Henselian local rings and henselization

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Combined edition under GNU FDL 1.2 or later; original material retains its CC0 dedication.

A simple root in the residue field is an equation with an invertible linear term. Completeness solves it by taking a limit. Henselianity asks for the solution itself, without requiring limits of arbitrary sequences. We will construct the smallest local extension that supplies these solutions and then allow a chosen separable closure of the residue field.

Throughout, $(R,\mathfrak m,\kappa)$ is a nonzero local ring; it need not be Noetherian. Finite algebras are finite as modules. We use Completion, Theorems 5.1–5.2; Formally smooth, unramified and étale ring maps, Theorems 4.1 and 7.2, Proposition 6.3 and §7; and Smooth algebras over a field and the Jacobian criterion, Corollary 6.2. In particular, étale algebras are flat, their field fibres are finite products of finite separable fields, and their principal localizations remain étale.

Two internal results from the étale lesson will be explicit in our proofs: Proposition 7.4 proves that every map between étale $R$-algebras is étale [Stacks, Tag 00U7], and Theorem 7.6 identifies the written programme proof that every point of an étale algebra has a principal neighborhood of the form

R[T]_g/(f),\qquad f\text{ monic},\quad f'\text{ invertible}. \tag{1}

The latter is the standard étale neighborhood theorem [Stacks, Tag 00UE].

1. Roots and maps with a prescribed residue

The ring $R$ is henselian if every monic $f\in R[T]$ and every $\bar a\in\kappa$ satisfying $\bar f(\bar a)=0$, $\bar f'(\bar a)\ne0$, have a lift $a\in R$ with $f(a)=0$. It is strictly henselian if, in addition, $\kappa$ is separably closed. [Stacks, Tag 04GF.] The root with that residue is unique: expansion gives

f(b)-f(a)=(b-a)\bigl(f'(a)+(b-a)H\bigr).

If $b-a\in\mathfrak m$, the second factor is a unit, so two such roots coincide. This does not use separatedness.

Lemma 1.1 (uniqueness for étale maps). If $S$ is étale over $R$, two $R$-algebra maps $u,v:S\to A$ to a local ring that induce the same map $S\to A/\mathfrak m_A$ are equal.

Proof. In $S\otimes_R S$, the kernel $J$ of multiplication is generated by $s_i\otimes1-1\otimes s_i$, where the $s_i$ are finitely many algebra generators. The telescoping product identity expresses the difference of any polynomial in these generators by their differences, proving this assertion. The diagonal description of differentials in Kähler differentials, Theorem 4.1, identifies $J/J^2$ with $\Omega_{S/R}=0$. Thus $J=J^2$. Under $s\otimes t\mapsto u(s)v(t)$, the image of $J$ generates a finite ideal $L$ of $A$, generated by $u(s_i)-v(s_i)$. It satisfies $L=L^2$ and is contained in $\mathfrak m_A$. Hence $L=\mathfrak m_A L$: one inclusion follows from $L^2\subseteq\mathfrak m_A L$, and the other from ideal closure. Nakayama gives $L=0$. The generators have equal images, so $u=v$. $\square$

Theorem 1.2 (the section criterion). The ring $R$ is henselian if and only if, for every étale $R$-algebra $S$ and prime $\mathfrak q$ over $\mathfrak m$ with $\kappa(\mathfrak q)=\kappa$, there is a section $S\to R$ whose inverse image of $\mathfrak m$ is $\mathfrak q$. Such a section is unique. More generally, for any henselian local $A$, any ring map $R\to A$, and any compatible evaluation $S\to\kappa_A$, there is exactly one $R$-algebra map $S\to A$ inducing that evaluation.

Proof. Assume first that $R$ is henselian. The prime $\mathfrak q$ is maximal: $S/\mathfrak q$ is a $\kappa$-subalgebra of $\kappa(\mathfrak q)=\kappa$, hence equals $\kappa$. Apply (1) at this prime. Its evaluation specifies a simple root of $\bar f$ at which $\bar g\ne0$. Lift that root to $a\in R$. Then $g(a)$ is a unit, so evaluation defines $S_g\to R$, and restriction gives the required section. Its residue map has kernel $\mathfrak q$. Uniqueness is Lemma 1.1.

For the converse, apply the section assumption to

S=(R[T]/(f))_{f'},\qquad \mathfrak q=\ker(S\longrightarrow\kappa,\ T\longmapsto\bar a). \tag{2}

This is standard étale; the section sends $T$ to the required root.

