Artin approximation for polynomial equations and its desingularization proof

Written and self-checked by GPT-6.1 Sol (OpenAI), Codex, Ultra setting, 5 October 2026. This reconstruction draft includes the desingularization proof and its consumed foundations. It retains adapted GNU FDL 1.2 expression.

The purpose of approximation is to retain exact equations while replacing coefficients in a completed local ring by coefficients on an étale neighbourhood. A coefficient need not itself belong to the local ring, and approximating each coefficient separately generally destroys its equations. Instead, retain all equations in one finite system, including those that impose the desired congruence.

The only mathematical source used to construct the new desingularization material is the freely accessible pinned Stacks exposition and its supporting chapters. Earlier programme proofs are used where their exact statements apply. The source is material for the proof written here; a link to that source does not discharge a mathematical obligation. The original theorem below remains the target until all of its consumed dependencies are proved in this lesson or an earlier programme lesson.

1. The exact approximation statement

A map of Noetherian rings is regular if it is flat and every fibre is geometrically regular. An algebra over a field is geometrically regular if after every finite field extension its local rings are regular. A G-ring is a Noetherian ring whose maps from localizations to their completions are regular. The relevant implication for a local G-ring is therefore immediate from its definition: \(R\to\widehat R\) is regular. The broader local criterion and permanence statements, including the property for local rings essentially of finite type over \(\mathbb Z\), are supplied among the foundation arguments below.

Theorem 1.1 (Artin approximation in a pointed étale neighbourhood). Let \((R,\mathfrak m)\) be a Noetherian local G-ring. Let

For every positive integer \(N\), there exist a finitely presented étale \(R\)-algebra \(R'\), a maximal ideal \(\mathfrak m'\subset R'\) over \(\mathfrak m\), and a solution \(b\in(R')^n\), such that

Here \(\rho:R'\to\widehat R\) is the canonical lift of the specified residue-field point. This is a homomorphism; no assertion that the whole étale algebra embeds into the completion is required. The statement allows arbitrary characteristic, imperfect residue fields, nilpotent base elements, and nonregular local rings.

Lemma 1.2 (the canonical comparison map). A finitely presented étale \(R\)-algebra and a specified \(R\)-map \(R'\to R/\mathfrak m\) have a unique compatible map \(R'\to\widehat R\).

Proof. Starting with the specified map to \(R/\mathfrak m\), lift successively to \(R/\mathfrak m^r\). The kernel of \(R/\mathfrak m^{r+1}\to R/\mathfrak m^r\) is square zero for \(r\geq1\), so existence and uniqueness follow from formal étaleness. The lifts are compatible by uniqueness. Taking their inverse limit gives the desired map. A map into an inverse limit of rings is exactly a compatible collection of maps into the constituent rings, which also proves uniqueness. The equality \(\widehat R/\mathfrak m^r\widehat R=R/\mathfrak m^r\) used here is proved in AG-CA, Completion, Proposition 2.2 and Theorem 3.3. Formal étaleness of an étale algebra is the defining lifting property in AG-CA, Formally smooth, unramified and étale ring maps, §6. ∎

Lemma 1.3 (a smooth residue-field point has an étale lift). If \(B\) is a smooth finitely presented \(R\)-algebra and \(B\to\kappa=R/\mathfrak m\) is an \(R\)-map, there are a finitely presented étale \(R\)-algebra \(R'\), a point \(R'\to\kappa\), and an \(R\)-map \(B\to R'\) inducing the original point.

Proof. Use the local standard form proved in AG-CA, Formally smooth, unramified and étale ring maps, Theorem 5.1 and Solution 8.5. At the kernel of the given point, localize once to write

This presentation includes any localization variable and its equation. Choose \(r_{c+1},\ldots,r_s\in R\) lifting the point's coordinates in the remaining variables. Substitute these elements in the equations, keeping the first \(c\) variables. Localize the resulting algebra at the substituted \(\Delta\) and at any denominators needed for the original presentation. Its number of equations and variables agree, and its square Jacobian determinant is a unit. The direct Newton calculation proving a standard étale presentation is étale appears in the same earlier lesson, Theorem 4.1 and §6: a square-zero error vector is corrected uniquely by the inverse Jacobian matrix. Thus the resulting \(R'\) is finitely presented étale. The original residue coordinates define its map to \(\kappa\), since the inverted elements have nonzero residues. The kernel of that map is maximal and has residue field \(\kappa\), because the structural map from \(R\) already surjects onto \(\kappa\). Substitution gives \(B_g\to R'\), hence \(B\to R'\). ∎

Proof of Theorem 1.1. The equality of finite-order quotients of a Noetherian local ring and its completion is proved at the preceding Completion locators. Choose \(c_i\in R\) representing \(\widehat a_i\) modulo \(\mathfrak m^N\widehat R\), and choose finitely many generators \(d_1,\ldots,d_M\) of \(\mathfrak m^N\). Write

Put

The tuple \(\widehat u\) gives an \(R\)-map \(A\to\widehat R\). Since \(R\to\widehat R\) is regular, the desingularization theorem proved by the sequence ending at Theorem D.38, and the elementary factorization criterion proved at the corresponding foundation lemma, give a factorization

with \(B\) smooth and finitely presented over \(R\). Compose with the residue map \(\widehat R\to\kappa\), then apply Lemma 1.3. The images \(u_{il}\in R'\) solve the \(g_j\) exactly. Set \(b_i=c_i+\sum_l d_lu_{il}\). Substitution in (1.2) gives \(f_j(b)=0\) exactly. Under the canonical map of Lemma 1.2,

This proves the prescribed congruence without demanding that the auxiliary \(u_{il}\) themselves be close to \(\widehat u_{il}\). The exact written foundations and the earlier programme proofs used in this argument are registered in §6. ∎

Corollary 1.4 (henselian formulation). If \(R\) is henselian, the solution can be chosen in \(R\). In general it gives a solution in the henselization \(R^h\), descending to the pointed étale neighbourhood just constructed.

Proof. The section criterion proved in AG-CA, Henselian local rings and henselization, Theorem 1.2 gives \(R'\to R\) with the specified residue map. Its composition with \(R\to\widehat R\) is the map \(\rho\) by Lemma 1.2, so (1.1) is preserved. For \(R^h\), use the pointed étale colimit and its universal property proved in that lesson, Lemma 4.1 and Theorem 4.2. Compatibility with the map into the complete henselian ring follows from the same uniqueness. Here the principal monic chart consumed by that section-criterion proof is supplied by the full earlier structural étale proof, Lemma D2.2, whose integral closure input is the affine conductor theorem, Theorem C4.3. Its polynomial normality input is proved in the preliminary coefficient argument, Corollary B1.2. To spell out the actual section step, localize the selected étale algebra to $(R[T]/(F))_h$, with $F$ monic and $F'$ nonzero at the given residue root $\bar a$. The henselian simple-root property gives $a\in R$ with $F(a)=0$ and $a\bmod\mathfrak m=\bar a$. Every denominator has nonzero residue at $\bar a$, hence is a unit at $a$, and evaluation defines the section. The resulting map $R'\to R$ induces the prescribed residue map; formal étale uniqueness along the quotients $R/\mathfrak m^n$ proves its composition with completion is $\rho$. Thus this corollary uses these written earlier proofs rather than an external conductor proof tag. ∎

2. Returning from a local model to the original base

Lemma 2.1 (spreading finite presentation data). Let \(R_0\) be Noetherian and \(\mathfrak p\subset R_0\). A pointed étale algebra, a finite tuple, and finitely many equations over \((R_0)_{\mathfrak p}\), as in Theorem 1.1, spread to \((R_0)_s\) for some \(s\notin\mathfrak p\). Their solution therefore gives a pointed étale neighbourhood over \(R_0\).

Proof. In the construction of Lemma 1.3, every polynomial coefficient, every fixed nonpivot coordinate, and every chosen inverse has a denominator outside \(\mathfrak p\). Invert their product in \(R_0\). The finitely many equations for the tuple and the finitely many witnesses verifying them likewise hold after multiplying by finitely many additional denominators; invert their product too. The standard étale presentation spreads with its invertible Jacobian determinant, so its spread algebra is étale by the same direct Newton calculation. Regard it as an \(R_0\)-algebra through the étale localization \(R_0\to(R_0)_s\); composing two formally étale, finitely presented maps remains étale because a square-zero lifting problem is solved uniquely in two successive steps. The chosen residue map survives and gives a point over \(\mathfrak p\) with unchanged residue field. The defining equations remain exact, and localization at the selected point gives the original local comparison. ∎

This proof uses the specific standard étale algebra produced above and therefore does not invoke an external spreading theorem for an arbitrary presentation. Pulling the resulting neighbourhood back to the original scheme preserves every defining equation and the specified fibre point.

3. The finite system for an embedded torus

The approximation step in the torus lesson begins after the formal deformation and effectivity argument has produced an affine torus \(\widehat T\) over the completed local model and a closed homomorphism \(\widehat T\hookrightarrow G_{\widehat R}\). The present lesson proves the approximation input; it does not claim to supply that preceding effectivity argument.

Lemma 3.A (finite splitting over a complete local ring). Let \((A,\mathfrak m,\kappa)\) be a local ring with \[ A\simeq\varprojlim_{n\geq1}A/\mathfrak m^n. \] No Noetherian, reducedness, normality or perfection hypothesis is imposed. Suppose a torus \(T/A\) has an affine etale splitting chart \(U\to\operatorname{Spec}A\) and a point of its closed fibre with finite separable residue field \(K/\kappa\). Then a finite free, faithfully flat, etale algebra \(E/A\) splits \(T\). We may choose \(E\) local and complete, with residue field a finite Galois extension \(L/\kappa\) containing \(K\), and with an action of \(\Gamma=\operatorname{Gal}(L/\kappa)\) for which \[ E\otimes_AE\xrightarrow{\sim}\prod_{\sigma\in\Gamma}E, \qquad x\otimes y\longmapsto\bigl(x\sigma(y)\bigr)_\sigma. \tag{3.2} \] Here a torus is etale locally a product of copies of \(\mathbf G_m\). Starting instead with the fpqc multiplicative-type definition gives this same condition by AG-GS-05, Theorem 7.7; its finite torsion-free character-group step is proved in AG-RG-S04, Lemma P0.10. Every torus therefore has such a chart and residue point: choose an affine finite-presentation part of an etale splitting cover meeting the closed fibre, and a closed point of that fibre. Its residue field is finite over \(\kappa\) by the finite-type field-algebra argument in AG-RG-S04, Lemma P0.9. The quotient sequence for differentials gives \(\Omega_{K/\kappa}=0\), because the chart is formally unramified. The finite-field differential calculation in AG-CA-17, Lemma 7.1 then proves separability.

Furthermore, a torus \(T_0/\kappa\), with character lattice split by \(L\), has a torus lift over \(A\) split by this same \(E\). Its character lattice and descent action reduce compatibly modulo every \(\mathfrak m^n\).

Proof. A finite separable extension embeds in a finite normal separable extension: adjoin all the roots of the minimal polynomials of a finite generating list. There are finitely many roots and each adjunction has finite separable degree. Every embedding permutes the root sets, so the resulting extension \(L/\kappa\) is normal and separable. Counting embeddings successively gives \(|\Gamma|=[L:\kappa]\). The primitive-element argument of AG-RG-S02, Lemma 4.2, applies also to finite fields, and gives \(L=\kappa(\alpha)\). Let \(\bar f\in\kappa[X]\) be the monic minimal polynomial of \(\alpha\), and lift its coefficients to a monic \(f\in A[X]\). Put \[ E=A[t]=A[X]/(f),\qquad d=\deg(f). \] Successively subtracting leading multiples of the monic polynomial \(f\) reduces every polynomial to degree less than \(d\). A nonzero multiple \(qf\) has degree \(\deg(q)+d\), since its leading coefficient is the nonzero leading coefficient of \(q\). Thus the remainder is unique, and \(1,t,\ldots,t^{d-1}\) is a basis making \(E\) free of positive finite rank over \(A\); it is faithfully flat because tensoring a module with this free module gives a direct sum of \(d\) copies. Its quotient by \(\mathfrak mE\) is \(L\). If \(e\in E\) has nonzero residue, multiplication by \(e\) on this basis is invertible modulo \(\mathfrak m\). Its determinant is a unit of \(A\), and the cofactor calculation proved in Lemma 3.B gives an inverse to multiplication by \(e\). Thus \(e\) is a unit, \(E\) is local with maximal ideal \(\mathfrak mE\), and coordinatewise completeness gives \[ E\simeq\varprojlim_n E/\mathfrak m^nE. \] Separability makes \(f'(t)\) a unit by the same argument. A root in a quotient by a square-zero ideal lifts uniquely by replacing any lift \(a\) with \(a-f'(a)^{-1}f(a)\). Polynomial expansion proves existence, and \(f(b)-f(a)=(b-a)(f'(a)+(b-a)H)\) proves uniqueness. Iteration gives the corresponding assertion for every nilpotent ideal. This proves formal etaleness of \(E/A\); its finite presentation makes it etale.

The residue point of \(U\) gives an algebra map \(\mathcal O(U)\to K\to L\). Formal etaleness of \(U/A\) lifts this map uniquely through \(E/\mathfrak m^{n+1}E\to E/\mathfrak m^nE\), whose kernel is square-zero for \(n\geq1\). Taking the compatible limit gives a map \(\mathcal O(U)\to E\), hence a section of \(U_E\to\operatorname{Spec}E\). Pulling back the splitting on \(U\) therefore splits \(T_E\).

We now construct (3.2), including its automorphisms. For every \(\sigma\in\Gamma\), the residue \(\sigma(\alpha)\in L\) is a simple root of \(\bar f\). In the complete local ring \(E\), Newton's recursion \[ a_{j+1}=a_j-f'(a_j)^{-1}f(a_j) \] starting from any lift of this residue gives \(f(a_j)\in(\mathfrak mE)^{2^j}\) and \(a_{j+1}-a_j\in(\mathfrak mE)^{2^j}\). Completeness and separatedness give a root \(a_\sigma\), and the difference formula above gives its uniqueness with that residue. Distinct \(a_\sigma\) have unit differences. If a polynomial \(P\) vanishes at all these roots, monic division by \(X-a_1\) gives \(P=(X-a_1)Q\), since the remainder is \(P(a_1)=0\). At each other root \(a_\sigma\), the factor \(a_\sigma-a_1\) is a unit, so \(Q(a_\sigma)=0\). Repeating the division shows that \(\prod_\sigma(X-a_\sigma)\) divides \(P\). If \(\deg(P)<d\), monicity forces \(P=0\). The difference between \(f\) and this product has degree less than \(d\) and vanishes at every root, so \[ f(X)=\prod_{\sigma\in\Gamma}(X-a_\sigma). \] The map \(t\mapsto a_\sigma\) defines an \(A\)-algebra endomorphism \(g_\sigma\) of \(E\). The two roots \(g_\sigma g_\tau(t)\) and \(a_{\sigma\tau}\) have the same residue, so uniqueness gives \(g_\sigma g_\tau=g_{\sigma\tau}\). Also \(a_1=t\); hence these maps are automorphisms and lift the Galois action on \(L\).

The isomorphism \(E\otimes_AE=E[X]/(f)\) and evaluation at the \(a_\sigma\) now give (3.2). Explicitly, the polynomials \[ \ell_\sigma(X)=\prod_{\tau\ne\sigma} \frac{X-a_\tau}{a_\sigma-a_\tau} \] give the inverse to evaluation: a tuple \((b_\sigma)_\sigma\) is represented by \(\sum_\sigma b_\sigma\ell_\sigma(X)\). Thus the isomorphism uses no division by \(|\Gamma|\). Its compatibility on triple products is exactly \(g_\sigma g_\tau=g_{\sigma\tau}\). This is the required finite etale Galois torsor.

For the final assertion choose the finite Galois splitting field of \(T_0\) supplied by AG-GS-05, Lemma 5.1, and let \(M\) be its free finite-rank character lattice, with the resulting \(\Gamma\)-action. On \(E[M]\) put \[ \sigma(ae^m)=g_\sigma(a)e^{\sigma m}. \tag{3.3} \] The group law and (3.2) make this a descent datum. Multiplication and every Hopf operation commute with it, since addition in \(M\) commutes with its action. The actual module and algebra effectivity proof in AG-RG-S04, Lemmas A1.1--A1.2, gives the invariant Hopf algebra \[ C=E[M]^\Gamma,\qquad E\otimes_AC\simeq E[M]. \tag{3.4} \] Tensor products and Hopf identities descend by this isomorphism and faithful flatness. Finite presentation descends here by a finite-generator argument. Express the finitely many Laurent algebra generators over \(E\) as finite \(E\)-linear combinations of elements of \(C\). Those elements generate \(C\), since the cokernel of their generated subalgebra becomes zero over \(E\). For a resulting finite polynomial surjection \(A[Z]\to C\), its kernel becomes a finite ideal after tensoring with \(E\), because \(E[M]\) is finitely presented. To see that finiteness for this particular surjection, compare it with a finite presentation of \(E[M]\): express each presentation's generators in the other, and substitute these expressions in the finitely many relations and composite-generator identities. They generate the new kernel. Write a finite list of its generators as finite sums of elements of the original kernel with coefficients in \(E\). Their finitely many kernel components generate the original kernel by faithful flatness. Thus \(C\) is finitely presented. It defines a torus \(T_A\), because (3.4) splits it on the finite etale faithfully flat cover.

Put \(A_n=A/\mathfrak m^n\) and \(E_n=E\otimes_AA_n\). Tensoring (3.2)--(3.4) with \(A_n\) gives the same torsor and the same character action over \(E_n\): \[ E_n\otimes_{A_n}(C\otimes_AA_n)\simeq E_n[M]. \tag{3.5} \] The algebra on the left descends the datum (3.3) modulo \(\mathfrak m^n\); uniqueness in Lemma A1.2 proves that it is its effective descent. This proves compatibility for every \(n\), even though \(A\to A_n\) need not be flat. It does not assume that taking invariants commutes with an arbitrary quotient. At \(n=1\), the descent datum is precisely the given one on \(L[M]\), so its descended torus is \(T_0\).

More generally, any compatible character data split by these \(E_n\) and identified with the same lattice have no additional nilpotent transition parameters. Comparing coefficients in \(\Delta(\sum c_me^m)=(\sum c_me^m)\otimes(\sum c_me^m)\) gives \(c_m^2=c_m\), \(c_mc_{m'}=0\) for \(m\ne m'\), and \(\sum c_m=1\). Each \(E_n\) is local, so exactly one coefficient is \(1\). A character is therefore one monomial; a torus automorphism is one integral lattice automorphism. A compatible automorphism reducing to a fixed one is that same one. Hence a compatible lattice descent action is the fixed action in (3.3), and (3.5) gives its actual compatible descent. This completes both assertions. \(\square\)

Choose a finite presentation of the Hopf algebra of \(\widehat T\). Lemma 3.A supplies a finite free, etale, faithfully flat splitting algebra \(E/\widehat R\); choose the resulting splitting isomorphism

Lemma 3.B (perfect trace, over an arbitrary base). Let $A$ be a commutative ring and $C$ a commutative $A$-algebra free of finite rank $d$. If the determinant of the pairing $(c,z)\mapsto\operatorname{Tr}_{C/A}(cz)$ is a unit, then $C/A$ is finitely presented and formally etale, hence etale. If $d>0$, it is faithfully flat. These assertions require no Noetherian, reducedness or characteristic hypothesis.

Proof. First, the elementary matrix identities used here hold over every commutative ring. Define a determinant by the signed permutation sum and a cofactor by the corresponding smaller signed determinant. Expanding along a row or column gives the determinant. A matrix with two equal rows or columns has determinant zero: pair each permutation with the permutation interchanging those positions; their contributions have opposite signs, including in characteristic two. Consequently, if $M$ is square, multiplying it on either side by its transposed cofactor matrix gives $(\det M)I$: a diagonal entry is the row or column expansion, and an off-diagonal entry is the expansion of a determinant with two equal rows or columns. A unit determinant therefore gives the inverse $(\det M)^{-1}\operatorname{adj}(M)$. For size zero the determinant is one. To verify the converse implication as well, expand $\det(UV)$ by multilinearity in its columns. The expansion chooses, for each column of $UV$, one column of $U$ with its coefficient from $V$. A tuple repeating a column of $U$ contributes zero by the cancellation just proved. Every remaining tuple is a permutation of all the columns of $U$, whose determinant is its permutation sign times $\det(U)$. Summing its coefficients with these signs is precisely $\det(V)$. Thus $\det(UV)=\det(U)\det(V)$, including for size zero. If $UV=I$, this equation gives $\det(U)\det(V)=1$, so an invertible matrix has a unit determinant. For square matrices $U,V$, the finite double sum also gives $\operatorname{tr}(UV)=\operatorname{tr}(VU)$. Thus conjugating a multiplication matrix by an invertible change of basis preserves its trace. Multiplication matrices, their traces and their determinants commute with every scalar extension, entry by entry.

Suppose first that $d>0$, and choose a basis $u_1,\ldots,u_d$. Write $u_i u_j=\sum_k m_{ijk}u_k$ and $1=\sum_i a_i u_i$. There is an explicit finite algebra presentation $$ C=A[X_1,\ldots,X_d]/\bigl(X_iX_j-\sum_k m_{ijk}X_k,\;1-\sum_i a_iX_i\bigr). $$ Indeed, repeatedly replace a product of two variables by the displayed linear combination, reducing each monomial to degree at most one. Replace its constant term by that constant times $\sum_i a_iX_i$. Every quotient element is therefore represented by $\sum_i b_iX_i$. The map to $C$ is surjective, and if this representative maps to zero, linear independence of the $u_i$ makes every $b_i$ zero. This proves the presentation without an assumption on ideals of $A$.

The inverse pairing matrix supplies a trace-dual basis $v_1,\ldots,v_d$, characterized by $\operatorname{Tr}(u_jv_i)=\delta_{ij}$. In particular $x=\sum_i u_i\operatorname{Tr}(v_ix)$ and $x=\sum_i\operatorname{Tr}(u_ix)v_i$. To check the second equality, pair both sides with each $u_j$ and use invertibility of the pairing; the first is the ordinary basis-coordinate equality. Put $$ e=\sum_i u_i\otimes v_i\in C\otimes_A C. $$ The $A$-linear map $$ \Phi:C\otimes_A C\longrightarrow\operatorname{End}_A(C),\qquad \Phi(a\otimes b)(x)=a\operatorname{Tr}(bx) $$ is an isomorphism, with inverse $F\mapsto\sum_i F(u_i)\otimes v_i$. The first coordinate equality proves $\Phi\Phi^{-1}(F)=F$; the second proves $\Phi^{-1}\Phi(a\otimes b)=a\otimes b$, and pure tensors generate. Since $\Phi(e)$ is the identity, both $(c\otimes1)e$ and $(1\otimes c)e$ have image multiplication by $c$. Thus $$ (c\otimes1)e=(1\otimes c)e\qquad(c\in C). \tag{3.6} $$ Let $s=\sum_i u_iv_i$. The diagonal coefficient of multiplication by $c$ in position $i$ is $\operatorname{Tr}(v_i c u_i)$. Summing those coefficients gives $\operatorname{Tr}(c)=\operatorname{Tr}(cs)$ for every $c$. Invertibility of the pairing now gives $$ \sum_i u_iv_i=1. \tag{3.7} $$ No division by $d$, or assertion that $d$ is a unit, occurs.

We prove the lifting property directly. Let $J\subset D$ be a square-zero ideal of a commutative $A$-algebra, and let $\bar f:C\to D/J$ be an $A$-algebra map. The free-module lifting calculation in AG-RG-S01, Lemma 1.0 gives an $A$-linear lift $l:C\to D$. Its value $l(1)$ is $1+j$ with $j\in J$, and has inverse $1-j$. Multiplying the whole lift by that inverse makes $l(1)=1$ while preserving $A$-linearity and its reduction. Give $J$ its $C$-module action via $\bar f$: multiply by any lift of $\bar f(c)$, which is independent of that lift because $J^2=0$.

The defect $b(c,z)=l(cz)-l(c)l(z)$ lies in $J$, is $A$-bilinear and has $b(c,1)=b(1,c)=0$. Associativity in $C$ and $D$ gives, by expanding the four terms, $$ c\,b(z,t)-b(cz,t)+b(c,zt)-b(c,z)\,t=0. \tag{3.8} $$ Define the $A$-linear correction $h(c)=\sum_i u_i b(v_i,c)$ in $J$. Apply the $A$-linear map $a\otimes z\mapsto a b(z,t)$ to (3.6); it gives $c h(t)=\sum_i u_i b(cv_i,t)$. Using (3.8) with $(v_i,c,t)$ and then (3.7), we obtain $$ c h(t)-h(ct)+h(c)t =\sum_i u_i\bigl(b(v_ic,t)-b(v_i,ct)+b(v_i,c)t\bigr) =\sum_i u_iv_i b(c,t)=b(c,t). $$ Consequently $f=l+h$ is multiplicative: the defect of this new map is $b(c,t)+h(ct)-c h(t)-h(c)t=0$, since $h(c)h(t)=0$. Also $h(1)=0$. Thus $f$ is a unital $A$-algebra lift of $\bar f$.

For uniqueness, the difference $\delta$ of two such lifts takes values in $J$ and satisfies $\delta(ct)=c\delta(t)+\delta(c)t$; expand their products and use $J^2=0$. Apply $a\otimes z\mapsto\delta(a)z$ to (3.6). The difference between its two sides is $\delta(c)\sum_i u_iv_i=\delta(c)$, by that identity and (3.7). It is zero, so every lift is unique. If $J^N=0$, lift successively through $D/J^{r+1}\to D/J^r$, for $1\le r<N$. Its kernel is square-zero because $2r\ge r+1$. Existence and uniqueness at each step prove formal etaleness for every nilpotent ideal. Together with the explicit finite presentation, this is etaleness by the programme definition.

For $d=0$, $C$ is the zero algebra with presentation $A/(1)$. If a unital map $C\to D/J$ exists and $J$ is nilpotent, then $D/J=0$, so $1\in J$ and nilpotence forces $D=0$. There is exactly one lift. Finally, for positive $d$, tensoring an $A$-module with $C$ gives $d$ copies of that module; this is exact and detects whether it is zero. This proves faithful flatness. $\square$

This is the finite-free case of the discriminant criterion in the freely accessible Stacks Project, Lemma 49.3.1, Tag 0BJF. The complete lifting proof above supplies the result used here.

Return to the splitting algebra $E/\widehat R$ from Lemma 3.A. Its trace determinant is a unit, as we now check rather than adding an unsupported inverse variable. By (3.2), after scalar extension from $\widehat R$ to $E$ the algebra becomes $\prod_{\sigma\in\Gamma}E$. Multiplication in this product is diagonal in the coordinate idempotent basis. Its trace pairing matrix is the identity, hence invertible. The preceding trace and basis-change calculations imply that the original pairing matrix becomes invertible over $E$, so its determinant $\Delta$ becomes a unit there. Then $(\widehat R/\Delta\widehat R)\otimes_{\widehat R}E=0$. Faithful flatness from Lemma 3.A detects zero modules, giving $\widehat R/\Delta\widehat R=0$, which means that $\Delta$ is already a unit of $\widehat R$.

Fix the finite free basis of $E$ supplied by Lemma 3.A. Its multiplication table, unit and multiplication matrices define finitely many coefficient variables; associativity, commutativity and the unit laws are finite polynomial equations. Add a variable $w$ and the equation $w\Delta=1$ for the trace determinant. The original completed coefficient tuple satisfies this equation by the preceding argument. After approximation it still satisfies the same equation exactly. Lemma 3.B makes the resulting finite free algebra etale and, since its unchanged rank is positive, faithfully flat.

Add coefficients for the finite presentations, Hopf maps, the isomorphism (3.1), its inverse, and the homomorphism to \(G\). Laurent expressions can be written polynomially by adding inverse variables and relations \(Z_iW_i=1\). The maps need to be specified on finite generating sets. Relations between their images need finitely many witnesses in the finitely generated relation ideals. The two composites of the splitting isomorphism and its inverse are tested on finite generators; equality of these composites with the identity is recorded by their witnesses. The same method records comultiplication, counit, inverse, and the homomorphism identities on a finite list of generators. Coassociativity and every needed descent identity are likewise checked on finite generators, using the finite multiplication tables on the finite overlap algebras. Each expression and witness has finite support.

Retain the closed immersion in this same system. The pullback \(\mathcal O(G_{\widehat R})\to\mathcal O(\widehat T)\) is surjective. Choose preimages of a finite list of algebra generators of \(\mathcal O(\widehat T)\), and include the equations asserting that their images are those generators. After approximation these equations still prove surjectivity; consequently the resulting homomorphism is a closed immersion.

We have obtained one finite polynomial system over the local model, solved by the original completed coefficient tuple. Apply Theorem 1.1 at order one. The defining equations hold exactly after approximation, while all the chosen coefficients retain their original residues. The preserved splitting maps identify the source Hopf algebra with a Laurent polynomial Hopf algebra on the finite étale faithfully flat cover. Thus it is a torus, with the specified special fibre and closed embedding. If a prescribed higher jet is needed, use that order instead. Lemma 2.1 spreads the data and the neighbourhood back to an open part of the finite-type model; base change returns them to the original base.

