which is defined because $\delta_i\in D'$ has nonzero residue under the faithfully flat local map $D'\to\bar\Lambda$, so is a unit. The relation follows on multiplying $(\pi_i\delta_i)^{2N}=\sum_j a_j\lambda_{ij}$ by $t_i^{2N}/\delta_i^{2N}$. The image of $x_i$ is $\pi_it_i$, giving the specified base algebra map; evaluation $t_i\mapsto\delta_i$ sends $z_{ij}$ back to $\lambda_{ij}$. Thus both triangles commute. $\square$

The main theorem

In this section we wrap up the discussion.

Theorem. General Néron desingularization (Popescu)

Any regular homomorphism of Noetherian rings is a filtered colimit of smooth ring maps.

Proof. By Lemma Reducing desingularization to field bases it suffices to prove this for $k \to \Lambda$ where $\Lambda$ is Noetherian and geometrically regular over $k$. Let $k \to A \to \Lambda$ be a factorization with $A$ a finite type $k$-algebra. It suffices to construct a factorization $A \to B \to \Lambda$ with $B$ of finite type such that $\mathfrak h_B = \Lambda$, see Lemma Solving the strict-standard Jacobian relations. Hence we may perform Noetherian induction on the ideal $\mathfrak h_A$. Pick a prime $\mathfrak q \supset \mathfrak h_A$ such that $\mathfrak q$ is minimal over $\mathfrak h_A$. It now suffices to resolve $k \to A \to \Lambda \supset \mathfrak q$ (as defined in the text following Situation The local desingularization problem). If the characteristic of $k$ is zero, this follows from Lemma Desingularization with separable residue fields. If the characteristic of $k$ is $p > 0$, this follows from Lemma Desingularization with inseparable residue fields. $\square$

The approximation property for G-rings

Let $R$ be a Noetherian local ring. In this case $R$ is a G-ring if and only if the ring map $R \to R^\wedge$ is regular, see More on Algebra, Lemma Testing the G-ring property at maximal ideals. In this case it is true that the henselization $R^h$ and the strict henselization $R^{sh}$ of $R$ are G-rings, see More on Algebra, Lemma The G-ring property under henselization. Moreover, any algebra essentially of finite type over a field, over a complete local ring, over $\mathbf{Z}$, or over a characteristic zero Dedekind ring is a G-ring, see More on Algebra, Proposition Examples and permanence of G-rings. This gives an ample supply of rings to which the result below applies.

Let $R$ be a ring. Let $f_1, \ldots, f_m \in R[x_1, \ldots, x_n]$. Let $S$ be an $R$-algebra. In this situation we say a vector $(a_1, \ldots, a_n) \in S^n$ is a solution in $S$ if and only if $$f_j(a_1, \ldots, a_n) = 0 \text{ in } S, \text{ for } j = 1, \ldots, m$$ Of course an important question in algebraic geometry is to see when systems of polynomial equations have solutions. The following theorem tells us that having solutions in the completion of a local Noetherian ring is often enough to show there exist solutions in the henselization of the ring.

Theorem. Artin approximation for henselian local G-rings

Let $R$ be a Noetherian local ring. Let $f_1, \ldots, f_m \in R[x_1, \ldots, x_n]$. Suppose that $(a_1, \ldots, a_n) \in (R^\wedge)^n$ is a solution in $R^\wedge$. If $R$ is a henselian G-ring, then for every integer $N$ there exists a solution $(b_1, \ldots, b_n) \in R^n$ in $R$ such that $a_i - b_i \in \mathfrak m^NR^\wedge$.

Proof. Let $c_i \in R$ be an element such that $a_i - c_i \in \mathfrak m^N$. Choose generators $\mathfrak m^N = (d_1, \ldots, d_M)$. Write $a_i = c_i + \sum a_{i, l} d_l$. Consider the polynomial ring $R[x_{i, l}]$ and the elements $$g_j = f_j(c_1 + \sum x_{1, l} d_l , \ldots, c_n + \sum x_{n, l} d_l) \in R[x_{i, l}]$$ The system of equations $g_j = 0$ has the solution $(a_{i, l})$. Suppose that we can show that the system of equations $g_j = 0$ has a solution $(b_{i, l})$ in $R$. Then it follows that $b_i = c_i + \sum b_{i, l}d_l$ is a solution of $f_j = 0$ which is congruent to $a_i$ modulo $\mathfrak m^N$. Thus it suffices to show that solvability over $R^\wedge$ implies solvability over $R$.

Let $A \subset R^\wedge$ be the $R$-subalgebra generated by $a_1, \ldots, a_n$. Since we've assumed $R$ is a G-ring, i.e., that $R \to R^\wedge$ is regular, we see that there exists a factorization $$A \to B \to R^\wedge$$ with $B$ smooth over $R$, see Theorem General Néron desingularization. Denote $\kappa = R/\mathfrak m$ the residue field. It is also the residue field of $R^\wedge$, so we get a commutative diagram $$\begin{gathered}\begin{matrix}B & R' \\ R & \kappa\end{matrix} \\[6pt] \begin{aligned}B & \longrightarrow \kappa \\ B & \cdots\!\!\rightarrow R' \\ R' & \cdots\!\!\rightarrow \kappa \\ R & \longrightarrow \kappa \\ R & \longrightarrow B\end{aligned}\end{gathered}$$ Since the vertical arrow is smooth, More on Algebra, Lemma Lifting a section of a smooth morphism implies that there exists an étale ring map $R \to R'$ which induces an isomorphism $R/\mathfrak m \to R'/\mathfrak mR'$ and an $R$-algebra map $B \to R'$ making the diagram above commute. Since $R$ is henselian we see that $R \to R'$ has a section, see Algebra, Lemma Characterizations of henselian local rings. Let $b_i \in R$ be the image of $a_i$ under the ring maps $A \to B \to R' \to R$. Since all of these maps are $R$-algebra maps, we see that $(b_1, \ldots, b_n)$ is a solution in $R$. $\square$

Given a Noetherian local ring $(R, \mathfrak m)$, an étale ring map $R \to R'$, and a maximal ideal $\mathfrak m' \subset R'$ lying over $\mathfrak m$ with $\kappa(\mathfrak m) = \kappa(\mathfrak m')$, then we have inclusions $$R \subset R_{\mathfrak m'} \subset R^h \subset R^\wedge,$$ by Algebra, Lemma Extending a henselian lifting problem to a finite algebra and More on Algebra, Lemma Noetherianity of a henselization.

Theorem. Artin approximation in an étale neighbourhood

Let $R$ be a Noetherian local ring. Let $f_1, \ldots, f_m \in R[x_1, \ldots, x_n]$. Suppose that $(a_1, \ldots, a_n) \in (R^\wedge)^n$ is a solution. If $R$ is a G-ring, then for every integer $N$ there exist

1. an étale ring map $R \to R'$,

2. a maximal ideal $\mathfrak m' \subset R'$ lying over $\mathfrak m$

3. a solution $(b_1, \ldots, b_n) \in (R')^n$ in $R'$

such that $\kappa(\mathfrak m) = \kappa(\mathfrak m')$ and $a_i - b_i \in (\mathfrak m')^NR^\wedge$.

Proof. We could deduce this theorem from Theorem Artin approximation for henselian local G-rings using that the henselization $R^h$ is a G-ring by More on Algebra, Lemma The G-ring property under henselization and writing $R^h$ as a directed colimit of étale extension $R'$. Instead we prove this by redoing the proof of the previous theorem in this case.

Let $c_i \in R$ be an element such that $a_i - c_i \in \mathfrak m^N$. Choose generators $\mathfrak m^N = (d_1, \ldots, d_M)$. Write $a_i = c_i + \sum a_{i, l} d_l$. Consider the polynomial ring $R[x_{i, l}]$ and the elements $$g_j = f_j(c_1 + \sum x_{1, l} d_l , \ldots, c_n + \sum x_{n, l} d_l) \in R[x_{i, l}]$$ The system of equations $g_j = 0$ has the solution $(a_{i, l})$. Suppose that we can show that the system of equations $g_j = 0$ has a solution $(b_{i, l})$ in $R'$ for some étale ring map $R \to R'$ endowed with a maximal ideal $\mathfrak m'$ such that $\kappa(\mathfrak m) = \kappa(\mathfrak m')$. Then it follows that $b_i = c_i + \sum b_{i, l}d_l$ is a solution of $f_j = 0$ which is congruent to $a_i$ modulo $(\mathfrak m')^N$. Thus it suffices to show that solvability over $R^\wedge$ implies solvability over some étale ring extension which induces a trivial residue field extension at some prime over $\mathfrak m$.

Let $A \subset R^\wedge$ be the $R$-subalgebra generated by $a_1, \ldots, a_n$. Since we've assumed $R$ is a G-ring, i.e., that $R \to R^\wedge$ is regular, we see that there exists a factorization $$A \to B \to R^\wedge$$ with $B$ smooth over $R$, see Theorem General Néron desingularization. Denote $\kappa = R/\mathfrak m$ the residue field. It is also the residue field of $R^\wedge$, so we get a commutative diagram $$\begin{gathered}\begin{matrix}B & R' \\ R & \kappa\end{matrix} \\[6pt] \begin{aligned}B & \longrightarrow \kappa \\ B & \cdots\!\!\rightarrow R' \\ R' & \cdots\!\!\rightarrow \kappa \\ R & \longrightarrow \kappa \\ R & \longrightarrow B\end{aligned}\end{gathered}$$ Since the vertical arrow is smooth, More on Algebra, Lemma Lifting a section of a smooth morphism implies that there exists an étale ring map $R \to R'$ which induces an isomorphism $R/\mathfrak m \to R'/\mathfrak mR'$ and an $R$-algebra map $B \to R'$ making the diagram above commute. Let $b_i \in R'$ be the image of $a_i$ under the ring maps $A \to B \to R'$. Since all of these maps are $R$-algebra maps, we see that $(b_1, \ldots, b_n)$ is a solution in $R'$. $\square$

Here is another variant of the main theorem of this section.

Lemma. Approximation after localization at a prime

Let $R$ be a Noetherian ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $f_1, \ldots, f_m \in R[x_1, \ldots, x_n]$. Suppose that $(a_1, \ldots, a_n) \in ((R_\mathfrak p)^\wedge)^n$ is a solution. If $R_\mathfrak p$ is a G-ring, then for every integer $N$ there exist

1. an étale ring map $R \to R'$,

2. a prime ideal $\mathfrak p' \subset R'$ lying over $\mathfrak p$

3. a solution $(b_1, \ldots, b_n) \in (R')^n$ in $R'$

such that $\kappa(\mathfrak p) = \kappa(\mathfrak p')$ and $a_i - b_i \in (\mathfrak p')^N(R'_{\mathfrak p'})^\wedge$.

Proof. By Theorem Artin approximation in an étale neighbourhood we can find a solution $(b'_1, \ldots, b'_n)$ in some ring $R''$ étale over $R_\mathfrak p$ which comes with a prime ideal $\mathfrak p''$ lying over $\mathfrak p$ such that $\kappa(\mathfrak p) = \kappa(\mathfrak p'')$ and $a_i - b'_i \in (\mathfrak p'')^N(R''_{\mathfrak p''})^\wedge$. We can write $R'' = R' \otimes_R R_\mathfrak p$ for some étale $R$-algebra $R'$ (see Algebra, Lemma Étale morphisms). After replacing $R'$ by a principal localization if necessary we may assume $(b'_1, \ldots, b'_n)$ come from a solution $(b_1, \ldots, b_n)$ in $R'$. Setting $\mathfrak p' = R' \cap \mathfrak p''$ we see that $R''_{\mathfrak p''} = R'_{\mathfrak p'}$ which finishes the proof. $\square$

B.10. Predecessor constructions used in the proof

These are the complete selected statements and proofs from the native Stacks source files. Their source labels are retained for precise cross-references. Remaining lower dependencies are listed in §12; not all of them are proved in this lesson.

B.10.1. More on Algebra

Lemma. Localizing an algebra while preserving its closed fibre

Let $A \to B$ be a ring map and $J \subset B$ an ideal. If $A \to B$ is étale at every prime of $V(J)$, then there exists a $g \in B$ mapping to an invertible element of $B/J$ such that $A' = B_g$ is étale over $A$.

Proof. The set of points of $\operatorname{Spec}(B)$ where $A \to B$ is not étale is a closed subset of $\operatorname{Spec}(B)$, see Algebra, Definition Étale ring maps. Write this as $V(J')$ for some ideal $J' \subset B$. Then $V(J') \cap V(J) = \emptyset$ hence $J + J' = B$ by Algebra, Lemma The Zariski topology on an affine spectrum. Write $1 = f + g$ with $f \in J$ and $g \in J'$. Then $g$ works. $\square$

Lemma. Lifting a monic polynomial factorization

Let $A$ be a ring, let $I \subset A$ be an ideal. Let $f \in A[x]$ be a monic polynomial. Let $\overline{f} = \overline{g} \overline{h}$ be a factorization of $f$ in $A/I[x]$ such that $\overline{g}$ and $\overline{h}$ are monic and generate the unit ideal in $A/I[x]$. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a factorization $f = g' h'$ in $A'[x]$ with $g'$, $h'$ monic lifting the given factorization over $A/I$.

Proof. We will deduce this from results on the universal factorization proved earlier; however, we encourage the reader to find their own proof not using this trick. Say $\deg(\overline{g}) = n$ and $\deg(\overline{h}) = m$ so that $\deg(f) = n + m$. Write $f = x^{n + m} + \sum \alpha_i x^{n + m - i}$ for some $\alpha_1, \ldots, \alpha_{n + m} \in A$. Consider the ring map $$R = \mathbf{Z}[a_1, \ldots, a_{n + m}] \longrightarrow S = \mathbf{Z}[b_1, \ldots, b_n, c_1, \ldots, c_m]$$ of Algebra, Example Étale algebras from polynomial factorizations. Let $R \to A$ be the ring map which sends $a_i$ to $\alpha_i$. Set $$B = A \otimes_R S$$ By construction the image $f_B$ of $f$ in $B[x]$ factors, say $f_B = g_B h_B$ with $g_B = x^n + \sum (1 \otimes b_i) x^{n - i}$ and similarly for $h_B$. Write $\overline{g} = x^n + \sum \overline{\beta}_i x^{n - i}$ and $\overline{h} = x^m + \sum \overline{\gamma}_i x^{m - i}$. The $A$-algebra map $$B \longrightarrow A/I, \quad 1 \otimes b_i \mapsto \overline{\beta}_i, \quad 1 \otimes c_i \mapsto \overline{\gamma}_i$$ maps $g_B$ and $h_B$ to $\overline{g}$ and $\overline{h}$ in $A/I[x]$. The displayed map is surjective; denote $J \subset B$ its kernel. From the discussion in Algebra, Example Étale algebras from polynomial factorizations it is clear that $A \to B$ is etale at all points of $V(J) \subset \operatorname{Spec}(B)$. Choose $g \in B$ as in Lemma Localizing an algebra while preserving its closed fibre and consider the $A$-algebra $B_g$. Since $g$ maps to a unit in $B/J = A/I$ we obtain also a map $B_g/I B_g \to A/I$ of $A/I$-algebras. Since $A/I \to B_g/I B_g$ is étale, also $B_g/IB_g \to A/I$ is étale (Algebra, Lemma Morphisms between étale algebras). Hence there exists an idempotent $e \in B_g/I B_g$ such that $A/I = (B_g/I B_g)_e$ (Algebra, Lemma Finite presentation and flatness). Choose a lift $h \in B_g$ of $e$. Then $A \to A' = (B_g)_h$ with factorization given by the image of the factorization $f_B = g_B h_B$ in $A'$ is a solution to the problem posed by the lemma. $\square$

Lemma. Lifting a coprime factorization

Let $A$ be a ring, let $I \subset A$ be an ideal. Let $f \in A[x]$ be a monic polynomial. Let $\overline{f} = \overline{g} \overline{h}$ be a factorization of $f$ in $A/I[x]$ and assume

1. the leading coefficient of $\overline{g}$ is an invertible element of $A/I$, and

2. $\overline{g}$, $\overline{h}$ generate the unit ideal in $A/I[x]$.

Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a factorization $f = g' h'$ in $A'[x]$ lifting the given factorization over $A/I$.

Proof. Applying Lemma Lifting a unit we may assume that the leading coefficient of $\overline{g}$ is the reduction of an invertible element $u \in A$. Then we may replace $\overline{g}$ by $\overline{u}^{-1}\overline{g}$ and $\overline{h}$ by $\overline{u}\overline{h}$. Thus we may assume that $\overline{g}$ is monic. Since $f$ is monic we conclude that $\overline{h}$ is monic too. In this case the result follows from Lemma Lifting a monic polynomial factorization. $\square$

Lemma. Separating a closed image from another closed subset

Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal of $R$ and let $J \subset S$ be an ideal of $S$. If the closure of the image of $V(J)$ in $\operatorname{Spec}(R)$ is disjoint from $V(I)$, then there exists an element $f \in R$ which maps to $1$ in $R/I$ and to an element of $J$ in $S$.

Proof. Let $I' \subset R$ be an ideal such that $V(I')$ is the closure of the image of $V(J)$. Then $V(I) \cap V(I') = \emptyset$ by assumption and hence $I + I' = R$ by Algebra, Lemma The Zariski topology on an affine spectrum. Write $1 = g + f$ with $g \in I$ and $f \in I'$. We have $V(f') \supset V(J)$ where $f'$ is the image of $f$ in $S$. Hence $(f')^n \in J$ for some $n$, see Algebra, Lemma The Zariski topology on an affine spectrum. Replacing $f$ by $f^n$ we win. $\square$

Lemma. Integral elements compatible with a lifted factorization

Let $I$ be an ideal of a ring $A$. Let $A \to B$ be an integral ring map. Let $b \in B$ map to an idempotent in $B/IB$. Then there exists a monic $f \in A[x]$ with $f(b) = 0$ and $f \bmod I = x^d(x - 1)^d$ for some $d \geq 1$.

Proof. Observe that $z = b^2 - b$ is an element of $IB$. By Algebra, Lemma Integral extensions there exist a monic polynomial $g(x) = x^d + \sum a_j x^j$ of degree $d$ with $a_j \in I$ such that $g(z) = 0$ in $B$. Hence $f(x) = g(x^2 - x) \in A[x]$ is a monic polynomial such that $f(x) \equiv x^d(x - 1)^d \bmod I$ and such that $f(b) = 0$ in $B$. $\square$

Lemma. Lifting an idempotent after localization

Let $A$ be a ring, let $I \subset A$ be an ideal. Let $A \to B$ be an integral ring map. Let $\overline{e} \in B/IB$ be an idempotent. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and an idempotent $e' \in B \otimes_A A'$ lifting $\overline{e}$.

Proof. Choose an element $y \in B$ lifting $\overline{e}$. Choose $f \in A[x]$ as in Lemma Integral elements compatible with a lifted factorization for $y$. By Lemma Lifting a coprime factorization we can find an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and such that $f = gh$ in $A[x]$ with $g(x) = x^d \bmod IA'$ and $h(x) = (x - 1)^d \bmod IA'$. After replacing $A$ by $A'$ we may assume that the factorization is defined over $A$. In that case we see that $b_1 = g(y) \in B$ is a lift of $\overline{e}^d = \overline{e}$ and $b_2 = h(y) \in B$ is a lift of $(\overline{e} - 1)^d = (-1)^d (1 - \overline{e})^d = (-1)^d(1 - \overline{e})$ and moreover $b_1b_2 = 0$. Thus $(b_1, b_2)B/IB = B/IB$ and $V(b_1, b_2) \subset \operatorname{Spec}(B)$ is disjoint from $V(IB)$. Since $\operatorname{Spec}(B) \to \operatorname{Spec}(A)$ is closed (see Algebra, Lemmas Going up for integral ring maps and Going up and closed maps of spectra) we can find an $a \in A$ which maps to an invertible element of $A/I$ whose image in $B$ lies in $(b_1, b_2)$, see Lemma Separating a closed image from another closed subset. After replacing $A$ by the localization $A_a$ we get that $(b_1, b_2) = B$. Then $\operatorname{Spec}(B) = D(b_1) \amalg D(b_2)$; disjoint union because $b_1b_2 = 0$ and covers $\operatorname{Spec}(B)$ because $(b_1, b_2) = B$. Let $e \in B$ be the idempotent corresponding to the open and closed subset $D(b_1)$, see Algebra, Lemma Product decompositions from disjoint closed subsets. Since $b_1$ is a lift of $\overline{e}$ and $b_2$ is a lift of $\pm (1 - \overline{e})$ we conclude that $e$ is a lift of $\overline{e}$ by the uniqueness statement in Algebra, Lemma Product decompositions from disjoint closed subsets. $\square$

Lemma. Lifting a finite projective module

Let $A$ be a ring, let $I \subset A$ be an ideal. Let $\overline{P}$ be a finite projective $A/I$-module. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a finite projective $A'$-module $P'$ lifting $\overline{P}$.