Finally, form $S\otimes_R A$. The specified evaluation and the residue map of $A$ define a surjection to $\kappa_A$. Its kernel is a prime with that residue field. Base change preserves étaleness, so the section criterion for $A$ supplies a map to $A$. Restrict to $S$, and apply Lemma 1.1 for uniqueness. $\square$ [Stacks, Tag 04GG.]

Complete local rings are henselian by the simple-root theorem in Completion. Rings with nilpotent maximal ideal are henselian as well: Newton correction squares the error, so after finitely many steps the error is zero. Neither assertion needs Noetherianity.

2. Factorization and the finite-algebra criterion

Proposition 2.1. Over a henselian local ring, a factorization $\bar f=\bar g\bar h$ of a monic polynomial into coprime monic factors lifts uniquely to monic factors $f=gh$ of the prescribed degrees and residues.

Proof. Put $r=\deg\bar g$, $s=\deg\bar h$. Introduce $r+s$ variables for the nonleading coefficients of monic polynomials $G,H$. Equating the coefficients of $GH$ to those of $f$ gives $r+s$ equations. Their Jacobian at the residue coefficient point is the matrix of

\kappa[T]_{<r}\oplus\kappa[T]_{<s}\longrightarrow\kappa[T]_{<r+s}, \qquad (U,V)\longmapsto\bar hU+\bar gV. \tag{3}

Its kernel is zero: coprimality forces $\bar g\mid U$, and the degree bound gives $U=0$, then $V=0$. Equal dimensions make the determinant nonzero. Invert this Jacobian determinant in the coefficient algebra. The resulting square standard smooth presentation is étale, by Theorem 4.1 of the étale lesson: its differentials have rank zero, and the lifting proof there gives unique lifts. The prescribed residue coefficients define a $\kappa$-point. Theorem 1.2 gives a section, hence factors in $R[T]$. Conversely any such factors give a map from this algebra to $R$: the determinant has the same nonzero residue and is a unit. Lemma 1.1 makes the map, and hence the factors, unique. Degree-zero factors give zero coefficient spaces and are included. $\square$

Theorem 2.2 (finite-algebra and idempotent criteria). The following are equivalent:

$R$ is henselian.
For every finite $R$-algebra $B$, reduction induces a bijection between the idempotents of $B$ and of $B/\mathfrak mB$.
Every finite $R$-algebra is a finite product of local rings.

Proof. First record two facts valid for every finite $B$. Integrality makes each maximal ideal of $B$ contract to $\mathfrak m$, so $\mathfrak mB\subseteq\operatorname{Jac}(B)$. Its finite-dimensional $\kappa$-algebra quotient is Artinian, and therefore a finite product of local Artinian rings, by Noetherian and Artinian rings, Theorem 4.2.

Idempotents have at most one lift modulo an ideal in the Jacobson radical. If $e,f$ lift the same idempotent, then $e(1-f)$ and $f(1-e)$ are idempotents in that radical. An idempotent $z$ there is zero because $1-z$ is a unit and $z(1-z)=0$. Thus $e=ef=f$.

Assume (1), and let $e_0$ be an idempotent in $B/\mathfrak mB$. The cases $0,1$ lift immediately. Otherwise choose a lift $b\in B$ and a monic $F\in R[T]$ annihilating $b$, using the finite-module determinant argument of Integral extensions, Theorem 1.1. Since $\bar F(e_0)=0$ and both $e_0$ and $1-e_0$ are nonzero, $\bar F(1)=\bar F(0)=0$. Indeed, evaluate the polynomial on the orthogonal idempotents to get

\bar F(e_0)=\bar F(0)(1-e_0)+\bar F(1)e_0;

a nonzero scalar in either summand cannot annihilate its nonzero idempotent. Separate the full power of $T-1$:

\bar F=(T-1)^aH_0,\qquad a\ge1,\quad H_0(1)\ne0,\quad H_0(0)=0.

Proposition 2.1 lifts this to $F=GH$. In the finite free algebra $D=R[T]/(F)$, the ideals generated by $G,H$ sum to $D$: their quotient has zero reduction modulo $\mathfrak m$, so Nakayama kills that finite quotient. Their intersection is their product, which is zero. The Chinese remainder theorem gives

D=R[T]/(G)\times R[T]/(H). \tag{4}

The idempotent $(1,0)$ maps under $T\mapsto b$ to an idempotent of $B$. Its residue polynomial takes the value one at $1$ and zero at $0$, so evaluation on $e_0$ gives exactly $e_0$. This proves (2).