The local model is essentially of finite type over \(\mathbb Z\). The G-ring permanence argument in §5 is therefore a consumed part of this application, including its coefficient-ring inputs. It must be proved before the approximation step is declared complete. An excellence label or an external Stacks link does not replace this work.

Scope of the supporting algebra. The approximation theorem and the desingularization theorem concern Noetherian rings. Every base ring of the finite-type algebra diagrams in their proof is Noetherian. In the foundational complete-intersection arguments below we state and prove the exact Noetherian forms consumed there; arguments explicitly stated over arbitrary rings retain that wider scope. Polynomial presentations with arbitrarily many variables occur only in the two-term cotangent comparison, whose elementary presentation and filtered-colimit proofs are written without Noetherian assumptions. This distinction preserves arbitrary characteristic, imperfect residue fields, and the full Noetherian local G-ring statement.

4. Desingularization proof sequence

The following sequence contains the actual lifting, presentation, local descent, separable, and inseparable arguments. It is adapted from the pinned free source. The source’s historical bibliography is not reproduced. Its statements use the conormal definition of smoothness reproduced among the foundation statements below. The displayed arrow lists specify the maps of the original commutative diagrams.

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 D.1: singular ideal

Adapted from the Stacks project, smoothing.tex, lines 124–132.

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

This makes sense because the set of primes where $R \to A$ is smooth is open, see Algebra, Definition F.7. In order to find an explicit set of generators for the singular ideal we first prove the following lemma.

Lemma D.2: find strictly standard

Adapted from the Stacks project, smoothing.tex, lines 141–181.

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

for some $a' \in A$ and

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 F.11. We will use the formation of the naive cotangent complex commutes with localization, see Algebra, Section the cotangent-complex context, especially Algebra, Lemma F.65. By Algebra, Definitions F.6 and F.7 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 $U = V = \{1, \ldots, c\}$). ∎

We will use the notion of a strictly standard element in $A$ over $R$. 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 D.15.

Definition D.3: strictly standard

Adapted from the Stacks project, smoothing.tex, lines 191–222.

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

for some $a' \in A$ and

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

for some $a_I \in A$ and

for $j = 1, \ldots, m - c$.

The following lemma is useful to find implications of (D.3s1: strict Jacobian-minor condition).

Lemma D.4: parse equation strictly standard one

Adapted from the Stacks project, smoothing.tex, lines 228–250.

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 (D.3s1: strict Jacobian-minor condition) 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

is multiplication by $a$. Conversely, if such a $\psi$ exists, then $a^c$ satisfies (D.3s1: strict Jacobian-minor condition). The relation condition D.3s2 is a separate hypothesis. This includes $c=0$: the empty minor is $1$ and the module map is the unique map between zero modules.

Proof. Let $M$ be the $n\times c$ differential matrix, with entry $M_{ij}=\partial f_j/\partial x_i$. The $c\times c$ minor on rows indexed by $U$ has the determinant in D.3s1: transposing a square matrix preserves its determinant by reindexing the signed permutation expansion. Thus D.3s1 says that $a$ belongs to the ideal of these minors. Lemma F.70, proved below over arbitrary commutative rings, supplies a $c\times n$ matrix $B$ with $BM=a1_c$. This is the required map $\psi$. Conversely the displayed composite being $a1_c$ says $BM=a1_c$, and the same lemma puts $a^c$ in the minor ideal, which is exactly D.3s1 for $a^c$. No relation among the remaining generators is used in this matrix assertion. ∎

Lemma D.5: elkik

Adapted from the Stacks project, smoothing.tex, lines 252–283.

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 D.3. 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 (D.3s2: strict relation condition) implies that $(I/I^2)_a$ is generated by the classes of $f_1, \ldots, f_c$. Assumption (D.3s1: strict Jacobian-minor condition) implies that the differential $(I/I^2)_a \to \bigoplus A_a\text{d}x_i$ has a left inverse, see Lemma D.4. Hence $R \to A_a$ is smooth by definition and Algebra, Lemma F.65.

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 D.2. ∎

Example D.6: not quasi compact

Adapted from the Stacks project, smoothing.tex, lines 285–292.

The set of points where a finitely presented ring map is smooth needn't be a quasi-compact open. For example, let $k$ be a field and set $$ R=k[x,y_1,y_2,y_3,\ldots]/(xy_i\mid i\geq1),\qquad A=R/(x). $$ The distinguished opens below are in $\operatorname{Spec}(A)$. The smooth locus of $R\to A$ is $$ U=\bigcup_{i\geq1}D(y_i)=\operatorname{Spec}(A)\setminus\{\mathfrak m\}, \qquad \mathfrak m=(y_1,y_2,\ldots), $$ and this open is not quasi-compact.

Proof. The quotient identifies $A$ with $k[y_1,y_2,\ldots]$. The map $R\to A$ is of finite presentation: it is the quotient by the single generator $x$, with no algebra variables required. The base ring $R$ is not assumed Noetherian. Evaluation at all $y_i=0$ gives $A/\mathfrak m=k$, so $\mathfrak m$ is maximal and is the only prime containing every $y_i$.

Here is the ideal calculation. Setting all $y_i$ to zero gives a retraction $R\to k[x]$ of the natural inclusion $k[x]\to R$, so that inclusion is injective. For $I=(x)\subset R$, the relations $xy_i=0$ give $xr=xr(x,0,0,\ldots)$ for every $r\in R$. Therefore $$ I=xk[x],\qquad I^2=x^2k[x],\qquad I/I^2=k\,\bar x\simeq A/\mathfrak m. $$ In the last identification every $y_i$ annihilates $\bar x$. For the zero-variable polynomial presentation $R\twoheadrightarrow A$, the conormal complex is $[I/I^2\to0]$. The presentation comparison in Lemma F.11 identifies it with the naive cotangent complex used in Definition F.6.

First suppose $y_i\notin\mathfrak q$. Inverting $y_i$ in $R$ makes $x=0$, hence $A_{y_i}=R_{y_i}$. As an $R$-algebra this is $$ R[t]/(h),\qquad h=y_it-1. $$ The element $h$ is a nonzerodivisor over this arbitrary base ring. Indeed, if $h\sum_{n=0}^Nb_nt^n=0$, its constant coefficient gives $b_0=0$, and its successive coefficients $y_ib_{n-1}-b_n=0$ give every $b_n=0$. Consequently $(h)/(h^2)$ is free on $\bar h$ over $R[t]/(h)$: multiplication by $h$ gives a surjection onto it, and cancellation of the nonzerodivisor $h$ shows that its kernel is exactly $(h)$. An $R$-derivation of $R[t]$ is determined by its value on $t$; the product rule gives $d(t^n)=nt^{n-1}dt$. Thus $\Omega_{R[t]/R}=R[t]dt$, and this presentation has conormal complex $$ A_{y_i}\,\bar h\xrightarrow{\ y_i\ }A_{y_i}\,dt. $$ Its differential is an isomorphism, since $y_i$ is a unit. Both first homology and the differential module are zero, and the zero module is finite projective. The displayed presentation is finite, so Definition F.6 proves $R\to A_{y_i}$ smooth. Definition F.7 proves smoothness at every point of $D(y_i)$.

We must also exclude the exceptional point $\mathfrak m$ from every smooth neighbourhood. Let $g\in A\setminus\mathfrak m$. Its constant coefficient $c=g(0,0,\ldots)$ is nonzero. Regard its polynomial expression as a lift $\widetilde g\in R$. A presentation of $A_g$ over $R$ is $$ P=R[t]\twoheadrightarrow A_g,\qquad J=\ker(P\to A_g)=(x,\widetilde g t-1). $$ The class of $x$ in $J/J^2$ has relative differential zero, because $x$ belongs to the base ring $R$. This class is nonzero. To see this, put $C=(R/(x^2))_{\widetilde g}$. The $R$-algebra map $P\to C$ sending $t$ to $\widetilde g^{-1}$ maps $J$ into the square-zero ideal $(x)C$, and maps $J^2$ to zero. It therefore maps the class of $x$ in $J/J^2$ to $x\in C$. The assignments $x\mapsto\epsilon$ and $y_i\mapsto0$ define $$ R/(x^2)\longrightarrow k[\epsilon]/(\epsilon^2). $$ They respect all the relations, and send $\widetilde g$ to the unit $c$. The map extends to $C$ and sends $x$ to the nonzero element $\epsilon$; its nonzeroness follows directly from the degree bound for a multiple of $\epsilon^2$. Hence $x\ne0$ in $C$, and the class of $x$ is a nonzero kernel element in the conormal complex $[J/J^2\to A_gdt]$. By the same presentation comparison, $H_1(\operatorname{NL}_{A_g/R})\ne0$. Definition F.6 shows that $R\to A_g$ is not smooth. This holds for every $g\notin\mathfrak m$, so Definition F.7 shows that $R\to A$ is not smooth at $\mathfrak m$.

Every other prime omits some $y_i$, proving that the smooth locus is exactly the displayed open $U$. In the notation of Definition D.1, this also gives $H_{A/R}=\mathfrak m$.

Finally the open cover $U=\bigcup_{i\geq1}D(y_i)$ has no finite subcover. Given a finite set $F$ of indices, choose $j\notin F$. Evaluation $y_j\mapsto1$ and $y_i\mapsto0$ for $i\ne j$ defines a surjection $A\to k$ with prime kernel $\mathfrak q_j$. This point lies in $D(y_j)\subset U$ but in none of the $D(y_i)$ with $i\in F$. Thus no finite subfamily covers $U$, and $U$ is not quasi-compact. $\square$

Lemma D.7: strictly standard base change

Adapted from the Stacks project, smoothing.tex, lines 294–309.

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. ∎

Lemma D.8: final solve

Adapted from the Stacks project, smoothing.tex, lines 311–327.

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 $A_{f_i}$. Explicitly, in

the equality $\sum_i f_it_i=1$ makes the opens $D(f_i)$ cover $\operatorname{Spec}(B)$. On $D(f_i)$ the defining equation uniquely eliminates $t_i$, giving

Each of these algebras is smooth over $R$, because $A_{f_i}$ is smooth. Smoothness is local on the target, so $B$ is smooth. The given relation $\sum_i f_i\lambda_i=1$ defines the map $B\to\Lambda$, $t_i\mapsto\lambda_i$, and the factorization. ∎

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 D.9: improve presentation

Adapted from the Stacks project, smoothing.tex, lines 341–439.

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 F.94) 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 $A=R[x_1,\ldots,x_n]/I$ with $I=(f_1,\ldots,f_m)$, and set $M=I/I^2$. Define $K$ by

Put $C=\operatorname{Sym}_A(M)$. Projection onto degree zero is the required retraction. The surjection $R[x_1,\ldots,x_n,y_1,\ldots,y_m]\to C$ sends $y_j$ to the class of $f_j$. Its kernel $J$ is generated by the $f_j$ and the linear forms $\sum h_jy_j$ whose coefficient classes $(h_j\bmod I)$ belong to $K$. This follows directly from the universal property of the symmetric algebra. In particular $C$ is of finite type over $A$.

Let $h\in R[x_1,\ldots,x_n]$ have image $a\in A$. Use the presentations

where $I'=(hx_0-1,I)$ and $J'=(hx_0-1,J)$. The principal-localization conormal calculation F.72 gives

Now assume that $A_a$ is a local complete intersection over $R$. The Koszul-regular local equation lists make $M_a$ locally free. Indeed, if $(g_i)$ is such a list, a conormal relation $\sum b_i g_i\in (g)^2$ can first be corrected by coefficients in $(g)$ to an exact relation. Its coefficients are then Koszul boundaries, by vanishing of first Koszul homology, and hence all belong to $(g)$. Thus the equation classes form a free conormal basis. Choose a finite principal cover trivializing $M_a$, using quasi-compactness of its spectrum. On each chart the kernel $K_a$ of the finite free surjection onto $M_a$ is a finite direct summand. Clear denominators of these finitely many chart generators and collect them in $K_a$; their cokernel vanishes on the cover, so they generate $K_a$ globally. Consequently $M_a$ is finitely presented. It is flat, because each ideal tensor map is injective on that same free cover and local detection proves its global injectivity. The finite-presentation projectivity theorem AG-CA-07, Theorem 5.3 therefore makes $M_a$ finite projective. The displayed defining sequence of $K$ splits after localization, so $K_a$ is finite projective and $K_a\oplus M_a\cong A_a^{\oplus m}$. Only at this point do we apply the symmetric-algebra conormal identity: the finite-projective quotient hypothesis in F.102 identifies the conormal module of $A_a[y_j]\to C_a$ with $K_a\otimes_{A_a}C_a$. Also $C_a=\operatorname{Sym}_{A_a}(M_a)$ is smooth over $A_a$ by F.128. On principal neighborhoods trivializing $K_a$ and $M_a$, the coordinates in F.102 make the kernel of $A_a[y_j]\to C_a$ an ideal generated by polynomial variables. That is a regular sequence, so its first Koszul homology vanishes by the exterior-complex cone calculation. Thus the exact conormal injectivity theorem F.130 applies to this particular presentation. The transitivity conormal sequence F.42 for the localized presentations, with left injectivity supplied by F.130, is therefore

The right module is projective, so this sequence splits and gives

This proves (1), including the asserted smoothness over $A_a$.

If $A_a$ is smooth over $R$, its conormal sequence for the original $n$ variables is split exact by the smoothness definition F.6. It makes $M_a$ finite projective and gives $M_a\oplus\Omega_{A_a/R}\cong A_a^{\oplus n}$ independently of (1). Now F.128 and F.102 apply: $\Omega_{C_a/A_a}=M_a\otimes_{A_a}C_a$ and $H_1(L_{C_a/A_a})=0$. Thus F.42 gives

The right module is projective, so this sequence splits. Combining it with the preceding split conormal identity gives

which proves (2). ∎

Proposition D.10: lift smooth

Adapted from the Stacks project, smoothing.tex, lines 447–557.

Let $R \to R_0$ be a surjective map of Noetherian rings with kernel $I$. This is the auxiliary lifting scope consumed by the Noetherian desingularization argument below.

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 F.129). 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 F.88). By Lemma D.9 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 F.55 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

By construction $C_0 = C/IC$. By Algebra, Lemma F.66 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 F.117 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 F.128). Picture

The identities of F.128 give

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 F.32 and F.83). The question is local on $\operatorname{Spec}(D)$, hence we may assume that $D$ is a relative global complete intersection (Algebra, Lemma F.87). 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 F.66 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 F.45). Thus we can replace $A$ by a localization to obtain the result we want. ∎

We know that any syntomic ring map $R \to A$ is locally a relative global complete intersection, see Algebra, Lemma F.87. The next lemma says that a vector bundle over $\operatorname{Spec}(A)$ is a relative global complete intersection.

Lemma D.11: syntomic complete intersection

Adapted from the Stacks project, smoothing.tex, lines 566–587.

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.,

flat over $R$ and all fibres of dimension $n - c$.

Proof. Apply Lemma D.9 to get $A \to C$. By Algebra, Lemma F.55 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 F.57 (the fibres are local complete intersections, so the lemma applies). ∎

Lemma D.12: smooth standard smooth

Adapted from the Stacks project, smoothing.tex, lines 589–672.

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.,

and $\det(\partial f_j/\partial x_i)_{i, j = 1, \ldots, c}$ is invertible in $B$.

Proof. Apply Lemma D.11 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

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

The $R$-algebra map $C \to B'$ has a retraction given by mapping $y_j$ to zero. We claim that the map

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

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

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 F.8. 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 F.9 and F.18. By the crit`ere de platitude par fibre (Algebra, Lemma F.36) we conclude that $R[z_1, \ldots, z_n] \to B'_g$ is flat. Finally, Algebra, Lemma F.16 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 F.41 we can write

with $\det(\partial g_j/\partial w_i)$ invertible in $B$. This proves the lemma. ∎

Lemma D.13: colimit standard smooth

Adapted from the Stacks project, smoothing.tex, lines 674–691.

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 F.90 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 D.12. Then the desired factorization is $A \to B \to C \to B \to \Lambda$. ∎

Lemma D.14: standard smooth include generators

Adapted from the Stacks project, smoothing.tex, lines 693–716.

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

and set $c = n + d$. ∎

Lemma D.15: compare standard

Adapted from the Stacks project, smoothing.tex, lines 718–850.

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

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

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

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

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

6. (f) (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 F.85. Part (b) is clear from Definition D.3.

Proof of (c). Choose a presentation $A = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ such that (D.3e1: elementary Jacobian-minor condition) and (D.3e2: elementary relation condition) hold. Choose $h \in R[x_1, \ldots, x_n]$ mapping to $a$. Then

Write $J = (x_0h - 1, f_1, \ldots, f_m)$. By (D.3e2: elementary relation condition) 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 F.55, we can choose a $g \in 1 + J$ such that

At this point (D.3e1: elementary Jacobian-minor condition) 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 (D.3s1: strict Jacobian-minor condition) and (D.3s2: strict relation condition) hold. Write $I = (f_1, \ldots, f_m)$. We already know that $A_a$ is smooth over $R$, see Lemma D.5. By Lemma D.4 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

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 (D.3s2: strict relation condition) 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

is multiplication by $a^{e_0}$ for some $e_0 \geq 1$. By Lemma D.4 we see (D.3s1: strict Jacobian-minor condition) 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 D.14. 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

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 $J = \ker(\Psi)$. 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 (D.3e1: elementary Jacobian-minor condition) and (D.3e2: elementary relation condition) hold for $a^e$ and all large enough $e$ as desired. ∎

The lifting problem

The goal in this section is to prove (Proposition D.18) 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 § presentations.

Lemma D.16: lift once

Adapted from the Stacks project, smoothing.tex, lines 1237–1305.

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

where $B$ is a smooth $R$-algebra and $J \subset IB$ is a finitely generated ideal.

Proof. Choose a factorization

with $\bar B$ standard smooth over $R/I$; this is possible by assumption and Lemma D.13. Write

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

and consider the map

We have

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

is smooth over $R$. Since smooth ring maps are formally smooth (Algebra, Proposition AG-CA-17, Theorem 3.1 and the split conormal definition of smoothness in this draft) 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 F.10). Thus $A' = B/J$ with $J \subset IB$ finitely generated (see Algebra, Lemma F.44). ∎

Lemma D.17: lift twice

Adapted from the Stacks project, smoothing.tex, lines 1307–1371.

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

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 AG-CA-07, Theorem 5.1) 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

with $C \to \Lambda$ given by $\varphi$ and $x_j \mapsto \lambda_j$. Choose a factorization

as in Lemma D.16. 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

for some $\xi_j \in B'$ and $\theta_i \in J' \subset IB'$. Hence we see that

because of the relations above and the fact that $I^2 = 0$. ∎

Proposition D.18: lift

Adapted from the Stacks project, smoothing.tex, lines 1373–1407.

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 F.90. By Lemma D.16 we may assume that $A = B/J$ with $B$ smooth over $R$ and $J \subset IB$ a finitely generated ideal. By Lemma D.17 we can find a commutative diagram

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. ∎

The lifting lemma

The next lemma constructs a finite presentation lifting the smooth locus.

Lemma D.19: lifting

Adapted from the Stacks project, smoothing.tex, lines 1420–1651.

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

with the following properties

1. (a) $D$ is of finite presentation,

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

3. (c) $R \to D$ 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. (d) $\bar C/\pi \bar C \to D/\pi D$ 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

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 F.136. Applying Lemma D.2 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. (3) $(\bar I_k)_{a_k} = (\bar I)_{a_k}$.

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

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

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

Here $j \in \{1, \ldots, m\}$, $k \in \{1, \ldots, r\}$, $\ell \in \{1, \ldots, m\}$, and

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

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

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

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

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 F.66. Hence $R \to D_k$ is smooth at $\mathfrak q_k$, see Algebra, Lemma F.45.

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 $(D_k)_{\mathfrak q_k}$. Reducing relations (4) modulo $\pi^2$ kills their $\pi^2g_{k,\ell}$ term. They therefore give $a_k\bar f_\ell=\sum_{j\in E_k}\bar h_{k,\ell}^j\bar f_j$ in $\bar I/\bar I^2$; this assertion is made in the reduced conormal module, not 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

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

is in $I_k + \pi^2R[x_1, \ldots, x_n]$. Computing modulo $\pi$ we have

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

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. ∎

The desingularization lemma

The next construction improves the singular ideal.

Lemma D.20: desingularize

Adapted from the Stacks project, smoothing.tex, lines 1664–1809.

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

and $0 \leq c \leq \min(n, m)$ such that (D.3s1: strict Jacobian-minor condition) holds for $\pi$ and such that

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

This implies that the constant term of $\partial f_j/\partial x_i$ is $r_{ji}$. Apply $\rho$ to (D.3s1: strict Jacobian-minor condition) for $\pi$ and we see that

for some $r_I \in R$. Thus we have

for some $u \in 1 + \pi^3R$. By Algebra, Lemma F.70 this implies there exists a $n \times c$ matrix $(s_{ik})$ such that

(Kronecker delta). We introduce auxiliary variables $v_1, \ldots, v_c, w_1, \ldots, w_n$ and we set

In the following we will use that

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

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 $Q/(h_1,\ldots,h_n)$, where

We set

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 D.5.

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 F.66 and F.45 (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 (D.20). 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 AG-CA-03, Theorem 6.1. As $R \to (B')_{\mathfrak q'}$ is flat we see that

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. ∎

Lemma D.21: desingularize strictly standard

Adapted from the Stacks project, smoothing.tex, lines 1811–1859.

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 D.20 to

and the image of $\pi$ in $D$. By Lemma D.7 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 D.20 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

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 F.29). ∎

Lemma D.22: desingularize lifting apply

Adapted from the Stacks project, smoothing.tex, lines 1861–1894.

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

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

Proof. Apply Lemma D.19 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 D.21 we can find a finitely presented $R$-algebra $B$ and factorizations $A \to B \to \Lambda$ and $D \to B \to \Lambda$ such that $H_{D/R}B\subset H_{B/R}$. To verify the assertion, first note that equality of the annihilators of $\pi$ and $\pi^2$ implies equality for every positive power: if $\pi^{r+1}x=0$, successive applications of that equality give $\pi x=0$. Thus the lifting lemma applies with $\pi^4$ in place of $\pi$.

At a prime not containing $\pi$, $R\to D$ is smooth; the singular-ideal inclusion then makes $R\to B$ smooth there too. Hence $B[1/\pi]$ is smooth. Let $\mathfrak l$ be a prime of $\Lambda$ containing $H_{B/R}\Lambda$, and let $\mathfrak b$ and $\mathfrak d$ be its contractions to $B$ and $D$. The preceding observation implies $\pi\in\mathfrak l$. The inclusion also implies $H_{D/R}\subset\mathfrak d$, so $D$ is not smooth at $\mathfrak d$. By the lifting lemma's property (c), the corresponding prime of $\bar C/\pi^4\bar C$ cannot come from a smooth prime of $\bar C$ over $R/\pi^8R$. Therefore every element of $H_{\bar C/(R/\pi^8R)}$ maps into $\mathfrak l/\pi^8\Lambda$. Intersect over all such $\mathfrak l$. This gives exactly the stated radical inclusion. ∎

Warmup: reduction to a base field

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 D.26.

Situation D.23: global

Adapted from the Stacks project, smoothing.tex, lines 1915–1918.

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

Let $R \to \Lambda$ be as in Situation D.23. We say PT holds for $R \to \Lambda$ if $\Lambda$ is a filtered colimit of smooth $R$-algebras.

Lemma D.24: product

Adapted from the Stacks project, smoothing.tex, lines 1925–1934.

Let $R_i \to \Lambda_i$, $i = 1, 2$ be as in Situation D.23. 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 $\Lambda_i=\operatorname*{colim}_{a\in I_i} B_{i,a}$ with $I_i$ filtered and every $B_{i,a}$ smooth over $R_i$. The category $I_1\times I_2$ is filtered: nonemptiness, a common successor for two objects, and an equalizer for two parallel arrows are constructed in the two coordinates. The $R_1\times R_2$-algebra $B_{1,a}\times B_{2,b}$ is smooth, because its spectrum is the disjoint union of the two corresponding smooth spectra. The natural map

is surjective by choosing representatives of the two components, and injective because any equality of pairs becomes an equality at successors in each coordinate. It is a ring isomorphism, proving the assertion. ∎

Lemma D.25: delocalize base

Adapted from the Stacks project, smoothing.tex, lines 1936–2000.

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 D.3) in $B$ over $R$.

Proof. We first apply Lemma D.12 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 D.14. 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

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

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

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

which divides $s$ as desired. ∎

Lemma D.26: reduce to field

Adapted from the Stacks project, smoothing.tex, lines 2002–2063.

If for every Situation D.23 where $R$ is a field PT holds, then PT holds in general.

Proof. Assume PT holds for any Situation D.23 where $R$ is a field. Let $R \to \Lambda$ be as in Situation D.23 arbitrary. Note that $R/I \to \Lambda/I\Lambda$ is another regular ring map of Noetherian rings, see More on Algebra, Lemma F.123. Consider the set of ideals

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 D.18 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 F.90.

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 F.89 and AG-CA-03, Proposition 2.2. By Lemma D.24 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 D.25 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 D.19 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 D.21 we can find $A \to B \to \Lambda$ with $B$ smooth over $R$ which finishes the proof. ∎

Local tricks

Situation D.27: local

Adapted from the Stacks project, smoothing.tex, lines 2078–2084.

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 D.27. 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 D.28: lift solution

Adapted from the Stacks project, smoothing.tex, lines 2096–2145.

Let $R \to A \to \Lambda \supset \mathfrak q$ be as in Situation D.27. 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

and

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

Then, if

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 D.22. 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 D.7. The induction resolves the situation modulo $\pi_1^8$; the already proved case $r=1$, applied to $\pi_1$, lifts this resolution to the original situation. ∎

Lemma D.29: delocalize weak

Adapted from the Stacks project, smoothing.tex, lines 2147–2227.

Let $R \to A \to \Lambda \supset \mathfrak q$ be as in Situation D.27. 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 D.8 we may assume that $C$ is smooth over $R_\mathfrak p$. By Lemma D.12 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 D.14. 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

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

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

which is smooth over $R$ by construction. Since $b$ does not map to an element of $\mathfrak q$, we win. ∎

Lemma D.30: delocalize height zero

Adapted from the Stacks project, smoothing.tex, lines 2229–2274.

Let $R \to A \to \Lambda \supset \mathfrak q$ be as in Situation D.27. 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 D.29 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

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. ∎

Separable residue fields

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

Lemma D.31: ogoma

Adapted from the Stacks project, smoothing.tex, lines 2286–2303.

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 $\ker(\pi : S^{-1}M \to S^{-1}M) = \ker(\pi^2 : S^{-1}M \to S^{-1}M)$ then there exists an $s \in S$ such that for any $n > 0$ we have $\ker(s^n\pi : M \to M) = \ker((s^n\pi)^2 : M \to M)$.

Proof. Let $K = \ker(\pi : M \to M)$ 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 $\pi^2m=0$ after localization, so $m\in K'$. The equality $sQ=0$ gives $sm\in K$, hence $s\pi m=0$, and therefore $s^n\pi m=0$ for every $n>0$. The reverse kernel inclusion is immediate. Thus this single $s$ works for every positive power. ∎

Lemma D.32: find sequence

Adapted from the Stacks project, smoothing.tex, lines 2305–2349.

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

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 AG-CA-12, Theorem 6.1. 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 AG-CA-12, Proposition 1.3. 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 D.31. By induction on $t$ this produces a sequence satisfying (1), (2), and (3). ∎

Lemma D.33: resolve special

Adapted from the Stacks project, smoothing.tex, lines 2351–2430.

Let $k \to A \to \Lambda \supset \mathfrak q$ be as in Situation D.27 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. If $d=\dim(\Lambda_\mathfrak q)=0$, the regular local target is its residue field. The separability assumption in the sense of Definition F.5, together with Lemma F.23, writes this residue field as a filtered colimit of smooth $k$-algebras. Since $A$ is finitely presented over $k$, Lemma F.90 factors $A\to\Lambda_\mathfrak q$ through one such algebra. Lemma D.30 delocalizes the resulting height-zero solution. Skip the positive-length parameter and sequence-lifting steps. The remaining argument treats $d>0$.

Set $d = \dim \Lambda_\mathfrak q$. Set $R = k[x_1, \ldots, x_d]$. Put $I=H_{A/k}\Lambda$, so $\mathfrak h_A=\sqrt I$. Since $\mathfrak q$ is minimal over $\sqrt I$, the radical of $I\Lambda_\mathfrak q$ is the maximal ideal of the Noetherian local ring $\Lambda_\mathfrak q$. Choose finitely many generators of that maximal ideal. Each has a power in $I\Lambda_\mathfrak q$; expanding monomials of sufficiently high total degree gives an integer $n>0$ with $\mathfrak q^n\Lambda_\mathfrak q\subset I\Lambda_\mathfrak q$. Choose generators $a_1, \ldots, a_r$ of $H_{A/k}$. Set

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 D.9 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 D.15. Now choose $\pi_1, \ldots, \pi_d \in \Lambda$ as in Lemma D.32 where $n = n$, $e = 8c$, $\mathfrak q = \mathfrak q$ and $I = H_{A/k}\Lambda$. Thus $\pi_i^n$ belongs to the actual ideal generated by the images of $a_j$, rather than merely to its radical. 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

Now we apply Lemma D.28 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

for $e = 8c$. By Lemma D.30 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 F.46. Hence

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 F.23 and Proposition D.18. Existence of the desired solution follows from Algebra, Lemma F.90. ∎

Inseparable residue fields

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

Lemma D.34: helper

Adapted from the Stacks project, smoothing.tex, lines 2444–2549.