Proof. We can choose an integer $n$ and a direct sum decomposition $(A/I)^{\oplus n} = \overline{P} \oplus \overline{K}$ for some $A/I$-module $\overline{K}$. Choose a lift $\varphi : A^{\oplus n} \to A^{\oplus n}$ of the projector $\overline{p}$ associated to the direct summand $\overline{P}$. Let $f \in A[x]$ be the characteristic polynomial of $\varphi$. Set $B = A[x]/(f)$. By Cayley-Hamilton (Algebra, Lemma The characteristic polynomial) there is a map $B \to \text{End}_A(A^{\oplus n})$ mapping $x$ to $\varphi$. For every prime $\mathfrak p \supset I$ the image of $f$ in $\kappa(\mathfrak p)$ is $(x - 1)^rx^{n - r}$ where $r$ is the dimension of $\overline{P} \otimes_{A/I} \kappa(\mathfrak p)$. Hence $(x - 1)^nx^n$ maps to zero in $B \otimes_A \kappa(\mathfrak p)$ for all $\mathfrak p \supset I$. Thus $x(1 - x)$ is contained in every prime ideal of $B/IB$. Hence $x^N(1 - x)^N$ is contained in $IB$ for some $N \geq 1$. It follows that $x^N + (1 - x)^N$ is a unit in $B/IB$ and that $$\overline{e} = \text{image of }\frac{x^N}{x^N + (1 - x)^N}\text{ in }B/IB$$ is an idempotent as both assertions hold in $\mathbf{Z}[x]/(x^N(x - 1)^N)$. The image of $\overline{e}$ in $\text{End}_{A/I}((A/I)^{\oplus n})$ is $$\frac{\overline{p}^N}{\overline{p}^N + (1 - \overline{p})^N} = \overline{p}$$ as $\overline{p}$ is an idempotent. After replacing $A$ by an étale extension $A'$ as in the lemma, we may assume there exists an idempotent $e \in B$ which maps to $\overline{e}$ in $B/IB$, see Lemma Lifting an idempotent after localization. Then the image of $e$ under the map $$B = A[x]/(f) \longrightarrow \text{End}_A(A^{\oplus n}).$$ is an idempotent element $p$ which lifts $\overline{p}$. Setting $P = \operatorname{Im}(p)$ we win. $\square$

Lemma. The cotangent complex of a symmetric algebra

Let $A$ be a ring. Let $0 \to K \to A^{\oplus m} \to M \to 0$ be a sequence of $A$-modules. Consider the $A$-algebra $C = \text{Sym}^*_A(M)$ with its presentation $\alpha : A[y_1, \ldots, y_m] \to C$ coming from the surjection $A^{\oplus m} \to M$. Then $$\mathrm{NL}(\alpha) = (K \otimes_A C \to \bigoplus\nolimits_{j = 1, \ldots, m} C \text{d}y_j)$$ (see Algebra, Section The naive cotangent complex) in particular $\Omega_{C/A} = M \otimes_A C$.

Proof. Let $J = \operatorname{Ker}(\alpha)$. The lemma asserts that $J/J^2 \cong K \otimes_A C$. Note that $\alpha$ is a homomorphism of graded algebras. We will prove that in degree $d$ we have $(J/J^2)_d = K \otimes_A C_{d - 1}$. Note that $$J_d = \operatorname{Ker}(\text{Sym}^d_A(A^{\oplus m}) \to \text{Sym}^d_A(M)) = \operatorname{Im}(K \otimes_A \text{Sym}^{d - 1}_A(A^{\oplus m}) \to \text{Sym}^d_A(A^{\oplus m})),$$ see Algebra, Lemma Presentations of symmetric and exterior powers. It follows that $(J^2)_d = \sum_{a + b = d} J_a \cdot J_b$ is the image of $$K \otimes_A K \otimes_A \text{Sym}^{d - 2}_A(A^{\otimes m}) \to \text{Sym}^d_A(A^{\oplus m}).$$ The cokernel of the map $K \otimes_A \text{Sym}^{d - 2}_A(A^{\otimes m}) \to \text{Sym}^{d - 1}_A(A^{\oplus m})$ is $\text{Sym}^{d - 1}_A(M)$ by the lemma referenced above. Hence it is clear that $(J/J^2)_d = J_d/(J^2)_d$ is equal to $$\begin{aligned} \operatorname{Coker}( K \otimes_A K \otimes_A \text{Sym}^{d - 2}_A(A^{\otimes m}) \to K \otimes_A \text{Sym}^{d - 1}_A(A^{\otimes m})) & = K \otimes_A \text{Sym}^{d - 1}_A(M) \\ & = K \otimes_A C_{d -1} \end{aligned}$$ as desired. $\square$

Lemma. Smoothness of a symmetric algebra

Let $A$ be a ring. Let $M$ be an $A$-module. Then $C = \text{Sym}_A^*(M)$ is smooth over $A$ if and only if $M$ is a finite projective $A$-module.

Proof. Let $\sigma : C \to A$ be the projection onto the degree $0$ part of $C$. Then $J = \operatorname{Ker}(\sigma)$ is the part of degree $> 0$ and we see that $J/J^2 = M$ as an $A$-module. Hence if $A \to C$ is smooth then $M$ is a finite projective $A$-module by Algebra, Lemma Sections of smooth ring maps.

Conversely, assume that $M$ is finite projective and choose a surjection $A^{\oplus n} \to M$ with kernel $K$. Of course the sequence $0 \to K \to A^{\oplus n} \to M \to 0$ is split as $M$ is projective. In particular we see that $K$ is a finite $A$-module and hence $C$ is of finite presentation over $A$ as $C$ is a quotient of $A[x_1, \ldots, x_n]$ by the ideal generated by $K \subset \bigoplus Ax_i$. The computation of Lemma The cotangent complex of a symmetric algebra shows that $\mathrm{NL}_{C/A}$ is homotopy equivalent to $(K \to A^{\oplus n}) \otimes_A C$. Hence $\mathrm{NL}_{C/A}$ is quasi-isomorphic to $C \otimes_A M$ placed in degree $0$ which means that $C$ is smooth over $A$ by Algebra, Definition Smooth ring maps. $\square$

Lemma. Lifting a section of a smooth morphism

Let $A$ be a ring, let $I \subset A$ be an ideal. Consider a commutative diagram $$\begin{gathered}\begin{matrix}B \\ A & A/I\end{matrix} \\[6pt] \begin{aligned}B & \longrightarrow A/I \\ A & \longrightarrow B \\ A & \longrightarrow A/I\end{aligned}\end{gathered}$$ where $B$ is a smooth $A$-algebra. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and an $A$-algebra map $B \to A'$ lifting the ring map $B \to A/I$.

Proof. Let $J \subset B$ be the kernel of $B \to A/I$ so that $B/J = A/I$. By Algebra, Lemma Smoothness and the naive cotangent complex the sequence $$0 \to I/I^2 \to J/J^2 \to \Omega_{B/A} \otimes_B B/J \to 0$$ is split exact. Thus $\overline{P} = J/(J^2 + IB) = \Omega_{B/A} \otimes_B B/J$ is a finite projective $A/I$-module. Choose an integer $n$ and a direct sum decomposition $A/I^{\oplus n} = \overline{P} \oplus \overline{K}$. By Lemma Lifting a finite projective module we can find an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a finite projective $A$-module $K$ which lifts $\overline{K}$. We may and do replace $A$ by $A'$. Set $B' = B \otimes_A \text{Sym}_A^*(K)$. Since $A \to \text{Sym}_A^*(K)$ is smooth by Lemma Smoothness of a symmetric algebra we see that $B \to B'$ is smooth which in turn implies that $A \to B'$ is smooth (see Algebra, Lemmas Base change of smooth ring maps and Smooth morphisms and local algebra). Moreover the section $\text{Sym}^*_A(K) \to A$ determines a section $B' \to B$ and we let $B' \to A/I$ be the composition $B' \to B \to A/I$. Let $J' \subset B'$ be the kernel of $B' \to A/I$. We have $JB' \subset J'$ and $B \otimes_A K \subset J'$. These maps combine to give an isomorphism $$(A/I)^{\oplus n} \cong J/(J^2 + IB) \oplus \overline{K} \longrightarrow J'/((J')^2 + IB')$$ Thus, after replacing $B$ by $B'$ we may assume that $J/(J^2 + IB) = \Omega_{B/A} \otimes_B B/J$ is a free $A/I$-module of rank $n$.

In this case, choose $f_1, \ldots, f_n \in J$ which map to a basis of $J/(J^2 + IB)$. Consider the finitely presented $A$-algebra $C = B/(f_1, \ldots, f_n)$. Note that we have an exact sequence $$0 \to H_1(L_{C/A}) \to (f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2 \to \Omega_{B/A} \otimes_B C \to \Omega_{C/A} \to 0$$ see Algebra, Lemma The transitivity sequence for the naive cotangent complex (note that $H_1(L_{B/A}) = 0$ and that $\Omega_{B/A}$ is finite projective, in particular flat so the Tor group vanishes). For any prime $\mathfrak q \supset J$ of $B$ the module $\Omega_{B/A, \mathfrak q}$ is free of rank $n$ because $\Omega_{B/A}$ is finite projective and because $\Omega_{B/A} \otimes_B B/J$ is free of rank $n$ (see Algebra, Lemma Characterizations of finite projective modules). By our choice of $f_1, \ldots, f_n$ the map $$\left((f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2\right)_{\mathfrak q} \to \Omega_{B/A, \mathfrak q}$$ is surjective modulo $J$. Hence we see that this map of modules over the local ring $C_{\mathfrak q}$ has to be an isomorphism (by Algebra, Lemma Nakayama's lemma the map is surjective and for example by Algebra, Lemma Surjective endomorphisms of finite modules because $((f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2)_{\mathfrak q}$ is generated by $n$ elements the map is injective). Thus $H_1(L_{C/A})_{\mathfrak q} = 0$ and $\Omega_{C/A, \mathfrak q} = 0$. By Algebra, Lemma Smoothness at a point we see that $A \to C$ is smooth at the prime $\overline{\mathfrak q}$ of $C$ corresponding to $\mathfrak q$. Since $\Omega_{C/A, \mathfrak q} = 0$ it is actually étale at $\overline{\mathfrak q}$. Thus $A \to C$ is étale at all primes of $C$ containing $JC$. By Lemma Localizing an algebra while preserving its closed fibre we can find an $f \in C$ mapping to an invertible element of $C/JC$ such that $A \to C_f$ is étale. By our choice of $f$ it is still true that $C_f/JC_f = A/I$. The map $C_f/IC_f \to A/I$ is surjective and étale by Algebra, Lemma Morphisms between étale algebras. Hence $A/I$ is isomorphic to the localization of $C_f/IC_f$ at some element $g \in C$, see Algebra, Lemma Finite presentation and flatness. Set $A' = C_{fg}$ to conclude the proof. $\square$

Lemma. The conormal sequence for a first-homology regular sequence

Let $A$ be a ring. Let $I \subset J \subset A$ be ideals. Assume that $J/I \subset A/I$ is generated by an $H_1$-regular sequence. Then $I \cap J^2 = IJ$.

Proof. To prove this choose $g_1, \ldots, g_m \in J$ whose images in $A/I$ form a $H_1$-regular sequence which generates $J/I$. In particular $J = I + (g_1, \ldots, g_m)$. Suppose that $x \in I \cap J^2$. Because $x \in J^2$ can write $$x = \sum a_{ij} g_ig_j + \sum a_j g_j + a$$ with $a_{ij} \in A$, $a_j \in I$ and $a \in I^2$. Then $\sum a_{ij}g_ig_j \in I \cap (g_1, \ldots, g_m)$ hence by Lemma First cotangent homology after a regular quotient we see that $\sum a_{ij}g_ig_j \in I(g_1, \ldots, g_m)$. Thus $x \in IJ$ as desired. $\square$

Lemma. The conormal sequence for a first-homology regular ideal

Let $A$ be a ring. Let $I \subset J \subset A$ be ideals. Assume that $J/I \subset A/I$ is a $H_1$-regular ideal. Then $I \cap J^2 = IJ$.

Proof. Follows immediately from Lemma The conormal sequence for a first-homology regular sequence by localizing. $\square$

Lemma. Finite projectivity of a quasi-regular conormal module

Let $I \subset R$ be a quasi-regular ideal of a ring. Then $I/I^2$ is a finite projective $R/I$-module.

Proof. This follows from Algebra, Lemma Characterizations of finite projective modules and the definitions. $\square$

Lemma. Syntomic algebras and local complete intersections

Let $R \to S$ be a ring map. The following are equivalent

1. $R \to S$ is syntomic (Algebra, Definition Local complete intersections), and

2. $R \to S$ is flat and a local complete intersection.

Proof. Assume (1). Then $R \to S$ is flat by definition. By Algebra, Lemma Local criteria for a syntomic algebra and Lemma Locality of the complete-intersection condition we see that it suffices to show a relative global complete intersection is a local complete intersection homomorphism which is Lemma Koszul complexes of global complete intersections.

Assume (2). A local complete intersection is of finite presentation because a Koszul-regular ideal is finitely generated. Let $R \to k$ be a map to a field. It suffices to show that $S' = S \otimes_R k$ is a local complete intersection over $k$, see Algebra, Definition Complete intersections over a field. Choose a prime $\mathfrak q' \subset S'$. Write $S = R[x_1, \ldots, x_n]/I$. Then $S' = k[x_1, \ldots, x_n]/I'$ where $I' \subset k[x_1, \ldots, x_n]$ is the image of $I$. Let $\mathfrak p' \subset k[x_1, \ldots, x_n]$, $\mathfrak q \subset S$, and $\mathfrak p \subset R[x_1, \ldots, x_n]$ be the corresponding primes. By Definition Regular ideals exists an $g \in R[x_1, \ldots, x_n]$, $g \not \in \mathfrak p$ and $f_1, \ldots, f_r \in R[x_1, \ldots, x_n]_g$ which form a Koszul-regular sequence generating $I_g$. Since $S$ and hence $S_g$ is flat over $R$ we see that the images $f'_1, \ldots, f'_r$ in $k[x_1, \ldots, x_n]_g$ form a $H_1$-regular sequence generating $I'_g$, see Lemma Relative regular immersions in affine algebra. Thus $f'_1, \ldots, f'_r$ map to a regular sequence in $k[x_1, \ldots, x_n]_{\mathfrak p'}$ generating $I'_{\mathfrak p'}$ by Lemma Regularity conditions for finite ideals in Noetherian rings. Applying Algebra, Lemma Local criteria for complete intersections we conclude $S'_{gg'}$ for some $g' \in S$, $g' \not \in \mathfrak q'$ is a global complete intersection over $k$ as desired. $\square$

Lemma. Cotangent transitivity with a complete-intersection terminal map

Let $A \to B \to C$ be ring maps. Assume $B \to C$ is a local complete intersection homomorphism. Choose a presentation $\alpha : A[x_s, s \in S] \to B$ with kernel $I$. Choose a presentation $\beta : B[y_1, \ldots, y_m] \to C$ with kernel $J$. Let $\gamma : A[x_s, y_t] \to C$ be the induced presentation of $C$ with kernel $K$. Then we get a canonical commutative diagram $$\begin{gathered}\begin{matrix}0 & \Omega_{A[x_s]/A} \otimes C & \Omega_{A[x_s, y_t]/A} \otimes C & \Omega_{B[y_t]/B} \otimes C & 0 \\ 0 & I/I^2 \otimes C & K/K^2 & J/J^2 & 0\end{matrix} \\[6pt] \begin{aligned}0 & \longrightarrow \Omega_{A[x_s]/A} \otimes C \\ \Omega_{A[x_s]/A} \otimes C & \longrightarrow \Omega_{A[x_s, y_t]/A} \otimes C \\ \Omega_{A[x_s, y_t]/A} \otimes C & \longrightarrow \Omega_{B[y_t]/B} \otimes C \\ \Omega_{B[y_t]/B} \otimes C & \longrightarrow 0 \\ 0 & \longrightarrow I/I^2 \otimes C \\ I/I^2 \otimes C & \longrightarrow K/K^2 \\ I/I^2 \otimes C & \longrightarrow \Omega_{A[x_s]/A} \otimes C \\ K/K^2 & \longrightarrow J/J^2 \\ K/K^2 & \longrightarrow \Omega_{A[x_s, y_t]/A} \otimes C \\ J/J^2 & \longrightarrow 0 \\ J/J^2 & \longrightarrow \Omega_{B[y_t]/B} \otimes C\end{aligned}\end{gathered}$$ with exact rows. In particular, the six term exact sequence of Algebra, Lemma The transitivity sequence for the naive cotangent complex can be completed with a zero on the left, i.e., the sequence $$0 \to H_1(\mathrm{NL}_{B/A} \otimes_B C) \to H_1(L_{C/A}) \to H_1(L_{C/B}) \to \Omega_{B/A} \otimes_B C \to \Omega_{C/A} \to \Omega_{C/B} \to 0$$ is exact.

Proof. The only thing to prove is the injectivity of the map $I/I^2 \otimes C \to K/K^2$. By assumption the ideal $J$ is Koszul-regular. Hence we have $IA[x_s, y_j] \cap K^2 = IK$ by Lemma The conormal sequence for a first-homology regular ideal. This means that the kernel of $K/K^2 \to J/J^2$ is isomorphic to $IA[x_s, y_j]/IK$. Since $I/I^2 \otimes_B C = IA[x_s, y_j]/IK$ by right exactness of tensor product, this provides us with the desired injectivity of $I/I^2 \otimes_B C \to K/K^2$. $\square$

Lemma. Cotangent transitivity for filtered complete intersections

Let $A \to B \to C$ be ring maps. If $B \to C$ is a filtered colimit of local complete intersection homomorphisms then the conclusion of Lemma Cotangent transitivity with a complete-intersection terminal map remains valid.

Proof. Follows from Lemma Cotangent transitivity with a complete-intersection terminal map and Algebra, Lemma Filtered colimits of naive cotangent complexes. $\square$

Lemma. Cartier's equality for differentials (Cartier equality)

Let $K/k$ be a finitely generated field extension. Then $\Omega_{K/k}$ and $H_1(L_{K/k})$ are finite dimensional and $\text{trdeg}_k(K) = \dim_K \Omega_{K/k} - \dim_K H_1(L_{K/k})$.

Proof. We can find a global complete intersection $A = k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ over $k$ such that $K$ is isomorphic to the fraction field of $A$, see Algebra, Lemma Syntomic algebras in a filtered colimit and its proof. In this case we see that $\mathrm{NL}_{K/k}$ is homotopy equivalent to the complex $$\bigoplus\nolimits_{j = 1, \ldots, c} K \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, n} K\text{d}x_i$$ by Algebra, Lemmas Kähler differentials, Theorems 3.1–3.3, Proposition 3.4 and Theorem 7.1 and Localization of the naive cotangent complex. The transcendence degree of $K$ over $k$ is the dimension of $A$ (by Algebra, Lemma Prime ideals and dimension in a polynomial ring) which is $n - c$ and we win. $\square$

Lemma. Transitivity of first cotangent homology

Let $M/L/K$ be field extensions. Then the Jacobi-Zariski sequence $$0 \to H_1(L_{L/K}) \otimes_L M \to H_1(L_{M/K}) \to H_1(L_{M/L}) \to \Omega_{L/K} \otimes_L M \to \Omega_{M/K} \to \Omega_{M/L} \to 0$$ is exact.