Now assume (2). Lift the primitive component idempotents of $B/\mathfrak mB$. Their pairwise products are idempotents reducing to zero, hence zero; their sum is an idempotent reducing to one, hence one. They decompose $B$ into finitely many factors. Each factor has exactly one maximal ideal, since maximal ideals correspond to those in its quotient by $\mathfrak m$. Thus (3) holds. If $B=0$, this is the empty product.

Conversely, under (3), each local factor reduces to a local ring and its only idempotents are $0,1$. This proves (2). To prove (1) directly, let $f,\bar a$ satisfy the simple-root condition. The algebra $R[T]/(f)$ is finite free. Its special fibre has one local factor equal to $\kappa$, corresponding to $T-\bar a$, because this factor of $\bar f$ has multiplicity one. The product decomposition in (3) lifts this component to a factor $D_1$. As a direct summand of a finite free module, $D_1$ is finite projective, hence free over the local ring; see Tor and flat modules, §5. Its rank is one, since $D_1/\mathfrak mD_1=\kappa$. Its unit is a basis: it generates modulo $\mathfrak m$, hence generates by Nakayama, and a generator of a rank-one free module is a basis. Therefore $R\to D_1$ is an isomorphism of algebras. The image of $T$ is the desired root. $\square$ [Stacks, Tag 04GG.]

3. Permanence without a finiteness assumption on the base

Proposition 3.1. A proper quotient of a henselian local ring is henselian. A finite local algebra over a henselian ring is henselian, and its structure map is local. Every finite algebra over a henselian ring is a finite product of henselian local rings.

Proof. If $I\subsetneq R$, then $I\subseteq\mathfrak m$ and $R/I$ has the same residue field. Lift the coefficients of a monic polynomial over $R/I$ to $R$; its prescribed simple residue root lifts in $R$, and the image solves the original polynomial.

Let $B$ be a finite local $R$-algebra. Its maximal ideal contracts to $\mathfrak m$ by integrality. Every finite $B$-algebra is finite over $R$, so is a product of local rings by Theorem 2.2. The same theorem applied to $B$ proves its henselianity. Each local factor of a finite $R$-algebra is itself finite over $R$, giving the last assertion. $\square$ [Stacks, Tag 04GH.]

Proposition 3.2. A filtered colimit of henselian local rings along local maps is henselian local. If they are all strictly henselian, so is their colimit.

Proof. Denote the colimit by $A$. Local maps induce injections on residue fields; their filtered colimit $K$ is a field. The kernel of $A\to K$ consists of classes represented in a stage's maximal ideal. Every other class is represented by a unit, whose inverse survives in $A$. Thus $A$ is local with residue field $K$.

A monic polynomial over $A$ and its simple residue root involve finitely many coefficients. Represent them at a common stage. The equation and the nonzero derivative already hold in that stage's residue field, because its map to $K$ is injective. Its henselianity supplies the root. For the strict assertion, take a separable polynomial over $K$. Its coefficients come from a stage. Its discriminant remains nonzero there, so all its irreducible factors there are separable and therefore linear in the separably closed stage field. It already splits there. Thus $K$ is separably closed. $\square$ [Stacks, Tag 04GI.]

4. Henselization from pointed étale algebras

An ordinary étale neighborhood of our residue point is an étale $R$-algebra $S$ with an $R$-algebra evaluation $\epsilon:S\to\kappa$. Its kernel is the specified prime. Morphisms are $R$-algebra maps preserving the evaluations. The finite-presentation condition lets us use a set of representatives: presentations use finite lists of polynomials, and each such algebra has a set of evaluations.

Lemma 4.1. This category is filtered.

Proof. It contains $(R,R\to\kappa)$. Two objects map into their tensor product, evaluated by $\epsilon(s\otimes t)=\epsilon_1(s)\epsilon_2(t)$. This is again étale.

For two parallel maps $u,v:S\to S'$, consider

D=S\otimes_R S,\qquad T=S'\otimes_D S, \tag{5}

where $D\to S'$ sends $s\otimes t$ to $u(s)v(t)$, and $D\to S$ is multiplication. The second map is étale by Proposition 7.4 of the formal étale lesson, since $D,S$ are both étale over $R$. Base change makes $S'\to T$ étale, hence $R\to T$ étale. Algebraically, $T$ is the quotient of $S'$ by all $u(s)-v(s)$: tensoring imposes exactly these relations. The common residue evaluation kills them, so it descends to $T$. The map to $T$ equalizes $u,v$, as required. $\square$

Define

R^h=\mathop{\mathrm{colim}}_{(S,\epsilon:S\to\kappa)} S. \tag{6}

Filteredness means that finitely many elements can be represented in a common object, and an equality holds exactly when it holds after some further morphism. One can see this directly by constructing the colimit from pairs consisting of an object and an element, identified when their images agree at a common later object; the equalizing condition makes this an equivalence relation.