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 F.49) 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

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

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

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 F.40. ∎

Lemma D.34A: Artinian denominator enlargement

Let $k$ have characteristic $p>0$. Let $A\to B$ be a flat local map of Noetherian Artinian local $k$-algebras, with $\mathfrak m_AB=\mathfrak m_B$, and suppose $\mathfrak m_B^n=0$. Write $F$ and $K$ for their residue fields. For finitely many units $b_j\in B$, there is a factorization $A\to A'\to B$ by local maps such that $A\to A'$ is essentially smooth, $A'\to B$ is faithfully flat, $\mathfrak m_{A'}B=\mathfrak m_B$, and a $p$-power of each $b_j$ belongs to $A'$. If a residue $\bar b_j$ is transcendental over the current residue field, the construction adjoins $b_j$ itself. The residue extension $F'/F$ is a finite tower of simple transcendental and finite separable extensions.

Proof. Treat one unit $b$ at a time. If its residue $\beta$ is transcendental over $F$, use $A'=A[T]_{\mathfrak m_AA[T]}$ with $T\mapsto b$. Its residue field is $F(\beta)$: every nonzero polynomial in $F[T]$ is inverted. Each inverted element maps to a unit of $B$, so the map is defined and local. This is an essentially smooth Artinian local ring, with $\mathfrak m_{A'}=\mathfrak m_AA'$.

If $\beta$ is algebraic over $F$, choose a $p$-power $q_0$ for which $\beta^{q_0}$ is separable over $F$, by the inseparable exponent calculation already proved. Lift the monic separable minimal polynomial of $\beta^{q_0}$ from $F[T]$ to $h\in A[T]$ and put $A'=A[T]/(h)$. It is finite free over $A$. Its quotient modulo $\mathfrak m_A$ is the field $F(\beta^{q_0})$, and $\mathfrak m_AA'$ is nilpotent, so $A'$ is Artinian local with this maximal ideal. Bézout for the relatively prime residue polynomials $\bar h,\bar h'$ makes $h'$ a unit modulo $\mathfrak m_AA'$; lifting its inverse across a nilpotent ideal makes it a unit in $A'$. The square Jacobian calculation, AG-CA-17, Theorem 4.1 and Section 6, therefore makes $A'$ étale. The simple-root Hensel proof in AG-CA-19, Theorem 5.1, lifts the specified residue root in the Artinian local ring $B$ and gives $A'\to B$. This is exactly the finite étale lifting construction in F.54, written here in its consumed form. If $\beta\in F$, use instead $A'=A$, $q_0=1$.

In both cases $A'$ is flat over $A$, and $A'/\mathfrak m_AA'=F'$, its residue field. A free $A$-resolution of $F$ therefore becomes a free $A'$-resolution of $F'$ after tensoring. Tensor it further with $B$. Flatness of $B$ over $A$ gives

The Noetherian local criterion, AG-CA-08, Theorem 4.2, applied with finite $B$-module $B$, proves that $A'\to B$ is flat. It is local and hence faithfully flat, so in particular injective. Its maximal ideal still generates $\mathfrak m_B$.

In the algebraic case choose $x\in A'$ with residue $\beta^{-q_0}$. It is a unit. Then $xb^{q_0}-1\in\mathfrak m_B$. For a $p$-power $q_1\ge n$, Frobenius gives $(xb^{q_0})^{q_1}=1$, whence $b^{q_0q_1}=x^{-q_1}\in A'$. In the transcendental case $b$ is already present, so any $p$-power may be used. Repeating the finite construction proves every assertion.

We also record its differential effect, used immediately below. For a simple transcendental residue extension the polynomial presentation and localization give $\Omega_{F(\beta)/k}=(\Omega_{F/k}\otimes_FF(\beta))\oplus F(\beta)\,d\beta$. A finite separable residue extension gives the same old differential space after tensoring: its monic separable presentation has invertible derivative, so its relative conormal complex is acyclic. The field Jacobi–Zariski sequence F.131 likewise gives in both cases

These are natural identifications, and composition gives them for the finite tower. Thus the image of this first homology in $H_1(L_{K/k})$ is unchanged. ∎

Lemma D.35: solution modulo

Adapted from the Stacks project, smoothing.tex, lines 2551–2621.

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

where the first arrow is essentially smooth and the second is flat,

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

Proof. Put $L=\Lambda_{\mathfrak q}$, with maximal ideal $\mathfrak m$ and residue field $K$, and put $B=L/\mathfrak m^n$. Geometric regularity and F.133 give an injection $H_1(L_{K/k})\hookrightarrow\mathfrak m/\mathfrak m^2$, so this first homology is finite-dimensional. By D.34 choose an Artinian local subring $A\subset B$ containing $E$, essentially of finite type over $k$, such that $A\to B$ is faithfully flat and $\mathfrak m_AB=\mathfrak m_B$. Write $F$ for its residue field.

Here one must ensure global representatives; an arbitrary such $A$ does not supply them. Choose a basis of the finite-dimensional space $\Omega_{F/k}$ consisting of differentials $d\theta_j$ of elements of $F$, and choose lifts $\widetilde\theta_j\in A$. Since $L\twoheadrightarrow B$, represent each lift by a fraction $a_j/b_j$ with $a_j\in\Lambda$ and $b_j\in\Lambda\setminus\mathfrak q$.

By the field left injection F.131, $H_1(L_{F/k})\otimes_FK$ injects into $H_1(L_{K/k})$. Let $V$ be its image in $\mathfrak m/\mathfrak m^2$ under the preceding injection. If $n\ge2$, the equality $\mathfrak m_AB=\mathfrak m_B$ shows that the images of elements of $\mathfrak m_A$ span $\mathfrak m/\mathfrak m^2$ over $K$. Choose finitely many $\widetilde\eta_i\in\mathfrak m_A$ whose images form a complement to $V$. Lift each to a fraction $c_i/d_i\in L$ with $c_i\in\mathfrak q$ and $d_i\in\Lambda\setminus\mathfrak q$. If $n=1$, choose instead elements $c_i\in\mathfrak q$ whose classes form this complement; their images in $B=K$ are zero, hence already belong to $A$. Use $d_i=1$ in that case.

Apply D.34A to this finite list of denominators, as units of $B$. It enlarges $A$ to an Artinian local subring $A'\subset B$ by an essentially smooth map, retains faithful flatness to $B$ and the maximal-ideal equality, and puts $b_j^{q_j},d_i^{r_i}$ in $A'$ for suitable $p$-powers. Consequently the global elements

have images in $A'$: the first and third are the original fractional lifts multiplied by the indicated powers. Let $F'$ be its residue field. During the enlargement record also the global denominators whose residues were adjoined transcendently; those denominators themselves have images in $A'$.

The differentials of the residues of the first two displayed global elements, together with those recorded transcendental denominators, span $\Omega_{F'/k}$. Indeed the quotient identity $\theta_j=(a_jb_j^{q_j-1})/b_j^{q_j}$ expresses every old $d\theta_j$ in terms of their differentials. D.34A shows that only the recorded transcendental differentials must be added; a finite separable step adds none. Select a basis from this finite spanning list, and denote its global representatives by $\lambda_1,\ldots,\lambda_t\in\Lambda$. All their classes lie in $A'$. This argument does not assert that $d(b_j^{q_j})=0$ inside $\Omega_{F'/k}$: its root need not lie in $F'$.

The natural first-homology calculation at the end of D.34A gives $H_1(L_{F'/k})=H_1(L_{F/k})\otimes_FF'$. Hence its image in $\mathfrak m/\mathfrak m^2$ is still $V$. Define $P'=k[y_1,\ldots,y_t]\to\Lambda$ using these $\lambda_j$, and let $\mathfrak p'$ be the inverse image of $\mathfrak q$. Lemma F.109 proves that $P'_{\mathfrak p'}\to L$ is flat and that $L/\mathfrak p'L$ is regular. More precisely, the proof of that lemma identifies the image of $(\mathfrak p'/\mathfrak p'^2)\otimes K$ in $\mathfrak m/\mathfrak m^2$ with $V$. To check this identification, write $K_0=\kappa(\mathfrak p')$. The chosen differential basis makes $F'/K_0$ finite separable, by F.98 and the field separability calculation. Its acyclic relative conormal complex identifies $H_1(L_{K_0/k})\otimes_{K_0}K$ with $H_1(L_{F'/k})\otimes_{F'}K$. The natural conormal square in F.109 then gives exactly the image $V$.

Set $\lambda_{t+i}=c_id_i^{r_i-1}$. These are global, have classes in $A'$, and their cotangent classes are those of the original $\widetilde\eta_i$ multiplied by the nonzero residues $d_i^{r_i}$. They remain a complement to $V$, so they map to a regular system of parameters of $L/\mathfrak p'L$. Let $P=k[y_1,\ldots,y_m]$ append these variables and let $\mathfrak p$ be its inverse image of $\mathfrak q$. The local ring $P_{\mathfrak p}$ is obtained from $P'_{\mathfrak p'}[y_{t+1},\ldots,y_m]$ at the prime generated by $\mathfrak p'$ and the appended variables, since their residues are zero. Its regular parameters consist of the old parameters and these variables. Their images form a regular parameter list in $L$: the old list is regular by F.109 and the new list is regular on its regular quotient. The regular-base flatness lemma F.46 gives flatness of $P_{\mathfrak p}\to L$. Their classes form a basis of $\mathfrak m/\mathfrak m^2$, so Nakayama gives $\mathfrak pL=\mathfrak m$. This proves property (1).

Every global generator has its class in $A'$, so the polynomial map to $B$ factors through $A'$. The elements outside $\mathfrak p$ map to units of this local ring. Since $\mathfrak p^n B=\mathfrak m^nB=0$ and $A'\hookrightarrow B$, the factorization descends to

Its composite is flat by the flatness just proved and base change. Faithful-flat descent of module flatness, F.48, makes the first map flat. Also $\mathfrak pA'=\mathfrak m_{A'}$: both ideals become $\mathfrak m_B$ in $B$, and faithful flatness detects equality of ideals. The residue extension $F'/\kappa(\mathfrak p)=F'/K_0$ is finite separable as above. The Artinian length-filtration proof F.40 therefore makes $A'$ finite over $P_{\mathfrak p}/\mathfrak p^n$. The source is Noetherian, so this is a finitely presented algebra. Lemma F.16 now proves that the first map is étale. In particular it is essentially smooth. Taking $D=A'$ proves properties (2) and (3), since $A\subset A'$ retains $E$. ∎

Lemma D.36: enlarge solution modulo

Adapted from the Stacks project, smoothing.tex, lines 2623–2669.

Let $\varphi : k[y_1, \ldots, y_m] \to \Lambda$, $n$, $\mathfrak q$, $\mathfrak p$ and

be as in Lemma D.35. Then for any $\lambda \in \Lambda \setminus \mathfrak q$ there exists an integer $q > 0$ and a factorization

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. Put $B=\Lambda_{\mathfrak q}/\mathfrak q^n\Lambda_{\mathfrak q}$. The hypotheses of D.35 make $D\to B$ flat and local, hence faithfully flat. Its maximal ideal generates that of $B$: the maximal ideal of the polynomial Artinian source already generates it by property (1), and its image in $D$ is the maximal ideal because the first map in property (2) is local and essentially smooth. Its closed fibre is a localization of a smooth algebra over a field, hence regular by AG-CA-18, Theorem 2.1. It is Artinian local, so its regular dimension is zero; its cotangent space is zero and Nakayama makes its maximal ideal zero. Thus that fibre is a field, which proves the asserted maximal-ideal equality. The equality also holds directly in the étale construction of D.35. Apply D.34A to the unit image of $\lambda$. It gives the required essentially smooth Artinian extension $D\to D'$, flat final map, and a $p$-power $q$ with $\lambda^q\in D'$. Composing with the original polynomial source supplies the stated factorization. ∎

Lemma D.37: resolve general

Adapted from the Stacks project, smoothing.tex, lines 2671–3098.

Let $k \to A \to \Lambda \supset \mathfrak q$ be as in Situation D.27 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. 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 $H_{A/k}$.

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

4. 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 D.9. We also obtain a section $C \to B$.

6. 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 $n=\max\{N+dc,\ d(e-1)+1\}$. This choice is made before defining $E$, $D$, and all subsequent factorizations.

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

9. 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

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 $\mathfrak p \Lambda_\mathfrak q = \mathfrak q \Lambda_\mathfrak q$.

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

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

3. If necessary modify the choice of $\pi_i$ such that for $i = 1, \ldots, d$ we have

1. 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 $i = 1, \ldots, d$

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

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

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

5. 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 $C \to B$.

6. 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

by sending $x_i$ to $\gamma_i = \pi_i t_i$.

1. It suffices to resolve

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

2. It suffices to resolve

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

2. It suffices to resolve

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

2. It suffices to resolve

1. It suffices to resolve

1. It suffices to resolve

1. 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

where the second arrow sends $t_i$ to $\delta_i$ and induces the given homomorphism $D' \to \Lambda_\mathfrak q/\mathfrak q^n\Lambda_\mathfrak q$.

1. 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 (D.37 / power). This is possible as $\mathfrak q$ is minimal over $\mathfrak h_A = \sqrt{H_{A/k}\Lambda}$.

Ad (D.37 / strictly standard). 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 D.9 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 D.15.

Ad (D.37 / NP). This follows by applying Lemma D.35.

Ad (D.37 / choose pii). 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 AG-CA-11, Theorem 5.1. 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 (D.37 / modify pii). By Algebra, Lemma AG-CA-12, Theorem 6.1 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 F.75. 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 AG-CA-12, Proposition 1.3. Hence we see that

Thus we get

after replacing $\pi_{t + 1}$ by $s\pi_{t + 1}$ for some $s \in S$ by Lemma D.31. By induction on $t$ this produces the desired sequence.

Ad (D.37 / choose deltai). 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 (D.37 / choose deltai) 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 (D.37 / power). 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 AG-CA-08, Theorem 2.2. 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 $\Lambda$. Here are the exact denominator-clearing details. Let $u_j=\pi_{t+1}^N(c_j+c'_j)$ in $S^{-1}\Lambda$. Choose one $s_0\in S$ with $s_0^{2N}u_j=\widetilde\mu_j\in\Lambda$ for every $j$. The error $r=(s_0\pi_{t+1})^{2N}-\sum_j a_j\widetilde\mu_j$ vanishes in $S^{-1}\Lambda$, so some $v\in S$ annihilates $r$. Set $s=s_0v$ and $\mu_{t+1j}=v^{2N}\widetilde\mu_j$. Then the denominator identity still holds and the relation is exact because its error is $v^{2N}r=0$. Its residue is $s^{2N}\pi_{t+1}^Nd_j$, since $\pi_{t+1}^Nc'_j$ vanishes modulo $\mathfrak q^n$. We may further replace $s$ by a power and enlarge $D'$ such that $s$ maps to an element of $D'$. With these choices $\mu_{t + 1j}$ maps to $s^{2N}\pi_{t+1}^N d_j$ which is an element of $D'$. 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 AG-CA-12, Theorem 6.1. 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 AG-CA-12, Proposition 1.3. Thus we get

Hence we may apply Lemma D.31 to find an $s' \in S$ such that

for any $q > 0$. By Lemma D.36 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 (D.37 / first solve). 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

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 (D.37 / second resolve). If $d=0$, the parameter lists and the ideal $I$ are empty, and this step does nothing; the localized regular target is a field. For $d>0$, this follows on applying Lemma D.28 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 D.7. The other assumption of Lemma D.28 holds by steps (D.37 / modify pii) and (D.37 / choose deltai).

Ad (D.37 / third resolve). Apply Lemma D.30 to the situation in (D.37 / second resolve). 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 (D.37 / fifth resolve). 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

Then $C \otimes_{k[x_1, \ldots, x_d]} (R/I)_\mathfrak r \to T_\mathfrak r \to \Lambda_\mathfrak q/I\Lambda_\mathfrak q$ is a solution to the situation in (D.37 / third resolve).

Ad (D.37 / sixth resolve). Our corrected 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 (D.37 / fifth resolve) then we can take $T/JT$ as the solution for (D.37 / sixth resolve).

Ad (D.37 / seventh resolve). This is true because we have a section $C \to B$ in the category of $R$-algebras.

Ad (D.37 / eighth resolve). 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

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 (D.37 / done). 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

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 (D.37 / map B Lambda). 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 $D'[t_1, \ldots, t_d]$ and everything is clear. ∎

The main theorem

In this section we wrap up the discussion.

Theorem D.38: popescu

Adapted from the Stacks project, smoothing.tex, lines 3112–3135.

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

Proof. By Lemma D.26 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 D.8. 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 D.27). If the characteristic of $k$ is zero, this follows from Lemma D.33. If the characteristic of $k$ is $p > 0$, this follows from Lemma D.37. ∎

5. Foundation statements and their free-source proofs

All resolution comparisons, balance of Tor and connecting exact sequences used in the following flatness arguments are proved for arbitrary modules in the earlier AG-RG-S01, Lemma 1.A. This binds their homological inputs to a written programme proof before the approximation argument.

These proofs are included as mathematical text, rather than cited as substitutes for proofs. Their transitive dependencies are checked against the earlier programme locators and the open-obligation register in §6. Inclusion of a proof does not automatically close its unproved inputs.

Cotangent-complex context. For a polynomial presentation $P=R[X_s]\twoheadrightarrow S$ with kernel $I$, write $\operatorname{NL}(P\to S)$ for the two-term complex $[I/I^2\to\bigoplus_sS\,dX_s]$ in degrees $-1,0$. The differential takes the class of $f$ to $df$. The canonical presentation is $R[X_s\mid s\in S]\to S$, $X_s\mapsto s$. For a commuting square of ring maps $R\to R'$, $S\to S'$, a presentation map is a map of these polynomial algebras lifting $S\to S'$. The full comparison and homotopy calculations below show independence of the polynomial presentation. Localization is likewise the displayed two-term localization proved at the corresponding foundation lemma, not an import of the whole cotangent chapter.

The inseparable exponent calculation. Let $\alpha$ be algebraic over a field $F$ of characteristic $p$. Its irreducible polynomial is $f(T)=g(T^{p^s})$ for a maximal $s\geq0$. Such $s$ exists because $f$ has finite positive degree. Maximality gives $g'\ne0$. Any factorization of $g$ would give a factorization of $f$, so $g$ is irreducible, and its relatively prime derivative makes it separable. The element $\alpha^{p^s}$ is a root of $g$, hence is separable over $F$. This proves the field-theoretic exponent choice used in the enlargement argument.

Dimension notation. Krull dimension is the supremum of the lengths of finite chains of prime ideals; the height of a prime is the dimension of its localization. A system of parameters of a Noetherian local ring of dimension $d$ is a list of $d$ elements generating an ideal whose radical is the maximal ideal. Their existence is AG-CA-11, Theorem 2.1. Minimal primes, irreducible components and the dimension of polynomial local rings are supplied respectively by AG-CA-01, Theorems 4.1–4.3, AG-CA-09, Theorem 4.3, and AG-CA-11, Theorem 6.1. This paragraph fixes notation; it does not assert an additional dimension theorem. The separate dimension-in-a-neighbourhood and codimension arguments consumed below remain separately registered until matched or proved.

Definition F.1: lci

Adapted from the Stacks project, algebra.tex, lines 37874–37881.

A ring map $R \to S$ is called syntomic, or we say $S$ is a flat local complete intersection over $R$ if it is flat, of finite presentation, and if all of its fibre rings $S \otimes_R \kappa(\mathfrak p)$ are local complete intersections, see Definition F.2.

Definition F.2: lci field

Adapted from the Stacks project, algebra.tex, lines 37249–37264.

Let $k$ be a field. Let $S$ be a finite type $k$-algebra.

1. We say that $S$ is a global complete intersection over $k$ if there exists a presentation $S = k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ such that $\dim(S) = n - c$.

2. We say that $S$ is a local complete intersection over $k$ if there exists a covering $\operatorname{Spec}(S) = \bigcup D(g_i)$ such that each of the rings $S_{g_i}$ is a global complete intersection over $k$.

We will also use the convention that the zero ring is a global complete intersection over $k$.

Definition F.3: lci local ring

Adapted from the Stacks project, algebra.tex, lines 37446–37458.

Let $k$ be a field. Let $S$ be a local $k$-algebra essentially of finite type over $k$. We say $S$ is a complete intersection (over $k$) if there exists a local $k$-algebra $R$ and elements $f_1, \ldots, f_c \in \mathfrak m_R$ such that

1. $R$ is essentially of finite type over $k$,

2. $R$ is a regular local ring,

3. $f_1, \ldots, f_c$ form a regular sequence in $R$, and

4. $S \cong R/(f_1, \ldots, f_c)$ as $k$-algebras.

Definition F.4: relative global complete intersection

Adapted from the Stacks project, algebra.tex, lines 37943–37951.

Let $R \to S$ be a ring map. We say that $R \to S$ is a relative global complete intersection if there exists a presentation $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ and every nonempty fibre of $\operatorname{Spec}(S) \to \operatorname{Spec}(R)$ has dimension $n - c$. We will say ``let $S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_c)$ be a relative global complete intersection'' to indicate this situation.

Definition F.5: separable field extension

Adapted from the Stacks project, algebra.tex, lines 10135–10146.

Let $K/k$ be a field extension.

1. We say $K$ is separably generated over $k$ if there exists a transcendence basis $\{x_i; i \in I\}$ of $K/k$ such that the extension $K/k(x_i; i \in I)$ is a separable algebraic extension.

2. We say $K$ is separable over $k$ if for every subextension $k \subset K' \subset K$ with $K'$ finitely generated over $k$, the extension $K'/k$ is separably generated.

Definition F.6: smooth

Adapted from the Stacks project, algebra.tex, lines 38526–38533.

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

Definition F.7: smooth at prime

Adapted from the Stacks project, algebra.tex, lines 38953–38960.

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.

Definition F.8: unramified

Adapted from the Stacks project, algebra.tex, lines 42897–42912.

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 F.9: CM over regular flat

Adapted from the Stacks project, algebra.tex, lines 34036–34074.

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

the first equality by Lemma AG-CA-12, Lemma 5.4 and Theorem 5.5 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 F.80. 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 AG-CA-12, Theorem 4.2 and Corollary 4.3 and $S/xS$ is Cohen-Macaulay with $\dim(S/xS) = \dim(S) - 1$. By (1) and Lemma AG-CA-12, Theorem 6.1 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 AG-CA-08, Corollary 4.4 and equation (6) and the remark following it. ∎

Lemma F.10: NAK

Adapted from the Stacks project, algebra.tex, lines 3732–3813.

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 (F.10 / nakayama). 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 F.70 (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 F.34 an element of $1 + \text{rad}(R)$ is an invertible element of $R$. Hence we see that (F.10 / nakayama) 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. ∎

Lemma F.11: NL homotopy

Adapted from the Stacks project, algebra.tex, lines 36700–36815.

Let $R\to R'$ and $S\to S'$ be a commutative square of ring maps with structure maps $R\to S$ and $R'\to S'$. Let $\alpha : P \to S$ and $\alpha' : P' \to S'$ be presentations.

1. There exists a morphism of presentations from $\alpha$ to $\alpha'$.

2. Any two morphisms of presentations induce homotopic morphisms of complexes $\operatorname{NL}(\alpha) \to \operatorname{NL}(\alpha')$.

3. The construction is compatible with compositions of morphisms of presentations (see proof for exact statement).

4. If $R \to R'$ and $S \to S'$ are isomorphisms, then for any map $\varphi$ of presentations from $\alpha$ to $\alpha'$ the induced map $\operatorname{NL}(\alpha) \to \operatorname{NL}(\alpha')$ is a homotopy equivalence and a quasi-isomorphism.

In particular, comparing $\alpha$ to the canonical presentation described in the cotangent-complex context above we conclude there is a quasi-isomorphism $\operatorname{NL}(\alpha) \to \NL_{S/R}$ well defined up to homotopy and compatible with all functorialities (up to homotopy).

Proof. Since $P$ is a polynomial algebra over $R$ we can write $P = R[x_a, a \in A]$ for some set $A$. As $\alpha'$ is surjective, we can choose for every $a \in A$ an element $f_a \in P'$ such that $\alpha'(f_a) = \phi(\alpha(x_a))$. Let $\varphi : P = R[x_a, a \in A] \to P'$ be the unique $R$-algebra map such that $\varphi(x_a) = f_a$. This gives the morphism in (1).

Let $\varphi$ and $\varphi'$ be morphisms of presentations from $\alpha$ to $\alpha'$. Let $I = \ker(\alpha)$ and $I' = \ker(\alpha')$. We have to construct the diagonal map $h$ in the diagram

where the vertical maps are induced by $\varphi$ and $\varphi'$, and $h$ must satisfy

Consider the map $\varphi - \varphi' : P \to P'$. Since both $\varphi$ and $\varphi'$ are compatible with $\alpha$ and $\alpha'$ we obtain $\varphi - \varphi' : P \to I'$. This implies that $\varphi, \varphi' : P \to P'$ induce the same $P$-module structure on $I'/(I')^2$, since $\varphi(p)i' - \varphi'(p)i' = (\varphi - \varphi')(p)i' \in (I')^2$. Also $\varphi - \varphi'$ is $R$-linear and

Hence the induced map $D : P \to I'/(I')^2$ is an $R$-derivation. Thus we obtain a canonical map $h : \Omega_{P/R} \otimes_P S \to I'/(I')^2$ such that $D = h \circ \text{d}$. For $f\in I$, the definition of the universal derivation gives $h(df)=D(f)=\varphi(f)-\varphi'(f)\bmod (I')^2$, so the degree-one difference is $h\circ d$. On a basis element $dx_a$ of $\Omega_{P/R}\otimes_PS$, one has $d(h(dx_a))=d(\varphi(x_a)-\varphi'(x_a))$, exactly the difference of the degree-zero maps. These two identities verify the homotopy on generators, hence on the whole complexes.

Suppose that we have a commutative diagram

and that

1. $\alpha : P \to S$,

2. $\alpha' : P' \to S'$, and

3. $\alpha'' : P'' \to S''$

are presentations. Suppose that

1. $\varphi : P \to P'$ is a morphism of presentations from $\alpha$ to $\alpha'$ and

2. $\varphi' : P' \to P''$ is a morphism of presentations from $\alpha'$ to $\alpha''$.

Then it is immediate that $\varphi' \circ \varphi : P \to P''$ is a morphism of presentations from $\alpha$ to $\alpha''$ and that the induced map $\operatorname{NL}(\alpha) \to \operatorname{NL}(\alpha'')$ of naive cotangent complexes is the composition of the maps $\operatorname{NL}(\alpha) \to \operatorname{NL}(\alpha')$ and $\operatorname{NL}(\alpha') \to \operatorname{NL}(\alpha'')$ induced by $\varphi$ and $\varphi'$.

In the simple case of complexes with 2 terms a quasi-isomorphism is just a map that induces an isomorphism on both the cokernel and the kernel of the maps between the terms. Note that homotopic maps of 2 term complexes (as explained above) define the same maps on kernel and cokernel. Hence if $\varphi$ is a map from a presentation $\alpha$ of $S$ over $R$ to itself, then the induced map $\operatorname{NL}(\alpha) \to \operatorname{NL}(\alpha)$ is a quasi-isomorphism being homotopic to the identity by part (2). To prove (4) in full generality, consider a morphism $\varphi'$ from $\alpha'$ to $\alpha$ which exists by (1). The compositions $\operatorname{NL}(\alpha) \to \operatorname{NL}(\alpha') \to \operatorname{NL}(\alpha)$ and $\operatorname{NL}(\alpha') \to \operatorname{NL}(\alpha) \to \operatorname{NL}(\alpha')$ are homotopic to the identity maps by (2) and (3), hence these maps are homotopy equivalences by definition. It follows formally that both maps $\operatorname{NL}(\alpha) \to \operatorname{NL}(\alpha')$ and $\operatorname{NL}(\alpha') \to \operatorname{NL}(\alpha)$ are quasi-isomorphisms. Explicitly, if $uv$ and $vu$ are homotopic to the respective identities, their maps on kernel and cokernel are identities, since a homotopy contributes $hd$ in degree one and $dh$ in degree zero. The maps induced by $u$ and $v$ on both groups are therefore inverse isomorphisms. ∎

Lemma F.12: Noetherian field extension

Adapted from the Stacks project, algebra.tex, lines 6331–6347.

If $A$ is a Noetherian $k$-algebra and $K/k$ is a finitely generated field extension, then $A\otimes_kK$ is Noetherian.