Proof. Combine Lemma Cotangent transitivity for filtered complete intersections with Algebra, Lemma Syntomic algebras in a filtered colimit. $\square$

Lemma. Compatibility of cotangent homology with a quotient

Given a commutative diagram of fields $$\begin{gathered}\begin{matrix}K & K' \\ k & k'\end{matrix} \\[6pt] \begin{aligned}K & \longrightarrow K' \\ k & \longrightarrow K \\ k & \longrightarrow k' \\ k' & \longrightarrow K'\end{aligned}\end{gathered}$$ with $k'/k$ and $K'/K$ finitely generated field extensions the kernel and cokernel of the maps $$\alpha : \Omega_{K/k} \otimes_K K' \to \Omega_{K'/k'} \quad\text{and}\quad \beta : H_1(L_{K/k}) \otimes_K K' \to H_1(L_{K'/k'})$$ are finite dimensional and $$\dim \operatorname{Ker}(\alpha) - \dim \operatorname{Coker}(\alpha) -\dim \operatorname{Ker}(\beta) + \dim \operatorname{Coker}(\beta)

\text{trdeg}_k(k') - \text{trdeg}_K(K')$$

Proof. The Jacobi-Zariski sequences for $k \subset k' \subset K'$ and $k \subset K \subset K'$ are $$0 \to H_1(L_{k'/k}) \otimes K' \to H_1(L_{K'/k}) \to H_1(L_{K'/k'}) \to \Omega_{k'/k} \otimes K' \to \Omega_{K'/k} \to \Omega_{K'/k'} \to 0$$ and $$0 \to H_1(L_{K/k}) \otimes K' \to H_1(L_{K'/k}) \to H_1(L_{K'/K}) \to \Omega_{K/k} \otimes K' \to \Omega_{K'/k} \to \Omega_{K'/K} \to 0$$ By Lemma Cartier's equality for differentials the vector spaces $\Omega_{k'/k}$, $\Omega_{K'/K}$, $H_1(L_{K'/K})$, and $H_1(L_{k'/k})$ are finite dimensional and the alternating sum of their dimensions is $\text{trdeg}_k(k') - \text{trdeg}_K(K')$. The lemma follows. $\square$

Proposition. Characterizations of geometric regularity

Let $k$ be a field of characteristic $p > 0$. Let $(A, \mathfrak m, K)$ be a Noetherian local $k$-algebra. The following are equivalent

1. $A$ is geometrically regular over $k$,

2. for all $k \subset k' \subset k^{1/p}$ finite over $k$ the ring $A \otimes_k k'$ is regular,

3. $A$ is regular and the canonical map $H_1(L_{K/k}) \to \mathfrak m/\mathfrak m^2$ is injective, and

4. $A$ is regular and the map $\Omega_{k/\mathbf{F}_p} \otimes_k K \to \Omega_{A/\mathbf{F}_p} \otimes_A K$ is injective.

Proof. Proof of (3) $\Rightarrow$ (1). Assume (3). Let $k'/k$ be a finite purely inseparable extension. Set $A' = A \otimes_k k'$. This is a local ring with maximal ideal $\mathfrak m'$. Set $K' = A'/\mathfrak m'$. We get a commutative diagram $$\begin{gathered}\begin{matrix}0 & H_1(L_{K/k}) \otimes K' & \mathfrak m/\mathfrak m^2 \otimes K' & \Omega_{A/k} \otimes_A K' & \Omega_{K/k} \otimes K' & 0 \\ \phantom{X} & H_1(L_{K'/k'}) & \mathfrak m'/(\mathfrak m')^2 & \Omega_{A'/k'} \otimes_{A'} K' & \Omega_{K'/k'} & 0\end{matrix} \\[6pt] \begin{aligned}0 & \longrightarrow H_1(L_{K/k}) \otimes K' \\ H_1(L_{K/k}) \otimes K' & \longrightarrow \mathfrak m/\mathfrak m^2 \otimes K' \\ H_1(L_{K/k}) \otimes K' & \xrightarrow{\beta} H_1(L_{K'/k'}) \\ \mathfrak m/\mathfrak m^2 \otimes K' & \longrightarrow \Omega_{A/k} \otimes_A K' \\ \mathfrak m/\mathfrak m^2 \otimes K' & \longrightarrow \mathfrak m'/(\mathfrak m')^2 \\ \Omega_{A/k} \otimes_A K' & \longrightarrow \Omega_{K/k} \otimes K' \\ \Omega_{A/k} \otimes_A K' & \xrightarrow{\cong} \Omega_{A'/k'} \otimes_{A'} K' \\ \Omega_{K/k} \otimes K' & \longrightarrow 0 \\ \Omega_{K/k} \otimes K' & \xrightarrow{\alpha} \Omega_{K'/k'} \\ H_1(L_{K'/k'}) & \longrightarrow \mathfrak m'/(\mathfrak m')^2 \\ \mathfrak m'/(\mathfrak m')^2 & \longrightarrow \Omega_{A'/k'} \otimes_{A'} K' \\ \Omega_{A'/k'} \otimes_{A'} K' & \longrightarrow \Omega_{K'/k'} \\ \Omega_{K'/k'} & \longrightarrow 0\end{aligned}\end{gathered}$$ with exact rows. The third vertical arrow is an isomorphism by base change for modules of differentials (Algebra, Lemma Base change of Kähler differentials). Thus $\alpha$ is surjective. By Lemma Compatibility of cotangent homology with a quotient we have $$\dim \operatorname{Ker}(\alpha) - \dim \operatorname{Ker}(\beta) + \dim \operatorname{Coker}(\beta) = 0$$ (and these dimensions are all finite). A diagram chase shows that $\dim \mathfrak m'/(\mathfrak m')^2 \leq \dim \mathfrak m/\mathfrak m^2$. However, since $A \to A'$ is finite flat we see that $\dim(A) = \dim(A')$, see Algebra, Lemma Dimensions of a base, fibre and total space. Hence $A'$ is regular by definition.

Equivalence of (3) and (4). Consider the Jacobi-Zariski sequences for rows of the commutative diagram $$\begin{gathered}\begin{matrix}\mathbf{F}_p & A & K \\ \mathbf{F}_p & k & K\end{matrix} \\[6pt] \begin{aligned}\mathbf{F}_p & \longrightarrow A \\ A & \longrightarrow K \\ \mathbf{F}_p & \longrightarrow k \\ \mathbf{F}_p & \longrightarrow \mathbf{F}_p \\ k & \longrightarrow K \\ k & \longrightarrow A \\ K & \longrightarrow K\end{aligned}\end{gathered}$$ to get a commutative diagram $$\begin{gathered}\begin{matrix}0 & \mathfrak m/\mathfrak m^2 & \Omega_{A/\mathbf{F}_p} \otimes_A K & \Omega_{K/\mathbf{F}_p} & 0 & \phantom{X} \\ 0 & H_1(L_{K/k}) & \Omega_{k/\mathbf{F}_p} \otimes_k K & \Omega_{K/\mathbf{F}_p} & \Omega_{K/k} & 0\end{matrix} \\[6pt] \begin{aligned}0 & \longrightarrow \mathfrak m/\mathfrak m^2 \\ \mathfrak m/\mathfrak m^2 & \longrightarrow \Omega_{A/\mathbf{F}_p} \otimes_A K \\ \Omega_{A/\mathbf{F}_p} \otimes_A K & \longrightarrow \Omega_{K/\mathbf{F}_p} \\ \Omega_{K/\mathbf{F}_p} & \longrightarrow 0 \\ 0 & \longrightarrow H_1(L_{K/k}) \\ H_1(L_{K/k}) & \longrightarrow \Omega_{k/\mathbf{F}_p} \otimes_k K \\ H_1(L_{K/k}) & \longrightarrow \mathfrak m/\mathfrak m^2 \\ \Omega_{k/\mathbf{F}_p} \otimes_k K & \longrightarrow \Omega_{K/\mathbf{F}_p} \\ \Omega_{k/\mathbf{F}_p} \otimes_k K & \longrightarrow \Omega_{A/\mathbf{F}_p} \otimes_A K \\ \Omega_{K/\mathbf{F}_p} & \longrightarrow \Omega_{K/k} \\ \Omega_{K/\mathbf{F}_p} & \longrightarrow \Omega_{K/\mathbf{F}_p} \\ \Omega_{K/k} & \longrightarrow 0 \\ \Omega_{K/k} & \longrightarrow 0\end{aligned}\end{gathered}$$ with exact rows. We have used that $H_1(L_{K/A}) = \mathfrak m/\mathfrak m^2$ and that $H_1(L_{K/\mathbf{F}_p}) = 0$ as $K/\mathbf{F}_p$ is separable, see Algebra, Proposition Characterizations of separable field extensions. Thus it is clear that the kernels of $H_1(L_{K/k}) \to \mathfrak m/\mathfrak m^2$ and $\Omega_{k/\mathbf{F}_p} \otimes_k K \to \Omega_{A/\mathbf{F}_p} \otimes_A K$ have the same dimension.

Proof of (2) $\Rightarrow$ (4) following Faltings, see the original source citation Faltings-einfacher. Let $a_1, \ldots, a_n \in k$ be elements such that $\text{d}a_1, \ldots, \text{d}a_n$ are linearly independent in $\Omega_{k/\mathbf{F}_p}$. Consider the field extension $k' = k(a_1^{1/p}, \ldots, a_n^{1/p})$. By Algebra, Lemma Degrees of extensions obtained by adjoining p-th roots we see that $k' = k[x_1, \ldots, x_n]/(x_1^p - a_1, \ldots, x_n^p - a_n)$. In particular we see that the naive cotangent complex of $k'/k$ is homotopic to the complex $\bigoplus_{j = 1, \ldots, n} k' \rightarrow \bigoplus_{i = 1, \ldots, n} k'$ with the zero differential as $\text{d}(x_j^p - a_j) = 0$ in $\Omega_{k[x_1, \ldots, x_n]/k}$. Set $A' = A \otimes_k k'$ and $K' = A'/\mathfrak m'$ as above. By Algebra, Lemma Base change of the naive cotangent complex we see that $\mathrm{NL}_{A'/A}$ is homotopy equivalent to the complex $\bigoplus_{j = 1, \ldots, n} A' \rightarrow \bigoplus_{i = 1, \ldots, n} A'$ with the zero differential, i.e., $H_1(L_{A'/A})$ and $\Omega_{A'/A}$ are free of rank $n$. The Jacobi-Zariski sequence for $\mathbf{F}_p \to A \to A'$ is $$H_1(L_{A'/A}) \to \Omega_{A/\mathbf{F}_p} \otimes_A A' \to \Omega_{A'/\mathbf{F}_p} \to \Omega_{A'/A} \to 0$$ Using the presentation $A[x_1, \ldots, x_n] \to A'$ with kernel $(x_j^p - a_j)$ we see, unwinding the maps in Algebra, Lemma The transitivity sequence for the naive cotangent complex, that the $j$th basis vector of $H_1(L_{A'/A})$ maps to $\text{d}a_j \otimes 1$ in $\Omega_{A/\mathbf{F}_p} \otimes A'$. As $\Omega_{A'/A}$ is free (hence flat) we get on tensoring with $K'$ an exact sequence $$K'^{\oplus n} \to \Omega_{A/\mathbf{F}_p} \otimes_A K' \xrightarrow{\beta} \Omega_{A'/\mathbf{F}_p} \otimes_{A'} K' \to K'^{\oplus n} \to 0$$ We conclude that the elements $\text{d}a_j \otimes 1$ generate $\operatorname{Ker}(\beta)$ and we have to show that are linearly independent, i.e., we have to show $\dim(\operatorname{Ker}(\beta)) = n$. Consider the following big diagram $$\begin{gathered}\begin{matrix}0 & \mathfrak m'/(\mathfrak m')^2 & \Omega_{A'/\mathbf{F}_p} \otimes K' & \Omega_{K'/\mathbf{F}_p} & 0 \\ 0 & \mathfrak m/\mathfrak m^2 \otimes K' & \Omega_{A/\mathbf{F}_p} \otimes K' & \Omega_{K/\mathbf{F}_p} \otimes K' & 0\end{matrix} \\[6pt] \begin{aligned}0 & \longrightarrow \mathfrak m'/(\mathfrak m')^2 \\ \mathfrak m'/(\mathfrak m')^2 & \longrightarrow \Omega_{A'/\mathbf{F}_p} \otimes K' \\ \Omega_{A'/\mathbf{F}_p} \otimes K' & \longrightarrow \Omega_{K'/\mathbf{F}_p} \\ \Omega_{K'/\mathbf{F}_p} & \longrightarrow 0 \\ 0 & \longrightarrow \mathfrak m/\mathfrak m^2 \otimes K' \\ \mathfrak m/\mathfrak m^2 \otimes K' & \longrightarrow \Omega_{A/\mathbf{F}_p} \otimes K' \\ \mathfrak m/\mathfrak m^2 \otimes K' & \xrightarrow{\alpha} \mathfrak m'/(\mathfrak m')^2 \\ \Omega_{A/\mathbf{F}_p} \otimes K' & \longrightarrow \Omega_{K/\mathbf{F}_p} \otimes K' \\ \Omega_{A/\mathbf{F}_p} \otimes K' & \xrightarrow{\beta} \Omega_{A'/\mathbf{F}_p} \otimes K' \\ \Omega_{K/\mathbf{F}_p} \otimes K' & \longrightarrow 0 \\ \Omega_{K/\mathbf{F}_p} \otimes K' & \xrightarrow{\gamma} \Omega_{K'/\mathbf{F}_p}\end{aligned}\end{gathered}$$ By Lemma Cartier's equality for differentials and the Jacobi-Zariski sequence for $\mathbf{F}_p \to K \to K'$ we see that the kernel and cokernel of $\gamma$ have the same finite dimension. By assumption $A'$ is regular (and of the same dimension as $A$, see above) hence the kernel and cokernel of $\alpha$ have the same dimension. It follows that the kernel and cokernel of $\beta$ have the same dimension which is what we wanted to show.

The implication (1) $\Rightarrow$ (2) is trivial. This finishes the proof of the proposition. $\square$

Lemma. Geometric regularity over a field

Let $k$ be a field of characteristic $p > 0$. Let $(A, \mathfrak m, K)$ be a Noetherian local $k$-algebra. Assume $A$ is geometrically regular over $k$. Let $K/F/k$ be a finitely generated subextension. Let $\varphi : k[y_1, \ldots, y_m] \to A$ be a $k$-algebra map such that $y_i$ maps to an element of $F$ in $K$ and such that $\text{d}y_1, \ldots, \text{d}y_m$ map to a basis of $\Omega_{F/k}$. Set $\mathfrak p = \varphi^{-1}(\mathfrak m)$. Then $$k[y_1, \ldots, y_m]_\mathfrak p \to A$$ is flat and $A/\mathfrak pA$ is regular.

Proof. Set $A_0 = k[y_1, \ldots, y_m]_\mathfrak p$ with maximal ideal $\mathfrak m_0$ and residue field $K_0$. Note that $\Omega_{A_0/k}$ is free of rank $m$ and $\Omega_{A_0/k} \otimes K_0 \to \Omega_{K_0/k}$ is an isomorphism. It is clear that $A_0$ is geometrically regular over $k$. Hence $H_1(L_{K_0/k}) \to \mathfrak m_0/\mathfrak m_0^2$ is an isomorphism, see Proposition Characterizations of geometric regularity. Now consider $$\begin{gathered}\begin{matrix}H_1(L_{K_0/k}) \otimes K & \mathfrak m_0/\mathfrak m_0^2 \otimes K \\ H_1(L_{K/k}) & \mathfrak m/\mathfrak m^2\end{matrix} \\[6pt] \begin{aligned}H_1(L_{K_0/k}) \otimes K & \longrightarrow H_1(L_{K/k}) \\ H_1(L_{K_0/k}) \otimes K & \longrightarrow \mathfrak m_0/\mathfrak m_0^2 \otimes K \\ \mathfrak m_0/\mathfrak m_0^2 \otimes K & \longrightarrow \mathfrak m/\mathfrak m^2 \\ H_1(L_{K/k}) & \longrightarrow \mathfrak m/\mathfrak m^2\end{aligned}\end{gathered}$$ Since the left vertical arrow is injective by Lemma Transitivity of first cotangent homology and the lower horizontal by Proposition Characterizations of geometric regularity we conclude that the right vertical one is too. Hence a regular system of parameters in $A_0$ maps to part of a regular system of parameters in $A$. We win by Algebra, Lemmas Flatness over a regular local ring and Regular rings are Cohen–Macaulay. $\square$

Lemma. Base change of regular ring maps (Regular maps and base change)

Let $R \to \Lambda$ be a regular ring map. For any finite type ring map $R \to R'$ the base change $R' \to \Lambda \otimes_R R'$ is regular too.

Proof. Flatness is preserved under any base change, see Algebra, Lemma Base change of flat modules. Consider a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p \subset R$. The residue field extension $\kappa(\mathfrak p')/\kappa(\mathfrak p)$ is finitely generated as $R'$ is of finite type over $R$. Hence the fibre ring $$(\Lambda \otimes_R R') \otimes_{R'} \kappa(\mathfrak p') = \Lambda \otimes_R \kappa(\mathfrak p) \otimes_{\kappa(\mathfrak p)} \kappa(\mathfrak p')$$ is Noetherian by Algebra, Lemma Noetherianity under extension of the ground field and the assumption on the fibre rings of $R \to \Lambda$. Geometric regularity of the fibres is preserved by Algebra, Lemma Criteria for geometric regularity. $\square$

Lemma. Composition of regular ring maps (Composition of regular maps)

Let $A \to B$ and $B \to C$ be regular ring maps. If the fibre rings of $A \to C$ are Noetherian, then $A \to C$ is regular.

Proof. Let $\mathfrak p \subset A$ be a prime. Let $\kappa(\mathfrak p) \subset k$ be a finite purely inseparable extension. We have to show that $C \otimes_A k$ is regular. By Lemma Base change of regular ring maps we may assume that $A = k$ and we reduce to proving that $C$ is regular. The assumption is that $B$ is regular and that $B \to C$ is flat with regular fibres. Then $C$ is regular by Algebra, Lemma Regularity over a regular base with regular fibre. Here are the details of the reduction. First localize at the source prime, then quotient by it and make a finite purely inseparable extension $k$ of its residue field. Base change preserves flatness and the regular-fibre condition. The intermediate ring $B\otimes_A k$ is Noetherian and regular, by the regular-map fibre hypothesis and finite field extension. The target $C\otimes_A k$ is Noetherian by the assumed Noetherian composite fibres, and $B\otimes_A k\to C\otimes_A k$ has regular fibres. The flat regular-source regular-fibre criterion proves that this target is regular. These are all geometric fibres of the composite. The composite is flat because both maps are flat. $\square$

Lemma. Permanence of regular ring maps

Let $A \to B \to C$ be ring maps. If $A \to C$ is regular and $B \to C$ is flat and surjective on spectra, then $A \to B$ is regular.

Proof. By Algebra, Lemma Permanence of flat ring maps we see that $A \to B$ is flat. Let $\mathfrak p \subset A$ be a prime. The ring map $B \otimes_A \kappa(\mathfrak p) \to C \otimes_A \kappa(\mathfrak p)$ is flat and surjective on spectra. Hence $B \otimes_A \kappa(\mathfrak p)$ is geometrically regular by Algebra, Lemma Descent of geometric regularity. $\square$

B.10.2. Commutative Algebra

Lemma. A left inverse for a matrix

Let $R$ be a ring. Let $n \geq m$. Let $A$ be an $n \times m$ matrix with coefficients in $R$. Let $J \subset R$ be the ideal generated by the $m \times m$ minors of $A$.

1. For any $f \in J$ there exists a $m \times n$ matrix $B$ such that $BA = f 1_{m \times m}$.

2. If $f \in R$ and $BA = f 1_{m \times m}$ for some $m \times n$ matrix $B$, then $f^m \in J$.

Proof. For $I \subset \{1, \ldots, n\}$ with $|I| = m$, we denote by $E_I$ the $m \times n$ matrix of the projection $$R^{\oplus n} = \bigoplus\nolimits_{i \in \{1, \ldots, n\}} R \longrightarrow \bigoplus\nolimits_{i \in I} R$$ and set $A_I = E_I A$, i.e., $A_I$ is the $m \times m$ matrix whose rows are the rows of $A$ with indices in $I$. Let $B_I$ be the adjugate (transpose of cofactor) matrix to $A_I$, i.e., such that $A_I B_I = B_I A_I = \det(A_I) 1_{m \times m}$. The $m \times m$ minors of $A$ are the determinants $\det A_I$ for all the $I \subset \{1, \ldots, n\}$ with $|I| = m$. If $f \in J$ then we can write $f = \sum c_I \det(A_I)$ for some $c_I \in R$. Set $B = \sum c_I B_I E_I$ to see that (1) holds.