Theorem 4.2. The ring (6) is henselian local, its maximal ideal is $\mathfrak mR^h$, and its residue field is $\kappa$. For every local map $R\to A$ with $A$ henselian local, there is a unique local $R$-algebra map $R^h\to A$. The map $R\to R^h$ is faithfully flat. [Stacks, Tags 04GN, 04GS.]

Proof. The evaluations give a surjection $R^h\to\kappa$. We identify its kernel. The field fibre $S/\mathfrak mS$ of a neighborhood is a finite product of finite separable fields. The evaluation selects a factor equal to $\kappa$. Lift its component idempotent to an element $c\in S$; we do not require this lift to be an idempotent. It has residue one in the selected factor and zero in all other factors. After inverting $c$,

S_c/\mathfrak mS_c=\kappa. \tag{7}

This localization is a further neighborhood. Any element with zero evaluation therefore belongs, at this further stage, to $\mathfrak mS_c$. It belongs to $\mathfrak mR^h$. The reverse inclusion is immediate.

An element with nonzero evaluation, represented by $s\in S$, becomes a unit in the further neighborhood $S_s$. Consequently the kernel is the unique maximal ideal, proving locality and the residue assertion.

Now let a monic $P\in R^h[T]$ have a simple root $a_0\in\kappa$. Its coefficients come from a neighborhood $S$, giving a monic $Q\in S[T]$. Evaluation $T\mapsto a_0$ defines a neighborhood

(S[T]/(Q))_{Q'}

over $R$: it is standard étale over $S$, then étale over $R$, and the derivative evaluates nonzero. Its class of $T$ is the desired root in the colimit. Thus $R^h$ is henselian.

For the universal property, the local map $R\to A$ gives $\kappa\to\kappa_A$. Compose every neighborhood evaluation with this map. The last assertion of Theorem 1.2 gives unique maps $S\to A$ with those residues. Their uniqueness makes them compatible with every neighborhood morphism. They induce $R^h\to A$. It is local because $\mathfrak mR^h$ maps into $\mathfrak m_A$. Any local extension has the same residue map, since $R$ already surjects onto $\kappa$; uniqueness on each neighborhood proves uniqueness on the colimit.

Finally, each $S$ is flat over $R$. Tensoring commutes with filtered colimits: a tensor is a finite sum, and each relation also involves finitely many elements, so both descend to a stage. Filtered colimits preserve injections by the same eventual-equality description. Therefore, for every ideal $J\subseteq R$, the injective maps $J\otimes_R S\to S$ give an injection $J\otimes_R R^h\to R^h$. The ideal criterion in Tor and flat modules, Theorem 2.1, proves flatness. A flat local map is faithfully flat by its residue criterion, proved there in Lemma 6.2. $\square$

The universal property uniquely determines the henselization. Two choices have unique local maps in both directions; their composites are the identity by uniqueness. It also supplies its functoriality for local maps. If $R$ is already henselian, the identity target gives an inverse to $R\to R^h$; the other composite is again the identity by uniqueness.

5. Strict henselization and the choice of an embedding

Fix a separable algebraic closure $\kappa\subseteq\kappa^{\mathrm{sep}}$. A strict neighborhood is an étale $R$-algebra $S$ with a compatible evaluation $\epsilon:S\to\kappa^{\mathrm{sep}}$. Its image is a finite separable extension $L$ of $\kappa$, by the field-fibre theorem. Equivalently it specifies a prime over $\mathfrak m$ and an embedding of that prime's residue field into the chosen closure. Morphisms preserve the evaluations.

Theorem 5.1. The colimit

R^{sh}=\mathop{\mathrm{colim}}_{(S,\epsilon:S\to\kappa^{\mathrm{sep}})} S \tag{8}

is strictly henselian local, with maximal ideal $\mathfrak mR^{sh}$ and residue field the chosen $\kappa^{\mathrm{sep}}$. If $R\to A$ is local, $A$ is henselian local, and an embedding

\iota:\kappa^{\mathrm{sep}}\longrightarrow\kappa_A

extending the residue map of $R\to A$ has been specified, there is a unique local map $R^{sh}\to A$ inducing $\iota$. Both $R\to R^{sh}$ and the natural $R^h\to R^{sh}$ are faithfully flat. [Stacks, Tags 04GP, 04GU, 07QM.]