Proof. Choose finitely many field generators and let $B\subset K$ be the $k$-algebra they generate. Its fraction field is $K$, so $K=(B\setminus\{0\})^{-1}B$. A finite polynomial presentation of $B$ identifies $A\otimes_kB$ with a quotient of a polynomial algebra in finitely many variables over $A$. Hilbert basis AG-CA-03, Theorem 2.1, makes it Noetherian. Tensor and the fraction universal property identify $A\otimes_kK$ with its localization at the images of the nonzero elements of $B$. The same Noetherian permanence proof in that earlier theorem makes this localization Noetherian. In the special case of a finite field extension, one can alternatively use its finite field basis to see a finite free $A$-module, whose ideals are finite submodules. This proves the full finitely generated form consumed in the regularity tests. ∎

Lemma F.13: another variant local criterion flatness

Adapted from the Stacks project, algebra.tex, lines 24942–24990.

For a local square of Noetherian rings $R\to S$, $R'\to S'$ with $S'$ a localization of $S\otimes_RR'$, a finite $S$-module $M$, and a proper ideal $I\subset R$, assume $M/IM$ is flat over $R/I$ and the map $\operatorname{Tor}_1^R(M,R/I)\to\operatorname{Tor}_1^{R'}(M\otimes_SS',R'/IR')$ is zero. Then $M\otimes_SS'$ is $R'$-flat.

Proof. Write a free presentation $0\to K\to F\to M\to0$ over $R$, allowing an infinite free $F$. Put $T=\operatorname{Tor}_1^R(M,R/I)$ and let $\bar K$ be the image of $K/IK$ in $F/IF$. The two sequences $0\to T\to K/IK\to\bar K\to0$ and $0\to\bar K\to F/IF\to M/IM\to0$ are exact. The assumed flatness of the last quotient makes $\bar K$ flat by AG-CA-07, Theorem 6.1. Tensor these sequences with any $R/I$-module $N$; their quotient flatness preserves their left injections and identifies $$\operatorname{Tor}_1^R(M,N)=T\otimes_{R/I}N.$$ This identity follows from the same free-presentation kernel formula for Tor, so does not assume that $K$ is flat.

After base change to $R'$, let $K'$ be the image of $K\otimes_RR'$ in $F\otimes_RR'$. The map onto $K'$ is surjective. Tensor it with $R'/IR'$; every cycle in $\ker(K'\otimes R'/IR'\to F\otimes R'/IR')$ lifts to an element of $K\otimes_RR'/IR'$, and that lift is automatically a cycle because the two free-module targets are identical. Therefore $\operatorname{Tor}_1^R(M,R'/IR')\to\operatorname{Tor}_1^{R'}(M\otimes_RR',R'/IR')$ is surjective. By the preceding identity its source is $T\otimes_RR'$. Localizing at the stipulated prime in the algebra commutes with this kernel calculation by exact localization. Thus $T\otimes_RR'$ surjects onto the Tor group of $M'=M\otimes_SS'$. The assumed zero map annihilates this surjection, so that Tor group is zero.

Base change of flatness makes $M'/IR'M'$ flat over $R'/IR'$. AG-CA-08, Corollary 4.4, is the proved proper-ideal local criterion under these Noetherian local and finite-over-$S'$ hypotheses; it now makes $M'$ flat over $R'$. This proves the transition criterion rather than invoking an unproved change-of-rings spectral sequence. ∎

Lemma F.14: base change relative global complete intersection

Adapted from the Stacks project, algebra.tex, lines 38076–38106.

Relative global complete-intersection presentations are preserved by base change, principal localization, and replacing the base by a localization already inverted in the algebra.

Proof. First record the required field-dimension argument. For a finite-type algebra over a field and any extension field, dimension is unchanged. Quotient by its finitely many minimal primes; their intersection is nilpotent by AG-CA-03, Proposition 2.2. This remains nilpotent after field extension. For each resulting domain, Noether normalization AG-CA-09, Corollary 3.2, gives an injective finite polynomial subalgebra. Field tensor is exact, so the injection and finiteness persist. Dimension is preserved by an injective integral map, AG-CA-05, Theorem 5.1. Thus each component and their maximum retain dimension. This proof applies to an arbitrary, possibly infinite, field extension.

The fibre after any base change is a field extension of the fibre over the contracted base prime. Consequently its dimension stays $n-c$. For localization, each original fibre is equidimensional of dimension $n-c$: the minimal-prime height calculation in F.77, applied just to the polynomial algebra over the fibre field, proves this without any base Noetherianity. A nonempty principal open in an irreducible finite-type field scheme has unchanged dimension, since its coordinate domain has the same fraction field and AG-CA-09, Theorem 4.2, identifies dimension with transcendence degree. Therefore every nonempty localized fibre still has dimension $n-c$. The presentation $S_g=R[X,Z]/(f_1,\ldots,f_c,hZ-1)$ has one additional variable and equation, hence the same expected dimension. Finally if a base element is already inverted in $S$, tensoring with that base localization gives back $S$, proving the third assertion. ∎

Lemma F.15: change base NL

Adapted from the Stacks project, algebra.tex, lines 36963–36981.

For a flat base change $R\to R'$, the two-term conormal complex of a polynomial presentation $P\twoheadrightarrow S$ base changes to the corresponding complex for $P\otimes_RR'\twoheadrightarrow S\otimes_RR'$.

Proof. Let $I$ be the presentation kernel. Flatness identifies the new kernel with $I\otimes_RR'$, since it preserves the exact sequence $0\to I\to P\to S\to0$. Its square is the image of $I^2\otimes_RR'$, by writing products of generators. The cokernel description therefore identifies the conormal module with $(I/I^2)\otimes_RR'$. The differential module of the polynomial ring is free on the same $dX_s$ before and after base change. Both differentials send the class of a polynomial $f$ to $df$, so these are an isomorphism of complexes. Comparison of polynomial presentations and its explicit homotopies, proved above, give the canonical homotopy equivalence. ∎

Lemma F.16: characterize etale

Adapted from the Stacks project, algebra.tex, lines 40942–40975.

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. Put $k=\kappa(\mathfrak p)$ and $L=\kappa(\mathfrak q)$. The local fibre ring is $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}=L$ by (3). Since $L/k$ is finite, F.56 supplies $g\notin\mathfrak q$ for which $\mathfrak q$ is the only point of the fibre of $S_g$. Replace $S$ by $S_g$; finite presentation is preserved by F.35. The fibre $B=S\otimes_R k$ has a unique prime, so is local: every element outside its unique maximal ideal is a unit. Its localization there is $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$, hence $B=L$.

Here is the finite separable fibre calculation, including its finite presentation. Choose a finite generating tower $k=E_0\subset E_1\subset\cdots\subset E_t=L$, with $E_i=E_{i-1}(\alpha_i)$. A finite $k$-basis provides such a generating list. Let $p_i(T)$ be the monic minimal polynomial of $\alpha_i$ over $E_{i-1}$. Polynomial division gives $E_i=E_{i-1}[T]/(p_i)$; the quotient is a field by polynomial Bezout. These full field calculations are proved in AG-RG-S02, Lemma 4.A. Represent the coefficients of $p_i$ by polynomials in the preceding generators, obtaining $P_i(T_1,\ldots,T_i)$. Successively taking these monic quotients gives the finite presentation $$ L=k[T_1,\ldots,T_t]/(P_1,\ldots,P_t). $$ Put $I=(P_1,\ldots,P_t)\subset k[T_1,\ldots,T_t]$. The minimal polynomial of $\alpha_i$ over $E_{i-1}$ divides its separable minimal polynomial over $k$. A repeated root of the former would be a repeated root of the latter, so $p_i'(\alpha_i)\ne0$, as also proved in Lemma 4.A. The Jacobian of this presentation over $L$ is triangular with these nonzero diagonal entries, hence invertible. The map $L^t\to I/I^2$ given by the equation classes is surjective, and its composite with the conormal differential to $L^t$ is this invertible matrix. Its kernel is therefore zero, and the conormal differential is an isomorphism. Thus $H_1(\NL_{L/k})=0$ and $\Omega_{L/k}=0$, proving that $L/k$ is smooth by Definition F.6. The empty tower gives the same assertion for $L=k$.

The stated local flatness and finite presentation now satisfy F.45, so another principal neighbourhood of $\mathfrak q$ is smooth over $R$. By the arbitrary-base standard-chart assertion of F.83, shrink once more to a standard smooth presentation $$ S=R[X_1,\ldots,X_n]/(f_1,\ldots,f_c). $$ Every element inverted in these shrinkings has nonzero image in the field $L$, so the fibre remains $L$. By F.85(2), $\Omega_{S/R}$ is free on $dX_{c+1},\ldots,dX_n$. The complete base-change calculation in AG-CA-16, Theorem 2.1, identifies its tensor with $L$ with $\Omega_{L/k}=0$. Hence $L^{n-c}=0$; a coordinate vector in a nonzero field module shows that this forces $n=c$.

For clarity, the square Jacobian gives étaleness directly. Given an $R$-map $S\to D/J$ with $J^2=0$, lift its variable images to a tuple $a\in D^n$. Its error vector $f(a)$ lies in $J^n$. The Jacobian matrix $A(a)$ is invertible: its determinant is a unit modulo $J$, and a lift of a modular inverse gives $1+j$, whose inverse is $1-j$; the cofactor identity of F.70 then gives the matrix inverse. Put $b=a-A(a)^{-1}f(a)$. Polynomial expansion gives $f(b)=f(a)+A(a)(b-a)=0$, since terms with two corrections vanish in $J^2$. Any other lifting root is $b+\epsilon$ with $\epsilon\in J^n$; the same expansion gives $A(b)\epsilon=0$, so $\epsilon=0$. Thus every square-zero lifting test has a unique solution. Finite presentation makes this algebra étale by the programme definition, so the original map is étale at $\mathfrak q$. Empty variable lists use determinant one and the unique empty tuple. No Noetherian or perfect-field assumption was used. ∎

Lemma F.17: characterize formally smooth field extension

Adapted from the Stacks project, algebra.tex, lines 45959–45969.

For a field extension $K/k$, formal smoothness is equivalent to $H_1(\operatorname{NL}_{K/k})=0$.

Proof. Use any polynomial presentation, including an infinite one. Its first homology is the kernel of the conormal differential by definition. If it vanishes, the conormal sequence is short exact and its quotient $\Omega_{K/k}$ is a vector space, hence free and projective; choosing and lifting a basis splits it. AG-CA-17, Theorem 3.1, with arbitrary variable sets then proves formal smoothness. Conversely that theorem gives split exactness and hence zero kernel. Polynomial comparison already proved identifies this homology with the presentation-independent $H_1(L_{K/k})$ used in the statement. ∎

Lemma F.18: characterize smooth kbar

Adapted from the Stacks project, algebra.tex, lines 39962–40064.

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 AG-CA-18, Proposition 3.1 and Definition AG-CA-14, Section 1, definition.

Assume that $S$ is smooth at $\mathfrak m$. By Lemma F.83 we see that $S_g$ is standard smooth over $k$ for a suitable $g \in S$, $g \not \in \mathfrak m$. Hence by Lemma F.85 we see that $\Omega_{S_g/k}$ is free of rank $\dim(S_g)$. Hence by Lemma AG-CA-18, Proposition 3.1 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

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

By our choice of $J$ the exact sequence

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 AG-CA-14, Section 1, definition 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 F.57. 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

see Lemma F.77. By Lemma F.78 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 AG-CA-18, Proposition 3.1 the matrix $A \bmod \mathfrak m$ has rank $c$. Thus we win. ∎

Lemma F.19: charpoly

Adapted from the Stacks project, algebra.tex, lines 2856–2879.

If $M$ is an $n\times n$ matrix over a commutative ring, its characteristic polynomial $P(T)=\det(TI-M)$ satisfies $P(M)=0$.

Proof. The cofactor expansion proves $(TI-M)\operatorname{adj}(TI-M)=P(T)I$ over the polynomial ring. Write the adjugate as $\sum_{j=0}^{n-1}C_jT^j$ and $P(T)=\sum_{j=0}^n p_jT^j$. Comparing coefficients gives $C_{n-1}=I$, $C_{j-1}-MC_j=p_jI$ for $1\le j<n$, and $-MC_0=p_0I$. Multiplying these equations respectively by $M^j$ and adding telescopes to $\sum_{j=0}^n p_jM^j=0$. The case $n=0$ is the identity on the zero module and causes no exception. ∎

Lemma F.20: codimension

Adapted from the Stacks project, algebra.tex, lines 29578–29596.

For a surjection of finite-type $k$-algebras $S'\to S$ and corresponding primes, $\dim_{x'}\operatorname{Spec}S'-\dim_x\operatorname{Spec}S=\operatorname{height}(\mathfrak p')-\operatorname{height}(\mathfrak p)$.

Proof. AG-CA-09, Theorem 6.1, gives point dimension as local dimension plus the transcendence degree of the residue field. The two residue fields here are equal, since the map is a surjection and the primes correspond. The local dimensions are the heights of those primes, by their chain definition. Subtracting the two identities cancels exactly the equal transcendence degrees, giving the asserted formula. ∎

Lemma F.21: colimit eventually flat

Adapted from the Stacks project, algebra.tex, lines 34105–34145.

In the local finite-presentation models constructed above, if the colimit module $M$ is flat over $R$, then some model $M_\mu$ is flat over $R_\mu$.

Proof. Fix one stage $\lambda$ and set $T=\operatorname{Tor}_1^{R_\lambda}(M_\lambda,R_\lambda/\mathfrak m_\lambda)$. The free-ideal kernel formula identifies it with $\ker(\mathfrak m_\lambda\otimes_{R_\lambda}M_\lambda\to M_\lambda)$. It is finite over $S_\lambda$: the ideal and module are finite, and kernels of maps of finite modules over that Noetherian algebra are finite. Choose finitely many generators. In the limit their images in $(\mathfrak m_\lambda R)\otimes_RM$ vanish, because multiplication into $M$ is injective by its $R$-flatness. Tensor and filtered colimits commute by their finite-expression universal properties. Consequently each such vanishing holds at a later stage; a common successor $\mu$ kills all generators in $(\mathfrak m_\lambda R_\mu)\otimes_{R_\mu}M_\mu$.

Thus the transition map from $T$ to $\operatorname{Tor}_1^{R_\mu}(M_\mu,R_\mu/\mathfrak m_\lambda R_\mu)$ is zero. The module $M_\lambda/\mathfrak m_\lambda M_\lambda$ is flat over the field $R_\lambda/\mathfrak m_\lambda$. The preceding transition criterion applies to the local square $R_\lambda\to S_\lambda$, $R_\mu\to S_\mu$, whose localized tensor description was proved in the model construction, with ideal $I=\mathfrak m_\lambda$. It makes $M_\mu$ flat over $R_\mu$, as required. ∎

Lemma F.22: colimit formally etale

Adapted from the Stacks project, algebra.tex, lines 42732–42747.

A filtered colimit of formally etale $R$-algebras is formally etale over $R$.

Proof. Given a map from the colimit to $A/J$ with $J^2=0$, restrict to each stage. Formal etaleness gives a unique lift from that stage to $A$. Uniqueness makes these maps compatible with every transition arrow, so their colimit supplies a lift. Two such lifts coincide on every stage and hence on the colimit. No finite-presentation assumption is used. ∎

Lemma F.23: colimit syntomic

Adapted from the Stacks project, algebra.tex, lines 46090–46123.

Every field extension $K/k$ is a filtered union of global complete-intersection $k$-subalgebras. If it is separable, it is a filtered union of smooth subalgebras.

Proof. Given a finite subset $E\subset K$, form its finitely generated subfield $L/k$. Choose a transcendence basis $T_1,\ldots,T_d$ and a finite algebraic generator tower $y_1,\ldots,y_r$. At the $i$th step the monic minimal polynomial has coefficients in the preceding field, which is spanned over $k(T)$ by the standard monomials in the preceding generators. Represent those coefficients by polynomials in the preceding $y_j$ with rational-function coefficients in $T$. Clearing a common denominator $h(T)$ puts all of these monic polynomials $P_i$ in $k[T,1/h,Y]$. Their successive quotients are free over the preceding coefficient ring on $1,Y_i,\ldots,Y_i^{\deg P_i-1}$, by monic division. In particular each monic polynomial is a nonzerodivisor in the relevant polynomial ring: its highest nonzero coefficient in a putative product cannot disappear. The list is therefore regular.

The resulting algebra $C$ is finite free over $k[T,1/h]$. After passing to $k(T)$ it is exactly the field $L$, by the minimal-polynomial tower. Since the coefficient ring is a domain and $C$ is free, $C\to C\otimes k(T)=L$ is injective. Each member of $E$ is a linear combination of the finite standard monomial basis with coefficients in $k(T)$; enlarging $h$ clears those finitely many denominators and puts $E$ in $C$. This finite injective extension has dimension $d$ by AG-CA-05, Theorem 5.1, and AG-CA-09, Theorem 4.2. Adjoining the localization variable with $hZ-1$ gives a polynomial presentation with $d+r+1$ variables and $r+1$ equations and dimension $d$, so it is a global complete intersection. The collection of these subalgebras is directed: to dominate two, apply the construction to their finite generator lists. Every field element is included, proving that the union is $K$.

In the separable case every finitely generated subfield remains separable, since the $p$-basis monomial independence criterion restricts to subfields (and characteristic zero is automatic). Use its smooth domain model proved above and clear the finitely many fraction denominators for $E$ by principal localization. This remains smooth and gives a containing subalgebra. The same directedness and union argument applies. ∎

Lemma F.24: colimits NL

Adapted from the Stacks project, algebra.tex, lines 36983–36999.

The canonical two-term conormal complex commutes with filtered colimits of ring maps.

Proof. Use the canonical polynomial presentation with one variable for each element of the target ring. Every polynomial is a finite sum in finitely many variables and coefficients, so it occurs at one stage of the filtered system. Its being in the kernel is a finite equality in the target, hence holds at a later stage. An element in the square of the kernel is a finite sum of products of such kernel elements, and that equality also holds at a common later stage. These facts prove that the polynomial rings, their kernels, their squared kernels, and their conormal quotients have the indicated filtered colimits. Polynomial differentials are the direct sums on the same variables, and their derivatives are finite expressions, so the differentials commute with these identifications. Here is the exactness calculation for the module colimits being used. For a compatible system of short exact sequences, an element from the left whose image is zero in the middle colimit has zero image at a later stage; injectivity at that stage makes its left representative zero as well. A middle representative mapping to zero in the right colimit maps to zero at a later stage, where exactness lifts it from the left module. Every right representative lifts from the middle module at its own stage by surjectivity. Addition, scalar multiplication and comparison of representatives are checked at a common stage. Thus filtered module colimits preserve the entire short exact sequence even when transition maps are not injective. Applied to the kernels and quotients just described, this gives the asserted equality of two-term complexes. ∎

Lemma F.25: complete local Noetherian domain finite over regular

Adapted from the Stacks project, algebra.tex, lines 46654–46712.

A complete Noetherian local domain $A$ is finite over a regular complete local subring $A_0$ with the same residue field. One may take $A_0$ to be a power-series ring over a field or a Cohen ring.

Proof. The coefficient-ring existence proof is AG-CA-19S, Theorem 5.1. Since $A$ is a domain its coefficient image is a field in equal characteristic, and is an embedded Cohen DVR in mixed characteristic. Let $D=\dim A$. In the field case choose $D$ parameters $x_i$, by AG-CA-11, Theorem 2.1, and set $A_0=k[[X_1,\ldots,X_D]]$. In the mixed case $p$ is a nonzerodivisor; AG-CA-11, Theorem 3.2, gives $\dim A/pA=D-1$. Choose parameters $x_1,\ldots,x_{D-1}$ of that quotient and set $A_0=C[[X_1,\ldots,X_{D-1}]]$. Here $C$ is its coefficient Cohen ring. In either case the image $I$ of the maximal ideal of $A_0$ is primary to the maximal ideal of $A$. A power of the latter lies in $I$ by AG-CA-03, Proposition 2.2, making the two filtrations cofinal. Thus $A$ is $I$-adically complete and separated. Evaluation of a power series at the $x_i$ is well defined, by the same convergence and multiplication proof as AG-CA-19S, Theorem 6.1.

Choose lifts $a_1,\ldots,a_s$ of a basis of the finite-dimensional residue quotient $A/I$ over $k$. Every $a\in A$ can be expressed modulo $I$ as a coefficient combination of them. Express the error in $I$ as a sum of its chosen generators times elements of $A$, and repeat the same residue decomposition on those elements. After $n$ steps this constructs coefficients $c_{jn}\in A_0$, compatible modulo its $n$-th maximal-ideal power, such that $a-\sum_jc_{jn}a_j\in I^n$. Completeness of $A_0$ gives limits $c_j$ and separatedness gives $a=\sum_jc_ja_j$. Hence $A$ is finite as an $A_0$-module. The quotient $A/I$ is finite-dimensional because its maximal ideal is nilpotent and its finitely many successive maximal-ideal layers are finite-dimensional, by AG-CA-03, Theorem 4.2.

The power-series ring $A_0$ is the maximal-ideal completion of the regular polynomial local ring over its coefficient field or DVR, as is checked directly on the finite quotients. Regularity of the polynomial local ring is AG-CA-14, Proposition 3.3, with the DVR base regular of dimension one; its dimension is the number of displayed parameters, including $p$ in mixed characteristic, by AG-CA-11, Theorem 5.1. Completion preserves dimension and regularity by AG-CA-19, Theorem 4.1. Thus $A_0$ is a regular local domain of dimension $D$. If the evaluation kernel contained $0\ne f$, the quotient $A_0/(f)$ would have dimension $D-1$ by AG-CA-11, Theorem 3.2. The injective finite integral map $A_0/\ker\to A$ preserves dimension by AG-CA-05, Theorem 5.1, but $\dim(A_0/\ker)\le D-1$ would contradict $\dim A=D$. The evaluation is therefore injective. Its residue map is the coefficient-field identification, completing the proof. For $D=0$ the domain is a field and the assertion takes $A_0=A$. ∎

Lemma F.26: completion at quasi finite prime

Adapted from the Stacks project, algebra.tex, lines 32412–32464.

Let $R$ be Noetherian, $R\to S$ finite type, and $q$ a quasi-finite prime over $p$. Then $\widehat{R_p}\otimes_RS\cong\widehat{S_q}\times D$, with its first projection the natural completed local map.

Proof. The earlier full affine conductor theorem Zariski Main, Theorem C4.3 gives an integral subalgebra $C\subset S$ and $g\in C\setminus q$ with $C_g=S_g$. Choose finitely many numerators from $C$ that, together with $g^{-1}$, generate $S_g$ over $R$. Let $S'\subset C$ be generated by them and $g$. Its generators are integral, hence it is a finite $R$-module algebra, and $S'_g=S_g$. Put $q'=q\cap S'$. Localizing this equality again gives $S'_{q'}=S_q$, so their maximal-adic completions agree.

The finite-extension completion argument of F.27 gives a product $$\widehat{R_p}\otimes_RS'=A\times D',\qquad A=\widehat{S'_{q'}}=\widehat{S_q}.$$ Let $e$ be the central idempotent of this selected factor. Its image splits $T=\widehat{R_p}\otimes_RS$ as $eT\times(1-e)T$. The element $g$ maps to a unit in $A$, because it is outside $q'$, so it is a unit in $eT$ through the map $A\to eT$. Consequently $eT=(eT)_g$. But localizing the tensor map at $g$ is an isomorphism, since $S'_g=S_g$. Its selected $e$-factor therefore identifies $(eT)_g$ with $A_g=A$. This proves $eT=A$ and the desired product decomposition, including its adic factor.

To check its first projection is the natural map, it agrees on $\widehat{R_p}$ and on $S'$ by the finite-extension completion construction. For any $s\in S$, equality after localizing at $g$ supplies an exponent $a$ with $g^as\in S'$: write its fraction using $S'_g$, then multiply by one further power to kill the localization error in $S$. Since $g$ is a unit in $\widehat{S_q}$, agreement on $g^as$ and on $g$ forces agreement on $s$. Pure tensors generate $T$, so its projection is precisely the natural completed-local map. This is an explicit adic comparison, not an identification of finite scheme completion with ring completion. ∎

Lemma F.27: completion finite extension

Adapted from the Stacks project, algebra.tex, lines 24144–24184.

For a finite map $R\to S$ of Noetherian rings and a prime $\mathfrak p$ of $R$, with primes $\mathfrak q_i$ above it, one has $\widehat{R_{\mathfrak p}}\otimes_RS=\prod_i\widehat{S_{\mathfrak q_i}}$.

Proof. Localize the base at $\mathfrak p$. All maximal ideals of the finite algebra lie above this maximal ideal, by AG-CA-05, Theorem 3.3, and there are finitely many by Theorem 3.4. For each $n$, the finite algebra $S/\mathfrak p^nS$ is a finite module over the Artinian ring $R/\mathfrak p^n$, hence Artinian by AG-CA-03, Theorem 3.3 and its finite maximal-ideal-layer proof. The product theorem 4.2 of that lesson gives $S/\mathfrak p^nS=\prod_iS_{\mathfrak q_i}/\mathfrak p^nS_{\mathfrak q_i}$. On each factor $\mathfrak pS_{\mathfrak q_i}$ has radical $\mathfrak q_iS_{\mathfrak q_i}$. A power of the latter ideal lies in the former, by AG-CA-03, Proposition 2.2; their adic filtrations are cofinal, so their inverse limits are the same. Taking the finite product of the inverse limits now gives the asserted product of local completions. Finally the completion-tensor equality for the finite $R_{\mathfrak p}$-module $S_{\mathfrak p}$ is AG-CA-19, Theorem 3.1. ∎

Lemma F.28: compose finite type

Adapted from the Stacks project, algebra.tex, lines 547–571.

Finite-type and finitely presented ring maps compose. If $R\to S'\to S$ and $S$ is finite type over $R$, it is finite type over $S'$. If additionally $S/R$ is finitely presented and $S'/R$ finite type, then $S/S'$ is finitely presented.

Proof. Generators of $S'/R$, followed by generators of $S/S'$, generate the composite. For presentations $S'=R[Y]/(a)$ and $S=S'[X]/(b)$, lift the finitely many coefficients of $b$ to $R[Y]$ to obtain the finite composite presentation $R[Y,X]/(a,\widetilde b)$. The same generators of $S/R$ generate it over $S'$. Finally write $S=R[X]/(f_1,\ldots,f_m)$ and $S'=R[Y]/I$, with finitely many $Y_i$ mapping to polynomials $h_i(X)$ in $S$. Then $S=S'[X]/(f_j,Y_i-h_i(X))$. To check this equality, substitute $Y_i=h_i(X)$; every relation in $I$ vanishes in $S$ and so belongs to the ideal of the $f_j$. This proves both inverse maps directly without assuming $I$ finitely generated. ∎

Lemma F.29: compose smooth

Adapted from the Stacks project, algebra.tex, lines 39039–39053.

A composition of smooth ring maps is smooth.

Proof. Let $R\to S\to T$ be smooth maps. Lemma F.28 gives finite presentation of $T/R$. The Jacobi–Zariski sequence F.42 places $H_1(\operatorname{NL}_{T/R})$ between $H_1(\operatorname{NL}_{S/R}\otimes_ST)$ and $H_1(\operatorname{NL}_{T/S})$. The latter is zero by smoothness. For the former, the last assertion of F.42 identifies it with $H_1(\operatorname{NL}_{S/R})\otimes_ST=0$: its two Tor hypotheses hold because $\Omega_{S/R}$ is projective. Hence $H_1(\operatorname{NL}_{T/R})=0$. The same sequence gives

The last module is projective, so this sequence splits. The first is finite projective because a finite direct-summand presentation for $\Omega_{S/R}$ remains such a presentation after tensoring with $T$; the last is finite projective by smoothness of $T/S$. Their direct sum is therefore finite projective. Definition F.6 proves smoothness of $T/R$. ∎

Lemma F.30: compose standard smooth

Adapted from the Stacks project, algebra.tex, lines 38815–38878.

A composite of standard smooth ring maps is standard smooth.

Proof. Write $S=R[X_1,\ldots,X_n]/(f_1,\ldots,f_c)$ and $T=S[Y_1,\ldots,Y_m]/(g_1,\ldots,g_d)$, with their selected Jacobian minors invertible. Lift the coefficients of the $g_i$ to $R[X]$. The resulting $R$-presentation of $T$ has equations $f_i,g_j$ and pivot variables the $c$ selected $X$ variables followed by the $d$ selected $Y$ variables. Its Jacobian matrix is block lower triangular, since $\partial f_i/\partial Y_j=0$. Its determinant is the product of the two given unit determinants. This is the required standard smooth presentation. Any principal localization is encoded by one more variable $z$ and equation $hz-1$, whose additional diagonal Jacobian entry $h$ is a unit. Thus the assertion covers the localized presentations as well. ∎

Lemma F.31: composition essentially of finite type

Adapted from the Stacks project, algebra.tex, lines 13129–13138.

Essentially finite type, and essentially finite presentation, are preserved by composition.