If $f 1_{m \times m} = BA$ then by the Cauchy-Binet formula (Commutative algebra) we have $f^m = \sum b_I \det(A_I)$ where $b_I$ is the determinant of the $m \times m$ matrix whose columns are the columns of $B$ with indices in $I$. $\square$

Lemma. Flatness over a regular local ring

Let $R \to S$ be a homomorphism of Noetherian local rings. Assume that $R$ is a regular local ring and that a regular system of parameters maps to a regular sequence in $S$. Then $R \to S$ is flat.

Proof. Suppose that $x_1, \ldots, x_d$ are a system of parameters of $R$ which map to a regular sequence in $S$. Note that $S/(x_1, \ldots, x_d)S$ is flat over $R/(x_1, \ldots, x_d)$ as the latter is a field. Then $x_d$ is a nonzerodivisor in $S/(x_1, \ldots, x_{d - 1})S$ hence $S/(x_1, \ldots, x_{d - 1})S$ is flat over $R/(x_1, \ldots, x_{d - 1})$ by the local criterion of flatness (see Lemma A variant of the local criterion for flatness and remarks following). Then $x_{d - 1}$ is a nonzerodivisor in $S/(x_1, \ldots, x_{d - 2})S$ hence $S/(x_1, \ldots, x_{d - 2})S$ is flat over $R/(x_1, \ldots, x_{d - 2})$ by the local criterion of flatness (see Lemma A variant of the local criterion for flatness and remarks following). Continue till one reaches the conclusion that $S$ is flat over $R$. $\square$

Lemma. Regular sequences in a polynomial ring

Let $R$ be a ring. Let $f_1, \ldots, f_r \in R$ which do not generate the unit ideal. The following are equivalent:

1. any permutation of $f_1, \ldots, f_r$ is a regular sequence,

2. any subsequence of $f_1, \ldots, f_r$ (in the given order) is a regular sequence, and

3. $f_1x_1, \ldots, f_rx_r$ is a regular sequence in the polynomial ring $R[x_1, \ldots, x_r]$.

Proof. It is clear that (1) implies (2). We prove (2) implies (1) by induction on $r$. The case $r = 1$ is trivial. The case $r = 2$ says that if $a, b \in R$ are a regular sequence and $b$ is a nonzerodivisor, then $b, a$ is a regular sequence. This is clear because the kernel of $a : R/(b) \to R/(b)$ is isomorphic to the kernel of $b : R/(a) \to R/(a)$ if both $a$ and $b$ are nonzerodivisors. The case $r > 2$. Assume (2) holds and say we want to prove $f_{\sigma(1)}, \ldots, f_{\sigma(r)}$ is a regular sequence for some permutation $\sigma$. We already know that $f_{\sigma(1)}, \ldots, f_{\sigma(r - 1)}$ is a regular sequence by induction. Hence it suffices to show that $f_s$ where $s = \sigma(r)$ is a nonzerodivisor modulo $f_1, \ldots, \hat f_s, \ldots, f_r$. If $s = r$ we are done. If $s < r$, then note that $f_s$ and $f_r$ are both nonzerodivisors in the ring $R/(f_1, \ldots, \hat f_s, \ldots, f_{r - 1})$ (by induction hypothesis again). Since we know $f_s, f_r$ is a regular sequence in that ring we conclude by the case of sequence of length $2$ that $f_r, f_s$ is too.

Note that $R[x_1, \ldots, x_r]/(f_1x_1, \ldots, f_ix_i)$ as an $R$-module is a direct sum of the modules $$R/I_E \cdot x_1^{e_1} \ldots x_r^{e_r}$$ indexed by multi-indices $E = (e_1, \ldots, e_r)$ where $I_E$ is the ideal generated by $f_j$ for $1 \leq j \leq i$ with $e_j > 0$. Hence $f_{i + 1}x_{i + 1}$ is a nonzerodivisor on this if and only if $f_{i + 1}$ is a nonzerodivisor on $R/I_E$ for all $E$. Taking $E$ with all positive entries, we see that $f_{i + 1}$ is a nonzerodivisor on $R/(f_1, \ldots, f_i)$. Thus (3) implies (2). Conversely, if (2) holds, then any subsequence of $f_1, \ldots, f_i, f_{i + 1}$ is a regular sequence in particular $f_{i + 1}$ is a nonzerodivisor on all $R/I_E$. In this way we see that (2) implies (3). $\square$

Lemma. Finite generation in an Artinian local target

Let $R \to S$ be a ring map. Assume $S$ is an Artinian local ring with maximal ideal $\mathfrak m$. Then

1. $R \to S$ is finite if and only if $R \to S/\mathfrak m$ is finite,

2. $R \to S$ is of finite type if and only if $R \to S/\mathfrak m$ is of finite type.

3. $R \to S$ is essentially of finite type if and only if the composition $R \to S/\mathfrak m$ is essentially of finite type.

Proof. If $R \to S$ is finite, then $R \to S/\mathfrak m$ is finite by Lemma Transitivity of finite ring extensions. Conversely, assume $R \to S/\mathfrak m$ is finite. As $S$ has finite length over itself (Lemma Finite length over an Artinian ring) we can choose a filtration $$0 \subset I_1 \subset \ldots \subset I_n = S$$ by ideals such that $I_i/I_{i - 1} \cong S/\mathfrak m$ as $S$-modules. Thus $S$ has a filtration by $R$-submodules $I_i$ such that each successive quotient is a finite $R$-module. Thus $S$ is a finite $R$-module by Lemma Commutative algebra.

If $R \to S$ is of finite type, then $R \to S/\mathfrak m$ is of finite type by Lemma Composition of finite-type ring maps. Conversely, assume that $R \to S/\mathfrak m$ is of finite type. Choose $f_1, \ldots, f_n \in S$ which map to generators of $S/\mathfrak m$. Then $A = R[x_1, \ldots, x_n] \to S$, $x_i \mapsto f_i$ is a ring map such that $A \to S/\mathfrak m$ is surjective (in particular finite). Hence $A \to S$ is finite by part (1) and we see that $R \to S$ is of finite type by Lemma Composition of finite-type ring maps.

If $R \to S$ is essentially of finite type, then $R \to S/\mathfrak m$ is essentially of finite type by Lemma Composition of essentially finite-type ring maps. Conversely, assume that $R \to S/\mathfrak m$ is essentially of finite type. Suppose $S/\mathfrak m$ is the localization of $R[x_1, \ldots, x_n]/I$. Choose $f_1, \ldots, f_n \in S$ whose congruence classes modulo $\mathfrak m$ correspond to the congruence classes of $x_1, \ldots, x_n$ modulo $I$. Consider the map $R[x_1, \ldots, x_n] \to S$, $x_i \mapsto f_i$ with kernel $J$. Set $A = R[x_1, \ldots, x_n]/J \subset S$ and $\mathfrak p = A \cap \mathfrak m$. Note that $A/\mathfrak p \subset S/\mathfrak m$ is equal to the image of $R[x_1, \ldots, x_n]/I$ in $S/\mathfrak m$. Hence $\kappa(\mathfrak p) = S/\mathfrak m$. Thus $A_\mathfrak p \to S$ is finite by part (1). We conclude that $S$ is essentially of finite type by Lemma Composition of essentially finite-type ring maps. $\square$

Lemma. Syntomic algebras in a filtered colimit

Let $K/k$ be a field extension. Then $K$ is a filtered colimit of global complete intersection algebras over $k$. If $K/k$ is separable, then $K$ is a filtered colimit of smooth algebras over $k$.

Proof. Suppose that $E \subset K$ is a finite subset. It suffices to show that there exists a $k$ subalgebra $A \subset K$ which contains $E$ and which is a global complete intersection (resp. smooth) over $k$. The separable/smooth case follows from Lemma Smooth localizations of separable extensions. In general let $L \subset K$ be the subfield generated by $E$. Pick a transcendence basis $x_1, \ldots, x_d \in L$ over $k$. The extension $L/k(x_1, \ldots, x_d)$ is finite. Say $L = k(x_1, \ldots, x_d)[y_1, \ldots, y_r]$. Pick inductively polynomials $P_i \in k(x_1, \ldots, x_d)[Y_1, \ldots, Y_r]$ such that $P_i = P_i(Y_1, \ldots, Y_i)$ is monic in $Y_i$ over $k(x_1, \ldots, x_d)[Y_1, \ldots, Y_{i - 1}]$ and maps to the minimal polynomial of $y_i$ in $k(x_1, \ldots, x_d)[y_1, \ldots, y_{i - 1}][Y_i]$. Then it is clear that $P_1, \ldots, P_r$ are a regular sequence in $k(x_1, \ldots, x_d)[Y_1, \ldots, Y_r]$ and that $L = k(x_1, \ldots, x_d)[Y_1, \ldots, Y_r]/(P_1, \ldots, P_r)$. If $h \in k[x_1, \ldots, x_d]$ is a polynomial such that $P_i \in k[x_1, \ldots, x_d, 1/h, Y_1, \ldots, Y_r]$, then we see that $P_1, \ldots, P_r$ is a regular sequence in $k[x_1, \ldots, x_d, 1/h, Y_1, \ldots, Y_r]$ and $A = k[x_1, \ldots, x_d, 1/h, Y_1, \ldots, Y_r]/(P_1, \ldots, P_r)$ is a global complete intersection. After adjusting our choice of $h$ we may assume $E \subset A$ and we win. $\square$

B.11. The additional G-ring permanence argument

The definition, local criterion, complete-local and henselization results, and full essentially-finite-type permanence proof are retained here. This explicitly supplies the theorem used on the local polynomial algebra in Lesson 7. Its exact unintegrated lower supports remain visible in §12.

G-rings

Let $A$ be a Noetherian local ring. In Section Completion and formal smoothness we have seen that some but not all properties of $A$ are reflected in the completion $A^\wedge$ of $A$. To study this further we introduce some terminology. For a prime $\mathfrak q$ of $A$ the fibre ring $$A^\wedge \otimes_A \kappa(\mathfrak q) = (A^\wedge)_\mathfrak q/\mathfrak q(A^\wedge)_\mathfrak q = (A/\mathfrak q)^\wedge \otimes_{A/\mathfrak q} \kappa(\mathfrak q)$$ is called a formal fibre of $A$. We think of the formal fibre as an algebra over $\kappa(\mathfrak q)$. Thus $A \to A^\wedge$ is a regular ring homomorphism if and only if all the formal fibres are geometrically regular algebras.

Definition. G-rings

A ring $R$ is called a G-ring if $R$ is Noetherian and for every prime $\mathfrak p$ of $R$ the ring map $R_\mathfrak p \to (R_\mathfrak p)^\wedge$ is regular.

By the discussion above we see that $R$ is a G-ring if and only if every local ring $R_\mathfrak p$ has geometrically regular formal fibres. Note that if $\mathbf{Q} \subset R$, then it suffices to check the formal fibres are regular. Another way to express the G-ring condition is described in the following lemma.

Lemma. Recognizing a G-ring from a completion

Let $R$ be a Noetherian ring. Then $R$ is a G-ring if and only if for every pair of primes $\mathfrak q \subset \mathfrak p \subset R$ the algebra $$(R/\mathfrak q)_\mathfrak p^\wedge \otimes_{R/\mathfrak q} \kappa(\mathfrak q)$$ is geometrically regular over $\kappa(\mathfrak q)$.

Proof. This follows from the fact that $$R_\mathfrak p^\wedge \otimes_R \kappa(\mathfrak q) = (R/\mathfrak q)_\mathfrak p^\wedge \otimes_{R/\mathfrak q} \kappa(\mathfrak q)$$ as algebras over $\kappa(\mathfrak q)$. $\square$

Lemma. G-rings under quasi-finite extensions

Let $R \to R'$ be a finite type map of Noetherian rings and let $$\begin{gathered}\begin{matrix}\mathfrak q' & \mathfrak p' & R' \\ \mathfrak q & \mathfrak p & R\end{matrix} \\[6pt] \begin{aligned}\mathfrak q' & \longrightarrow \mathfrak p' \\ \mathfrak p' & \longrightarrow R' \\ \mathfrak q & \longrightarrow \mathfrak p \\ \mathfrak q & \mathrel{-} \mathfrak q' \\ \mathfrak p & \longrightarrow R \\ \mathfrak p & \mathrel{-} \mathfrak p' \\ R & \longrightarrow R'\end{aligned}\end{gathered}$$ be primes. Assume $R \to R'$ is quasi-finite at $\mathfrak p'$.

1. If the formal fibre $R_\mathfrak p^\wedge \otimes_R \kappa(\mathfrak q)$ is geometrically regular over $\kappa(\mathfrak q)$, then the formal fibre $(R'_{\mathfrak p'})^\wedge \otimes_{R'} \kappa(\mathfrak q')$ is geometrically regular over $\kappa(\mathfrak q')$.

2. If the formal fibres of $R_\mathfrak p$ are geometrically regular, then the formal fibres of $R'_{\mathfrak p'}$ are geometrically regular.

3. If $R \to R'$ is quasi-finite and $R$ is a G-ring, then $R'$ is a G-ring.

Proof. It is clear that (1) \(\Rightarrow\) (2) \(\Rightarrow\) (3). Assume \(R_\mathfrak p^\wedge \otimes_R \kappa(\mathfrak q)\) is geometrically regular over \(\kappa(\mathfrak q)\). By Algebra, Lemma Completion at a quasi-finite prime we see that

for some \(R_\mathfrak p^\wedge\)-algebra \(B\). Hence \(R'_{\mathfrak p'} \to (R'_{\mathfrak p'})^\wedge\) is a factor of a base change of the map \(R_\mathfrak p \to R_\mathfrak p^\wedge\). It follows that \((R'_{\mathfrak p'})^\wedge \otimes_{R'} \kappa(\mathfrak q')\) is a factor of

Thus the result follows as extension of base field preserves geometric regularity, see Algebra, Lemma Criteria for geometric regularity. \(\square\)

Lemma. Testing geometric regularity of formal fibres

Let $R$ be a Noetherian ring. Then $R$ is a G-ring if and only if for every finite free ring map $R \to S$ the formal fibres of $S$ are regular rings.

Proof. Assume that for any finite free ring map \(R \to S\) the ring \(S\) has regular formal fibres. Let \(\mathfrak q \subset \mathfrak p \subset R\) be primes and let \(\kappa(\mathfrak q) \subset L\) be a finite purely inseparable extension. To show that \(R\) is a G-ring it suffices to show that

is a regular ring. Choose a finite free extension \(R \to R'\) such that \(\mathfrak q' = \mathfrak qR'\) is a prime and such that \(\kappa(\mathfrak q')\) is isomorphic to \(L\) over \(\kappa(\mathfrak q)\), see Algebra, Lemma Finite free algebras with a prescribed residue extension. By Algebra, Lemma Completion of a finite ring extension we have

where \(\mathfrak p_i'\) are the primes of \(R'\) lying over \(\mathfrak p\). Thus we have

Our assumption is that the rings on the right are regular, hence the ring on the left is regular too. Thus \(R\) is a G-ring. The converse follows from Lemma G-rings under quasi-finite extensions. \(\square\)

Lemma. Geometric regularity of generic formal fibres in positive characteristic

Let $k$ be a field of characteristic $p$. Let $A = k[[x_1, \ldots, x_n]][y_1, \ldots, y_m]$ and denote $K$ the fraction field of $A$. Let $\mathfrak p \subset A$ be a prime. Then $A_\mathfrak p^\wedge \otimes_A K$ is geometrically regular over $K$.

Proof. Let $L/K$ be a finite purely inseparable field extension. We will show by induction on $[L : K]$ that $A_\mathfrak p^\wedge \otimes L$ is regular. The base case is $L = K$: as $A$ is regular, $A_\mathfrak p^\wedge$ is regular (Lemma Regularity and completion), hence the localization $A_\mathfrak p^\wedge \otimes K$ is regular. Let $K \subset M \subset L$ be a subfield such that $L$ is a degree $p$ extension of $M$ obtained by adjoining a $p$th root of an element $f \in M$. Let $B$ be a finite $A$-subalgebra of $M$ with fraction field $M$. Clearing denominators, we may and do assume $f \in B$. Set $C = B[z]/(z^p -f)$ and note that $B \subset C$ is finite and that the fraction field of $C$ is $L$. Since $A \subset B \subset C$ are finite and $L/M/K$ are purely inseparable we see that for every element of $B$ or $C$ some power of it lies in $A$. Hence there is a unique prime $\mathfrak r \subset B$, resp. $\mathfrak q \subset C$ lying over $\mathfrak p$. Note that $$A_\mathfrak p^\wedge \otimes_A M = B_\mathfrak r^\wedge \otimes_B M$$ see Algebra, Lemma Completion of a finite ring extension. By induction we know that this ring is regular. In the same manner we have $$A_\mathfrak p^\wedge \otimes_A L = C_\mathfrak q^\wedge \otimes_C L = B_\mathfrak r^\wedge \otimes_B M[z]/(z^p - f)$$ the last equality because the completion of $C = B[z]/(z^p - f)$ equals $B_\mathfrak r^\wedge[z]/(z^p -f)$. By Lemma Derivations of formal power series in positive characteristic we know there exists a derivation $D : B \to B$ such that $D(f) \not = 0$. In other words, $g = D(f)$ is a unit in $M$! By Lemma Extending a derivation $D$ extends to a derivation of $B_\mathfrak r$, $B_\mathfrak r^\wedge$ and $B_\mathfrak r^\wedge \otimes_B M$ (successively extending through a localization, a completion, and a localization). Since it is an extension we end up with a derivation of $B_\mathfrak r^\wedge \otimes_B M$ which maps $f$ to $g$ and $g$ is a unit of the ring $B_\mathfrak r^\wedge \otimes_B M$. Hence $A_\mathfrak p^\wedge \otimes_A L$ is regular by Lemma Regularity after an extension of degree p and we win. $\square$

Proposition. Complete Noetherian rings are G-rings

A Noetherian complete local ring is a G-ring.

Proof. Let $A$ be a Noetherian complete local ring. By Lemma Recognizing a G-ring from a completion it suffices to check that $B = A/\mathfrak q$ has geometrically regular formal fibres over the minimal prime $(0)$ of $B$. Thus we may assume that $A$ is a domain and it suffices to check the condition for the formal fibres over the minimal prime $(0)$ of $A$. Let $K$ be the fraction field of $A$.

We can choose a subring $A_0 \subset A$ which is a regular complete local ring such that $A$ is finite over $A_0$, see Algebra, Lemma A complete local domain finite over a regular ring. Moreover, we may assume that $A_0$ is a power series ring over a field or a Cohen ring. By Lemma G-rings under quasi-finite extensions we see that it suffices to prove the result for $A_0$.

Assume that $A$ is a power series ring over a field or a Cohen ring. Since $A$ is regular the localizations $A_\mathfrak p$ are regular (see Algebra, Definition Regular Noetherian rings and the discussion preceding it). Hence the completions $A_\mathfrak p^\wedge$ are regular, see Lemma Regularity and completion. Hence the fibre $A_{\mathfrak p}^\wedge \otimes_A K$ is, as a localization of $A_\mathfrak p^\wedge$, also regular. Thus we are done if the characteristic of $K$ is $0$. The positive characteristic case is the case $A = k[[x_1, \ldots, x_d]]$ which is a special case of Lemma Geometric regularity of generic formal fibres in positive characteristic. $\square$

Lemma. Testing the G-ring property at maximal ideals

Let $R$ be a Noetherian ring. Then $R$ is a G-ring if and only if $R_\mathfrak m$ has geometrically regular formal fibres for every maximal ideal $\mathfrak m$ of $R$.