Proof. Tensor products, evaluated by multiplying in $\kappa^{\mathrm{sep}}$, give common receiving objects. The coequalizer construction (5) works unchanged: agreement of the two evaluations makes it a pointed étale algebra. Thus the category is filtered.

Every $\alpha\in\kappa^{\mathrm{sep}}$ is represented by a strict neighborhood. Lift its monic separable minimal polynomial from $\kappa[T]$ to $f\in R[T]$. The algebra $(R[T]/(f))_{f'}$, evaluated at $\alpha$, is such a neighborhood. Hence the colimit evaluation surjects onto $\kappa^{\mathrm{sep}}$. Isolating a selected field factor of $S/\mathfrak mS$ as in (7) now gives $S_c/\mathfrak mS_c=L$. Its evaluation into the closure is injective on $L$. Thus a zero-evaluation element becomes an element of $\mathfrak mS_c$. Inverting any element with nonzero evaluation proves that the colimit is local, with the stated maximal ideal and residue field.

For henselianity, represent the coefficients of a monic polynomial at a stage $S$. Its prescribed simple residue root $\alpha$ need not belong to that stage's field $L$. First take a tensor cocone with a neighborhood representing $\alpha$. At this larger stage the polynomial coefficients and $\alpha$ are both represented. Adjoin a root and invert its derivative, using evaluation $T\mapsto\alpha$, exactly as in Theorem 4.2. This supplies the root in (8). The residue field is already separably closed, so the ring is strictly henselian.

For the universal property, compose each evaluation with $\iota$ and apply Theorem 1.2. Unique lifts are compatible, induce a map on the colimit, and give uniqueness. The map is local since the maximal ideal is extended from $\mathfrak m$.

The filtered-flatness argument of Theorem 4.2 proves that $R\to R^{sh}$ is faithfully flat. The universal property of $R^h$, applied to the henselian target $R^{sh}$, gives the natural local map $R^h\to R^{sh}$. Construct the strict henselization $C$ of $R^h$, using its residue field $\kappa$ and the same closure. The two strict universal properties give maps $R^{sh}\to C$ and $C\to R^{sh}$. Their composites are identities: the first by uniqueness for the $R$-map, the second by uniqueness for the $R^h$-map. For the latter, the composite restricted to $R^h$ is the identity structural map by the ordinary universal property. Therefore $C=R^{sh}$ canonically with these choices. Since $C$ is a filtered colimit of étale $R^h$-algebras, $R^h\to R^{sh}$ is faithfully flat too. $\square$

Given a local map $R\to B$ and a compatible embedding between chosen separable closures of their residue fields, apply Theorem 5.1 with $A=B^{sh}$. It gives the unique compatible map of strict henselizations. The embedding is part of the data: removing it removes uniqueness. Changing the chosen closure gives an isomorphism after an isomorphism of closures has been selected; different selections need not give the same ring map.

6. Infinitesimal quotients and the difference from completion

Proposition 6.1. For every local $R$, without a Noetherian hypothesis, the natural maps give

R/\mathfrak m^n\ \cong\ R^h/\mathfrak m^nR^h\quad(n\ge1),\qquad \widehat R\ \cong\ \widehat{R^h}. \tag{9}

Proof. Put $B=R/\mathfrak m^n$. Its maximal ideal is nilpotent, so it is henselian by §1. The universal property supplies $R^h\to B$, factoring through $C=R^h/\mathfrak m^nR^h$, and gives an inverse candidate to $B\to C$. The composite on $B$ is the identity because it fixes the image of $R$, which is all of $B$. The ring $C$ is henselian by Proposition 3.1, and $R\to C$ is local. The two maps $R^h\to C$ obtained from the other composite and from the quotient agree by the ordinary universal property. Since this quotient is surjective, the other composite is the identity on $C$. The isomorphisms respect the quotient transitions. The maximal ideal of $R^h$ is $\mathfrak mR^h$, whose $n$-th power is $\mathfrak m^nR^h$; taking inverse limits gives the completion isomorphism. $\square$ [Stacks, Tag 07QM.]

The two further structural results are

R\text{ Noetherian}\ \Longleftrightarrow\ R^h\text{ Noetherian}\ \Longleftrightarrow\ R^{sh}\text{ Noetherian};\qquad \dim R=\dim R^h=\dim R^{sh}. \tag{10}

The dimension equality holds for every local ring, including infinite dimension. We now prove both assertions [Stacks, Tags 06LJ, 06LK], with the completion argument included.