Proof. Write the first algebra as $U^{-1}R[Y]/(a)$, and the second before its final localization as a polynomial algebra over it modulo finitely many chosen relations in the finite-presentation case. Only finitely many denominators from $U$ occur in those relations; absorb them in one element $u\in U$. This finite algebra then descends to $(R[Y]/(a))_u$, and its further localization gives the original composite. The composite is therefore a localization of a finite-type $R$-algebra, or of a finitely presented one in the second case. The remaining elements of $U$ and the second localization set are simply included in the final localization set. The finite-type argument only needs the finite generator list and uses the same construction with the possibly infinite relation ideal. ∎

Lemma F.32: composition syntomic

Adapted from the Stacks project, algebra.tex, lines 38404–38453.

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

1. If $R$ is Noetherian and $R \to S$ 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 F.4. Then

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 AG-CA-11, Theorem 5.1 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 $R$ Noetherian and $R \to S$ and $S \to S'$ are syntomic. Their finite presentation makes $S$ and $S'$ Noetherian by the earlier Hilbert basis theorem, so both applications of F.87 have their stated base hypothesis. Let $\mathfrak q' \subset S'$ be a prime ideal lying over $\mathfrak q \subset S$. By Lemma F.87 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 F.14 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 F.87 and F.62 we see that $R \to S'$ is syntomic (this also uses that the spectrum of $S'$ is quasi-compact, see Lemma AG-CA-01, Proposition 2.3). ∎

Lemma F.33: conormal module localize

Adapted from the Stacks project, algebra.tex, lines 37203–37225.

For finite polynomial presentations $R[X_1,\ldots,X_n]\twoheadrightarrow S$ and $R[Y_1,\ldots,Y_m]\twoheadrightarrow S_g$ with kernels $I,J$, there is an isomorphism $(I/I^2)_g\oplus S_g^m\cong J/J^2\oplus S_g^n$.

Proof. Put $C=S_g$, $P=R[X_1,\ldots,X_n]$, and $Q=R[Y_1,\ldots,Y_m]$. Choose $f\in P$ representing $g$. Then $P_f\twoheadrightarrow C$ has kernel $I_f$, by exact localization. Choose $u_j\in P_f$ lifting the images of the $Y_j$ in $C$, and choose $v_i\in Q$ lifting the images of the $X_i$. The latter lifts exist because $Q\twoheadrightarrow C$ is surjective.

Map $W=P[Y_1,\ldots,Y_m]$ to $C$ using both specified presentations. Its kernel is $L=(J,X_i-v_i(Y))$: quotienting by this ideal eliminates the $X_i$ and gives exactly $Q/J=C$. Translation makes $W=Q[s_1,\ldots,s_n]$, with $s_i=X_i-v_i(Y)$, and $L=(J,s)$. Expanding a polynomial by its $s$ degree proves $$ L/L^2\cong J/J^2\oplus C^n. $$ Its constant part is read modulo $J^2$, its linear coefficients modulo $J$, and all terms of degree at least two belong to $L^2$. These rules define inverse maps, and justify the formula over every base ring.

The image of $f$ in $C$ is the unit $g$, so localization leaves this conormal module unchanged. Ideal products and quotients localize by their fraction formulas, giving $L_f/L_f^2=L/L^2$. On the other hand $W_f=P_f[Y]$, and its kernel for the same map to $C$ is $(I_f,Y_j-u_j)$. Eliminating the $Y_j$ gives $P_f/I_f=C$, which proves this kernel equality. Translating these variables and applying the identical constant/linear calculation gives $$ L_f/L_f^2\cong (I_f/I_f^2)\oplus C^m \cong (I/I^2)_g\oplus C^m. $$ Combine the two displayed isomorphisms. No free summand is cancelled and no flatness of $C$ is assumed. The formulas also hold for empty variable lists and $C=0$. ∎

Lemma F.34: contained in radical

Adapted from the Stacks project, algebra.tex, lines 3648–3678.

An ideal $I$ is contained in the Jacobson radical exactly when every element of $1+I$ is a unit. In that case units modulo $I$ lift to units.

Proof. If $I$ lies in every maximal ideal, $1+i$ lies in none and therefore is a unit: a nonunit generates a proper ideal, which lies in a maximal ideal by Zorn's lemma. Conversely, if $I\not\subset\mathfrak m$, write $1=i+x$ with $i\in I$, $x\in\mathfrak m$; then $x=1-i$ is a nonunit, contradiction. If $ab=1+i$ after lifting a modular inverse, its unit status gives an inverse $b(ab)^{-1}$ for $a$. ∎

Lemma F.35: cover upstairs

Adapted from the Stacks project, algebra.tex, lines 4346–4405.

Let $g_1,\ldots,g_r\in S$ generate the unit ideal. Finite type and finite presentation of $S/R$ can be checked on the finitely many $S_{g_i}$.

Proof. First record the module detection calculation on a finite principal cover. Let $t_1,\ldots,t_r$ generate one in a ring $T$, say $\sum c_it_i=1$, and suppose a $T$-module $M$ has $M_{t_i}=0$ for each $i$. For any $m\in M$, choose positive integers $e_i$ with $t_i^{e_i}m=0$, using the equality criterion for localization. Expand $$ 1=\left(\sum c_it_i\right)^N,\qquad N=1+\sum_i(e_i-1). $$ Every monomial contains some $t_i^{e_i}$, so multiplying by $m$ gives $m=0$. Thus $M=0$, with no finiteness hypothesis on $M$. An empty cover forces $T=0$, when this assertion also holds.

Choose $h_i\in S$ with $\sum h_ig_i=1$. For finite type, let $S_0\subset S$ be the $R$-subalgebra generated by all $h_i,g_i$ and the numerators of finite $R$-algebra generator lists for the $S_{g_i}$. The maps $(S_0)_{g_i}\to S_{g_i}$ are injective by the localization equality criterion and surjective by these chosen generators. The $g_i$ generate one in $S_0$ as well. The cokernel $S/S_0$, viewed as an $S_0$-module, vanishes on this principal cover, so the preceding calculation gives $S=S_0$.

For finite presentation first use finite type to write $S=P/J$, $P=R[X_1,\ldots,X_n]$. Lift $g_i,h_i$ to $u_i,v_i\in P$. The surjection $P[Z]\to S_{g_i}$ has kernel $JP[Z]+(u_iZ-1)$, since its quotient is $S[Z]/(g_iZ-1)=S_{g_i}$ by the fraction universal property. This kernel is finitely generated because $S_{g_i}$ is finitely presented. Here is the precise presentation-independence calculation: given $C=R[Y_1,\ldots,Y_m]/(a_1,\ldots,a_s)$ and a finite polynomial surjection $R[W_1,\ldots,W_t]\to C$, choose polynomials $U_j(W)$ representing the images of $Y_j$ and $V_l(Y)$ representing the images of $W_l$. The ideal $$ \big(a_j(U(W)),\ W_l-V_l(U(W))\big)_{j,l} $$ is its kernel. Indeed substitution $Y_j\mapsto U_j(W)$ defines a map from $C$ to the displayed quotient, inverse to its map to $C$: the first list makes substitution well-defined, the second makes the composite on the $W_l$ the identity, and the chosen images make the other composite on the $Y_j$ the identity.

Expand each finite kernel generator for $P[Z]\to S_{g_i}$ as a finite combination of elements of $J$ and $u_iZ-1$. Collect the finitely many elements $a_{ij}\in J$ that occur. These elements together with $u_iZ-1$ generate exactly that kernel: they belong to it and generate each of its chosen generators. Set $J_0=(\sum v_iu_i-1,a_{ij})\subset J$. The classes of the $u_i$ generate one in $P/J_0$. The equality of the preceding kernels shows that every map $(P/J_0)_{u_i}\to(P/J)_{u_i}$ is an isomorphism. Its global kernel $J/J_0$ therefore vanishes by the module detection calculation, so $J=J_0$ is finitely generated.

Conversely, if $S$ is of finite type, then $S_{g_i}$ is generated by its finite algebra generators and $g_i^{-1}$. If $S=P/(a_1,\ldots,a_s)$ is finitely presented and $u_i\in P$ lifts $g_i$, then $S_{g_i}=P[Z]/(a_1,\ldots,a_s,u_iZ-1)$ is finitely presented. This proves both directions, including zero rings and empty generator lists. ∎

Lemma F.36: criterion flatness fibre

Adapted from the Stacks project, algebra.tex, lines 34374–34496.

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 F.59 we may first write $R = \operatorname*{colim} R_\lambda$ 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

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

Note that this also implies that

Choose a presentation

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 F.21 that for all $\lambda$ big enough the module $M_\lambda$ is flat over $R_\lambda$. Moreover, note that $\mathfrak m = \operatorname*{colim} \mathfrak p_\lambda$, $S/\mathfrak mS = \operatorname*{colim} S_\lambda/\mathfrak p_\lambda S_\lambda$, $S'/\mathfrak mS' = \operatorname*{colim} S'_\lambda/\mathfrak p_\lambda S'_\lambda$, and $M/\mathfrak mM = \operatorname*{colim} M_\lambda/\mathfrak p_\lambda M_\lambda$. Also, for each $\lambda_3 \leq \lambda \leq \mu$ we see (from the properties listed above) that

is a localization, and the map

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 F.60 as well. We may apply Lemma F.21 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 F.37. 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). ∎

Lemma F.37: criterion flatness fibre Noetherian

Adapted from the Stacks project, algebra.tex, lines 24999–25045.

Let $R\to S\to T$ be local maps of Noetherian local rings, and $M\ne0$ finite over $T$. If $M$ is $R$-flat and $M/\mathfrak m_RM$ is flat over $S/\mathfrak m_RS$, then $M$ is faithfully flat over $S$ and $S$ is flat over $R$.

Proof. Put $I=\mathfrak m_RS$. The map $\mathfrak m_R\otimes_RM\to I\otimes_SM$ is surjective, because the images of $\mathfrak m_R$ generate $I$. Its composite to $M$ is injective by $R$-flatness, so both maps are injective. Thus $\operatorname{Tor}_1^S(S/I,M)=0$. The ideal form of the local flatness criterion, AG-CA-08, Corollary 4.4, applied to the proper ideal $I$ gives $S$-flatness. Nakayama gives $M/\mathfrak m_TM\ne0$, hence $M/\mathfrak m_SM\ne0$; the faithful-flatness criterion for modules, AG-CA-08, Theorem 1.1, gives faithfulness over $S$.

Tensor the exact sequence $0\to\operatorname{Tor}_1^R(R/\mathfrak m_R,S)\to\mathfrak m_R\otimes_RS\to I\to0$ with this flat $S$-module. Its first term maps to the kernel of the already injective map $\mathfrak m_R\otimes_RM\to I\otimes_SM$, hence is zero. Faithfulness kills the original Tor module. AG-CA-08, Theorem 4.2, then gives $R$-flatness of $S$. All modules to which the local criterion is applied are finite over the stipulated Noetherian local overring. ∎

Lemma F.38: descent regular

Adapted from the Stacks project, algebra.tex, lines 48353–48371.

Regularity descends along faithfully flat maps of Noetherian rings.

Proof. Fix a source prime and choose a target prime above it, which exists by faithful flatness. The localized map $A\to B$ is flat local, hence faithfully flat by AG-CA-08, Theorem 2.1. The target is regular local. Take a free resolution of the source residue field $k$ over $A$. After tensoring with $B$ it remains a free resolution of $k\otimes_AB$, by flatness. Consequently $$\operatorname{Tor}_i^A(k,k)\otimes_AB=\operatorname{Tor}_i^B(k\otimes_AB,k\otimes_AB).$$ AG-CA-14, Theorem 2.2, gives finite global dimension of $B$, so these groups vanish for all sufficiently large $i$. Faithfulness gives vanishing of the original source Tor groups. The minimal-free-resolution calculation AG-CA-13, Theorem 2.2, makes the source residue field have finite projective dimension; AG-CA-14, Theorem 2.2, then makes $A$ regular. Doing this at every prime proves the assertion. ∎

Lemma F.39: dimension fibres bounded open upstairs

Adapted from the Stacks project, algebra.tex, lines 32674–32690.

For a finite-type map of Noetherian rings $R\to S$, if the fibre dimension at a point $q$ is $n$, some principal neighbourhood of $q$ has fibre point-dimension at most $n$ everywhere. This is the exact form used here.

Proof. Let $p=q\cap R$. Choose a principal neighbourhood whose $\kappa(p)$-fibre has dimension exactly $n$, using the point-dimension definition and the finite-type field component formula. Apply Noether normalization to that fibre. The coordinate changes in AG-CA-09, Lemma 2.1 and Corollary 3.2, use polynomials with integer coefficients in the algebra generators: at each step replace a generator by it minus a sufficiently high power of the last generator, so one selected relation has a constant nonzero leading coefficient; dividing that coefficient makes the last generator integral. Inducting yields $n$ such parameters and finiteness over their polynomial algebra. Thus these parameters lift to elements $t_i\in S$, and $R[T_1,\ldots,T_n]\to S$, $T_i\mapsto t_i$, has a finite fibre over $p$. At the chosen point it is quasi-finite: its further fibre over the contracted polynomial prime is a finite algebra over that prime's residue field.

The earlier affine conductor proof Zariski Main, Theorem C5.1 proves the open quasi-finite locus from its full Theorem C4.3. Shrink once more around $q$ so this polynomial map is quasi-finite everywhere. In a fibre over any base prime $p'$, its algebra $U$ is quasi-finite over $\kappa(p')[T_1,\ldots,T_n]$. At a minimal prime $\eta$ of $U$, the residue field is finite over that of its contracted polynomial prime, by the isolated-fibre criterion proved above. Its transcendence degree over $\kappa(p')$ is therefore at most $n$. AG-CA-09, Theorem 4.2, gives $\dim(U/\eta)$ equal to that transcendence degree. Taking the maximum over the finitely many minimal primes gives $\dim U\le n$. Every fibre point-dimension is at most that fibre dimension. This proves the claimed neighbourhood bound. Its earlier conductor input includes the polynomial-normality proof in Zariski Main over an arbitrary base, Corollary B1.2, so no old external proof tag is being substituted. ∎

Lemma F.40: essentially of finite type into artinian local

Adapted from the Stacks project, algebra.tex, lines 13151–13207.

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 AG-CA-05, Theorem 1.2. Conversely, assume $R \to S/\mathfrak m$ is finite. As $S$ has finite length over itself (Lemma AG-CA-03, Theorem 4.2) we can choose a filtration

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 F.43.

If $R \to S$ is of finite type, then $R \to S/\mathfrak m$ is of finite type by Lemma F.28. 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 F.28.

If $R \to S$ is essentially of finite type, then $R \to S/\mathfrak m$ is essentially of finite type by Lemma F.31. 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 F.31. ∎

Lemma F.41: etale standard smooth

Adapted from the Stacks project, algebra.tex, lines 40737–40760.

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

is an isomorphism, in particular $I/I^2$ is a free $S$-module. Thus by Lemma F.55 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 $S$. ∎

Lemma F.42: exact sequence NL

Adapted from the Stacks project, algebra.tex, lines 36838–36913.

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

with exact rows. We get the following exact sequence of homology groups

of $C$-modules extending the sequence of Lemma AG-CA-16, Theorem 3.1. If $\text{Tor}_1^B(\Omega_{B/A}, C) = 0$ and $\text{Tor}_2^B(\Omega_{B/A}, C) = 0$, then $H_1(\NL_{B/A} \otimes_B C) = H_1(L_{B/A}) \otimes_B C$.

Proof. On differentials, the first map sends $dx_s$ to $dx_s$, the second sends every $dx_s$ to zero and $dy_t$ to $dy_t$. On conormal modules the first map sends $(f\bmod I^2)\otimes c$ to $cf\bmod K^2$, using any lift of $c$ to $A[x_s,y_t]$; the ambiguity changes the answer by $K^2$. The second sends $g\bmod K^2$ to its image in $J/J^2$. Derivatives of these representatives verify every commutative square. The connecting homology map is obtained by lifting a cycle from $J/J^2$ to $K/K^2$, taking its derivative (which lies in the image of the left differential module), and reducing modulo the image of $I/I^2\otimes C$. Changing the lift changes this by a boundary. The same lifting calculation proves exactness at each homology term. 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

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 F.11 to identify the terms as homology groups of the naive cotangent complexes.

The final assertion is a statement in homological algebra. Recall that $\NL_{B/A} = (N^{-1} \to N^0)$ 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

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

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. ∎

Lemma F.43: extension

Adapted from the Stacks project, algebra.tex, lines 398–483.

For $0\to M_1\to M_2\to M_3\to0$, finite generation is preserved under extensions and quotients; finite presentation is preserved under extensions; a quotient of a finitely presented module by a finite submodule is finitely presented; and if $M_3$ is finitely presented and $M_2$ finite, then $M_1$ is finite.

Proof. Generators of $M_1$ and lifts of generators of $M_3$ generate $M_2$; images of generators generate any quotient. For the last assertion take a finite free surjection $F\to M_2$. Its composite $F\to M_3$ has finite kernel $K$: to see presentation independence, compare it with a fixed finite presentation of $M_3$ by lifting each of the two free bases to the other free module; the finite defining relations and the finite substitution differences generate the kernel. The image of $K$ in $M_2$ is precisely $M_1$, which is consequently finite.

If $M_2=F/K$ has finite generators and relations and $M_1$ is finite, lift its finite generator list to $F$; adjoining those lifts to the finite generator list of $K$ presents $M_3$. Finally for an extension of two finitely presented modules take finite free surjections $F_i\to M_i$ for $i=1,3$ and lift the second basis to $M_2$. Then $F_1\oplus F_3\to M_2$ is surjective. Its kernel fits into a short exact sequence of the two presentation kernels: projection to $F_3$ gives the map onto the third kernel, and subtracting a suitable $F_1$ lift proves surjectivity; its kernel is the first presentation kernel. Both outer kernels are finite, so the generator argument at the start makes the middle kernel finite. This proves finite presentation of $M_2$. ∎

Lemma F.44: finite presentation independent

Adapted from the Stacks project, algebra.tex, lines 573–589.

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. ∎

Lemma F.45: flat fibre smooth

Adapted from the Stacks project, algebra.tex, lines 39115–39149.

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. We prove the stated arbitrary-ring assertion. First replace $S$ by the given $S_s$, with $s\notin\mathfrak q$, and keep the corresponding prime denoted by $\mathfrak q$. Choose a finite presentation $$ P=R[X_1,\ldots,X_n]\longrightarrow S=P/I,\qquad I=(a_1,\ldots,a_m). $$ Let $\mathfrak r\subset P$ be the inverse image of $\mathfrak q$, put $A=R_{\mathfrak p}$, $Q=P_{\mathfrak r}$, and $B=S_{\mathfrak q}$. The exact sequence $0\to I_{\mathfrak r}\to Q\to B\to0$ is a sequence of $A$-modules, and its quotient $B$ is $A$-flat by hypothesis. We will use only finite generation of $I$, not a chain condition on any of these rings.

Here is the tensor-injection calculation needed for that sequence. More generally, suppose $0\to K\to M\to F\to0$ is exact over a ring $A$ and $F$ is flat; $M$ need not be flat or free. For an arbitrary $A$-module $N$, choose a free presentation $F_1\xrightarrow{d}F_0\to N\to0$, and put $T=\ker d$ and $L=\operatorname{im}d$. The free presentations, direct-sum exactness and right-exact tensor calculations are written in AG-RG-S01, Lemma 1.0. If $z\in K\otimes_A N$ maps to zero in $M\otimes_A N$, lift it to $x\in K\otimes_A F_0$. Right exactness gives $y\in M\otimes_A F_1$ whose image in $M\otimes_A F_0$ is the image of $x$. The image of $y$ in $F\otimes_A F_1$ is killed by $1\otimes d$. Tensoring $0\to T\to F_1\to L\to0$ with the flat $F$, and using the injection $L\to F_0$, shows that this kernel is the image of $F\otimes_A T$. Lift such a preimage to $M\otimes_A T$, using the surjection $M\to F$ and right exactness, and subtract its image from $y$. The resulting $y'$ still maps to the image of $x$, and its image in $F\otimes_A F_1$ is zero. Since $F_1$ is free, coordinatewise exactness lifts $y'$ from $K\otimes_A F_1$. Its image in $K\otimes_A F_0$ is $x$: this follows from the injection $K\otimes_A F_0\to M\otimes_A F_0$, again coordinatewise because $F_0$ is free. Thus $x$ comes from $K\otimes_A F_1$, so $z=0$. We have proved the tensor injection with no Tor assertion.

Set $k=\kappa(\mathfrak p)=A/\mathfrak pA$. Apply that calculation to $N=k$. Right exactness and the quotient-tensor formula identify $$ J:=\ker(Q/\mathfrak pQ\to B/\mathfrak pB) \simeq I_{\mathfrak r}/\mathfrak pI_{\mathfrak r}. $$ Write $\overline Q=Q/\mathfrak pQ$ and $\overline B=B/\mathfrak pB$. Then $$ \overline Q=k[X_1,\ldots,X_n]_{\overline{\mathfrak r}}, \qquad \overline B=\overline Q/J. $$ Indeed localization commutes with quotient and coefficient change; every polynomial over $k$ lifts coefficientwise to $A$, and being outside $\overline{\mathfrak r}$ is exactly being outside the prime before reduction. The ring $\overline Q$ is local, $J$ is contained in its maximal ideal, and $\overline B$ is the local ring of the fibre at the specified point. In particular $J$ is finite, generated by the images of the $a_j$.

The smooth-fibre hypothesis makes $\overline B$ formally smooth over $k$. Here the equivalence of the definitions is used at its proved scope: Definition F.6 and the presentation comparison in Lemma F.11 give, on a smooth fibre neighbourhood, an injective conormal map with projective differential cokernel. Projectivity splits that quotient, and the full conormal criterion AG-CA-17, Theorem 3.1 gives formal smoothness. Its Theorem 1.2 proves that arbitrary further localizations preserve formal smoothness: a lift on the unlocalized algebra sends the inverted elements to units, since units lift across a square-zero ideal, and hence extends to the localization. Thus this step does not assume that a local ring is itself finitely presented.

We obtain the needed conormal splitting in this same localized polynomial presentation, without changing presentations. Formal smoothness lifts the identity of $\overline B$ to a $k$-algebra section $$ \sigma:\overline B\longrightarrow\overline Q/J^2 $$ of its square-zero quotient. The map $$ D(q)=q\bmod J^2-\sigma(q\bmod J) \quad:\quad\overline Q\longrightarrow J/J^2 $$ is a $k$-derivation: writing $q=\sigma(\overline q)+D(q)$, and discarding products of two elements of $J/J^2$, gives $D(qq')=\overline qD(q')+\overline q'D(q)$. For $j\in J$, it has $D(j)=j\bmod J^2$. The universal differential construction therefore supplies a left inverse to $$ J/J^2\longrightarrow \Omega_{\overline Q/k}\otimes_{\overline Q}\overline B =\overline B^{\,n},\qquad [j]\longmapsto\left(\frac{\partial j}{\partial X_\nu}\right)_\nu. $$ The free differential basis and its localization here are precisely AG-CA-16, Theorems 5.1 and 2.2; localization extends a derivation by $D(q/t)=t^{-1}D(q)-qt^{-2}D(t)$. Thus $J/J^2$ is a finite direct summand of $\overline B^{\,n}$.

We spell out the finite local-module argument, including the required Nakayama uses. Over any local ring $(D,\mathfrak n)$, a finite module $M$ with $M=HM$ for $H\subset\mathfrak n$ is zero: choose generators and write $m_i=\sum_j h_{ij}m_j$ with $h_{ij}\in H$. The cofactor identity of Lemma 3.B applied to $I-(h_{ij})$ shows that its determinant annihilates every generator. That determinant is congruent to one modulo $\mathfrak n$, hence is a unit. An element outside the unique maximal ideal is a unit, because a nonunit generates a proper ideal, which lies in a maximal ideal by the proper-ideal Zorn argument. Therefore $M=0$. It follows also that lifts of a basis of $M/\mathfrak nM$ generate a finite $M$, by applying this assertion to their quotient.

Choose from the finite list $a_1,\ldots,a_m$ elements $f_1,\ldots,f_c$ whose conormal classes give a basis of $(J/J^2)\otimes_{\overline B}\kappa(\mathfrak q)$; a finite spanning list over a field contains a basis by successively adjoining independent members. Nakayama over $\overline B$ makes these classes generate $J/J^2$. They are in fact a free basis. To see this, the surjection $\overline B^c\to J/J^2$ splits: lift the projection $\overline B^n\to J/J^2$ through it by choosing lifts of the $n$ coordinate images, and restrict that lift to the direct summand $J/J^2$. Its kernel is then a finite direct summand of $\overline B^c$, generated by the projections of its coordinate vectors. Modulo the maximal ideal the surjection is an isomorphism by our basis choice, so this kernel has zero residue quotient. The just-proved Nakayama argument kills it.

Let $H$ be the $c$ by $n$ equation-by-variable matrix $(\partial f_i/\partial X_\nu)$, with coefficients reduced to $\overline B$. The split conormal injection has matrix $H^{\mathsf T}$ and a left inverse $V$, so $VH^{\mathsf T}=I_c$. We write the minor argument here. Expanding $\det(VH^{\mathsf T})$ by multilinearity in its columns chooses $c$ columns of $V$ with their coefficients from $H^{\mathsf T}$. Repeated choices have determinant zero by the signed-permutation cancellation. Grouping the remaining choices by their distinct index set $U$, in increasing order, gives $$ 1=\sum_{|U|=c}\det(V^U)\det\bigl((H^{\mathsf T})_U\bigr). $$ The signed-permutation formula also gives $\det(W^{\mathsf T})=\det W$, by replacing a permutation with its inverse. If $c>n$ every choice repeats, contradicting $1\ne0$ in the local ring $\overline B$. Thus $c\le n$, and at least one $c$ by $c$ minor of $H$ is a unit of $\overline B$: if all belonged to its maximal ideal the displayed sum could not be one. Choose its variable indices $U$. The corresponding polynomial $$ \Delta=\det\left(\frac{\partial f_i}{\partial X_\nu}\right)_ {1\le i\le c,\ \nu\in U}\in P $$ is outside $\mathfrak r$, since its image in $\kappa(\mathfrak q)$ is nonzero. For $c=0$ this argument uses the empty determinant $\Delta=1$ and the empty generator list.

The chosen $f_i$ generate the whole fibre ideal $J$ in $\overline Q$, not merely its conormal quotient. Indeed $J=(\overline f_1,\ldots,\overline f_c)+J^2$. Its finite quotient by the displayed ideal is equal to $J$ times that quotient. Since $J$ is contained in the maximal ideal of the local ring $\overline Q$, Nakayama makes the quotient zero. Now put $M=I_{\mathfrak r}/(f_1,\ldots,f_c)Q$, a finite $Q$-module. The tensor-injection identification above gives $$ M/\mathfrak pM =J/(\overline f_1,\ldots,\overline f_c)=0. $$ The ideal $\mathfrak pQ$ lies in the maximal ideal of $Q$, so another application of finite Nakayama gives $M=0$. This is the step where flatness of the local quotient $B$ is essential.

For each of the finitely many original generators $a_j$ of $I$, this equality expresses $a_j/1$ as a combination of the $f_i/1$ in $P_{\mathfrak r}$. Clear the finitely many coefficient denominators and also any multiplier outside $\mathfrak r$ witnessing each equality in that localization. Their finite product $v\notin\mathfrak r$ gives $I_v=(f_1,\ldots,f_c)P_v$. This keeps the killed-cross-difference condition and does not cancel a possible zero divisor. Consequently a neighbourhood of $\mathfrak q$ is $$ S_{v\Delta} =R[X_1,\ldots,X_n]_{v\Delta}/(f_1,\ldots,f_c). $$ It is finitely presented over $R$: add one variable $Y$ with the relation $(v\Delta)Y-1$, together with the $c$ equations. These further principal localizations of the earlier $S_s$ give a principal neighbourhood in the original $\operatorname{Spec}S$: write their elements with numerators in the original $S$ and denominators powers of $s$, and invert $s$ and those finitely many numerators.

Finally we prove formal smoothness of this neighbourhood by the square-zero calculation. Given an $R$-algebra $D$, an ideal $K$ with $K^2=0$, and a map from the displayed algebra to $D/K$, lift the variable images to $b_\nu\in D$. The errors $r_i=f_i(b)$ lie in $K$. The selected square Jacobian block $H_U(b)$ has unit determinant, because its determinant is $\Delta(b)$ and its image modulo $K$ is a unit. The cofactor calculation in Lemma 3.B gives its inverse. Change only the selected coordinates by $$ (b_\nu')_{\nu\in U} =(b_\nu)_{\nu\in U}-H_U(b)^{-1}(r_i)_{1\le i\le c}, $$ leaving the others fixed. All corrections lie in $K$. Expanding each polynomial monomial shows $f_i(b')=f_i(b)+\sum_\nu(\partial f_i/\partial X_\nu)(b)(b_\nu'-b_\nu)$, because every term with two corrections lies in $K^2$. Thus all $f_i(b')$ are zero. The image of $v\Delta(b')$ modulo $K$ is unchanged and is a unit; it is a unit in $D$ as well. Explicitly, if a lift of its inverse gives a product $1+j$ with $j\in K$, multiply that lift by $1-j$. Evaluation therefore extends through the quotient and localization to an $R$-algebra lift. For $c=0$ there are no errors to correct; choosing arbitrary variable lifts and using the same unit argument suffices. We have proved formal smoothness for every square-zero test. Together with the finite presentation it is smooth, by the written conormal criterion and its agreement with Definition F.6. The neighbourhood contains $\mathfrak q$, proving the conclusion. No Noetherian hypothesis has been used. ∎

Lemma F.46: flat over regular

Adapted from the Stacks project, algebra.tex, lines 34076–34098.