Proof. Assume $R_\mathfrak m \to R_\mathfrak m^\wedge$ is regular for every maximal ideal $\mathfrak m$ of $R$. Let $\mathfrak p$ be a prime of $R$ and choose a maximal ideal $\mathfrak p \subset \mathfrak m$. Since $R_\mathfrak m \to R_\mathfrak m^\wedge$ is faithfully flat we can choose a prime $\mathfrak p'$ in $R_\mathfrak m^\wedge$ lying over $\mathfrak pR_\mathfrak m$. Consider the commutative diagram $$\begin{gathered}\begin{matrix}R_\mathfrak m^\wedge & (R_\mathfrak m^\wedge)_{\mathfrak p'} & (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge \\ R_\mathfrak m & R_\mathfrak p & R_\mathfrak p^\wedge\end{matrix} \\[6pt] \begin{aligned}R_\mathfrak m^\wedge & \longrightarrow (R_\mathfrak m^\wedge)_{\mathfrak p'} \\ (R_\mathfrak m^\wedge)_{\mathfrak p'} & \longrightarrow (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge \\ R_\mathfrak m & \longrightarrow R_\mathfrak m^\wedge \\ R_\mathfrak m & \longrightarrow R_\mathfrak p \\ R_\mathfrak p & \longrightarrow (R_\mathfrak m^\wedge)_{\mathfrak p'} \\ R_\mathfrak p & \longrightarrow R_\mathfrak p^\wedge \\ R_\mathfrak p^\wedge & \longrightarrow (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge\end{aligned}\end{gathered}$$ By assumption the ring map $R_\mathfrak m \to R_\mathfrak m^\wedge$ is regular. By Proposition Complete Noetherian rings are G-rings $(R_\mathfrak m^\wedge)_{\mathfrak p'} \to (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge$ is regular. The localization $R_\mathfrak m^\wedge \to (R_\mathfrak m^\wedge)_{\mathfrak p'}$ is regular. Hence $R_\mathfrak m \to (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge$ is regular by Lemma Composition of regular ring maps. Since it factors through the localization $R_\mathfrak p$, also the ring map $R_\mathfrak p \to (R_\mathfrak m^\wedge)_{\mathfrak p'}^\wedge$ is regular. Thus we may apply Lemma Permanence of regular ring maps to see that $R_\mathfrak p \to R_\mathfrak p^\wedge$ is regular. $\square$

Lemma. The G-ring property under henselization

Let $R$ be a Noetherian local ring which is a G-ring. Then the henselization $R^h$ and the strict henselization $R^{sh}$ are G-rings.

Proof. We will use the criterion of Lemma Testing the G-ring property at maximal ideals. Let $\mathfrak q \subset R^h$ be a prime and set $\mathfrak p = R \cap \mathfrak q$. Set $\mathfrak q_1 = \mathfrak q$ and let $\mathfrak q_2, \ldots, \mathfrak q_t$ be the other primes of $R^h$ lying over $\mathfrak p$, so that $R^h \otimes_R \kappa(\mathfrak p) = \prod\nolimits_{i = 1, \ldots, t} \kappa(\mathfrak q_i)$, see Lemma Fibres of henselization maps. Using that $(R^h)^\wedge = R^\wedge$ (Lemma Noetherianity of a henselization) we see $$\prod\nolimits_{i = 1, \ldots, t} (R^h)^\wedge \otimes_{R^h} \kappa(\mathfrak q_i) = (R^h)^\wedge \otimes_{R^h} (R^h \otimes_R \kappa(\mathfrak p)) = R^\wedge \otimes_R \kappa(\mathfrak p)$$ Hence $(R^h)^\wedge \otimes_{R^h} \kappa(\mathfrak q_i)$ is geometrically regular over $\kappa(\mathfrak p)$ by assumption. Since $\kappa(\mathfrak q_i)$ is separable algebraic over $\kappa(\mathfrak p)$ it follows from Algebra, Lemma Geometric regularity under separable algebraic extensions that $(R^h)^\wedge \otimes_{R^h} \kappa(\mathfrak q_i)$ is geometrically regular over $\kappa(\mathfrak q_i)$.

Let $\mathfrak r \subset R^{sh}$ be a prime and set $\mathfrak p = R \cap \mathfrak r$. Set $\mathfrak r_1 = \mathfrak r$ and let $\mathfrak r_2, \ldots, \mathfrak r_s$ be the other primes of $R^{sh}$ lying over $\mathfrak p$, so that $R^{sh} \otimes_R \kappa(\mathfrak p) = \prod\nolimits_{i = 1, \ldots, s} \kappa(\mathfrak r_i)$, see Lemma Fibres of henselization maps. Then we see that $$\prod\nolimits_{i = 1, \ldots, s} (R^{sh})^\wedge \otimes_{R^{sh}} \kappa(\mathfrak r_i) = (R^{sh})^\wedge \otimes_{R^{sh}} (R^{sh} \otimes_R \kappa(\mathfrak p)) = (R^{sh})^\wedge \otimes_R \kappa(\mathfrak p)$$ Note that $R^\wedge \to (R^{sh})^\wedge$ is formally smooth in the $\mathfrak m_{(R^{sh})^\wedge}$-adic topology, see Lemma Noetherianity of a henselization. Hence $R^\wedge \to (R^{sh})^\wedge$ is regular by Proposition Formal smoothness and regularity. We conclude that $(R^{sh})^\wedge \otimes_{R^{sh}} \kappa(\mathfrak r_i)$ is regular over $\kappa(\mathfrak p)$ by Lemma Composition of regular ring maps as $R^\wedge \otimes_R \kappa(\mathfrak p)$ is regular over $\kappa(\mathfrak p)$ by assumption. Since $\kappa(\mathfrak r_i)$ is separable algebraic over $\kappa(\mathfrak p)$ it follows from Algebra, Lemma Geometric regularity under separable algebraic extensions that $(R^{sh})^\wedge \otimes_{R^{sh}} \kappa(\mathfrak r_i)$ is geometrically regular over $\kappa(\mathfrak r_i)$. $\square$

Lemma. Geometric regularity of polynomial formal fibres in positive characteristic

Let $p$ be a prime number. Let $A$ be a Noetherian complete local domain with fraction field $K$ of characteristic $p$. Let $\mathfrak q \subset A[x]$ be a maximal ideal lying over the maximal ideal of $A$ and let $(0) \not = \mathfrak r \subset \mathfrak q$ be a prime lying over $(0) \subset A$. Then $A[x]_\mathfrak q^\wedge \otimes_{A[x]} \kappa(\mathfrak r)$ is geometrically regular over $\kappa(\mathfrak r)$.

Proof. Note that $K \subset \kappa(\mathfrak r)$ is finite. Hence, given a finite purely inseparable extension $L/\kappa(\mathfrak r)$ there exists a finite extension of Noetherian complete local domains $A \subset B$ such that $\kappa(\mathfrak r) \otimes_A B$ surjects onto $L$. Namely, you take $B \subset L$ a finite $A$-subalgebra whose field of fractions is $L$. Denote $\mathfrak r' \subset B[x]$ the kernel of the map $B[x] = A[x] \otimes_A B \to \kappa(\mathfrak r) \otimes_A B \to L$ so that $\kappa(\mathfrak r') = L$. Then $$A[x]_\mathfrak q^\wedge \otimes_{A[x]} L = A[x]_\mathfrak q^\wedge \otimes_{A[x]} B[x] \otimes_{B[x]} \kappa(\mathfrak r') = \prod B[x]_{\mathfrak q_i}^\wedge \otimes_{B[x]} \kappa(\mathfrak r')$$ where $\mathfrak q_1, \ldots, \mathfrak q_t$ are the primes of $B[x]$ lying over $\mathfrak q$, see Algebra, Lemma Completion of a finite ring extension. Thus we see that it suffices to prove the rings $B[x]_{\mathfrak q_i}^\wedge \otimes_{B[x]} \kappa(\mathfrak r')$ are regular. This reduces us to showing that $A[x]_\mathfrak q^\wedge \otimes_{A[x]} \kappa(\mathfrak r)$ is regular in the special case that $K = \kappa(\mathfrak r)$.

Assume \(K = \kappa(\mathfrak r)\). In this case we see that \(\mathfrak r K[x]\) is generated by \(x - f\) for some \(f \in K\) and

The derivation \(D = \text{d}/\text{d}x\) of \(A[x]\) extends to \(K[x]\) and maps \(x - f\) to a unit of \(K[x]\). Moreover \(D\) extends to \(A[x]_\mathfrak q^\wedge \otimes_A K\) by Lemma Extending a derivation. As \(A \to A[x]_\mathfrak q^\wedge\) is formally smooth (see Lemmas Formal smoothness and smooth morphisms and Formal smoothness and completion) the ring \(A[x]_\mathfrak q^\wedge \otimes_A K\) is regular by Proposition Formal smoothness and regularity (the arguments of the proof of that proposition simplify significantly in this particular case). We conclude by Lemma Regularity of a quotient. \(\square\)

Proposition. The G-ring property for finite-type algebras

Let $R$ be a G-ring. If $R \to S$ is essentially of finite type then $S$ is a G-ring.

Proof. Since being a G-ring is a property of the local rings it is clear that a localization of a G-ring is a G-ring. Conversely, if every localization at a prime is a G-ring, then the ring is a G-ring. Thus it suffices to show that $S_\mathfrak q$ is a G-ring for every finite type $R$-algebra $S$ and every prime $\mathfrak q$ of $S$. Writing $S$ as a quotient of $R[x_1, \ldots, x_n]$ we see from Lemma G-rings under quasi-finite extensions that it suffices to prove that $R[x_1, \ldots, x_n]$ is a G-ring. By induction on $n$ it suffices to prove that $R[x]$ is a G-ring. Let $\mathfrak q \subset R[x]$ be a maximal ideal. By Lemma Testing the G-ring property at maximal ideals it suffices to show that $$R[x]_\mathfrak q \longrightarrow R[x]_\mathfrak q^\wedge$$ is regular. If $\mathfrak q$ lies over $\mathfrak p \subset R$, then we may replace $R$ by $R_\mathfrak p$. Hence we may assume that $R$ is a Noetherian local G-ring with maximal ideal $\mathfrak m$ and that $\mathfrak q \subset R[x]$ lies over $\mathfrak m$. Note that there is a unique prime $\mathfrak q' \subset R^\wedge[x]$ lying over $\mathfrak q$. Consider the diagram $$\begin{gathered}\begin{matrix}R[x]_\mathfrak q^\wedge & (R^\wedge[x]_{\mathfrak q'})^\wedge \\ R[x]_\mathfrak q & R^\wedge[x]_{\mathfrak q'}\end{matrix} \\[6pt] \begin{aligned}R[x]_\mathfrak q^\wedge & \longrightarrow (R^\wedge[x]_{\mathfrak q'})^\wedge \\ R[x]_\mathfrak q & \longrightarrow R^\wedge[x]_{\mathfrak q'} \\ R[x]_\mathfrak q & \longrightarrow R[x]_\mathfrak q^\wedge \\ R^\wedge[x]_{\mathfrak q'} & \longrightarrow (R^\wedge[x]_{\mathfrak q'})^\wedge\end{aligned}\end{gathered}$$ Since $R$ is a G-ring the lower horizontal arrow is regular (as a localization of a base change of the regular ring map $R \to R^\wedge$). Suppose we can prove the right vertical arrow is regular. Then it follows that the composition $R[x]_\mathfrak q \to (R^\wedge[x]_{\mathfrak q'})^\wedge$ is regular, and hence the left vertical arrow is regular by Lemma Permanence of regular ring maps. Hence we see that we may assume $R$ is a Noetherian complete local ring and $\mathfrak q$ a prime lying over the maximal ideal of $R$.

Let $R$ be a Noetherian complete local ring and let $\mathfrak q \subset R[x]$ be a maximal ideal lying over the maximal ideal of $R$. Let $\mathfrak r \subset \mathfrak q$ be a prime ideal. We want to show that $R[x]_\mathfrak q^\wedge \otimes_{R[x]} \kappa(\mathfrak r)$ is a geometrically regular algebra over $\kappa(\mathfrak r)$. Set $\mathfrak p = R \cap \mathfrak r$. Then we can replace $R$ by $R/\mathfrak p$ and $\mathfrak q$ and $\mathfrak r$ by their images in $R/\mathfrak p[x]$, see Lemma Recognizing a G-ring from a completion. Hence we may assume that $R$ is a domain and that $\mathfrak r \cap R = (0)$.

By Algebra, Lemma A complete local domain finite over a regular ring we can find $R_0 \subset R$ which is regular and such that $R$ is finite over $R_0$. Applying Lemma G-rings under quasi-finite extensions we see that it suffices to prove $R[x]_\mathfrak q^\wedge \otimes_{R[x]} \kappa(\mathfrak r)$ is geometrically regular over $\kappa(\mathfrak r)$ when, in addition to the above, $R$ is a regular complete local ring.

Now $R$ is a regular complete local ring, we have $\mathfrak r \subset \mathfrak q \subset R[x]$, we have $(0) = R \cap \mathfrak r$ and $\mathfrak q$ is a maximal ideal lying over the maximal ideal of $R$. Since $R$ is regular the ring $R[x]$ is regular (Algebra, Lemma Regularity ascends along a regular ring map). Hence the localization $R[x]_\mathfrak q$ is regular. Hence the completions $R[x]_\mathfrak q^\wedge$ are regular, see Lemma Regularity and completion. Hence the fibre $R[x]_{\mathfrak q}^\wedge \otimes_{R[x]} \kappa(\mathfrak r)$ is, as a localization of $R[x]_\mathfrak q^\wedge$, also regular. Thus we are done if the characteristic of the fraction field of $R$ is $0$.

If the characteristic of $R$ is positive, then $R = k[[x_1, \ldots, x_n]]$. In this case we split the argument in two subcases:

1. The case $\mathfrak r = (0)$. The result is a direct consequence of Lemma Geometric regularity of generic formal fibres in positive characteristic.

2. The case $\mathfrak r \not = (0)$. This is Lemma Geometric regularity of polynomial formal fibres in positive characteristic.

$\square$

Remark. Failure of the G-ring property under completion

Let $R$ be a G-ring and let $I \subset R$ be an ideal. In general it is not the case that the $I$-adic completion $R^\wedge$ is a G-ring. An example was given by Nishimura in the original source citation Nishimura. A generalization and, in some sense, clarification of this example can be found in the last section of the original source citation Dumitrescu.

Proposition. Examples and permanence of G-rings

The following types of rings are G-rings:

1. fields,

2. Noetherian complete local rings,

3. $\mathbf{Z}$,

4. Dedekind domains with fraction field of characteristic zero,

5. finite type ring extensions of any of the above.

Proof. For fields, $\mathbf{Z}$ and Dedekind domains of characteristic zero this follows immediately from the definition and the fact that the completion of a discrete valuation ring is a discrete valuation ring. A Noetherian complete local ring is a G-ring by Proposition Complete Noetherian rings are G-rings. The statement on finite type overrings is Proposition The G-ring property for finite-type algebras. $\square$

Additional proofs of the supporting constructions

These statements and full proofs supply local support for the preceding arguments. They reuse the same Stacks Project edition. A remaining genuine prerequisite is marked explicitly rather than treated as proved.

Versality and algebraicity criteria

Lemma. Approximation of a marked family with its associated graded algebra

Let $S$ be a locally Noetherian scheme. Let $p : \mathcal{X} \to (\mathrm{Sch}/S)_{fppf}$ be a category fibred in groupoids. Let $x$ be an object of $\mathcal{X}$ lying over $\operatorname{Spec}(R)$ where $R$ is a Noetherian complete local ring with residue field $k$ of finite type over $S$. Let $s \in S$ be the image of $\operatorname{Spec}(k) \to S$. Assume that (a) $\mathcal{O}_{S, s}$ is a G-ring and (b) $p$ is limit preserving on objects. Then for every integer $N \geq 1$ there exist

1. a finite type $S$-algebra $A$,

2. a maximal ideal $\mathfrak m_A \subset A$,

3. an object $x_A$ of $\mathcal{X}$ over $\operatorname{Spec}(A)$,

4. an $S$-isomorphism $R/\mathfrak m_R^N \cong A/\mathfrak m_A^N$,

5. an isomorphism $x|_{\operatorname{Spec}(R/\mathfrak m_R^N)} \cong x_A|_{\operatorname{Spec}(A/\mathfrak m_A^N)}$ compatible with (4), and

6. an isomorphism $\text{Gr}_{\mathfrak m_R}(R) \cong \text{Gr}_{\mathfrak m_A}(A)$ of graded $k$-algebras.

Proof. Choose an affine open $\operatorname{Spec}(\Lambda) \subset S$ such that $k$ is a finite $\Lambda$-algebra, see Morphisms, Lemma Points of finite type (uncovered prerequisite). We may and do replace $S$ by $\operatorname{Spec}(\Lambda)$.

We may write $R$ as a directed colimit $R = \mathop{\operatorname{colim}} C_j$ where each $C_j$ is a finite type $\Lambda$-algebra (see Algebra, Lemma Filtered limits and finite presentation). By assumption (b) the object $x$ is isomorphic to the restriction of an object over one of the $C_j$. Hence we may choose a finite type $\Lambda$-algebra $C$, a $\Lambda$-algebra map $C \to R$, and an object $x_C$ of $\mathcal{X}$ over $\operatorname{Spec}(C)$ such that $x = x_C|_{\operatorname{Spec}(R)}$. The choice of $C$ is a bookkeeping device and could be avoided. For later use, let us write $C = \Lambda[y_1, \ldots, y_u]/(f_1, \ldots, f_v)$ and we denote $\overline{a}_i \in R$ the image of $y_i$ under the map $C \to R$. Set $\mathfrak m_C = C \cap \mathfrak m_R$.

Choose a $\Lambda$-algebra surjection $\Lambda[x_1, \ldots, x_s] \to k$ and denote by $\mathfrak m'$ the kernel. By the universal property of polynomial rings we may lift this to a $\Lambda$-algebra map $\Lambda[x_1, \ldots, x_s] \to R$. We add some variables (i.e., we increase $s$ a bit) mapping to generators of $\mathfrak m_R$. Having done this we see that $\Lambda[x_1, \ldots, x_s] \to R/\mathfrak m_R^2$ is surjective. Then we see that

$$P = \Lambda[x_1, \ldots, x_s]_{\mathfrak m'}^\wedge \longrightarrow R$$ is a surjective map of Noetherian complete local rings, see for example Formal Deformation Theory, Lemma Surjectivity on cotangent spaces.

Choose lifts $a_i \in P$ of $\overline{a}_i$ we found above. Choose generators $b_1, \ldots, b_r \in P$ for the kernel of (the displayed identity). Choose $c_{ji} \in P$ such that $$f_j(a_1, \ldots, a_u) = \sum c_{ji} b_i$$ in $P$ which is possible by the choices made so far. Choose generators $$k_1, \ldots, k_t \in \operatorname{Ker}(P^{\oplus r} \xrightarrow{(b_1, \ldots, b_r)} P)$$ and write $k_i = (k_{i1}, \ldots, k_{ir})$ and $K = (k_{ij})$ so that $$P^{\oplus t} \xrightarrow{K} P^{\oplus r} \xrightarrow{(b_1, \ldots, b_r)} P \to R \to 0$$ is an exact sequence of $P$-modules. In particular we have $\sum k_{ij} b_j = 0$. After possibly increasing $N$ we may assume $N - 1$ works in the Artin-Rees lemma for the first two maps of this exact sequence (see More on Algebra, Section Artin–Rees constants for finite module maps for terminology).

By assumption $\mathcal{O}_{S, s} = \Lambda_{\Lambda \cap \mathfrak m'}$ is a G-ring. Hence by More on Algebra, Proposition The G-ring property for finite-type algebras the ring $\Lambda[x_1, \ldots, x_s]_{\mathfrak m'}$ is a $G$-ring. Hence by Smoothing Ring Maps, Theorem Artin approximation in an étale neighbourhood there exist an étale ring map $$\Lambda[x_1, \ldots, x_s]_{\mathfrak m'} \to B,$$ a maximal ideal $\mathfrak m_B$ of $B$ lying over $\mathfrak m'$, and elements $a'_i, b'_i, c'_{ij}, k'_{ij} \in B$ such that

1. $\kappa(\mathfrak m') = \kappa(\mathfrak m_B)$ which implies that $\Lambda[x_1, \ldots, x_s]_{\mathfrak m'} \subset B_{\mathfrak m_B} \subset P$ and $P$ is identified with the completion of $B$ at $\mathfrak m_B$, see remark preceding Smoothing Ring Maps, Theorem Artin approximation in an étale neighbourhood,

2. $a_i - a'_i, b_i - b'_i, c_{ij} - c'_{ij}, k_{ij} - k'_{ij} \in (\mathfrak m')^N P$, and

3. $f_j(a'_1, \ldots, a'_u) = \sum c'_{ji} b'_i$ and $\sum k'_{ij}b'_j = 0$.