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 AG-CA-08, Corollary 4.4 and equation (6) 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 AG-CA-08, Corollary 4.4 and equation (6) and remarks following). Continue till one reaches the conclusion that $S$ is flat over $R$. ∎

Lemma F.47: flat over regular with regular fibre

Adapted from the Stacks project, algebra.tex, lines 28729–28751.

A flat local map of Noetherian local rings with regular source and regular closed fibre has regular target.

Proof. Let the source dimension be $d$ and fibre dimension $e$. Choose $d$ source parameters and lift $e$ fibre parameters. Together they generate the target maximal ideal: the fibre generators generate it modulo the source maximal ideal, which the source parameters already generate. The flat local dimension formula AG-CA-11, Theorem 5.1, gives target dimension $d+e$. Its embedding dimension is at most the number $d+e$ of these generators, and at least its dimension by the height theorem AG-CA-11, Theorem 3.1. Equality is the definition of a regular local ring. ∎

Lemma F.48: flatness descends more general

Adapted from the Stacks project, algebra.tex, lines 9135–9161.

Let $R$ be a ring. Let $S \to S'$ be a flat map of $R$-algebras. Let $M$ be a module over $S$, and set $M' = S' \otimes_S M$.

1. If $M$ is flat over $R$, then $M'$ is flat over $R$.

2. If $S \to S'$ is faithfully flat, then $M$ is flat over $R$ if and only if $M'$ is flat over $R$.

Proof. Let $N \to N'$ be an injection of $R$-modules. By the flatness of $S \to S'$ we have

If $M$ is flat over $R$, then the left hand side is zero and we find that $M'$ is flat over $R$ by the second characterization of flatness in Lemma AG-CA-07, Theorem 2.1. If $M'$ is flat over $R$ then we have the vanishing of the right hand side and if in addition $S \to S'$ is faithfully flat, this implies that $\ker(N \otimes_R M \to N' \otimes_R M)$ is zero which in turn shows that $M$ is flat over $R$. ∎

Lemma F.49: formally smooth extensions easy

Adapted from the Stacks project, algebra.tex, lines 45971–46012.

Let $K/k$ be an extension of fields.

1. If $K$ is purely transcendental over $k$, then $K$ is formally smooth over $k$.

2. If $K$ is separable algebraic over $k$, then $K$ is formally smooth over $k$.

3. If $K$ is separable over $k$, then $K$ is formally smooth over $k$.

Proof. For (1) write $K = k(x_j; j \in J)$. Suppose that $A$ is a $k$-algebra, and $I \subset A$ is an ideal of square zero. Let $\varphi : K \to A/I$ be a $k$-algebra map. Let $a_j \in A$ be an element such that $a_j \mod I = \varphi(x_j)$. Then it is easy to see that there is a unique $k$-algebra map $K \to A$ which maps $x_j$ to $a_j$ and which reduces to $\varphi$ mod $I$. Hence $k \subset K$ is formally smooth.

In case (2) we see that $k \subset K$ is a colimit of étale ring extensions. An étale ring map is formally étale (Lemma AG-CA-17, Theorem 2.2 and Proposition 6.1). Hence this case follows from Lemma F.22 and the trivial observation that a formally étale ring map is formally smooth.

In case (3), write $K = \operatorname*{colim} K_i$ as the filtered colimit of its finitely generated $k$-subextensions. By Definition F.5 each $K_i$ is separable algebraic over a purely transcendental extension of $k$. Hence $K_i/k$ is formally smooth by cases (1) and (2) and Lemma AG-CA-17, Theorem 1.2. Thus $H_1(L_{K_i/k}) = 0$ by Lemma F.17. Hence $H_1(L_{K/k}) = 0$ by Lemma F.24. Hence $K/k$ is formally smooth by Lemma F.17 again. ∎

Lemma F.50: geometrically regular

Adapted from the Stacks project, algebra.tex, lines 48753–48798.

For a Noetherian $k$-algebra, regularity after every finitely generated field extension is equivalent to regularity after every finite purely inseparable field extension.

Proof. One implication is immediate. For the other, given a finitely generated $K/k$, choose $k'/k$ and $K'/K$ by the explicit separabilization proof F.69. The assumed $A\otimes_k k'$ is regular and Noetherian. Write $K'$ as the localization of the smooth finite-type $k'$-domain constructed in the field lemma. Base change gives a smooth algebra over $A\otimes_k k'$. Smooth maps are flat by AG-CA-18, Theorem 6.1, and their residue-field fibres are regular by AG-CA-18, Theorem 2.1; regularity ascent proved above makes this base-changed algebra regular. Localization then makes $A\otimes_k K'$ regular. The map $A\otimes_k K\to A\otimes_kK'$ is faithfully flat, so regularity descent proved above makes $A\otimes_k K$ regular. These rings are Noetherian because finitely generated field tensor is a localization of a finite-type polynomial extension, the Noetherian field-tensor argument already written in this lesson. ∎

Lemma F.51: geometrically regular descent

Adapted from the Stacks project, algebra.tex, lines 48815–48834.

Geometric regularity descends through faithfully flat maps of Noetherian $k$-algebras.

Proof. For a finite purely inseparable $k'/k$, field tensor preserves faithful flatness by AG-CA-08, Proposition 2.3. The target tensor ring is regular by hypothesis, so the regularity descent proof above makes the source tensor ring regular. The finite purely inseparable criterion just proved gives geometric regularity of the source. No additional fibre assertion is assumed. ∎

Lemma F.52: geometrically regular over subfields

Adapted from the Stacks project, algebra.tex, lines 48851–48866.

If $A$ is Noetherian and geometrically regular over $k$, then $A\otimes_kE$ is regular for any field extension $E/k$ for which this tensor product is Noetherian. If $A\otimes_kE$ is Noetherian, it is geometrically regular over $E$. Also if $k$ is a directed union of subfields $k_i$ and $A$ is geometrically regular over every $k_i$, then it is geometrically regular over $k$.

Proof. Write $E$ as the directed union of its finitely generated subfields $E_i$. Fix a local ring $T=(A\otimes_kE)_{\mathfrak q}$. Its maximal ideal is finite, and a finite generating list lifts from one $A\otimes_k E_i$ after including the finitely many coefficients in the numerators and denominators. Put $B_i=(A\otimes_kE_i)_{\mathfrak q_i}$ at the contracted prime. The map $B_i\to T$ is flat local: field extension is flat, and the localizations preserve flatness. The extended ideal $\mathfrak q_iT$ contains the chosen maximal-ideal generators and lies in that maximal ideal, hence equals it. Its closed fibre is therefore the residue field of $T$, a regular ring. The source $B_i$ is regular by geometric regularity and Noetherian by the finite-type field argument. The already proved flat local ascent makes $T$ regular. This works at every prime.

For any finite purely inseparable $F/E$, the ring $(A\otimes_kE)\otimes_EF=A\otimes_kF$ is Noetherian, being finite over the given Noetherian ring. The argument just proved makes it regular. The finite purely inseparable criterion above now proves geometric regularity over $E$.

For the last assertion let $k'/k$ be finite purely inseparable. Choose its finite generator tower and the finite coefficients of the monic root equations, all in some $k_i$. Enlarge that index to include coefficients expressing products of a fixed $k$-basis of $k'$ and generators in that basis. Linear independence of that finite basis persists over every subfield; the product expressions give a field $k_i'\subset k'$ with the same basis over $k_i$, and $k\otimes_{k_i}k_i'=k'$. The extension $k_i'/k_i$ is finite purely inseparable because the generator powers lie in $k_i$. Therefore $A\otimes_kk'=A\otimes_{k_i}k_i'$ is regular by the hypothesis over $k_i$. The finite purely inseparable criterion proves geometric regularity over $k$. ∎

Lemma F.53: grothendieck regular sequence

Adapted from the Stacks project, algebra.tex, lines 24513–24526.

For a flat local map of Noetherian local rings $R\to P$, a list whose reduction is regular in the closed fibre is regular in $P$, and every successive quotient remains flat over $R$.

Proof. For the first element apply AG-CA-08, Lemma 5.1, to multiplication $P\xrightarrow{f_1}P$. The target is $R$-flat, both modules are finite over the Noetherian local ring $P$, and the map on the closed fibre is injective. That exact proved lemma gives injectivity and flatness of $P/(f_1)$. Repeat on the successive local quotient, whose closed fibre is the corresponding regular quotient. Local Nakayama ensures that every quotient remains nonzero, since the elements belong to the maximal ideal. This proves every assertion by induction. ∎

Lemma F.54: henselian cat finite etale

Adapted from the Stacks project, algebra.tex, lines 43976–44032.

For a henselian local ring $(R,\mathfrak m,k)$, reduction modulo $\mathfrak m$ gives an equivalence between finite etale $R$-algebras and finite products of finite separable extensions of $k$.

Proof. A finite algebra over $R$ is a product of local rings by AG-CA-20, Theorem 2.2. Each factor of a finite etale algebra is etale, by the lifting property restricted by its idempotent, and its residue is a finite separable field by AG-CA-17, Theorem 7.2. For full faithfulness it suffices to treat local factors $S_1,S_2$: a field map between their residues lifts uniquely to $S_1\to S_2$ by AG-CA-20, Theorem 1.2, since a finite local algebra over henselian $R$ is henselian by Proposition 3.1 there. Ring maps to a local factor select one source idempotent; its residue selects the same idempotent. Thus the local-factor statement proves full faithfulness for products.

For essential surjectivity, choose a finite separable tower $k=k_0\subset k_1\subset\cdots\subset k_t=L$ with each step generated by one element. Starting with $S_0=R$, lift the monic minimal polynomial to $S_{i-1}[T]$ and put $S_i=S_{i-1}[T]/(f_i)$. This ring is finite free over $S_{i-1}$ and has residue field $k_i$. Every maximal ideal contracts to the maximal ideal of $S_{i-1}$ by the integral maximal-ideal theorem, so $S_i$ is local with maximal ideal $\mathfrak mS_i$. The derivative $f_i'$ has nonzero residue in this field and hence is a unit. The one-equation Jacobian lift proves $S_i/S_{i-1}$ etale. Its finite etale composite over $R$ lifts $L$. Finite products lift the general object. This proof does not consume a decomposition theorem for arbitrary finite-type algebras over a henselian ring. ∎

Lemma F.55: huber

Adapted from the Stacks project, algebra.tex, lines 37957–37984.

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 F.44. Let $f_1, \ldots, f_c \in I$ be elements which map to a basis of $I/I^2$. By Nakayama's lemma (Lemma F.10) there exists a $g \in 1 + I$ such that

and $I_g \cong (f_1, \ldots, f_c)_g$. Hence we see that

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 F.72. ∎

Lemma F.56: isolated point fibre

Adapted from the Stacks project, algebra.tex, lines 31543–31578.

For a finite-type map $R\to S$ and a fibre point $q$, the following are equivalent: it is isolated in its fibre; its local fibre algebra is finite over the residue field; it is a closed fibre point with zero-dimensional local fibre ring; and one principal neighbourhood contains no other point of that fibre. These are also equivalent to zero dimension at that fibre point and to finite residue extension together with zero-dimensional local fibre ring.

Proof. Work in the finite-type field algebra $T=S\otimes_R\kappa(\mathfrak p)$. Its dimension-at-a-point formula AG-CA-09, Theorem 6.1, is $\dim_x\operatorname{Spec}T=\dim T_q+\operatorname{trdeg}\kappa(q)$. Both summands are nonnegative. Thus zero point-dimension is precisely local dimension zero and finite residue extension; the latter finiteness is equivalent to closedness by AG-CA-06, Theorem 2.1. A zero-dimensional neighbourhood exists by the definition of point-dimension and principal opens forming a basis. Its finite-type field algebra is Noetherian of dimension zero, hence Artinian by AG-CA-03, Theorem 4.2. Its finite product decomposition and finite residue fields imply it is finite-dimensional over the field: its finitely many maximal-ideal-power quotients have finite-dimensional successive layers because each ideal is finite. Its points are finitely many isolated points, and selecting the factor of $q$ gives a principal singleton neighbourhood.

Conversely an isolated point admits such a principal neighbourhood. Its only prime is $q$, so it is a zero-dimensional local finite-type algebra and the same Artinian-layer proof makes it finite-dimensional; localizing it again leaves it unchanged. If the local fibre ring is finite-dimensional, it is Artinian and its residue field is finite, giving the preceding zero-point-dimension criterion. Finally principal opens of $T$ lift to principal opens of $S$, after clearing the one coefficient denominator from the base. This identifies the last neighbourhood statement with the original fibre statement and proves every equivalence. ∎

Lemma F.57: lci

Adapted from the Stacks project, algebra.tex, lines 37341–37441.

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 F.20). 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 $R=k[x_1,\ldots,x_n]_{\mathfrak q'}$. Iterating AG-CA-14, Proposition 3.3, starting with $k$, makes this a regular local ring. The full proof of AG-CA-12, Theorem 6.1, makes it Cohen–Macaulay. Its dimension is $\operatorname{height}(\mathfrak q')$ by prime-localization correspondence. 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 F.10 we see that $f_1, \ldots, f_c$ generate $I_{\mathfrak q'}$. Since the dimensions work out, we conclude by Proposition AG-CA-12, Theorem 4.2 and Corollary 4.3 that $f_1, \ldots, f_c$ is a regular sequence in $R$. By Lemma F.74 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 the dimension notation. 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 F.20 this dimension is equal to $n - c$. Finally, set $g$ equal to the image of $g'g''$ in $S$. Then we see that

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 $S_g\cong k[y_1,\ldots,y_m]/J$, with $J=(f_1,\ldots,f_t)$ and $\dim S_g=m-t$, be a global complete-intersection presentation. Let $\mathfrak q''$ be the polynomial prime corresponding to $\mathfrak qS_g$. First justify the dimension at this particular point. Every minimal prime $\mathfrak a$ of $J$ has height at most $t$ by AG-CA-11, Theorem 3.1. Its component has dimension $m-\operatorname{height}(\mathfrak a)$ by AG-CA-09, Theorem 4.3. The maximum of these finitely many component dimensions is $\dim S_g=m-t$. Thus each such height is at least $t$ as well, and every component has dimension exactly $m-t$.

The complete point-dimension calculation in AG-CA-09, Theorem 6.1, now gives $\dim_{\mathfrak qS_g}\operatorname{Spec}S_g=m-t$: every component through this point has that dimension. The local rings $(S_g)_{\mathfrak qS_g}$ and $S_{\mathfrak q}$ are identical, so their heights agree. Applying F.20 to the polynomial presentation therefore gives $$ t=m-\dim_{\mathfrak qS_g}\operatorname{Spec}S_g =\operatorname{height}(\mathfrak q'')-\operatorname{height}(\mathfrak q). $$ The ambient polynomial local ring is regular by iterating AG-CA-14, Proposition 3.3, starting with the field $k$. It is Cohen–Macaulay by AG-CA-12, Theorem 6.1. Its quotient has dimension $\operatorname{height}(\mathfrak q)$, so the displayed equality and the expected-dimension criterion AG-CA-12, Theorem 4.2, prove that the actual chosen $f_1,\ldots,f_t$ form a regular sequence there. This includes the empty-list case. By Lemma F.74 we see that $(J/J^2)_{\mathfrak q}$ is free of rank $t$. By Lemma F.33 we have

Localize this isomorphism at $\mathfrak q$. Since its left side is finite free, $(I/I^2)_{\mathfrak q}$ is a finite direct summand of a finite free module. It is flat by restricting tensor injections and finitely presented because its complementary summand is finite. The complete local calculation in AG-RG-S01, Lemma 1.2 makes it finite free. Reduction to the residue field in this same isomorphism computes its rank. 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). ∎

Lemma F.58: lci at prime

Adapted from the Stacks project, algebra.tex, lines 37592–37613.

For a finite-type algebra $S$ over a field and a prime $\mathfrak q$, being a complete-intersection local ring is equivalent to a global complete-intersection principal neighbourhood and to its polynomial presentation kernel being generated locally by a regular sequence.

Proof. A polynomial local presentation $P_{\mathfrak q'}\twoheadrightarrow S_{\mathfrak q}$ has regular local ambient ring by AG-CA-14, Proposition 3.3 and Theorem 3.2. If its kernel is a regular list of length $c$, the conormal is free of rank $c$ by F.74. Conversely, if the conormal can be generated by $c=\dim P_{\mathfrak q'}-\dim S_{\mathfrak q}$ elements, Nakayama lifts them to generators of the ideal, and the CM expected-dimension criterion AG-CA-12, Theorem 4.2, makes them regular. Stable comparison of conormal modules under a change of polynomial presentation, proved above, preserves this assertion: the two ambient local dimensions differ by the difference of their variable counts, since AG-CA-09, Theorem 4.3, gives $\dim k[X]_{\mathfrak q'}=n-\operatorname{trdeg}_k\kappa(\mathfrak q)$. A stably free finite module over the local quotient is free, with the resulting prescribed rank.

For the definition allowing an arbitrary regular local ambient $Q$ essentially finite type over $k$, present $Q$ as a quotient of a polynomial local ring. Both ambient and quotient are regular. AG-CA-14, Proposition 1.4, proves that its kernel is generated by part of a regular parameter system. Lift the regular equations defining $S_{\mathfrak q}$ from $Q$ and append them to that system. By the definition of successive injectivity this concatenated list is regular. Hence the polynomial criterion applies; the reverse implication uses that polynomial ambient itself as $Q$.

To obtain a principal neighbourhood, clear the finitely many denominators in the local regular generator list. Kill the finite cokernel of its map onto the presentation ideal, then kill its finitely many positive Koszul homology modules by another element outside $\mathfrak q'$. At every prime on this neighbourhood containing the ideal the list is regular by the Noetherian $H_1$ criterion. Every minimal prime of the resulting equation ideal therefore has height exactly $c$: its quotient local ring has dimension zero, while each regular equation lowers dimension by one by AG-CA-12, Corollary 6.2. Each component has dimension $n-c$ by AG-CA-09, Theorem 4.3. Adjoining the localization variable gives a global presentation with $n+1$ variables and $c+1$ equations and this same dimension. This is a global complete intersection. A global complete-intersection presentation in turn has regular local equation lists by the minimal-prime height and CM argument proved above, completing the equivalence. ∎

Lemma F.59: limit essentially finite presentation

Adapted from the Stacks project, algebra.tex, lines 33722–33770.

A local essentially finitely presented map $R\to S$ is a directed colimit of local maps $R_\lambda\to S_\lambda$ essentially of finite type over $\mathbb Z$, with transition maps obtained by tensoring the base and localizing at the contracted prime.

Proof. Write $S=(R[X_1,\ldots,X_n]/(f_1,\ldots,f_c))_{\mathfrak q}$. For each finite subset of $R$ containing every equation coefficient, let $A_\lambda\subset R$ be the subring it generates over the image of $\mathbb Z$, and set $R_\lambda=(A_\lambda)_{\mathfrak m_R\cap A_\lambda}$. These are local subrings of $R$ with local inclusions. Their directed union is $R$, since each element is included in some finite subset. Form $C_\lambda=R_\lambda[X]/(f_i)$ and localize it at the inverse image $\mathfrak q_\lambda$ of the chosen target prime, giving $S_\lambda$. The maps are local and essentially finite type, and every stage is Noetherian by the Hilbert basis theorem.

Tensoring $C_\lambda$ with $R_\mu$ gives $C_\mu$. After initially inverting the complement of $\mathfrak q_\lambda$, localizing at the contracted prime inverts exactly the additional complement of $\mathfrak q_\mu$; the fraction universal property identifies this with $S_\mu$. Every fraction in $S$ has a numerator and denominator involving finitely many coefficients and hence occurs at one stage. If two fractions become equal, cross multiplication gives an equality in the polynomial quotient after one denominator outside $\mathfrak q$ is multiplied in. Its finite ideal-membership expression uses only finitely many coefficients, so the equality holds at a later stage. This proves both surjectivity and injectivity of $\operatorname{colim}S_\lambda\to S$, completing every claimed colimit identification. ∎

Lemma F.60: limit module essentially finite presentation

Adapted from the Stacks project, algebra.tex, lines 33804–33859.

A finitely presented module $M$ over the local algebra in the preceding approximation has finitely presented models $M_\lambda$ at sufficiently large stages, compatible by tensoring with $S_\mu$, with colimit $M$.

Proof. Choose a finite matrix presentation $S^a\xrightarrow{H}S^b\to M\to0$. Its finitely many entries are fractions and occur in one $S_{\lambda_0}$ by the preceding representative proof. Use their images to define $M_\lambda=\operatorname{coker}(S_\lambda^a\xrightarrow{H_\lambda}S_\lambda^b)$ at all later stages. Tensor is right exact, so $M_\lambda\otimes_{S_\lambda}S_\mu=M_\mu$. Taking the colimit of the finite free presentations gives the original cokernel: a vector represents zero precisely when it equals the image of a finite vector, and this finite equality holds at a common later stage. Thus the colimit is $M$, each model is finitely presented, and all ring-transition assertions remain those proved above. ∎

Lemma F.61: local dimension zero henselian

Adapted from the Stacks project, algebra.tex, lines 44097–44121.

Let $(R, \mathfrak m)$ be a local ring of dimension $0$. Then $R$ is henselian.

Proof. Let $R \to S$ be a finite ring map. By Lemma AG-CA-20, Theorem 1.2 and Theorem 2.2 it suffices to show that $S$ is a product of local rings. By Lemma AG-CA-05, Theorem 3.4 $S$ has finitely many primes $\mathfrak m_1, \ldots, \mathfrak m_r$ which all lie over $\mathfrak m$. There are no inclusions among these primes, see Lemma AG-CA-05, Theorem 3.3, hence they are all maximal. Every element of $\mathfrak m_1 \cap \ldots \cap \mathfrak m_r$ is nilpotent by Lemma AG-CA-01, Theorem 1.2 and Proposition 2.1. It follows $S$ is the product of the localizations of $S$ at the primes $\mathfrak m_i$ by Lemma F.73. ∎

Lemma F.62: local syntomic

Adapted from the Stacks project, algebra.tex, lines 37927–37941.

If finitely many principal opens cover $\operatorname{Spec}S$ and $R\to S_{g_i}$ is syntomic on each, then $R\to S$ is syntomic.

Proof. Finite presentation follows from F.35, including its explicit finite substitution calculation. For flatness, tensor any injection of $R$-modules with $S$ and localize its kernel at each $g_i$. Each kernel is zero by the assumed chart flatness; local detection on this principal cover makes the original kernel zero. In every residue-field fibre these same opens cover, and their local complete-intersection covers combine to give a cover by global complete-intersection neighbourhoods. This is precisely the fibre definition of syntomicness. No Noetherian hypothesis is required for this gluing step. ∎

Lemma F.63: localization colimit

Adapted from the Stacks project, algebra.tex, lines 1260–1278.

Localization of a module is the filtered colimit of copies of that module with transition maps given by multiplication by denominators. In particular a localization of a ring is the filtered colimit of its principal localizations.

Proof. For the first assertion, index the copies by finite products of elements of the multiplicative set, ordered with multiplication maps (or use the filtered divisibility category). Send an element $m$ in the copy indexed by $s$ to $m/s$. Every fraction has such a representative. Equality $m/s=n/t$ means $u(tm-sn)=0$ for some denominator $u$, precisely the equality at the common successor $ust$. Thus the map is bijective and respects addition and scalars. For rings use the principal subalgebras $R_s$: two such stages map to $R_{st}$; every element and every equality in the full localization is represented or becomes equal at a principal successor by the same fraction criterion. This proves the ring colimit assertion. ∎

Lemma F.64: localization smooth separable

Adapted from the Stacks project, algebra.tex, lines 46075–46088.

A finitely generated separably generated extension $K/k$ is the fraction field of a smooth finite-type $k$-domain. Conversely a fraction field of a smooth finite-type $k$-domain is separably generated.

Proof. For the forward implication choose a separating transcendence basis $T$, followed by a finite tower of separable simple algebraic generators. At each step its monic minimal polynomial has nonzero derivative at the generator. Clear finitely many coefficient denominators and invert that derivative and the needed denominator polynomials. These operations produce a finite-type domain inside $K$, with fraction field $K$, obtained from a polynomial algebra by a finite succession of standard étale extensions and principal localizations. AG-CA-17, Theorem 1.2 and Proposition 6.1, make it smooth.

For the converse localize the smooth algebra at its generic point; algebraic formal smoothness survives localization by AG-CA-17, Theorem 1.2. The field criterion proved below equates this with separability. Here is also why it supplies the finitely generated separating-basis convention. A standard smooth principal chart has an étale map from the polynomial algebra in the nonpivot variables: its square Jacobian and zero relative differentials prove this by AG-CA-17, Theorems 3.1 and 2.2. This map is flat by AG-CA-18, Corollary 6.2. A nonzero flat algebra over the polynomial domain has zero structural kernel, since tensoring multiplication by any nonzero element stays injective. Thus the nonpivot variables are algebraically independent in the domain. The dimension calculation AG-CA-18, Theorem 2.1, makes their number equal to the transcendence degree, so the fraction-field extension over them is finite algebraic. It is formally étale after localization, so its differentials vanish; the converse for finite fields in AG-CA-17, Lemma 7.1, makes it separable. They are the required separating transcendence basis. ∎

Lemma F.65: localize NL

Adapted from the Stacks project, algebra.tex, lines 37118–37133.

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

Proof. We have $S^{-1}B = \colim_{g \in S} B_g$ where we think of $S$ as a directed set (ordering by divisibility), see Lemma F.63. By Lemma F.72 each of the maps $\NL_{B/A} \otimes_B B_g \to \NL_{B_g/A}$ is a quasi-isomorphism. The lemma follows from Lemma F.24. ∎

Lemma F.66: localize relative complete intersection

Adapted from the Stacks project, algebra.tex, lines 38108–38158.

Let $R$ be Noetherian. 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

is a relative global complete intersection with a presentation as in Definition F.4 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 F.39 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 F.39 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 $S\otimes_R\kappa(\mathfrak p)$. Indeed, the image of $J$ generates the unit ideal in this fibre. Write a finite relation $1=\sum_j\bar a_j\bar g_j$ with $g_j\in J$. Choose a common denominator $s\in R\setminus\mathfrak p$ for the $\bar a_j$. Multiplication gives $g=\sum_j a_jg_j\in J$ whose fibre image is the nonzero scalar $s$, hence is a unit. Let $h \in R[x_1, \ldots, x_n]$ be any preimage of $g$.

Ad (3). By Lemma F.39 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 $g$.

In every case, a nonempty fibre of this principal localization has dimension at least $n-c$: its ambient polynomial algebra has $n+1$ variables and the displayed ideal has $c+1$ generators, so the height bound AG-CA-11, Theorem 3.1 and the finite-type field dimension formula give that lower bound. The upper bound already proved therefore gives dimension exactly $n-c$. This is precisely the relative global complete-intersection definition for the displayed presentation. ∎

Lemma F.67: locally smooth

Adapted from the Stacks project, algebra.tex, lines 39014–39037.

A ring map is smooth if and only if it is smooth on a principal neighbourhood of every target prime.

Proof. For the forward direction, take a finite polynomial presentation of the smooth algebra. Its conormal sequence is split and injective on the left by AG-CA-17, Theorem 3.1. Localization preserves this splitting and injection. By F.72 and the presentation comparison F.11, the conormal complex of a principal localization differs from the localized complex by a contractible summand. Thus that localization again has an injective conormal map and a projective differential module, giving formal smoothness by the same split-conormal criterion. Adding one variable and the relation $gZ-1$ preserves finite presentation by F.35, so the principal localization is smooth.

Conversely, choose for each target prime a principal neighbourhood on which the map is smooth. The elements of these principal opens generate one: otherwise the ideal they generate lies in a maximal ideal, which is an uncovered prime. A finite expression for one gives a finite subcover $D(g_1),\ldots,D(g_r)$. By F.35, $S$ is finitely presented over $R$. Choose $S=P/I$, where $P=R[X_1,\ldots,X_n]$ and $I$ is finitely generated. The conormal presentation $$ I/I^2\longrightarrow S^n\longrightarrow M=\Omega_{S/R}\longrightarrow0 $$ makes $M$ finitely presented: the differentials of a finite generating list of $I$ generate the kernel of $S^n\to M$. This is the conormal and polynomial-differential calculation of AG-CA-16, Section 3.