Set $A = B/(b'_1, \ldots, b'_r)$ and denote by $\mathfrak m_A$ the image of $\mathfrak m_B$ in $A$. (Note that $A$ is essentially of finite type over $\Lambda$; at the end of the proof we will show how to obtain an $A$ which is of finite type over $\Lambda$.) There is a ring map $C \to A$ sending $y_i \mapsto a'_i$ because the $a'_i$ satisfy the desired equations modulo $(b'_1, \ldots, b'_r)$. Note that $A/\mathfrak m_A^N = R/\mathfrak m_R^N$ as quotients of $P = B^\wedge$ by property (2) above. Set $x_A = x_C|_{\operatorname{Spec}(A)}$. Since the maps $$C \to A \to A/\mathfrak m_A^N \cong R/\mathfrak m_R^N \quad\text{and}\quad C \to R \to R/\mathfrak m_R^N$$ are equal we see that $x_A$ and $x$ agree modulo $\mathfrak m_R^N$ via the isomorphism $A/\mathfrak m_A^N = R/\mathfrak m_R^N$. At this point we have shown properties (1) -- (5) of the statement of the lemma. To see (6) note that $$P^{\oplus t} \xrightarrow{K} P^{\oplus r} \xrightarrow{(b_1, \ldots, b_r)} P \quad\text{and}\quad P^{\oplus t} \xrightarrow{K'} P^{\oplus r} \xrightarrow{(b'_1, \ldots, b'_r)} P$$ are two complexes of $P$-modules which are congruent modulo $(\mathfrak m')^N$ with the first one being exact. By our choice of $N$ above we see from More on Algebra, Lemma Lesson 7, Appendix A, Lemma A.2 that $R = P/(b_1, \ldots, b_r)$ and $P/(b'_1, \ldots, b'_r) = B^\wedge/(b'_1, \ldots, b'_r) = A^\wedge$ have isomorphic associated graded algebras, which is what we wanted to show.

This last paragraph of the proof serves to clean up the issue that $A$ is essentially of finite type over $S$ and not yet of finite type. The construction above gives $A = B/(b'_1, \ldots, b'_r)$ and $\mathfrak m_A \subset A$ with $B$ étale over $\Lambda[x_1, \ldots, x_s]_{\mathfrak m'}$. Hence $A$ is of finite type over the Noetherian ring $\Lambda[x_1, \ldots, x_s]_{\mathfrak m'}$. Thus we can write $A = (A_0)_{\mathfrak m'}$ for some finite type $\Lambda[x_1, \ldots, x_s]$ algebra $A_0$. Then $A = \mathop{\operatorname{colim}} (A_0)_f$ where $f \in \Lambda[x_1, \ldots, x_s] \setminus \mathfrak m'$, see Algebra, Lemma Localization as a filtered colimit. Because $p : \mathcal{X} \to (\mathrm{Sch}/S)_{fppf}$ is limit preserving on objects, we see that $x_A$ comes from some object $x_{(A_0)_f}$ over $\operatorname{Spec}((A_0)_f)$ for an $f$ as above. After replacing $A$ by $(A_0)_f$ and $x_A$ by $x_{(A_0)_f}$ and $\mathfrak m_A$ by $(A_0)_f \cap \mathfrak m_A$ the proof is finished. $\square$

Commutative algebra and regularity

Definition. Smoothness at a prime ideal

Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$. We say $R \to S$ is smooth at $\mathfrak q$ if there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is smooth.

Lemma. Localization of the naive cotangent complex

Let $A \to B$ be a ring map. Let $S \subset B$ be a multiplicative subset. The canonical map $\mathrm{NL}_{B/A} \otimes_B S^{-1}B \to \mathrm{NL}_{S^{-1}B/A}$ is a quasi-isomorphism.

Proof. We have $S^{-1}B = \mathop{\operatorname{colim}}_{g \in S} B_g$ where we think of $S$ as a directed set (ordering by divisibility), see Lemma Localization as a filtered colimit. By Lemma The cotangent complex of a principal localization each of the maps $\mathrm{NL}_{B/A} \otimes_B B_g \to \mathrm{NL}_{B_g/A}$ is a quasi-isomorphism. The lemma follows from Lemma Filtered colimits of naive cotangent complexes. $\square$

Definition. Smooth ring maps

A ring map $R \to S$ is smooth if it is of finite presentation and the naive cotangent complex $\mathrm{NL}_{S/R}$ is quasi-isomorphic to a finite projective $S$-module placed in degree $0$: this means that $H_1(\mathrm{NL}_{S/R}) = 0$ and that $\Omega_{S/R}$ is a finite projective $S$-module.

Lemma. The transitivity sequence for the naive cotangent complex (Jacobi-Zariski sequence)

Let $A \to B \to C$ be ring maps. Choose a presentation $\alpha : A[x_s, s \in S] \to B$ with kernel $I$. Choose a presentation $\beta : B[y_t, t \in T] \to C$ with kernel $J$. Let $\gamma : A[x_s, y_t] \to C$ be the induced presentation of $C$ with kernel $K$. Then we get a canonical commutative diagram $$\begin{gathered}\begin{matrix}0 & \Omega_{A[x_s]/A} \otimes C & \Omega_{A[x_s, y_t]/A} \otimes C & \Omega_{B[y_t]/B} \otimes C & 0 \\ \phantom{X} & I/I^2 \otimes C & K/K^2 & J/J^2 & 0\end{matrix} \\[6pt] \begin{aligned}0 & \longrightarrow \Omega_{A[x_s]/A} \otimes C \\ \Omega_{A[x_s]/A} \otimes C & \longrightarrow \Omega_{A[x_s, y_t]/A} \otimes C \\ \Omega_{A[x_s, y_t]/A} \otimes C & \longrightarrow \Omega_{B[y_t]/B} \otimes C \\ \Omega_{B[y_t]/B} \otimes C & \longrightarrow 0 \\ I/I^2 \otimes C & \longrightarrow K/K^2 \\ I/I^2 \otimes C & \longrightarrow \Omega_{A[x_s]/A} \otimes C \\ K/K^2 & \longrightarrow J/J^2 \\ K/K^2 & \longrightarrow \Omega_{A[x_s, y_t]/A} \otimes C \\ J/J^2 & \longrightarrow 0 \\ J/J^2 & \longrightarrow \Omega_{B[y_t]/B} \otimes C\end{aligned}\end{gathered},$$ with exact rows. We get the following exact sequence of homology groups $$H_1(\mathrm{NL}_{B/A} \otimes_B C) \to H_1(L_{C/A}) \to H_1(L_{C/B}) \to C \otimes_B \Omega_{B/A} \to \Omega_{C/A} \to \Omega_{C/B} \to 0$$ of $C$-modules extending the sequence of Lemma Kähler differentials, Theorems 3.1–3.3, Proposition 3.4 and Theorem 7.1. If $\text{Tor}_1^B(\Omega_{B/A}, C) = 0$ and $\text{Tor}_2^B(\Omega_{B/A}, C) = 0$, then $H_1(\mathrm{NL}_{B/A} \otimes_B C) = H_1(L_{B/A}) \otimes_B C$.

Proof. The precise definition of the maps is omitted. The exactness of the top row follows as the $\text{d}x_s$, $\text{d}y_t$ form a basis for the middle module. The map $\gamma$ factors $$A[x_s, y_t] \to B[y_t] \to C$$ with surjective first arrow and second arrow equal to $\beta$. Thus we see that $K \to J$ is surjective. Moreover, the kernel of the first displayed arrow is $IA[x_s, y_t]$. Hence $I/I^2 \otimes C$ surjects onto the kernel of $K/K^2 \to J/J^2$. Finally, we can use Lemma Kähler differentials, Theorems 3.1–3.3, Proposition 3.4 and Theorem 7.1 to identify the terms as homology groups of the naive cotangent complexes.

The final assertion is a statement in homological algebra. Recall that $\mathrm{NL}_{B/A} = (N^{-1} \to N^0)$ is a two term complex of $B$-modules with $N^0$ free and cohomology modules $H^0 = \Omega_{B/A}$ and $H^{-1} = H_1(L_{B/A})$. Write $M \subset N^0$ for the image of the differential. If $\text{Tor}_1^B(H^0, C) = 0$, then we have an exact sequence $$0 \to M \otimes_B C \to N^0 \otimes_B C \to H^0 \otimes_B C \to 0$$ Since $N^0$ is free, we also see that $\text{Tor}_2^B(H^0, C) = \text{Tor}_1^B(M, C)$. Hence if $\text{Tor}_2^B(H^0, C) = 0$ then we also have an exact sequence $$0 \to H^{-1} \otimes_B C \to N^{-1} \otimes_B C \to M \otimes_B C \to 0$$ Putting everything together we see that if $\text{Tor}_1^B(H^0, C) = 0$ and $\text{Tor}_2^B(H^0, C) = 0$, then $H^{-1} \otimes_B C$ is the kernel of $N^{-1} \otimes_B C \to N^0 \otimes_B C$ as desired. $\square$

Lemma. The conormal module of a syntomic presentation

Let $R$ be a ring. Let $S = R[x_1, \ldots, x_n]/I$ for some finitely generated ideal $I$. If $g \in S$ is such that $S_g$ is syntomic over $R$, then $(I/I^2)_g$ is a finite projective $S_g$-module.

Proof. By Lemma Local criteria for a syntomic algebra there exist finitely many elements $g_1, \ldots, g_m \in S$ which generate the unit ideal in $S_g$ such that each $S_{gg_j}$ is a relative global complete intersection over $R$. Since it suffices to prove that $(I/I^2)_{gg_j}$ is finite projective, see Lemma Characterizations of finite projective modules, we may assume that $S_g$ is a relative global complete intersection. In this case the result follows from Lemmas Localization of a conormal module and The conormal module of a global complete intersection. $\square$

Lemma. A presentation realizing a basis of the conormal module

Let $S$ be a finitely presented $R$-algebra which has a presentation $S = R[x_1, \ldots, x_n]/I$ such that $I/I^2$ is free over $S$. Then $S$ has a presentation $S = R[y_1, \ldots, y_m]/(f_1, \ldots, f_c)$ such that $(f_1, \ldots, f_c)/(f_1, \ldots, f_c)^2$ is free with basis given by the classes of $f_1, \ldots, f_c$.

Proof. Note that $I$ is a finitely generated ideal by Lemma Finite presentation and finite algebras. Let $f_1, \ldots, f_c \in I$ be elements which map to a basis of $I/I^2$. By Nakayama's lemma (Lemma Nakayama's lemma) there exists a $g \in 1 + I$ such that $$g \cdot I \subset (f_1, \ldots, f_c)$$ and $I_g \cong (f_1, \ldots, f_c)_g$. Hence we see that $$S \cong R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)[1/g] \cong R[x_1, \ldots, x_n, x_{n + 1}]/(f_1, \ldots, f_c, gx_{n + 1} - 1)$$ as desired. It follows that $f_1, \ldots, f_c,gx_{n + 1} - 1$ form a basis for $(f_1, \ldots, f_c, gx_{n + 1} - 1)/(f_1, \ldots, f_c, gx_{n + 1} - 1)^2$ for example by applying Lemma The cotangent complex of a principal localization. $\square$

Lemma. Localization of a relative complete intersection

Let $R$ be a ring. Let $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$. We will find $h \in R[x_1, \ldots, x_n]$ which maps to $g \in S$ such that $$S_g = R[x_1, \ldots, x_n, x_{n + 1}]/(f_1, \ldots, f_c, hx_{n + 1} - 1)$$ is a relative global complete intersection with a presentation as in Definition Relative global complete intersections in each of the following cases:

1. Let $I \subset R$ be an ideal. If the fibres of $\operatorname{Spec}(S/IS) \to \operatorname{Spec}(R/I)$ have dimension $n - c$, then we can find $(h, g)$ as above such that $g$ maps to $1 \in S/IS$.

2. Let $\mathfrak p \subset R$ be a prime. If $\dim(S \otimes_R \kappa(\mathfrak p)) = n - c$, then we can find $(h, g)$ as above such that $g$ maps to a unit of $S \otimes_R \kappa(\mathfrak p)$.

3. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. If $\dim_{\mathfrak q}(S/R) = n - c$, then we can find $(h, g)$ as above such that $g \not \in \mathfrak q$.

Proof. Ad (1). By Lemma An open neighbourhood with bounded fibre dimension there exists an open subset $W \subset \operatorname{Spec}(S)$ containing $V(IS)$ such that all fibres of $W \to \operatorname{Spec}(R)$ have dimension $\leq n - c$. Say $W = \operatorname{Spec}(S) \setminus V(J)$. Then $V(J) \cap V(IS) = \emptyset$ hence we can find a $g \in J$ which maps to $1 \in S/IS$. Let $h \in R[x_1, \ldots, x_n]$ be any preimage of $g$.

Ad (2). By Lemma An open neighbourhood with bounded fibre dimension there exists an open subset $W \subset \operatorname{Spec}(S)$ containing $\operatorname{Spec}(S \otimes_R \kappa(\mathfrak p))$ such that all fibres of $W \to \operatorname{Spec}(R)$ have dimension $\leq n - c$. Say $W = \operatorname{Spec}(S) \setminus V(J)$. Then $V(J \cdot S \otimes_R \kappa(\mathfrak p)) = \emptyset$. Hence we can find a $g \in J$ which maps to a unit in $S \otimes_R \kappa(\mathfrak p)$ (details omitted). Let $h \in R[x_1, \ldots, x_n]$ be any preimage of $g$.

Ad (3). By Lemma An open neighbourhood with bounded fibre dimension there exists a $g \in S$, $g \not \in \mathfrak q$ such that all nonempty fibres of $R \to S_g$ have dimension $\leq n - c$. Let $h \in R[x_1, \ldots, x_n]$ be any element that maps to $g$. $\square$

Lemma. Composition of syntomic ring maps

Let $R \to S$, $S \to S'$ be ring maps.

1. If $R \to S$ and $S \to S'$ are syntomic, then $R \to S'$ is syntomic.

2. If $R \to S$ and $S \to S'$ are relative global complete intersections, then $R \to S'$ is a relative global complete intersection.

Proof. Proof of (2). Say $R \to S$ and $S \to S'$ are relative global complete intersections and we have presentations $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ and $S' = S[y_1, \ldots, y_m]/(h_1, \ldots, h_d)$ as in Definition Relative global complete intersections. Then $$S' \cong R[x_1, \ldots, x_n, y_1, \ldots, y_m]/(f_1, \ldots, f_c, h'_1, \ldots, h'_d)$$ for some lifts $h_j' \in R[x_1, \ldots, x_n, y_1, \ldots, y_m]$ of the $h_j$. Hence it suffices to bound the dimensions of the fibre rings. Thus we may assume $R = k$ is a field. In this case we see that we have a ring, namely $S$, which is of finite type over $k$ and equidimensional of dimension $n - c$, and a finite type ring map $S \to S'$ all of whose nonempty fibre rings are equidimensional of dimension $m - d$. Then, by Lemma Dimensions of a base, fibre and total space for example applied to localizations at maximal ideals of $S'$, we see that $\dim(S') \leq n - c + m - d$ as desired.

We will reduce part (1) to part (2). Assume $R \to S$ and $S \to S'$ are syntomic. Let $\mathfrak q' \subset S'$ be a prime ideal lying over $\mathfrak q \subset S$. By Lemma Local criteria for a syntomic algebra there exists a $g' \in S'$, $g' \not \in \mathfrak q'$ such that $S \to S'_{g'}$ is a relative global complete intersection. Similarly, we find $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is a relative global complete intersection. By Lemma Base change of a global complete intersection the ring map $S_g \to S'_{gg'}$ is a relative global complete intersection. By part (2) we see that $R \to S'_{gg'}$ is a relative global complete intersection and $gg' \not \in \mathfrak q'$. Since $\mathfrak q'$ was arbitrary combining Lemmas Local criteria for a syntomic algebra and Locality of syntomic ring maps we see that $R \to S'$ is syntomic (this also uses that the spectrum of $S'$ is quasi-compact, see Lemma Quasi-compactness of an affine spectrum). $\square$

Lemma. Smooth algebras are syntomic

Let $R \to S$ be a smooth ring map. There exists an open covering of $\operatorname{Spec}(S)$ by standard opens $D(g)$ such that each $S_g$ is standard smooth over $R$. In particular $R \to S$ is syntomic.

Proof. Choose a presentation $\alpha : R[x_1, \ldots, x_n] \to S$ with kernel $I = (f_1, \ldots, f_m)$. For every subset $E \subset \{1, \ldots, m\}$ consider the open subset $U_E$ where the classes $f_e, e\in E$ freely generate the finite projective $S$-module $I/I^2$, see Lemma Flatness of a cokernel. We may cover $\operatorname{Spec}(S)$ by standard opens $D(g)$ each completely contained in one of the opens $U_E$. For such a $g$ we look at the presentation $$\beta : R[x_1, \ldots, x_n, x_{n + 1}] \longrightarrow S_g$$ mapping $x_{n + 1}$ to $1/g$. Setting $J = \operatorname{Ker}(\beta)$ we use Lemma The cotangent complex of a principal localization to see that $J/J^2 \cong (I/I^2)_g \oplus S_g$ is free. We may and do replace $S$ by $S_g$. Then using Lemma A presentation realizing a basis of the conormal module we may assume we have a presentation $\alpha : R[x_1, \ldots, x_n] \to S$ with kernel $I = (f_1, \ldots, f_c)$ such that $I/I^2$ is free on the classes of $f_1, \ldots, f_c$.

Using the presentation $\alpha$ obtained at the end of the previous paragraph, we more or less repeat this argument with the basis elements $\text{d}x_1, \ldots, \text{d}x_n$ of $\Omega_{R[x_1, \ldots, x_n]/R}$. Namely, for any subset $E \subset \{1, \ldots, n\}$ of cardinality $c$ we may consider the open subset $U_E$ of $\operatorname{Spec}(S)$ where the differential of $\mathrm{NL}(\alpha)$ composed with the projection $$S^{\oplus c} \cong I/I^2 \longrightarrow \Omega_{R[x_1, \ldots, x_n]/R} \otimes_{R[x_1, \ldots, x_n]} S \longrightarrow \bigoplus\nolimits_{i \in E} S\text{d}x_i$$ is an isomorphism. Again we may find a covering of $\operatorname{Spec}(S)$ by (finitely many) standard opens $D(g)$ such that each $D(g)$ is completely contained in one of the opens $U_E$. By renumbering, we may assume $E = \{1, \ldots, c\}$. For a $g$ with $D(g) \subset U_E$ we look at the presentation $$\beta : R[x_1, \ldots, x_n, x_{n + 1}] \to S_g$$ mapping $x_{n + 1}$ to $1/g$. Setting $J = \operatorname{Ker}(\beta)$ we conclude from Lemma The cotangent complex of a principal localization that $J = (f_1, \ldots, f_c, fx_{n + 1} - 1)$ where $\alpha(f) = g$ and that the composition $$J/J^2 \longrightarrow \Omega_{R[x_1, \ldots, x_{n + 1}]/R} \otimes_{R[x_1, \ldots, x_{n + 1}]} S_g \longrightarrow \bigoplus\nolimits_{i = 1}^c S_g\text{d}x_i \oplus S_g \text{d}x_{n + 1}$$ is an isomorphism. Reordering the coordinates as $x_1, \ldots, x_c, x_{n + 1}, x_{c + 1}, \ldots, x_n$ we conclude that $S_g$ is standard smooth over $R$ as desired.

This finishes the proof as standard smooth algebras are syntomic (Lemmas Standard smooth algebras and Criteria for global complete intersections) and being syntomic over $R$ is local on $S$ (Lemma Locality of syntomic ring maps). $\square$

Lemma. Local criteria for a syntomic algebra

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. The following are equivalent:

1. There exists an element $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is syntomic.

2. There exists an element $g \in S$, $g \not \in \mathfrak q$ such that $S_g$ is a relative global complete intersection over $R$.

3. There exists an element $g \in S$, $g \not \in \mathfrak q$, such that $R \to S_g$ is of finite presentation, the local ring map $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat, and the local ring $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is a complete intersection ring over $\kappa(\mathfrak p)$ (see Definition Complete-intersection local rings).

Proof. The implication (1) $\Rightarrow$ (3) is Lemma Complete intersections at a prime ideal. The implication (2) $\Rightarrow$ (1) is Lemma Criteria for global complete intersections. It remains to show that (3) implies (2).

Assume (3). After replacing $S$ by $S_g$ for some $g \in S$, $g\not\in \mathfrak q$, we may assume $S$ is finitely presented over $R$. Choose a presentation $S = R[x_1, \ldots, x_n]/I$. Let $\mathfrak q' \subset R[x_1, \ldots, x_n]$ be the prime corresponding to $\mathfrak q$. Write $\kappa(\mathfrak p) = k$. Note that $S \otimes_R k = k[x_1, \ldots, x_n]/\overline{I}$ where $\overline{I} \subset k[x_1, \ldots, x_n]$ is the ideal generated by the image of $I$. Let $\overline{\mathfrak q}' \subset k[x_1, \ldots, x_n]$ be the prime ideal generated by the image of $\mathfrak q'$. By Lemma Complete intersections at a prime ideal the equivalent conditions of Lemma Local criteria for complete intersections hold for $\overline{I}$ and $\overline{\mathfrak q}'$. Say the dimension of $\overline{I}_{\overline{\mathfrak q}'}/ \overline{\mathfrak q}'\overline{I}_{\overline{\mathfrak q}'}$ over $\kappa(\overline{\mathfrak q}')$ is $c$. Pick $f_1, \ldots, f_c \in I$ mapping to a basis of this vector space. The images $\overline{f}_j \in \overline{I}$ generate $\overline{I}_{\overline{\mathfrak q}'}$ (by Lemma Local criteria for complete intersections). Set $S' = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$. Let $J$ be the kernel of the surjection $S' \to S$. Since $S$ is of finite presentation $J$ is a finitely generated ideal (Lemma Composition of finite-type ring maps). Consider the short exact sequence $$0 \to J \to S' \to S \to 0.$$ As $S_\mathfrak q$ is flat over $R$ we see that $J_{\mathfrak q'} \otimes_R k \to S'_{\mathfrak q'} \otimes_R k$ is injective (Lemma Tor vanishing for a flat module). However, by construction $S'_{\mathfrak q'} \otimes_R k$ maps isomorphically to $S_\mathfrak q \otimes_R k$. Hence we conclude that $J_{\mathfrak q'} \otimes_R k = J_{\mathfrak q'}/\mathfrak pJ_{\mathfrak q'} = 0$. By Nakayama's lemma (Lemma Nakayama's lemma) we conclude that there exists a $g \in R[x_1, \ldots, x_n]$, $g \not \in \mathfrak q'$ such that $J_g = 0$. In other words $S'_g \cong S_g$. After further localizing we see that $S'$ (and hence $S$) becomes a relative global complete intersection by Lemma Localization of a relative complete intersection as desired. $\square$

Lemma. Smoothness from flatness and smooth fibres

Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. Assume

1. there exists a $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_g$ is of finite presentation,

2. the local ring homomorphism $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat,

3. the fibre $S \otimes_R \kappa(\mathfrak p)$ is smooth over $\kappa(\mathfrak p)$ at the prime corresponding to $\mathfrak q$.

Then $R \to S$ is smooth at $\mathfrak q$.

Proof. By Lemmas Local criteria for a syntomic algebra and Smooth algebras over a field and the Jacobian criterion, Theorems 5.1–6.1 and Sections 1–3 we see that there exists a $g \in S$, $g \not \in \mathfrak q$, such that $S_g$ is a relative global complete intersection. Replacing $S$ by $S_g$ we may assume $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ is a relative global complete intersection. For any subset $I \subset \{1, \ldots, n\}$ of cardinality $c$ consider the polynomial $g_I = \det (\partial f_j/\partial x_i)_{j = 1, \ldots, c, i \in I}$ of Lemma Smoothness of a global complete intersection. Note that the image $\overline{g}_I$ of $g_I$ in the polynomial ring $\kappa(\mathfrak p)[x_1, \ldots, x_n]$ is the determinant of the partial derivatives of the images $\overline{f}_j$ of the $f_j$ in the ring $\kappa(\mathfrak p)[x_1, \ldots, x_n]$. Thus the lemma follows by applying Lemma Smoothness of a global complete intersection both to $R \to S$ and to $\kappa(\mathfrak p) \to S \otimes_R \kappa(\mathfrak p)$. $\square$

Lemma. Local criteria for complete intersections

Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q$ be a prime of $S$. Choose any presentation $S = k[x_1, \ldots, x_n]/I$. Let $\mathfrak q'$ be the prime of $k[x_1, \ldots, x_n]$ corresponding to $\mathfrak q$. Set $c = \text{height}(\mathfrak q') - \text{height}(\mathfrak q)$, in other words $\dim_{\mathfrak q}(S) = n - c$ (see Lemma Dimension and codimension). The following are equivalent

1. There exists a $g \in S$, $g \not \in \mathfrak q$ such that $S_g$ is a global complete intersection over $k$.

2. The ideal $I_{\mathfrak q'} \subset k[x_1, \ldots, x_n]_{\mathfrak q'}$ can be generated by $c$ elements.

3. The conormal module $(I/I^2)_{\mathfrak q}$ can be generated by $c$ elements over $S_{\mathfrak q}$.

4. The conormal module $(I/I^2)_{\mathfrak q}$ is a free $S_{\mathfrak q}$-module of rank $c$.

5. The ideal $I_{\mathfrak q'}$ can be generated by a regular sequence in the regular local ring $k[x_1, \ldots, x_n]_{\mathfrak q'}$.

In this case any $c$ elements of $I_{\mathfrak q'}$ which generate $I_{\mathfrak q'}/\mathfrak q'I_{\mathfrak q'}$ form a regular sequence in the local ring $k[x_1, \ldots, x_n]_{\mathfrak q'}$.

Proof. Set $R = k[x_1, \ldots, x_n]_{\mathfrak q'}$. This is a Cohen-Macaulay local ring of dimension $\text{height}(\mathfrak q')$, see for example Lemma Complete intersections are Cohen–Macaulay. Moreover, $\overline{R} = R/IR = R/I_{\mathfrak q'} = S_{\mathfrak q}$ is a quotient of dimension $\text{height}(\mathfrak q)$. Let $f_1, \ldots, f_c \in I_{\mathfrak q'}$ be elements which generate $(I/I^2)_{\mathfrak q}$. By Lemma Nakayama's lemma we see that $f_1, \ldots, f_c$ generate $I_{\mathfrak q'}$. Since the dimensions work out, we conclude by Proposition Characterizations of Cohen–Macaulay modules that $f_1, \ldots, f_c$ is a regular sequence in $R$. By Lemma Regular sequences are quasi-regular we see that $(I/I^2)_{\mathfrak q}$ is free. These arguments show that (2), (3), (4) are equivalent and that they imply the last statement of the lemma, and therefore they imply (5).

If (5) holds, say $I_{\mathfrak q'}$ is generated by a regular sequence of length $e$, then $\text{height}(\mathfrak q) = \dim(S_{\mathfrak q}) = \dim(k[x_1, \ldots, x_n]_{\mathfrak q'}) - e = \text{height}(\mathfrak q') - e$ by dimension theory, see Section Dimension and codimension. We conclude that $e = c$. Thus (5) implies (2).

We continue with the notation introduced in the first paragraph. For each $f_i$ we may find $d_i \in k[x_1, \ldots, x_n]$, $d_i \not \in \mathfrak q'$ such that $f_i' = d_i f_i \in k[x_1, \ldots, x_n]$. Then it is still true that $I_{\mathfrak q'} = (f_1', \ldots, f_c')R$. Hence there exists a $g' \in k[x_1, \ldots, x_n]$, $g' \not \in \mathfrak q'$ such that $I_{g'} = (f_1', \ldots, f_c')$. Moreover, pick $g'' \in k[x_1, \ldots, x_n]$, $g'' \not \in \mathfrak q'$ such that $\dim(S_{g''}) = \dim_{\mathfrak q} \operatorname{Spec}(S)$. By Lemma Dimension and codimension this dimension is equal to $n - c$. Finally, set $g$ equal to the image of $g'g''$ in $S$. Then we see that $$S_g \cong k[x_1, \ldots, x_n, x_{n + 1}] / (f_1', \ldots, f_c', x_{n + 1}g'g'' - 1)$$ and by our choice of $g''$ this ring has dimension $n - c$. Therefore it is a global complete intersection. Thus each of (2), (3), and (4) implies (1).

Assume (1). Let $S_g \cong k[y_1, \ldots, y_m]/(f_1, \ldots, f_t)$ be a presentation of $S_g$ as a global complete intersection. Write $J = (f_1, \ldots, f_t)$. Let $\mathfrak q'' \subset k[y_1, \ldots, y_m]$ be the prime corresponding to $\mathfrak qS_g$. Note that $t = m - \dim(S_g) = \text{height}(\mathfrak q'') - \text{height}(\mathfrak q)$, see Lemma Dimension and codimension for the last equality. As seen in the proof of Lemma Complete intersections are Cohen–Macaulay (and also above) the elements $f_1, \ldots, f_t$ form a regular sequence in the local ring $k[y_1, \ldots, y_m]_{\mathfrak q''}$. By Lemma Regular sequences are quasi-regular we see that $(J/J^2)_{\mathfrak q}$ is free of rank $t$. By Lemma Localization of a conormal module we have $$J/J^2 \oplus S_g^n \cong (I/I^2)_g \oplus S_g^m$$ Thus $(I/I^2)_{\mathfrak q}$ is free of rank $t + n - m = m - \dim(S_g) + n - m = n - \dim(S_g) = \text{height}(\mathfrak q') - \text{height}(\mathfrak q) = c$. Thus we obtain (4). $\square$

Definition. Unramified ring maps

Let $R \to S$ be a ring map.

1. We say $R \to S$ is unramified if $R \to S$ is of finite type and $\Omega_{S/R} = 0$.

2. We say $R \to S$ is G-unramified if $R \to S$ is of finite presentation and $\Omega_{S/R} = 0$.

3. Given a prime $\mathfrak q$ of $S$ we say that $S$ is unramified at $\mathfrak q$ if there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is unramified.

4. Given a prime $\mathfrak q$ of $S$ we say that $S$ is G-unramified at $\mathfrak q$ if there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is G-unramified.

Lemma. Flatness from Cohen--Macaulayness over a regular base

Miracle flatness

Let $R \to S$ be a local homomorphism of Noetherian local rings. Assume

1. $R$ is regular,

2. $S$ is Cohen-Macaulay,

3. $\dim(S) = \dim(R) + \dim(S/\mathfrak m_R S)$.

Then $R \to S$ is flat.

Proof. By induction on $\dim(R)$. The case $\dim(R) = 0$ is trivial, because then $R$ is a field. Assume $\dim(R) > 0$. By (3) this implies that $\dim(S) > 0$. Let $\mathfrak q_1, \ldots, \mathfrak q_r$ be the minimal primes of $S$. Note that $\mathfrak q_i \not \supset \mathfrak m_R S$ since $$\dim(S/\mathfrak q_i) = \dim(S) > \dim(S/\mathfrak m_R S),$$ the first equality by Lemma Maximal prime chains in a Cohen–Macaulay ring and the inequality by (3). Thus $\mathfrak p_i = R \cap \mathfrak q_i$ is not equal to $\mathfrak m_R$. Pick $x \in \mathfrak m_R$, $x \not \in \mathfrak m_R^2$, and $x \not \in \mathfrak p_i$, see Lemma An elementary algebraic comparison. Hence we see that $x$ is not contained in any of the minimal primes of $S$. Hence $x$ is a nonzerodivisor on $S$ by (2), see Lemma Equivalent Cohen–Macaulay conditions and $S/xS$ is Cohen-Macaulay with $\dim(S/xS) = \dim(S) - 1$. By (1) and Lemma Regular rings are Cohen–Macaulay the ring $R/xR$ is regular with $\dim(R/xR) = \dim(R) - 1$. By induction we see that $R/xR \to S/xS$ is flat. Hence we conclude by Lemma A variant of the local criterion for flatness and the remark following it. $\square$

Lemma. Smoothness after an algebraic closure of the ground field

Let $k$ be an algebraically closed field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak m \subset S$ be a maximal ideal. The following are equivalent:

1. The ring $S_{\mathfrak m}$ is a regular local ring.

2. We have $\dim_{\kappa(\mathfrak m)} \Omega_{S/k} \otimes_S \kappa(\mathfrak m) \leq \dim(S_{\mathfrak m})$.

3. We have $\dim_{\kappa(\mathfrak m)} \Omega_{S/k} \otimes_S \kappa(\mathfrak m) = \dim(S_{\mathfrak m})$.

4. There exists a $g \in S$, $g \not \in \mathfrak m$ such that $S_g$ is smooth over $k$. In other words $S/k$ is smooth at $\mathfrak m$.

Proof. Note that (1), (2) and (3) are equivalent by Lemma The rank of Kähler differentials and Definition Regular Noetherian rings.

Assume that $S$ is smooth at $\mathfrak m$. By Lemma Smooth algebras are syntomic we see that $S_g$ is standard smooth over $k$ for a suitable $g \in S$, $g \not \in \mathfrak m$. Hence by Lemma Standard smooth algebras we see that $\Omega_{S_g/k}$ is free of rank $\dim(S_g)$. Hence by Lemma The rank of Kähler differentials we see that $\dim(S_{\mathfrak m}) = \dim (\mathfrak m/\mathfrak m^2)$ in other words $S_\mathfrak m$ is regular.

Conversely, suppose that $S_{\mathfrak m}$ is regular. Let $d = \dim(S_{\mathfrak m}) = \dim \mathfrak m/\mathfrak m^2$. Choose a presentation $S = k[x_1, \ldots, x_n]/I$ such that $x_i$ maps to an element of $\mathfrak m$ for all $i$. In other words, $\mathfrak m'' = (x_1, \ldots, x_n)$ is the corresponding maximal ideal of $k[x_1, \ldots, x_n]$. Note that we have a short exact sequence $$I/\mathfrak m''I \to \mathfrak m''/(\mathfrak m'')^2 \to \mathfrak m/(\mathfrak m)^2 \to 0$$ Pick $c = n - d$ elements $f_1, \ldots, f_c \in I$ such that their images in $\mathfrak m''/(\mathfrak m'')^2$ span the kernel of the map to $\mathfrak m/\mathfrak m^2$. This is clearly possible. Let $J = (f_1, \ldots, f_c)$. So $J \subset I$. Let $S' = k[x_1, \ldots, x_n]/J$ so there is a surjection $S' \to S$. Let $\mathfrak m' = \mathfrak m''S'$ be the corresponding maximal ideal of $S'$. Hence we have $$\begin{gathered}\begin{matrix}k[x_1, \ldots, x_n] & S' & S \\ \mathfrak m'' & \mathfrak m' & \mathfrak m\end{matrix} \\[6pt] \begin{aligned}k[x_1, \ldots, x_n] & \longrightarrow S' \\ S' & \longrightarrow S \\ \mathfrak m'' & \longrightarrow k[x_1, \ldots, x_n] \\ \mathfrak m'' & \longrightarrow \mathfrak m' \\ \mathfrak m' & \longrightarrow \mathfrak m \\ \mathfrak m' & \longrightarrow S' \\ \mathfrak m & \longrightarrow S\end{aligned}\end{gathered}$$ By our choice of $J$ the exact sequence $$J/\mathfrak m''J \to \mathfrak m''/(\mathfrak m'')^2 \to \mathfrak m'/(\mathfrak m')^2 \to 0$$ shows that $\dim( \mathfrak m'/(\mathfrak m')^2 ) = d$. Since $S'_{\mathfrak m'}$ surjects onto $S_{\mathfrak m}$ we see that $\dim(S'_{\mathfrak m'}) \geq d$. Hence by the discussion preceding Definition Regular local rings we conclude that $S'_{\mathfrak m'}$ is regular of dimension $d$ as well. Because $S'$ was cut out by $c = n - d$ equations we conclude that there exists a $g' \in S'$, $g' \not \in \mathfrak m'$ such that $S'_{g'}$ is a global complete intersection over $k$, see Lemma Local criteria for complete intersections. Also the map $S'_{\mathfrak m'} \to S_{\mathfrak m}$ is a surjection of Noetherian local domains of the same dimension and hence an isomorphism. Hence $S' \to S$ is surjective with finitely generated kernel and becomes an isomorphism after localizing at $\mathfrak m'$. Thus we can find $g' \in S'$, $g' \not \in \mathfrak m'$ such that $S'_{g'} \to S_{g'}$ is an isomorphism. All in all we conclude that after replacing $S$ by a principal localization we may assume that $S$ is a global complete intersection.

At this point we may write $S = k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ with $\dim S = n - c$. Recall that the naive cotangent complex of this algebra is given by $$\bigoplus S \cdot f_j \to \bigoplus S \cdot \text{d}x_i$$ see Lemma The conormal module of a global complete intersection. By Lemma Smoothness of a global complete intersection in order to show that $S$ is smooth at $\mathfrak m$ we have to show that one of the $c \times c$ minors $g_I$ of the matrix "$A$" giving the map above does not vanish at $\mathfrak m$. By Lemma The rank of Kähler differentials the matrix $A \bmod \mathfrak m$ has rank $c$. Thus we win. $\square$

Lemma. The fibrewise criterion for flatness (Critère de platitude par fibres)

Let $R$, $S$, $S'$ be local rings and let $R \to S \to S'$ be local ring homomorphisms. Let $M$ be an $S'$-module. Let $\mathfrak m \subset R$ be the maximal ideal. Assume

1. The ring maps $R \to S$ and $R \to S'$ are essentially of finite presentation.

2. The module $M$ is of finite presentation over $S'$.

3. The module $M$ is not zero.

4. The module $M/\mathfrak mM$ is a flat $S/\mathfrak mS$-module.

5. The module $M$ is a flat $R$-module.

Then $S$ is flat over $R$ and $M$ is a flat $S$-module.

Proof. As in the proof of Lemma Essentially finite presentations in a filtered limit we may first write $R = \mathop{\operatorname{colim}} R_\lambda$ as a directed colimit of local $\mathbf{Z}$-algebras which are essentially of finite type. Denote by $\mathfrak p_\lambda$ the maximal ideal of $R_\lambda$. Next, we may assume that for some $\lambda_1 \in \Lambda$ there exist $f_{j, \lambda_1} \in R_{\lambda_1}[x_1, \ldots, x_n]$ such that $$S = \mathop{\operatorname{colim}}_{\lambda \geq \lambda_1} S_\lambda, \text{ with } S_\lambda = (R_\lambda[x_1, \ldots, x_n]/ (f_{1, \lambda}, \ldots, f_{u, \lambda}))_{\mathfrak q_\lambda}$$ For some $\lambda_2 \in \Lambda$, $\lambda_2 \geq \lambda_1$ there exist $g_{j, \lambda_2} \in R_{\lambda_2}[x_1, \ldots, x_n, y_1, \ldots, y_m]$ with images $\overline{g}_{j, \lambda_2} \in S_{\lambda_2}[y_1, \ldots, y_m]$ such that $$S' = \mathop{\operatorname{colim}}_{\lambda \geq \lambda_2} S'_\lambda, \text{ with } S'_\lambda = (S_\lambda[y_1, \ldots, y_m]/ (\overline{g}_{1, \lambda}, \ldots, \overline{g}_{v, \lambda}))_{\overline{\mathfrak q}'_\lambda}$$ Note that this also implies that $$S'_\lambda = (R_\lambda[x_1, \ldots, x_n, y_1, \ldots, y_m]/ (f_{1, \lambda}, \ldots, f_{u, \lambda}, g_{1, \lambda}, \ldots, g_{v, \lambda}))_{\mathfrak q'_\lambda}$$ Choose a presentation $$(S')^{\oplus s} \to (S')^{\oplus t} \to M \to 0$$ of $M$ over $S'$. Let $A \in \text{Mat}(t \times s, S')$ be the matrix of the presentation. For some $\lambda_3 \in \Lambda$, $\lambda_3 \geq \lambda_2$ we can find a matrix $A_{\lambda_3} \in \text{Mat}(t \times s, S'_{\lambda_3})$ which maps to $A$. For all $\lambda \geq \lambda_3$ we let $M_\lambda = \operatorname{Coker}((S'_\lambda)^{\oplus s} \xrightarrow{A_\lambda} (S'_\lambda)^{\oplus t})$.