On each $D(g_i)$, F.72 and F.11 identify the first homology with that of the smooth chart, which is zero by AG-CA-17, Theorem 3.1. Exact localization commutes with this kernel, and the module detection calculation in F.35 makes the global first homology zero. On the same cover, the split-conormal criterion makes $M_{g_i}$ a direct summand of a finite free module. It is therefore flat: tensor injections remain injective after restricting to that summand. To prove $M$ flat over $S$, tensor any injection of $S$-modules with $M$. The kernel localizes to zero at every $g_i$, since the $M_{g_i}$ are flat; F.35 again makes the kernel zero.

The finitely presented flat module $M$ is finite projective by AG-CA-07, Theorem 5.3. Hence its conormal surjection splits, and the already proved vanishing of first homology makes the left conormal map injective. AG-CA-17, Theorem 3.1, gives formal smoothness. Together with finite presentation this is smoothness. If the target ring is zero, the empty presentation with the relation $1=0$ gives finite presentation, all these modules vanish, and a square-zero extension of the zero ring is zero because its unit is square-zero. Hence every lifting problem has the unique required map. Thus the empty-spectrum case is included. ∎

Lemma F.68: make etale map prescribed residue field

Adapted from the Stacks project, algebra.tex, lines 41287–41319.

For a prime $\mathfrak p$ of $R$ and a finite separable extension $L/\kappa(\mathfrak p)$, some finitely presented etale $R$-algebra has a point over $\mathfrak p$ with this prescribed residue field.

Proof. Choose finitely many elements generating $L$ as a field, and adjoin them one at a time. At a given step their monic minimal polynomial over the preceding residue field has nonzero derivative at the chosen element, since it is separable. Lift its finitely many coefficients to the local ring of the already constructed algebra at the chosen prime. Clear finitely many denominators to obtain that polynomial over one principal neighbourhood. Adjoin a root of this monic polynomial and invert its derivative. The square one-equation Jacobian calculation of AG-CA-17, Proposition 6.1, makes this algebra etale; its chosen residue-field evaluation maps the root to the required element. The kernel of this evaluation at the chosen point has exactly the enlarged residue field, since the localized base residue field and the root generate it. The polynomial coefficients and inverses involved at each of the finitely many steps descend from the local ring by finitely many denominator clearings. Composing the resulting finitely presented etale maps gives the asserted $R$-algebra. This construction uses a finite separable tower and does not assume a primitive element theorem. ∎

Lemma F.69: make separably generated

Adapted from the Stacks project, algebra.tex, lines 10182–10253.

For a finitely generated field extension $K/k$ there are finite purely inseparable extensions $k'/k$ and $K'/K$ with a commutative field square and $K'/k'$ separably generated.

Proof. Only characteristic $p>0$ needs proof. Choose a transcendence basis $T$; $K/k(T)$ is finite. For any finite extension, its separable elements form a finite separable subfield and the extension over it is purely inseparable: for each algebraic generator repeatedly write its irreducible polynomial as a polynomial in $X^p$ until its derivative is nonzero; the corresponding $p$th power of the generator is separable. Finite composita of separable simple extensions are separable, since they are generated by distinct-root polynomial towers. Thus sufficiently large $p$th powers of all generators lie in that separable subfield. Let $K_s$ be this subfield over $k(T)$ and induct on the finite $p$-power degree $[K:K_s]$.

If that degree exceeds one, choose $\beta\in K\setminus K_s$ with $\alpha=\beta^p\in K_s$. Write the monic separable minimal polynomial of $\alpha$ over $k(T)$ as $P(X)=X^d+\sum a_iX^i$. Each $a_i$ is a rational function involving finitely many coefficients of $k$. Adjoin their $p$th roots to $k$ to obtain a finite purely inseparable $k'/k$, and adjoin $T_j^{1/p}$ to the rational base. Every $a_i$ is now a $p$th power: take the $p$th roots of its numerator and denominator coefficientwise and replace each $T_j$ by $T_j^{1/p}$. Set $F'=k'(T_j^{1/p})$ and let $L=KF'$. It is finite purely inseparable over $K$.

Let $Q(X)$ be the polynomial obtained by taking those coefficient roots, so $Q(X)^p=P(X^p)$. The element $\beta$ is a root of $Q$, and $Q'(\beta)^p=P'(\alpha)\ne0$. Its minimal polynomial over $F'$ is therefore separable. The compositum $K_sF'$ is separable over $F'$, because separable polynomials remain separable under every field extension. Hence $K_s(\beta)F'$ lies in the separable subfield $L_s$ of $L/F'$. Since $[K:K_s(\beta)]=[K:K_s]/p$, the degree $[L:L_s]$ is at most this smaller degree: base change cannot increase finite vector-space degree. Induct on this strictly smaller integer, regarding $T_j^{1/p}$ as a transcendence basis over $k'$. The resulting finite purely inseparable extensions compose to the desired $k'/k$ and $K'/K$. At degree one the extension is already separably generated. This supplies the coefficient-root detail and the strict induction decrease. ∎

Lemma F.70: matrix left inverse

Adapted from the Stacks project, algebra.tex, lines 2709–2745.

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 $m=0$, the only minor is the empty determinant $1$, so $J=R$; both assertions hold for the unique matrix between the zero modules. Assume $m>0$. For $I\subset\{1,\ldots,n\}$ with $|I|=m$, let $E_I:R^n\to R^m$ select those coordinates in increasing order, and write $A_I=E_IA$.

For a square matrix $C$, its adjugate is the transpose of its signed cofactor matrix. The $(i,j)$ entry of $C\operatorname{adj}(C)$ is the cofactor expansion along row $j$, replacing that row by row $i$. It is $\det C$ when $i=j$ and zero when $i\ne j$, since then two rows coincide. The analogous column expansion gives $\operatorname{adj}(C)C=(\det C)1_m$ as well. The row and column expansions are obtained by grouping the signed permutation formula according to the chosen entry. Thus both identities hold over every commutative ring.

If $f=\sum_I c_I\det A_I$, set $B=\sum_I c_I\operatorname{adj}(A_I)E_I$. The identity just proved gives $BA=f1_m$, proving (1).

For (2), expand $\det(BA)$ by multilinearity in its $m$ columns, each a linear combination of columns of $B$. Terms selecting a repeated column vanish by alternation. Group the remaining terms by their $m$ distinct selected column indices $I$, in increasing order. The signed sum of the coefficients for each group is $\det A_I$, while its column determinant is $\det B^I$, where $B^I$ consists of columns indexed by $I$. Hence $$ \det(BA)=\sum_{|I|=m}\det(B^I)\det(A_I). $$ If $BA=f1_m$, its determinant is $f^m$, so the displayed identity gives $f^m\in J$. This proves (2) with no field, domain or reducedness hypothesis. ∎

Lemma F.71: presentation sym exterior

Adapted from the Stacks project, algebra.tex, lines 2282–2310.

For $M_2\to M_1\to M\to0$, the kernel of $\operatorname{Sym}^n M_1\to\operatorname{Sym}^nM$ is the image of $M_2\otimes\operatorname{Sym}^{n-1}M_1$; the analogous statement holds for exterior powers.

Proof. Put $N=\operatorname{im}(M_2\to M_1)$. The symmetric algebra is the tensor algebra modulo the pairwise commutation relations. Quotienting it further by the degree-one elements of $N$ gives the symmetric algebra of $M_1/N=M$, because maps from either quotient to a commutative algebra are exactly linear maps from $M_1$ annihilating $N$. In degree $n$ the ideal generated by $N$ consists precisely of finite sums $n_i u_i$ with $u_i$ of degree $n-1$; lifting each $n_i$ from $M_2$ gives the asserted image. Surjectivity follows from lifting every factor of a pure symmetric tensor. The exterior algebra is the tensor algebra modulo $x\otimes x$ for every degree-one element $x$; the same quotient and degree argument gives the exterior assertion, with product denoted by wedge. This includes $n=1$; for $n=0$ the kernel is zero. ∎

Lemma F.72: principal localization NL

Adapted from the Stacks project, algebra.tex, lines 37031–37116.

Let $A \to B$ be a ring map. Let $g \in B$. Suppose $\alpha : P \to B$ is a presentation with kernel $I$. Then a presentation of $B_g$ over $A$ is the map

extending $\alpha$ and sending $x$ to $1/g$. The kernel $J$ of $\beta$ is generated by $I$ and the element $f x - 1$ where $f \in P$ is an element mapped to $g \in B$ by $\alpha$. In this situation we have

1. $J/J^2 = (I/I^2)_g \oplus B_g (f x - 1)$,

2. $\Omega_{P[x]/A} \otimes_{P[x]} B_g = \Omega_{P/A} \otimes_P B_g \oplus B_g \text{d}x$,

3. $\operatorname{NL}(\beta) \cong \operatorname{NL}(\alpha) \otimes_B B_g \oplus (B_g \xrightarrow{g} B_g).$

Hence the canonical map $\NL_{B/A} \otimes_B B_g \to \NL_{B_g/A}$ is a homotopy equivalence.

Proof. Since $P[x]/(I, fx - 1) = B[x]/(gx - 1) = B_g$ we get the statement about $I$ and $fx - 1$ generating $J$. Consider the commutative diagram

with exact rows of Lemma F.42. The $B_g$-module $\Omega_{B[x]/B} \otimes_{B[x]} B_g$ is free of rank $1$ on $\text{d}x$. The element $\text{d}x$ in the $B_g$-module $\Omega_{P[x]/A} \otimes_{P[x]} B_g$ provides a splitting for the top row. The element $gx - 1 \in (gx - 1)/(gx - 1)^2$ is mapped to $g\text{d}x$ in $\Omega_{B[x]/B} \otimes_{B[x]} B_g$ and hence $(gx - 1)/(gx - 1)^2$ is free of rank $1$ over $B_g$. Here is a direct conormal calculation that also includes $B_g=0$. Put $q=gx-1$. Multiplication by $q$ on $B[x]$ is injective: for a nonzero polynomial, the coefficient of its lowest nonzero degree in its product with $q$ is the negative of that coefficient. The map $$ B[x]/(q)\longrightarrow(q)/(q^2),\qquad [a]\longmapsto[aq], $$ is surjective. If $aq=bq^2$, injectivity of multiplication by $q$ gives $a=bq$, so it is injective as well. Thus $(q)/(q^2)$ is the free rank-one $B_g$-module with generator $q$. When $g$ is nilpotent, $B_g=0$ and both modules are zero, so the same calculation applies without a proper regular-sequence hypothesis.

Let us prove $(I/I^2)_g \to J/J^2$ is injective. Consider the $P$-algebra map

sending $x$ to $1/f$. Since $J$ is generated by $I$ and $fx - 1$ we see that $\pi(J) \subset (I/I^2)_f = (I/I^2)_g$. Since this is an ideal of square zero we see that $\pi(J^2) = 0$. If $a \in I$ maps to an element of $J^2$ in $J$, then $\pi(a) = 0$, which implies that $a$ maps to zero in $I_f/I_f^2$. This proves the desired injectivity.

Thus we have a short exact sequence of two term complexes

For this sequence the following splittings commute with the differentials:

This works because $\text{d}(fx - 1) = x\text{d}f + f \text{d}x = g(g^{-2}\text{d}f + \text{d}x)$ in $\Omega_{P[x]/A} \otimes B_g$. The second summand is contractible: multiplication by $g^{-1}$ from degree zero to degree one is a homotopy, since its two composites with the differential $g$ are identity. Consequently inclusion and projection on the first summand are homotopy inverses. This proves the claimed homotopy equivalence. ∎

Lemma F.73: product local

Adapted from the Stacks project, algebra.tex, lines 13035–13062.

A ring with finitely many maximal ideals and locally nilpotent Jacobson radical is the product of its localizations at those ideals; every prime is maximal.

Proof. Write $I=\bigcap_i\mathfrak m_i$. Every element of $I$ is nilpotent, so every prime contains $I$, hence the product of the finitely many $\mathfrak m_i$, and therefore one maximal ideal. Thus these are all the primes. The Chinese remainder calculation gives $R/I=\prod_iR/\mathfrak m_i$: for two comaximal ideals choose $a+b=1$ with $a$ in the first and $b$ in the second and use $ub+va$ to lift a prescribed pair; induction gives the finite version. Idempotents lift uniquely across a nil ideal by AG-CA-01, Lemma 5.1. Lift the coordinate idempotents to $e_i$. Their products $e_ie_j$ and the difference $\sum e_i-1$ are idempotents in a nil ideal, hence zero. Consequently $R\to\prod_i e_iR$, $r\mapsto(e_ir)$, has inverse $(r_i)\mapsto\sum r_i$. The factor $e_iR$ has just the prime corresponding to $\mathfrak m_i$, so is local and is the localization $R_{\mathfrak m_i}$ by the fraction universal property. This proves the claimed product. ∎

Lemma F.74: regular quasi regular

Adapted from the Stacks project, algebra.tex, lines 17680–17766.

If $f_1,\ldots,f_c$ is a regular sequence in $A$, the conormal module of $I=(f_1,\ldots,f_c)$ is free over $A/I$ on the classes of the $f_i$. Only this consequence of quasi-regularity is used here.

Proof. First prove that every relation $\sum a_if_i=0$ is generated by the pairwise Koszul relations. For $c=1$, injectivity of $f_1$ gives $a_1=0$. For $c>1$, reduction modulo $(f_1,\ldots,f_{c-1})$ and injectivity of $f_c$ give $a_c=\sum_{i<c}b_if_i$. Subtract from the relation the sum of the relations $b_i(f_ie_c-f_ce_i)$, making its last coefficient zero. Apply induction to the remaining coefficients. In particular all coefficients of every relation lie in $I$. Now if $\sum a_if_i\in I^2$, subtract a displayed finite sum of products $f_if_j$, changing each $a_i$ by an element of $I$, to obtain a relation equal to zero. The preceding calculation puts its coefficients in $I$, hence puts the original $a_i$ in $I$. This proves injectivity of $(A/I)^c\to I/I^2$; generation proves surjectivity. ∎

Lemma F.75: regular sequence in polynomial ring

Adapted from the Stacks project, algebra.tex, lines 17576–17624.

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

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). ∎

Lemma F.76: relative global complete intersection

Adapted from the Stacks project, algebra.tex, lines 38280–38292.

A relative global complete intersection over a Noetherian base is syntomic.

Proof. It is finitely presented by its displayed finite equation list. Its fibres are global complete intersections by the dimension definition. Its flatness is the quotient-flatness conclusion of F.77, proved immediately below using F.53. These are exactly the three syntomic conditions. ∎

Lemma F.77: relative global complete intersection conormal

Adapted from the Stacks project, algebra.tex, lines 38204–38278.

Let $R$ be Noetherian and let $S=R[X_1,\ldots,X_n]/(f_1,\ldots,f_c)$ have all its nonempty fibres of dimension $n-c$. At every prime above the equation ideal, the $f_i$ form a regular sequence in the corresponding polynomial local ring, its successive quotients are $R$-flat, and $(f)/(f)^2$ is free over $S$ on their classes. This is the Noetherian form consumed in desingularization.

Proof. Fix a base prime $\mathfrak p$ and put $k=\kappa(\mathfrak p)$. In $k[X]$ every minimal prime of the equation ideal has height at most $c$, by AG-CA-11, Theorem 3.1. Its component has dimension $n$ minus that height, by AG-CA-09, Theorem 4.3. The asserted fibre dimension $n-c$ forces every such height to equal $c$. At a prime $\mathfrak r$ containing the equations, $A=k[X]_{\mathfrak r}$ is regular by iterating AG-CA-14, Proposition 3.3 from the field, hence CM by AG-CA-12, Theorem 6.1. For a minimal equation prime $\mathfrak a\subset\mathfrak r$, prime-localization correspondence preserves all primes below $\mathfrak a$, so $\dim A_{\mathfrak aA}=\operatorname{height}(\mathfrak a)=c$. The CM dimension formula AG-CA-12, Lemma 5.4, applied in $A$ at $\mathfrak aA$ therefore gives $\dim A/\mathfrak aA=\dim A-c$ for this component of the quotient. Taking the maximum gives that same quotient dimension. The expected-dimension criterion AG-CA-12, Theorem 4.2, makes the $c$ equations a regular sequence in this fibre local ring.

The polynomial local ring over $R_{\mathfrak p}$ is flat, since polynomial extension is free and localization is flat. Apply F.53, whose full Noetherian lifting proof uses AG-CA-08, Lemma 5.1, to obtain regularity before reduction and flatness of every successive quotient. The elementary regular-sequence relation calculation proved at F.74 makes its conormal module free on the classes of the $f_i$ at every prime of $S$. The global map $S^c\to(f)/(f)^2$ is surjective and has zero kernel after every prime localization, so AG-CA-02, Theorem 5.1, makes it an isomorphism. Flatness of the quotient is likewise detected locally by the ideal injection criterion and local detection. ∎

Lemma F.78: relative global complete intersection smooth

Adapted from the Stacks project, algebra.tex, lines 39071–39113.

For the preceding Noetherian presentation, smoothness at a prime is equivalent to one $c\times c$ Jacobian minor being outside that prime.

Proof. The preceding conormal calculation identifies the two-term complex with $S^c\xrightarrow{J}S^n$. If a chosen maximal minor is invertible, the inverse of that square block gives a left inverse to $J$; elimination identifies its cokernel with a free module of rank $n-c$. The split conormal criterion AG-CA-17, Theorem 3.1, gives smoothness on that localization. Conversely smoothness supplies a left inverse after localization by that same criterion. Reducing at the prime preserves the left inverse. Thus the matrix over the residue field has rank $c$, and some $c$-minor is nonzero. Principal localization of the conormal complex is the explicit contractible-summand computation proved at F.72. ∎

Lemma F.79: section smooth

Adapted from the Stacks project, algebra.tex, lines 39859–39914.

For a smooth ring map $R\to S$ with section $\sigma:S\to R$, put $I=\ker\sigma$. Then $I/I^2$ is finite projective over $R$. If it is free of rank $d$, the $I$-adic completion is $R[[T_1,\ldots,T_d]]$.

Proof. The map $s\mapsto s-\sigma(s)\bmod I^2$ is an $R$-derivation to $I/I^2$ and sends each $i\in I$ to its class. Universal differentials give the inverse to the natural map $I/I^2\to\Omega_{S/R}\otimes_SR$: both composites agree on the algebra generators and on the ideal classes. Smoothness makes $\Omega_{S/R}$ finite projective by the split conormal criterion AG-CA-17, Theorem 3.1, so its base change is finite projective. The ideal $I$ is finitely generated: for finite algebra generators $x_j$ of $S/R$, subtracting their section values shows that $x_j-\sigma(x_j)$ generate the section kernel.

Choose $t_i\in I$ representing a free conormal basis. Define a map $b:R[[T]]\to\widehat S$ by convergent evaluation $T_i\mapsto t_i$. It is surjective: start with the residue coefficient in $R$, then successively correct the error in $I^n/I^{n+1}$ by a homogeneous degree-$n$ polynomial in the $t_i$. These monomials generate that quotient because the $t_i$ generate $I/I^2$; expanding products proves this in every degree. The corrections converge in the completion.

The chosen conormal identification defines a first-order $R$-map $S\to R[[T]]/(T)^2$ sending $t_i$ to $T_i$ and reducing to $\sigma$. Formal smoothness of $S/R$, proved equivalent to the split conormal property in AG-CA-17, Theorem 3.1, lifts this map successively through the square-zero kernels $(T)^n/(T)^{n+1}$. Its compatible inverse limit sends $I$ into $(T)$, hence extends to a continuous map $a:\widehat S\to R[[T]]$. The composite $ab$ sends $T_i$ to $T_i$ plus terms of degree at least two. Such a substitution is an automorphism: recursively correct each inverse variable one homogeneous degree at a time; the linear term is identity, so the correction at each degree is uniquely its negative error, and the series converge. The same recursion proves that both composites with the resulting inverse are identity in every finite quotient and hence in the separated limit. Replace $a$ by $(ab)^{-1}a$. Then $ab=1$. Since $b$ is already surjective it is also injective and gives the required isomorphism. ∎

Lemma F.80: silly

Adapted from the Stacks project, algebra.tex, lines 2605–2636.

Let $I_1,\ldots,I_r,J$ be ideals of a ring with $J\not\subset I_i$ for every $i$, and at most two of the $I_i$ not prime. Then $J$ contains an element outside all $I_i$.

Proof. For two ideals, choose $x\in J\setminus I_1$ and $y\in J\setminus I_2$. If neither already works, then $x\in I_2$, $y\in I_1$, and $x+y$ works. Remove any $I_i$ contained in another listed ideal, since avoidance of the larger ideal implies avoidance of the smaller. Induct on the number of remaining ideals. For $r\ge3$ renumber so $I_r$ is prime. Choose $x$ avoiding the first $r-1$ by induction. If $x\in I_r$, choose $y\in JI_1\cdots I_{r-1}\setminus I_r$: such $y$ exists because none of these factors is contained in the prime $I_r$, and products of separately chosen elements outside a prime stay outside it. Then $y\in I_i$ for $i<r$, so $x+y$ still avoids them, and it avoids $I_r$ because $x\in I_r$ and $y\notin I_r$. ∎

Lemma F.81: size extension pth roots

Adapted from the Stacks project, algebra.tex, lines 45787–45812.

If $a_1,\ldots,a_n$ in a characteristic-$p$ field have independent absolute differentials, adjoining their $p$th roots has degree $p^n$.

Proof. First prove the absolute differential calculation used here. A maximal $p$-independent set is a $p$-basis, by the Zorn and degree-$p$ argument in AG-CA-19S, Lemma 1.1. In its monomial expansion define the partial derivation of one basis element by differentiating the corresponding monomial and killing all $p$th-power coefficients. The relations $T_b^p-b^p$ have derivative zero, so these are well-defined derivations. Their values on the basis elements show that their differentials are linearly independent; differentiating every expansion shows that they span. An expansion has all partial derivatives zero precisely when every exponent is zero, hence precisely when it lies in the subfield of $p$th powers. Thus $da=0$ if and only if $a\in k^p$.

For $n=1$ this proves $a_1$ is not a $p$th power. The monic polynomial $T^p-a_1$ is irreducible: any proper factor over $k$ is $(T-\alpha)^e$ in an algebraic closure, $1\le e<p$, and its coefficient $-e\alpha$ would put $\alpha\in k$. Suppose by induction the first $n-1$ roots give degree $p^{n-1}$, with their standard monomial basis. If $a_n$ became a $p$th power there, write a root in that basis and raise its expansion to $p$. Then $a_n=\sum_E\lambda_E^p\prod_{i<n}a_i^{e_i}$. Its differential belongs to the span of the preceding $da_i$, a contradiction. The same degree-$p$ irreducibility argument proves the last extension degree $p$, and multiplication of finite vector-space degrees gives $p^n$. ∎

Lemma F.82: smooth over field

Adapted from the Stacks project, algebra.tex, lines 38622–38676.

A smooth finite-type algebra over a field is locally a complete intersection.

Proof. At a given prime choose a standard smooth neighbourhood by AG-CA-17, Theorem 5.1. In its polynomial ambient local ring, the classes of its chosen equations in the maximal ideal modulo its square are linearly independent. Indeed, differentiating an element of the square gives zero after reduction to the residue field; a linear relation between those equation classes would therefore give the same relation between their differential rows, contradicting the invertible Jacobian minor. Extend the independent classes to a basis of the maximal ideal modulo its square. The ambient polynomial local ring is regular by iterating AG-CA-14, Proposition 3.3, starting with the field; therefore AG-CA-12, Theorem 6.1, makes this full minimal generator list a regular sequence; its initial equation list is regular. The complete-intersection neighbourhood criterion just proved gives a global complete-intersection chart around the prime. Such charts cover the spectrum, proving the local statement without any algebraic-closure reduction or omitted diagram calculation. ∎

Lemma F.83: smooth syntomic

Adapted from the Stacks project, algebra.tex, lines 38880–38951.

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 $R$. If $R$ is Noetherian, then $R \to S$ is syntomic.

Proof. If $S=0$, take the empty standard-open cover. The map $R\to0$ is finitely presented by $R/(1)$, its zero module is flat, and its fibres are empty, so the syntomic consequence holds as well. Assume $S\ne0$.

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 $I/I^2$. Here is the full basis-open calculation over an arbitrary base. Write $M=I/I^2$. The split conormal sequence of AG-CA-17, Theorem 3.1, makes $M$ a finite direct summand of $S^n$. Its complementary summand is finite, generated by the projections of the coordinate vectors, so this splitting is a finite presentation of $M$. Tensoring an injection with $M$ is injective by restricting the injection obtained with $S^n$. Thus $M$ is finitely presented and flat, and the complete proof of AG-RG-S01, Lemma 1.2 gives finite freeness over each local ring.

At a point $\mathfrak q$, choose a basis of $M\otimes_S\kappa(\mathfrak q)$ from the spanning classes of the finitely many $f_i$, and let $E$ be its indices, of cardinality $r$. Over $S_{\mathfrak q}$, Nakayama makes the resulting map $\varphi:S^r\to M$ surjective; its square coordinate matrix in a free local basis is invertible modulo the maximal ideal, so the cofactor identity makes $\varphi_{\mathfrak q}$ an isomorphism. The finite cokernel of $\varphi$ vanishes there. Clearing the finitely many denominators annihilating its generators gives $a\notin\mathfrak q$ for which $\varphi_a$ is surjective. Projectivity of $M_a$ splits this surjection, so its kernel is a finite direct summand of $S_a^r$. Its localization at $\mathfrak q$ is zero. Clear the finitely many annihilating denominators of its generators as well, taking their numerators in $S$, to obtain $b\notin\mathfrak q$ with $\varphi_{ab}$ an isomorphism. Hence $D(ab)\subset U_E$.

Apply the same argument at any point where a fixed selected list is a basis. It gives a principal neighbourhood contained in its basis locus, proving that every $U_E$ is open. Since a spanning list over a field contains a basis, these opens cover $\operatorname{Spec}S$. Empty products are one, and the argument also covers rank zero. Choose for each point a principal neighbourhood $D(g)$ contained in its selected $U_E$, using the basis-open calculation just proved. These opens cover $\operatorname{Spec}S$, and every chosen $S_g$ is nonzero because its open contains that point. For such a $g$ we look at the presentation

mapping $x_{n + 1}$ to $1/g$. Setting $J = \ker(\beta)$ we use Lemma F.72 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 F.55 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 $\operatorname{NL}(\alpha)$ composed with the projection

is an isomorphism. Here is the full minor-cover calculation. Let $A$ be the $n\times c$ matrix of the conormal differential in these bases. The split conormal sequence of AG-CA-17, Theorem 3.1, gives a $c\times n$ matrix $B$ with $BA=1_c$. Reduction at any maximal ideal gives an injection of a $c$-dimensional vector space into an $n$-dimensional one, hence $c\le n$. By F.70(2), the $c\times c$ minors $\Delta_E$ of $A$ generate the unit ideal. For a fixed $E$, the cofactor identity makes the selected square map invertible exactly where its determinant is a unit, so $U_E=D(\Delta_E)$. Thus these finitely many principal opens cover $\operatorname{Spec}S$, and we may use them as the required $D(g)$. For $c=0$, the empty determinant is one and the unique empty selected list covers the whole spectrum. By renumbering, we may assume $E = \{1, \ldots, c\}$. For a $g$ with $D(g) \subset U_E$ we look at the presentation

mapping $x_{n + 1}$ to $1/g$. Setting $J = \ker(\beta)$ we conclude from Lemma F.72 that $J = (f_1, \ldots, f_c, fx_{n + 1} - 1)$ where $\alpha(f) = g$ and that the composition

is an isomorphism: in the equation-class bases, its matrix is block triangular, with the selected square Jacobian block first and $g$ in the last diagonal slot. Its determinant is $g\Delta_E$, a unit in $S_g$, so the cofactor identity gives its inverse. 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.

For the final consequence, suppose $R$ is Noetherian. Standard smooth algebras are syntomic (Lemmas F.85 and F.76) and being syntomic over $R$ is local on $S$ (Lemma F.62). ∎

Lemma F.84: snake

Adapted from the Stacks project, algebra.tex, lines 311–343.

Given a commutative diagram

of abelian groups with exact rows, there is a canonical exact sequence

Moreover: if $X \to Y$ is injective, then the first map is injective; if $V \to W$ is surjective, then the last map is surjective.

Proof. The map $\partial : \ker(\gamma) \to \operatorname{coker}(\alpha)$ is defined as follows. Take $z \in \ker(\gamma)$. Choose $y \in Y$ mapping to $z$. Then $\beta(y) \in V$ maps to zero in $W$. Hence $\beta(y)$ is the image of some $u \in U$. Set $\partial z = \overline{u}$, the class of $u$ in the cokernel of $\alpha$. Changing the lift $y$ by the image of $x\in X$ changes $u$ by $\alpha(x)$, so the connecting map is well defined and additive. Exactness at $\ker\beta$ follows because a lift in $X$ of an element of $Y$ with zero image in $Z$ has $\alpha(x)=0$, using injectivity $U\to V$. Exactness at $\ker\gamma$ follows because $\partial z=0$ means $u=\alpha(x)$ for some $x$; then $y-x$ has zero image under $\beta$ and still maps to $z$. Conversely any such lift has connecting class zero. At $\operatorname{coker}\alpha$, if the image of $u$ in $V$ equals $\beta(y)$, its image $z$ in $Z$ is in $\ker\gamma$ and has $\partial z=[u]$; conversely this is the definition of $\partial$. At $\operatorname{coker}\beta$, if $v$ maps to $\gamma(z)$ in $W$, lift $z$ to $y\in Y$ and replace $v$ by $v-\beta(y)$. Exactness of the bottom row puts it in the image of $U$, so its class comes from $\operatorname{coker}\alpha$. The converse follows from the row being a complex. If $X\to Y$ is injective, so is its restriction to $\ker\alpha$. If $V\to W$ is surjective, every class modulo $\gamma(Z)$ has a representative lifted from $V$, proving the final surjectivity. ∎

Lemma F.85: standard smooth

Adapted from the Stacks project, algebra.tex, lines 38716–38780.

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

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.

For (4), represent $g$ by $h(x)$ and add a variable $z$ and the equation $hz-1$. The Jacobian block on $x_1,\ldots,x_c,z$ is block triangular, with determinant $h\det(\partial f_i/\partial x_j)$. Both factors are units in the localization, proving it standard smooth. For (6), use the same presentation over $R_f$ and the same Jacobian minor: it remains a unit, while $R_f[x]/(f_i)\cong S$ because $f$ was already a unit in $S$.

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 AG-CA-17, Section 4, equations (6) and (7) 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 $n - c$. Put $B=k[x_1,\ldots,x_n]/I\ne0$. It is locally a complete intersection by F.82. For a prime $\mathfrak q$ of $B$, with polynomial preimage $\mathfrak q'$, the complete equivalence in F.57 therefore identifies the free rank of $(I/I^2)_{\mathfrak q}$ with $\operatorname{height}(\mathfrak q')-\operatorname{height}(\mathfrak q)$. The standard-smooth conormal calculation above makes that rank exactly $c$; equality of free ranks is checked by reducing to the residue field.

In particular, if $\mathfrak q$ is a minimal prime of $B$, its height is zero, and its polynomial preimage is a minimal prime over $I$ of height $c$. By AG-CA-09, Theorem 4.3, this irreducible component has dimension $n-c$. There are finitely many such components, and their union is $\operatorname{Spec}B$, by AG-CA-03, Proposition 2.2. The finite-component dimension formula in AG-CA-09, Section 1, gives $\dim B=n-c$. This proves the assertion for every nonzero fibre, including empty equation lists and imperfect fields. Zero fibres impose no dimension condition. ∎

Lemma F.86: sum two terms

Adapted from the Stacks project, algebra.tex, lines 37135–37186.

Two homotopy-equivalent two-term complexes $A_1\xrightarrow{d_A}A_0$ and $B_1\xrightarrow{d_B}B_0$ satisfy $A_1\oplus B_0\cong B_1\oplus A_0$.

Proof. Let $\varphi,\psi$ be inverse comparison maps up to homotopies $h:A_0\to A_1$, $h':B_0\to B_1$, chosen with $1-\psi\varphi=hd_A$ in degree one and $d_Ah$ in degree zero, and the analogous equations on $B$. Form the block maps

Their product $UV$ has blocks $\psi_1\varphi_1+hd_A=1$, $-\psi_1h'+h\psi_0$, $-d_B\varphi_1+\varphi_0d_A=0$, and $d_Bh'+\varphi_0\psi_0=1$. Reversing the order gives blocks $1$, $\varphi_1h-h'\varphi_0$, $d_A\psi_1-\psi_0d_B=0$, and $1$. Both products are upper triangular with identity diagonal, hence invertible by negating their off-diagonal block. A map with an invertible product on each side has a left and a right inverse, which coincide: $(UV)^{-1}U$ and $U(VU)^{-1}$ are the inverses of $V$. Thus $U$ and $V$ are isomorphisms. ∎

Lemma F.87: syntomic

Adapted from the Stacks project, algebra.tex, lines 38315–38381.

Let $R \to S$ be a ring map with $R$ Noetherian. 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 F.3).

Proof. The implication (1) $\Rightarrow$ (3) is Lemma F.58. The implication (2) $\Rightarrow$ (1) is Lemma F.76. 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 F.58 the equivalent conditions of Lemma F.57 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 F.57). 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 F.28). Consider the short exact sequence

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 AG-CA-07, Theorem 2.1). 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 F.10) 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 F.66 as desired. ∎

Lemma F.88: syntomic presentation ideal mod squares

Adapted from the Stacks project, algebra.tex, lines 38383–38402.

Let $R$ be Noetherian. 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 F.87 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 AG-CA-07, Theorem 5.3, we may assume that $S_g$ is a relative global complete intersection. In this case the result follows from Lemmas F.33 and F.77. ∎

Lemma F.89: total ring fractions no embedded points

Adapted from the Stacks project, algebra.tex, lines 4748–4775.

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 F.80. Hence $\operatorname{Spec}(Q(R)) = \{\mathfrak q_1, \ldots, \mathfrak q_t\}$ (as subsets of $\operatorname{Spec}(R)$, see Lemma AG-CA-02, Theorem 3.1). 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 AG-CA-01, Theorem 5.2 and Corollary 5.3. Moreover $A_i$ is a local ring, which is a localization of $R$. Hence $A_i \cong R_{\mathfrak q_i}$. ∎

Lemma F.90: when colimit

Adapted from the Stacks project, algebra.tex, lines 33328–33357.

Let $R\to\Lambda$ be a ring map. Then $\Lambda$ is a filtered colimit of smooth $R$-algebras if and only if every map $A\to\Lambda$, with $A$ finitely presented over $R$, factors through a smooth $R$-algebra.

Proof. For the forward direction, choose finitely many generators and relations for $A$. Representatives of all generator images lie at one stage of the filtered diagram. The finitely many relations become zero at a common later stage, giving the factorization.

Conversely, consider the category of smooth finitely presented $R$-algebras equipped with a map to $\Lambda$, with morphisms commuting with that map. It is nonempty because it contains $R$. Given two objects $B_1,B_2$, their tensor product is finitely presented; factoring its map to $\Lambda$ through a smooth algebra gives a common successor. Given parallel arrows $u,v\colon B\rightrightarrows C$, choose finite algebra generators $b_i$ of $B$ and factor the finitely presented algebra $C/(u(b_i)-v(b_i))$ through a smooth algebra. The resulting arrow out of $C$ equalizes $u,v$. Thus the category is filtered. One may choose a set of representatives of finite presentations, so size causes no difficulty.

Its colimit maps onto $\Lambda$: for $\lambda\in\Lambda$, factor $R[T]\to\Lambda$, $T\mapsto\lambda$, through a smooth algebra. For injectivity, if an element represented by $b\in B$ maps to zero, factor $B/(b)\to\Lambda$ through a smooth algebra; then $b$ is already zero at a successor. Apply this to the difference of any two representatives after moving them to a common stage. This proves the ring map from the colimit to $\Lambda$ is an isomorphism. ∎

Proposition F.91: characterize separable field extensions

Adapted from the Stacks project, algebra.tex, lines 46037–46069.

For any field extension $K/k$, separability, geometric reducedness, algebraic formal smoothness, vanishing of $H_1(L_{K/k})$, and injectivity of $K\otimes_k\Omega_{k/P}\to\Omega_{K/P}$ are equivalent; here $P$ is the prime field. In characteristic $p$, separability means that a $p$-basis of $k$ remains $p$-independent over $K^p$. In characteristic zero all these conditions hold. For finitely generated extensions this agrees with having a separating transcendence basis.

Proof. Every field is formally smooth over its prime field by the full arbitrary-field proof AG-CA-19S, Lemma 1.1. Suppose the displayed differential map is injective. In a square-zero lifting test for $K/k$, first obtain a prime-field lift $u:K\to E$ by that lemma. Its restriction differs from the given map of $k$ by a derivation $D:k\to J$, where $J$ is a $K$-vector space through the map modulo $J$. The corresponding $K$-linear map $K\otimes_k\Omega_{k/P}\to J$ extends to $\Omega_{K/P}$ by choosing a basis complement. Subtract the resulting derivation from $u$. The square-zero product identity shows that this is still a ring map, and its restriction to $k$ is the prescribed one. Thus $K/k$ is formally smooth.

Conversely for any $P$-derivation $D:k\to V$, with $V$ a $K$-vector space, use the test algebra $K\oplus V$ with structural map $a\mapsto(a,D(a))$ and target quotient $K$. A lift of the identity gives a $P$-derivation of $K$ extending $D$. Hence every linear functional on $K\otimes_k\Omega_{k/P}$ extends to $\Omega_{K/P}$. If the displayed map had a nonzero kernel vector, a linear functional nonzero on that vector could not extend. It is therefore injective. The conormal/vector-space argument in the preceding formal-field lemma equates formal smoothness with the $H_1$ condition.

In characteristic zero choose a transcendence basis of $k/P$ and extend it to one of $K/P$. Algebraic extensions in characteristic zero are separable. The polynomial and separable algebraic differential formulas AG-CA-16, Theorem 6.2, identify their absolute differentials with the free spaces on those bases, proving the indicated injection. The same formula for arbitrary algebraic unions follows because each finite polynomial expression and relation occurs in a finite subextension. The lifting proof in the first paragraph consequently applies.

In characteristic $p$, the absolute differential computation in the preceding $p$-root degree proof says that a $p$-basis $B$ of $k$ has independent images of its differentials in $K$ precisely when it is $p$-independent over $K^p$. To verify this assertion directly, extend any maximal independent subset to a $p$-basis of $K$; its partial derivations detect linear independence. If a finite subset is $p$-dependent, take a minimal monomial relation over $K^p$ and differentiate it; a minimal relation has a nonzero partial derivative because its exponents are below $p$, giving differential dependence. This proves the equivalence with the stated separability condition.

That condition is equivalent to $K\otimes_k k^{1/p}$ being reduced. For a finite subset of $B$ the tensor product is $K[T_b]/(T_b^p-b)$, free on the standard monomials. It is a field exactly when those roots have degree $p^{\#B_0}$, which is exactly the indicated monomial independence. If the degree is smaller, its map to the field generated by the roots has a nonzero kernel; every element of that kernel has zero $p$th power, so it is not reduced. The whole tensor product is the directed union of these rings; these inclusions are injective by their monomial bases. Reducedness is therefore equivalent to independence of all finite subsets.

The independence persists at every root height. Indeed, a relation $\sum_a c_a^{p^r}B^a=0$ with $0\le a_b<p^r$ can be grouped by $a=pu+v$. Independence of the monomials $B^v$, $0\le v_b<p$, over $K^p$ makes each grouped coefficient, a $p$th power, zero. Taking its unique $p$th root and induction on $r$ makes every coefficient zero. Iterating the $p$-basis expansion proves that these are the bases for adjoining all $p^r$th roots of $k$. Consequently tensoring $K$ with any finite purely inseparable extension of $k$ gives a field: it injects into the corresponding full-root tensor field and is finite-dimensional over $K$.

For any finitely generated extension $E/k$, the explicit separabilization construction proved below provides finite purely inseparable $k'/k$, $E'/E$, with $E'/k'$ separably generated. Set $F=K\otimes_k k'$, a field by the last paragraph. Write $E'$ as a finite separable tower over $k'(T)$. Tensoring with $F$ gives a localization of a polynomial algebra over $F$, followed by finite étale algebras. These are regular by regularity ascent through flat maps with regular fibres, proved above, hence reduced. Thus $K\otimes_k E'=(K\otimes_k k')\otimes_{k'}E'$ is reduced. The map $K\otimes_kE\to K\otimes_kE'$ is faithfully flat, in particular injective, so the former is reduced. An arbitrary field extension is the directed union of its finitely generated subfields; all the corresponding tensor maps are injective, and a nilpotent element lies at one stage. This proves geometric reducedness for every field extension. Its converse follows by testing $k^{1/p}$.

Finally a finitely generated separable field is the fraction field of a smooth finite-type domain: choose any finite algebra model. In its polynomial local presentation at the generic point, formal smoothness splits the conormal differential; choose a conormal basis, lift it to the finite kernel, and use Nakayama to make these lifts generate the ideal locally. Their independent differential rows give a nonzero Jacobian minor. Clear the finitely many denominators for ideal generation and this minor, obtaining a principal standard smooth model by AG-CA-17, Theorem 4.1. The nonpivot-coordinate argument in the preceding lemma gives a separating transcendence basis. Conversely the polynomial and finite separable tower proves formal smoothness of a separably generated field. This completes all comparisons without imposing finite generation on the initial $K/k$. ∎

Definition F.92: G ring

Adapted from the Stacks project, more-algebra.tex, lines 12788–12793.

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.

Definition F.93: excellent

Adapted from the Stacks project, more-algebra.tex, lines 13961–13970.

Let $R$ be a ring.

1. We say $R$ is quasi-excellent if $R$ is Noetherian, a G-ring, and J-2.

2. We say $R$ is excellent if $R$ is quasi-excellent and universally catenary.

Definition F.94: local complete intersection

Adapted from the Stacks project, more-algebra.tex, lines 8947–8952.

A ring map $A \to B$ is called a local complete intersection if it is of finite type and for some (equivalently any) presentation $B = A[x_1, \ldots, x_n]/I$ the ideal $I$ is Koszul-regular.

Lemma F.94.A: arbitrary-ring presentation independence

Adapted from the freely accessible Stacks project, Tag 07CZ, with the finite-generation, change-of-basis and complex arguments written out below. This argument retains the lesson's GNU Free Documentation License terms.

Let $A$ be any ring and let $C$ have two finite polynomial presentations $$ P=A[T_1,\ldots,T_n]\twoheadrightarrow C=P/I,\qquad Q=A[X_1,\ldots,X_m]\twoheadrightarrow C=Q/J. $$ If $I$ is a Koszul-regular ideal in the sense of Definition F.95, then so is $J$. In particular the parenthetical assertion in Definition F.94 holds without a Noetherian hypothesis, and such a finite-type algebra is finitely presented.

Proof. First prove the required finite generation. At every prime containing $I$, a principal neighbourhood makes $I$ generated by a finite Koszul-regular list. The corresponding principal opens cover $V(I)$. If their elements are $a_\lambda$, no prime contains $I+(a_\lambda)$, so that ideal is the unit ideal: a proper ideal lies in a prime by the full prime-separation proof in AG-RG-S04, Lemma P0.1. Thus finitely many $a_1,\ldots,a_s$ suffice, and $$ 1=\sum_{i=1}^s b_i a_i+d,\qquad d\in I. $$ Choose in $I$ numerators of the finite generators of each $I_{a_i}$ and let $N$ be the ideal generated by those numerators and $d$. Then $N_{a_i}=I_{a_i}$. At a prime outside all $D(a_i)$, the displayed relation makes $d$ a unit, so $N$ and $I$ are both the unit ideal there. Every prime is in one of these cases. Local detection applied to $I/N$ gives $I=N$. Hence $I$ is finite and $C$ is finitely presented. Lemma F.44, proved for arbitrary rings, now makes $J$ finite too. If $C=0$, both kernels are unit ideals and their Koszul-regularity condition is vacuous; this also gives a finite presentation. We may henceforth fix a prime $\mathfrak q$ of the nonzero $C$.

Choose $p_j(T)\in P$ representing the images of $X_j$ and $q_i(X)\in Q$ representing the images of $T_i$. In $W=A[T,X]$ the common kernel is $$ L=(I,X_j-p_j(T))=(J,T_i-q_i(X)). $$ Quotienting either displayed ideal eliminates its added variables and gives precisely the specified map to $C$, proving both equalities. Let $\mathfrak r$ be the inverse image of $\mathfrak q$ in $W$ and put $s_i=T_i-q_i(X)$. Translation identifies $W$ with $Q[s_1,\ldots,s_n]$ and $L$ with $(J,s_1,\ldots,s_n)$. Expanding polynomials by their degree in the $s_i$ gives $$ L/L^2\simeq (J/J^2)\oplus C^{\oplus n}. $$ Indeed its constant part is $J/J^2$, its linear coefficients are read modulo $J$, and all terms with two $s$ factors lie in $L^2$. These descriptions define mutually inverse maps, so require no finiteness or flatness assumption.

The Koszul complex on $f_1,\ldots,f_c$ has free degree-$k$ basis indexed by the $k$-element subsets; its differential deletes one index with the alternating sign and multiplies by the corresponding $f_i$. The terms of its square cancel in pairs. If its first homology is zero, the classes of the $f_i$ form a free basis of $(f)/(f)^2$. To prove independence, correct a relation $\sum b_i f_i\in(f)^2$ by coefficients in $(f)$ so that its scalar product is exactly zero. First-homology vanishing expresses that relation as a sum of the skew Koszul boundaries, whose coefficients all belong to $(f)$. The original $b_i$ therefore belong to $(f)$ as well. Surjectivity follows because the $f_i$ generate the ideal.

By the hypothesis on $I$, choose $a\in P$ outside the contracted prime and a Koszul-regular list $f$ generating $I_a$. Polynomial extension and localization preserve its positive-homology vanishing: they are flat, and the exact flat-tensor homology calculation is proved in AG-RG-S01, Lemma 1.0. Append $u_j=X_j-p_j(T)$. Modulo $I$ these are successive polynomial variables translated by constants, and eliminating them leaves $P/I$. Multiplication by each such variable is injective, as comparison of polynomial coefficients shows, also after localization. Appending one element to a Koszul complex is the cone of its multiplication map; the connecting-homomorphism construction in AG-CA-06S, Section 2 then identifies its positive homology with the kernel and higher homology of that map. The higher groups are zero and the degree-zero multiplication is injective. Induction therefore makes the whole list $(f,u)$ Koszul-regular in $W_a$. Its conormal classes give a free basis at $\mathfrak r$ by the preceding relation calculation.

Consequently $(J/J^2)_{\mathfrak q}$ is a finite direct summand of that free conormal module. Such a summand is flat by restricting the free tensor injection, and is finitely presented because the complementary summand is generated by the projections of a finite basis. AG-RG-S01, Lemma 1.2 makes it finite free over the local ring $C_{\mathfrak q}$. Choose from a finite generating list of $J$ elements $g_1,\ldots,g_r$ whose conormal residue classes form a basis. Nakayama makes those classes generate the local conormal module; in a free basis their square coordinate matrix has unit determinant, so they form a basis there. They also generate $J$ in the local ring $Q_{\mathfrak q_Q}$, where $\mathfrak q_Q$ is the inverse image of $\mathfrak q$. In fact $J=(g)+J^2$ there; the finite module $J/(g)$ equals $J$ times itself, and $J$ is contained in that local maximal ideal. The same proved Nakayama calculation makes this quotient zero.

Thus $(g,s)$ and $(f,u)$ are two generator lists for $L_{\mathfrak r}$ whose conormal classes are bases. They have the same length, since reduction gives two bases of $L_{\mathfrak r}/\mathfrak rL_{\mathfrak r}$; here $L^2\subset\mathfrak rL$. Expressing one list in the other gives a square matrix invertible modulo the maximal ideal, hence invertible by the cofactor identity. Its exterior powers give an isomorphism of their Koszul complexes: the alternating deletion formula above checks commutation with the differential on every basis wedge, and the exterior powers of the inverse give the inverse maps. Permuting the list is the same argument with a permutation matrix. In particular the Koszul complex on $(s,g)$ has zero positive homology at this stage.

We need a principal neighbourhood, not only this stalk. Clear the finitely many denominators in both generator comparisons and in the invertible matrix, and invert its determinant. For equalities in a localization, also include an element outside $\mathfrak r$ killing each cross-difference; no zero divisor is cancelled. Together with $a$ and the finite generator comparison for $J$, their product $h\in W\setminus\mathfrak r$ makes the list $(s,g)$ generate $L_h$ and its Koszul complex isomorphic to the Koszul-regular complex on $(f,u)$ over $W_h$. Thus its positive homology vanishes over the actual neighbourhood $W_h$.

Finally set $v=h(q_i(X),X)\in Q$. It is outside $\mathfrak q_Q$, because evaluation modulo $(s)$ takes $\mathfrak r$ to $\mathfrak q_Q$. We have $$ W_h/(s_1,\ldots,s_n)=Q_v. $$ The Koszul complex on $s$ is a free resolution of this quotient: before localization it is the successive polynomial-variable complex, whose acyclicity follows from the same cone calculation; flat localization preserves it. Tensor this resolution with the finite free terms of the Koszul complex on $g$. The augmentation of the resulting total complex to the Koszul complex on $g$ over $Q_v$ is a homology isomorphism. Here is the finite elimination argument: include the quotient as augmented degree $-1$ in each column. Those columns are exact. In a total cycle choose its component with largest $g$ degree, lift its vertical cycle in that exact column, and subtract the total boundary. The largest degree disappears and only smaller degrees are changed. Repetition terminates and makes every cycle a boundary. This proves the augmentation assertion without a spectral-sequence or Noetherian assumption. The total complex is precisely the Koszul complex on $(s,g)$, so $g$ has zero positive Koszul homology over $Q_v$. The generator equality also descends to $J_v=(g)$. We have therefore produced the neighbourhood required by Definition F.95 at every prime. Empty lists obey the same formulas. This proves every assertion. ∎

Definition F.95: regular ideal

Adapted from the Stacks project, more-algebra.tex, lines 8710–8731.

Let $R$ be a ring and let $I \subset R$ be an ideal.

1. We say $I$ is a regular ideal if for every $\mathfrak p \in V(I)$ there exists a $g \in R$, $g \not \in \mathfrak p$ and a regular sequence $f_1, \ldots, f_r \in R_g$ such that $I_g$ is generated by $f_1, \ldots, f_r$.

2. We say $I$ is a Koszul-regular ideal if for every $\mathfrak p \in V(I)$ there exists a $g \in R$, $g \not \in \mathfrak p$ and a Koszul-regular sequence $f_1, \ldots, f_r \in R_g$ such that $I_g$ is generated by $f_1, \ldots, f_r$.

3. We say $I$ is a $H_1$-regular ideal if for every $\mathfrak p \in V(I)$ there exists a $g \in R$, $g \not \in \mathfrak p$ and an $H_1$-regular sequence $f_1, \ldots, f_r \in R_g$ such that $I_g$ is generated by $f_1, \ldots, f_r$.

4. We say $I$ is a quasi-regular ideal if for every $\mathfrak p \in V(I)$ there exists a $g \in R$, $g \not \in \mathfrak p$ and a quasi-regular sequence $f_1, \ldots, f_r \in R_g$ such that $I_g$ is generated by $f_1, \ldots, f_r$.

Lemma F.96: G ring goes up quasi finite

Adapted from the Stacks project, more-algebra.tex, lines 12823–12870.

Let $R \to R'$ be a finite type map of Noetherian rings and let

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 F.26 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 F.50. ∎

Lemma F.97: another helper G ring

Adapted from the Stacks project, more-algebra.tex, lines 13103–13159.

Let $R$ be any Noetherian complete local domain with fraction field $K$. First, for a prime $q$ of $R[X]$, $\widehat{R[X]_q}\otimes_RK$ is regular. If $q$ lies over the maximal ideal and $0\ne r\subset q$ lies over $(0)\subset R$, then $\widehat{R[X]_q}\otimes_{R[X]}\kappa(r)$ is geometrically regular over $\kappa(r)$.

Proof. Choose the already proved finite regular complete local subring $A_0\subset R$, and put $F=\operatorname{Frac}(A_0)$. Let $q_0$ contract $q$ to $A_0[X]$. The finite-extension completion decomposition gives $$\widehat{A_0[X]_{q_0}}\otimes_{A_0[X]}R[X]=\prod_i\widehat{R[X]_{q_i}},$$ with the finitely many primes over $q_0$, one of which is $q$. Tensor with $K$ over $R$. The left side becomes $\widehat{A_0[X]_{q_0}}\otimes_{A_0}K$: the finite domain $R$ over $A_0$ has $R\otimes_{A_0}F=K$, since a finite domain over a field is a field. The strong first assertion of the preceding helper makes $\widehat{A_0[X]_{q_0}}\otimes_{A_0}F$ geometrically regular over $F$. Its base change to the finite extension $K/F$ is regular by the field-extension proof. Each factor, including $\widehat{R[X]_q}\otimes_RK$, is regular. This proves the first assertion without any use of a general formal-smoothness/regularity equivalence.

For the second, $\kappa(r)/K$ is finite, because the nonzero prime $rK[X]$ is generated by an irreducible polynomial. Given a finite purely inseparable $L/\kappa(r)$, choose a finite integral $R$-subalgebra $B\subset L$ with fraction field $L$, by scaling finitely many algebraic generators into monic equations. It is complete by finite-module completion. It is local: a finite algebra over the complete henselian local ring decomposes into local factors by AG-CA-20, Theorem 2.2; this domain has no nontrivial idempotent and hence only one factor. Let $r'$ be the kernel of $B[X]\to L$ sending $X$ to its specified image from $\kappa(r)$; then $\kappa(r')=L=\operatorname{Frac}(B)$.

Finite-extension completion and tensor associativity identify the tested fibre with a product of $$\widehat{B[X]_{q_i}}\otimes_{B[X]}\kappa(r'),$$ over primes $q_i$ above $q$; factors not meeting this fibre are zero. In each remaining factor, $r'L[X]=(X-a)$ for some $a\in L$, so this is $$\big(\widehat{B[X]_{q_i}}\otimes_B L\big)/(X-a).$$ The parenthesized ring is regular by the first assertion, applied to the complete local domain $B$. The derivation $\partial/\partial X$ of $B[X]$ extends to its prime localization, completion and localization to $L$, kills $B$ and sends $X-a$ to one. The checked hypersurface quotient argument makes each factor regular. Thus every finite purely inseparable test extension $L$ gives a regular fibre, and the geometric-regularity criterion gives the assertion. This is the exact completion argument consumed by polynomial permanence of G-rings. ∎

Lemma F.98: cartier equality

Adapted from the Stacks project, more-algebra.tex, lines 9182–9206.

For a finitely generated field extension $K/k$, both $\Omega_{K/k}$ and $H_1(L_{K/k})$ are finite-dimensional and their dimension difference is $\operatorname{trdeg}_kK$.

Proof. Use the explicit finite global complete-intersection domain $C\subset K$ constructed in the monic-tower proof, with fraction field $K$. Write its polynomial presentation with $n$ variables and $c$ regular equations; the localization variable has already been counted, so $\dim C=n-c$. Its conormal module is free on the equation classes by the regular-relation calculation, and polynomial differentials are free on the variables. Localization and homotopy comparison give the two-term complex $K^c\to K^n$ for $K/k$. If the matrix rank is $r$, its homology dimensions are $c-r$ and $n-r$, respectively. Their difference is $n-c=\dim C=\operatorname{trdeg}_kK$, by AG-CA-09, Theorem 4.2. This also proves their finiteness. ∎

Lemma F.99: check G ring easy

Adapted from the Stacks project, more-algebra.tex, lines 12803–12821.

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

Proof. The completion map of each Noetherian local ring is flat by AG-CA-19, Theorem 3.2. Its fibre at a contracted prime $\mathfrak q$ is $\widehat{R_{\mathfrak p}}\otimes_R\kappa(\mathfrak q)$. Exactness of finite-module completion, AG-CA-19, Theorem 3.1, gives $\widehat{R_{\mathfrak p}}/\mathfrak q\widehat{R_{\mathfrak p}}=\widehat{(R/\mathfrak q)_{\mathfrak p}}$. Tensoring this equality with the fraction field of $R/\mathfrak q$ identifies the displayed fibre with the ring in the assertion. A flat map is regular precisely when all its residue-field fibres are geometrically regular, by the stated definition. Applying this to every local completion gives both directions. ∎

Lemma F.99A: completion of a flat local map

Let $(A,\mathfrak m)\to(B,\mathfrak n)$ be a flat local map of Noetherian local rings. Its induced map on maximal-ideal completions $\widehat A\to\widehat B$ is faithfully flat. The residue fields need not be equal.

Proof. A local map sends $\mathfrak m^r$ into $\mathfrak n^r$; passing to compatible finite quotients therefore defines the map of completions. It remains local because their maximal ideals are $\mathfrak m\widehat A$ and $\mathfrak n\widehat B$, and their residue fields are $A/\mathfrak m$ and $B/\mathfrak n$, respectively. These quotient and Noetherian assertions are AG-CA-19, Proposition 2.2 and Theorem 3.3.

Put $\kappa=A/\mathfrak m$. Choose a free $A$-resolution $F_\bullet\to\kappa$, as constructed in AG-CA-06S, Section 1. Completion is flat by AG-CA-19, Theorem 3.2, so $F_\bullet\otimes_A\widehat A$ is a free $\widehat A$-resolution of $\kappa\otimes_A\widehat A=\kappa$. The ring $\widehat B$ is flat over $B$ by the same theorem and hence flat over $A$ by composition. Consequently

Apply the proved local criterion for flatness, AG-CA-08, Theorem 4.2, to the Noetherian local map $\widehat A\to\widehat B$ and the finite $\widehat B$-module $\widehat B$. It proves flatness. A flat local map is faithfully flat by AG-CA-08, Theorem 2.1: its closed fibre is the nonzero residue-field extension, and the local faithful-flatness criterion applies. This proves the assertion. ∎

Lemma F.100: check G ring maximal ideals

Adapted from the Stacks project, more-algebra.tex, lines 13001–13037.