With these choices, we have for each $\lambda_3 \leq \lambda \leq \mu$ that $S_\lambda \otimes_{R_{\lambda}} R_\mu \to S_\mu$ is a localization, $S'_\lambda \otimes_{S_{\lambda}} S_\mu \to S'_\mu$ is a localization, and the map $M_\lambda \otimes_{S'_\lambda} S'_\mu \to M_\mu$ is an isomorphism. This also implies that $S'_\lambda \otimes_{R_{\lambda}} R_\mu \to S'_\mu$ is a localization. Thus, since $M$ is flat over $R$ we see by Lemma Eventual flatness in a filtered colimit that for all $\lambda$ big enough the module $M_\lambda$ is flat over $R_\lambda$. Moreover, note that $\mathfrak m = \mathop{\operatorname{colim}} \mathfrak p_\lambda$, $S/\mathfrak mS = \mathop{\operatorname{colim}} S_\lambda/\mathfrak p_\lambda S_\lambda$, $S'/\mathfrak mS' = \mathop{\operatorname{colim}} S'_\lambda/\mathfrak p_\lambda S'_\lambda$, and $M/\mathfrak mM = \mathop{\operatorname{colim}} M_\lambda/\mathfrak p_\lambda M_\lambda$. Also, for each $\lambda_3 \leq \lambda \leq \mu$ we see (from the properties listed above) that $$S'_\lambda/\mathfrak p_\lambda S'_\lambda \otimes_{S_{\lambda}/\mathfrak p_\lambda S_\lambda} S_\mu/\mathfrak p_\mu S_\mu \longrightarrow S'_\mu/\mathfrak p_\mu S'_\mu$$ is a localization, and the map $$M_\lambda / \mathfrak p_\lambda M_\lambda \otimes_{S'_\lambda/\mathfrak p_\lambda S'_\lambda} S'_\mu /\mathfrak p_\mu S'_\mu \longrightarrow M_\mu/\mathfrak p_\mu M_\mu$$ is an isomorphism. Hence the system $(S_\lambda/\mathfrak p_\lambda S_\lambda \to S'_\lambda/\mathfrak p_\lambda S'_\lambda, M_\lambda/\mathfrak p_\lambda M_\lambda)$ is a system as in Lemma Essentially finitely presented module models as well. We may apply Lemma Eventual flatness in a filtered colimit again because $M/\mathfrak m M$ is assumed flat over $S/\mathfrak mS$ and we see that $M_\lambda/\mathfrak p_\lambda M_\lambda$ is flat over $S_\lambda/\mathfrak p_\lambda S_\lambda$ for all $\lambda$ big enough. Thus for $\lambda$ big enough the data $R_\lambda \to S_\lambda \to S'_\lambda, M_\lambda$ satisfies the hypotheses of Lemma The Noetherian fibrewise criterion for flatness. Pick such a $\lambda$. Then $S$ is a localization of $S_\lambda \otimes_{R_\lambda} R$, hence is flat over $R$. Also $M$ is a localization of $M_\lambda \otimes_{S_\lambda} S$, hence is flat over $S$ (base change and localization preserve flatness). $\square$

Lemma. Characterizations of étale algebras

Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over a prime $\mathfrak p$ of $R$. If

1. $R \to S$ is of finite presentation,

2. $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat

3. $\mathfrak p S_{\mathfrak q}$ is the maximal ideal of the local ring $S_{\mathfrak q}$, and

4. the field extension $\kappa(\mathfrak q)/\kappa(\mathfrak p)$ is finite separable,

then $R \to S$ is étale at $\mathfrak q$.

Proof. Apply Lemma An isolated point of a fibre to find a $g \in S$, $g \not \in \mathfrak q$ such that $\mathfrak q$ is the only prime of $S_g$ lying over $\mathfrak p$. We may and do replace $S$ by $S_g$. Then $S \otimes_R \kappa(\mathfrak p)$ has a unique prime, hence is a local ring, hence is equal to $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} \cong \kappa(\mathfrak q)$. By Lemma Smoothness from flatness and smooth fibres there exists a $g \in S$, $g \not \in \mathfrak q$ such that $R \to S_g$ is smooth. Replacing $S$ by $S_g$ again, we may assume that $R \to S$ is smooth. By Lemma Smooth algebras are syntomic we may even assume that $R \to S$ is standard smooth, say $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$. Since $S \otimes_R \kappa(\mathfrak p) = \kappa(\mathfrak q)$ has dimension $0$ we conclude that $n = c$, i.e., $R \to S$ is étale. $\square$

Lemma. Étale algebras in standard smooth form

Any étale ring map is standard smooth. More precisely, if $R \to S$ is étale, then there exists a presentation $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_n)$ such that the image of $\det(\partial f_j/\partial x_i)$ is invertible in $S$.

Proof. Let $R \to S$ be étale. Choose a presentation $S = R[x_1, \ldots, x_n]/I$. As $R \to S$ is étale we know that $$\text{d} : I/I^2 \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, n} S\text{d}x_i$$ is an isomorphism, in particular $I/I^2$ is a free $S$-module. Thus by Lemma A presentation realizing a basis of the conormal module we may assume (after possibly changing the presentation), that $I = (f_1, \ldots, f_c)$ such that the classes $f_i \bmod I^2$ form a basis of $I/I^2$. It follows immediately from the fact that the displayed map above is an isomorphism that $c = n$ and that $\det(\partial f_j/\partial x_i)$ is invertible in $S$. $\square$

Lemma. Recognizing a filtered colimit of finite presentations

Let $R \to \Lambda$ be a ring map. Let $\mathcal{E}$ be a set of $R$-algebras such that each $A \in \mathcal{E}$ is of finite presentation over $R$. Then the following two statements are equivalent

1. $\Lambda$ is a filtered colimit of elements of $\mathcal{E}$, and

2. for any $R$-algebra map $A \to \Lambda$ with $A$ of finite presentation over $R$ we can find a factorization $A \to B \to \Lambda$ with $B \in \mathcal{E}$.

Proof. Suppose that $\mathcal{I} \to \mathcal{E}$, $i \mapsto A_i$ is a filtered diagram such that $\Lambda = \mathop{\operatorname{colim}}_i A_i$. Let $A \to \Lambda$ be an $R$-algebra map with $A$ of finite presentation over $R$. Then we get a factorization $A \to A_i \to \Lambda$ by applying Lemma Characterizations of finite presentation. Thus (1) implies (2).

Consider the category $\mathcal{I}$ of Lemma The filtered category of finite ring presentations. By Categories, Lemma Cofinal subcategories of filtered categories (uncovered prerequisite) the full subcategory $\mathcal{J}$ consisting of those $A \to \Lambda$ with $A \in \mathcal{E}$ is cofinal in $\mathcal{I}$ and is a filtered category. Then $\Lambda$ is also the colimit over $\mathcal{J}$ by Categories, Lemma Cofinality and colimits (uncovered prerequisite). $\square$

Lemma. Standard smooth algebras

Let $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c) = R[x_1, \ldots, x_n]/I$ be a standard smooth algebra. Then

1. the ring map $R \to S$ is smooth,

2. the $S$-module $\Omega_{S/R}$ is free on $\text{d}x_{c + 1}, \ldots, \text{d}x_n$,

3. the $S$-module $I/I^2$ is free on the classes of $f_1, \ldots, f_c$,

4. for any $g \in S$ the ring map $R \to S_g$ is standard smooth,

5. for any ring map $R \to R'$ the base change $R' \to R'\otimes_R S$ is standard smooth,

6. if $f \in R$ maps to an invertible element in $S$, then $R_f \to S$ is standard smooth, and

7. the ring $S$ is a relative global complete intersection over $R$.

Proof. Consider the naive cotangent complex of the given presentation $$(f_1, \ldots, f_c)/(f_1, \ldots, f_c)^2 \longrightarrow \bigoplus\nolimits_{i = 1}^n S \text{d}x_i.$$ Let us compose this map with the projection onto the first $c$ direct summands of the direct sum. According to the definition of a standard smooth algebra the classes $f_i \bmod (f_1, \ldots, f_c)^2$ map to a basis of $\bigoplus_{i = 1}^c S\text{d}x_i$. We conclude that $(f_1, \ldots, f_c)/(f_1, \ldots, f_c)^2$ is free of rank $c$ with a basis given by the elements $f_i \bmod (f_1, \ldots, f_c)^2$, and that the homology in degree $0$, i.e., $\Omega_{S/R}$, of the naive cotangent complex is a free $S$-module with basis the images of $\text{d}x_{c + j}$, $j = 1, \ldots, n - c$. In particular, this proves $R \to S$ is smooth.

The proofs of (4) and (6) are omitted. But see the example below and the proof of Lemma Base change of a global complete intersection.

Let $\varphi : R \to R'$ be any ring map. Set $S' = R'[x_1, \ldots, x_n]/(f_1^\varphi, \ldots, f_c^\varphi)$ where $f^\varphi$ is the polynomial obtained from $f \in R[x_1, \ldots, x_n]$ by applying $\varphi$ to all the coefficients. Then $S' \cong R' \otimes_R S$. Moreover, the determinant of Definition Standard smooth presentations for $S'/R'$ is equal to $g^\varphi$. Its image in $S'$ is therefore the image of $g$ via $R[x_1, \ldots, x_n] \to S \to S'$ and hence invertible. This proves (5).

To prove (7) it suffices to show that every nonzero fibre $S \otimes_R \kappa(\mathfrak p)$ has dimension $n - c$ for every prime $\mathfrak p \subset R$. By (5) it suffices to prove that any standard smooth algebra $k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ over a field $k$, if nonzero, has dimension $n - c$. We already know that $k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ is a local complete intersection by Lemma Smooth algebras over a field and the Jacobian criterion, Theorems 5.1–6.1 and Sections 1–3. Hence, since $I/I^2$ is free of rank $c$ we see that $k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ has dimension $n - c$, by Lemma Local criteria for complete intersections for example. $\square$

Proposition. Formal smoothness of smooth algebras

Let $R \to S$ be a ring map. The following are equivalent

1. $R \to S$ is of finite presentation and formally smooth,

2. $R \to S$ is smooth.

Proof. Follows from Proposition Characterizations of formal smoothness and Definition Smooth ring maps. (Note that $\Omega_{S/R}$ is a finitely presented $S$-module if $R \to S$ is of finite presentation, see Lemma Kähler differentials, Theorems 3.1–3.3, Proposition 3.4 and Theorem 7.1.) $\square$

Lemma. Nakayama's lemma

Source credit: the original source citation MatCA (1.M Lemma (NAK) page 11)

We quote from the original source citation MatCA: "This simple but important lemma is due to T. Nakayama, G. Azumaya and W. Krull. Priority is obscure, and although it is usually called the Lemma of Nakayama, late Prof. Nakayama did not like the name."

Let $R$ be a ring with Jacobson radical $\text{rad}(R)$. Let $M$ be an $R$-module. Let $I \subset R$ be an ideal.

1. If $IM = M$ and $M$ is finite, then there exists an $f \in 1 + I$ such that $fM = 0$.

2. If $IM = M$, $M$ is finite, and $I \subset \text{rad}(R)$, then $M = 0$.

3. If $N, N' \subset M$, $M = N + IN'$, and $N'$ is finite, then there exists an $f \in 1 + I$ such that $fM \subset N$ and $M_f = N_f$.

4. If $N, N' \subset M$, $M = N + IN'$, $N'$ is finite, and $I \subset \text{rad}(R)$, then $M = N$.

5. If $N \to M$ is a module map, $N/IN \to M/IM$ is surjective, and $M$ is finite, then there exists an $f \in 1 + I$ such that $N_f \to M_f$ is surjective.

6. If $N \to M$ is a module map, $N/IN \to M/IM$ is surjective, $M$ is finite, and $I \subset \text{rad}(R)$, then $N \to M$ is surjective.

7. If $x_1, \ldots, x_n \in M$ generate $M/IM$ and $M$ is finite, then there exists an $f \in 1 + I$ such that $x_1, \ldots, x_n$ generate $M_f$ over $R_f$.

8. If $x_1, \ldots, x_n \in M$ generate $M/IM$, $M$ is finite, and $I \subset \text{rad}(R)$, then $M$ is generated by $x_1, \ldots, x_n$.

9. If $IM = M$, $I$ is nilpotent, then $M = 0$.

10. If $N, N' \subset M$, $M = N + IN'$, and $I$ is nilpotent then $M = N$.

11. If $N \to M$ is a module map, $I$ is nilpotent, and $N/IN \to M/IM$ is surjective, then $N \to M$ is surjective.

12. If $\{x_\alpha\}_{\alpha \in A}$ is a set of elements of $M$ which generate $M/IM$ and $I$ is nilpotent, then $M$ is generated by the $x_\alpha$.

Proof. Proof of (the indicated step). Choose generators $y_1, \ldots, y_m$ of $M$ over $R$. For each $i$ we can write $y_i = \sum z_{ij} y_j$ with $z_{ij} \in I$ (since $M = IM$). In other words $\sum_j (\delta_{ij} - z_{ij})y_j = 0$. Let $f$ be the determinant of the $m \times m$ matrix $A = (\delta_{ij} - z_{ij})$. Note that $f \in 1 + I$ (since the matrix $A$ is entrywise congruent to the $m \times m$ identity matrix modulo $I$). By Lemma A left inverse for a matrix (1), there exists an $m \times m$ matrix $B$ such that $BA = f 1_{m \times m}$. Writing out we see that $\sum_{i} b_{hi} a_{ij} = f \delta_{hj}$ for all $h$ and $j$; hence, $\sum_{i, j} b_{hi} a_{ij} y_j = \sum_{j} f \delta_{hj} y_j = f y_h$ for every $h$. In other words, $0 = f y_h$ for every $h$ (since each $i$ satisfies $\sum_j a_{ij} y_j = 0$). This implies that $f$ annihilates $M$.

By Lemma Containment in the Jacobson radical an element of $1 + \text{rad}(R)$ is an invertible element of $R$. Hence we see that (the indicated step) implies (2). We obtain (3) by applying (1) to $M/N$ which is finite as $N'$ is finite. We obtain (4) by applying (2) to $M/N$ which is finite as $N'$ is finite. We obtain (5) by applying (3) to $M$ and the submodules $\operatorname{Im}(N \to M)$ and $M$. We obtain (6) by applying (4) to $M$ and the submodules $\operatorname{Im}(N \to M)$ and $M$. We obtain (7) by applying (5) to the map $R^{\oplus n} \to M$, $(a_1, \ldots, a_n) \mapsto a_1x_1 + \ldots + a_nx_n$. We obtain (8) by applying (6) to the map $R^{\oplus n} \to M$, $(a_1, \ldots, a_n) \mapsto a_1x_1 + \ldots + a_nx_n$.

Part (9) holds because if $M = IM$ then $M = I^nM$ for all $n \geq 0$ and $I$ being nilpotent means $I^n = 0$ for some $n \gg 0$. Parts (10), (11), and (12) follow from (9) by the arguments used above. $\square$

Lemma. Finite presentation and finite algebras

Let $R \to S$ be a ring map of finite presentation. For any surjection $\alpha : R[x_1, \ldots, x_n] \to S$ the kernel of $\alpha$ is a finitely generated ideal in $R[x_1, \ldots, x_n]$.

Proof. Write $S = R[y_1, \ldots, y_m]/(f_1, \ldots, f_k)$. Choose $g_i \in R[y_1, \ldots, y_m]$ which are lifts of $\alpha(x_i)$. Then we see that $S = R[x_i, y_j]/(f_l, x_i - g_i)$. Choose $h_j \in R[x_1, \ldots, x_n]$ such that $\alpha(h_j)$ corresponds to $y_j \bmod (f_1, \ldots, f_k)$. Consider the map $\psi : R[x_i, y_j] \to R[x_i]$, $x_i \mapsto x_i$, $y_j \mapsto h_j$. Then the kernel of $\alpha$ is the image of $(f_l, x_i - g_i)$ under $\psi$ and we win. $\square$

Lemma. The equational criterion for flatness (Equational criterion of flatness)

A module $M$ over $R$ is flat if and only if every relation in $M$ is trivial.

Proof. Assume $M$ is flat and let $\sum f_i x_i = 0$ be a relation in $M$. Let $I = (f_1, \ldots, f_n)$, and let $K = \operatorname{Ker}(R^n \to I, (a_1, \ldots, a_n) \mapsto \sum_i a_i f_i)$. So we have the short exact sequence $0 \to K \to R^n \to I \to 0$. Then $\sum f_i \otimes x_i$ is an element of $I \otimes_R M$ which maps to zero in $R \otimes_R M = M$. By flatness $\sum f_i \otimes x_i$ is zero in $I \otimes_R M$. Thus there exists an element of $K \otimes_R M$ mapping to $\sum e_i \otimes x_i \in R^n \otimes_R M$ where $e_i$ is the $i$th basis element of $R^n$. Write this element as $\sum k_j \otimes y_j$ and then write the image of $k_j$ in $R^n$ as $\sum a_{ij} e_i$ to get the result.

Assume every relation is trivial, let $I$ be a finitely generated ideal, and let $x = \sum f_i \otimes x_i$ be an element of $I \otimes_R M$ mapping to zero in $R \otimes_R M = M$. This just means exactly that $\sum f_i x_i$ is a relation in $M$. And the fact that it is trivial implies easily that $x$ is zero, because $$x

\sum f_i \otimes x_i

\sum f_i \otimes \left(\sum a_{ij}y_j\right)

\sum \left(\sum f_i a_{ij}\right) \otimes y_j

0$$ $\square$

Lemma. Krull's intersection theorem

Let $R$ be a Noetherian local ring. Let $I \subset R$ be a proper ideal. Let $M$ be a finite $R$-module. Then $\bigcap_{n \geq 0} I^nM = 0$.

Proof. Let $N = \bigcap_{n \geq 0} I^nM$. Then $N = I^nM \cap N$ for all $n \geq 0$. By the Artin-Rees Lemma The Artin–Rees lemma we see that $N = I^nM \cap N \subset IN$ for some suitably large $n$. By Nakayama's Lemma Nakayama's lemma we see that $N = 0$. $\square$

Lemma. Composition of smooth ring maps

A composition of smooth ring maps is smooth.

Proof. You can prove this in many different ways. One way is to use the snake lemma (Lemma The snake lemma), the Jacobi-Zariski sequence (Lemma The transitivity sequence for the naive cotangent complex), combined with the characterization of projective modules as being direct summands of free modules (Lemma Characterizations of projective modules). Another proof can be obtained by combining Lemmas Smooth algebras are syntomic, Composition of standard smooth presentations and Smooth morphisms and local algebra. $\square$

Lemma. Total rings of fractions without embedded primes

Let $R$ be a ring. Assume that $R$ has finitely many minimal primes $\mathfrak q_1, \ldots, \mathfrak q_t$, and that $\mathfrak q_1 \cup \ldots \cup \mathfrak q_t$ is the set of zerodivisors of $R$. Then the total ring of fractions $Q(R)$ is equal to $R_{\mathfrak q_1} \times \ldots \times R_{\mathfrak q_t}$.

Proof. There are natural maps $Q(R) \to R_{\mathfrak q_i}$ since any nonzerodivisor lies in $R \setminus \mathfrak q_i$. Hence a natural map $Q(R) \to R_{\mathfrak q_1} \times \ldots \times R_{\mathfrak q_t}$. For any nonminimal prime $\mathfrak p \subset R$ we see that $\mathfrak p \not \subset \mathfrak q_1 \cup \ldots \cup \mathfrak q_t$ by Lemma An elementary algebraic comparison. Hence $\operatorname{Spec}(Q(R)) = \{\mathfrak q_1, \ldots, \mathfrak q_t\}$ (as subsets of $\operatorname{Spec}(R)$, see Lemma The spectrum of a localization). Therefore $\operatorname{Spec}(Q(R))$ is a finite discrete set and it follows that $Q(R) = A_1 \times \ldots \times A_t$ with $\operatorname{Spec}(A_i) = \{\mathfrak{q}_i\}$, see Lemma A disjoint spectrum and a product of rings. Moreover $A_i$ is a local ring, which is a localization of $R$. Hence $A_i \cong R_{\mathfrak q_i}$. $\square$

Lemma. Irreducible components of a Noetherian spectrum

A Noetherian affine scheme has finitely many generic points.

If $R$ is a Noetherian ring then $\operatorname{Spec}(R)$ has finitely many irreducible components. In other words $R$ has finitely many minimal primes.

Proof. By Lemma The topology of a Noetherian spectrum and Topology, Lemma Noetherian topological spaces (programme binding) we see there are finitely many irreducible components. By Lemma Irreducibility of an affine spectrum these correspond to minimal primes of $R$. $\square$