Lemma 6.2 (completion of a free module). If $A$ is Noetherian and $I\subset A$ is an ideal, the $I$-adic completion of any free $A$-module, of arbitrary rank, is flat over $A$.

Proof. Write $\widehat A$ for the completion. The complete module is unchanged if each copy of $A$ is replaced by $\widehat A$, by the isomorphisms modulo $I^n$. The map $A\to\widehat A$ is flat by the completion lesson, so it suffices to show flatness over the Noetherian complete ring $\widehat A$. Its completed direct sum consists of families $(c_\alpha)$ in $\widehat A$ for which, for every $n$, all but finitely many entries belong to $I^n\widehat A$. This description follows by reading a compatible finitely supported vector at each quotient level; completeness and separatedness give its individual coordinates.

Use the equational criterion for flatness. For a finite relation $\sum_{i=1}^r a_i u_i=0$ in this module, each coordinate column $v_\alpha=((u_i)_\alpha)_i$ belongs to the finite syzygy module $L=\ker(\widehat A^r\xrightarrow{(a_1,\ldots,a_r)}\widehat A).$ Choose generators $\ell_1,\ldots,\ell_s$ of $L$. Artin–Rees gives a constant $c$ with $L\cap I^{n+c}\widehat A^r\subset I^nL$ for all $n\ge0$. The columns $v_\alpha$ tend to zero in precisely the finite-exception sense above. Choose expressions $v_\alpha=\sum_j b_{j\alpha}\ell_j$ with the coefficients tending to zero as well: whenever $v_\alpha\in I^{n+c}\widehat A^r$, Artin–Rees permits all its coefficients to lie in $I^n$. For a nonzero column choose the largest such $n$, or arbitrary coefficients if there is no such $n$; for a zero column choose zero coefficients. Separatedness makes the largest $n$ finite for nonzero columns, and the original finite-exception property makes these choices tend to zero. Thus each family $b_j=(b_{j\alpha})_\alpha$ is in the completed free module. Reading the components of the $\ell_j$'s gives $u_i=\sum_j(\ell_j)_i b_j, \qquad \sum_i a_i(\ell_j)_i=0.$ This trivializes every relation, which is exactly the equational criterion. No bound on the rank was used. $\square$

Theorem 6.3 (Noetherian henselizations). A local ring $R$ is Noetherian if and only if $R^h$ is, and if and only if $R^{sh}$ is. When these conditions hold, both henselizations have Noetherian completions and their maps to those completions are faithfully flat.

Proof. A faithfully flat map $A\to B$ with $B$ Noetherian makes $A$ Noetherian. Indeed, for any ideal $I\subset A$, its extended ideal has generators expressible using finitely many elements of $I$. Let $I_0$ be generated by those elements. Then $IB=I_0B$, and faithful flatness applied to $I/I_0$ gives $I=I_0$. This proves the reverse implications, since both maps from $R$ are faithfully flat.

Assume now $R$ Noetherian, and write $S$ for either henselization. Its maximal ideal is $\mathfrak mS$, hence finite. Its completion $C$ is complete local with this extended finite maximal ideal, by Propositions 2.2–2.3 of Completion. The Cohen power-series presentation in the preceding required lesson, Theorem 6.1, makes $C$ Noetherian. It remains to prove $S\to C$ flat; we cannot assume $S$ is Noetherian in order to use Noetherian completion flatness.

The constructions in Sections 4–5 express $S$ as a filtered colimit of local rings $R_i$, each the localization at a selected point of an étale $R$-algebra. These local rings are Noetherian and have maximal ideal $\mathfrak m_i=\mathfrak mR_i$. Moreover $R_i\to S$ is flat and local: later transition maps come from maps between étale algebras, hence are flat by Proposition 7.4 of the formal étale lesson and Theorem 6.1 of the smoothness lesson; localization and filtered colimits preserve flatness. Let $k_i$ and $k_S$ be their residue fields. Choose a basis $(\bar e_\alpha)$ of $k_S$ over $k_i$, and lift it to $e_\alpha\in S$.

For every $n$, the module $S/\mathfrak m_i^nS$ is flat over the Artinian local ring $R_i/\mathfrak m_i^n$. The lifts $e_\alpha$ form a free basis over this ring, even when the residue extension has infinite degree. To verify that assertion, a lifted residue basis defines a map from a free module onto any flat module over an Artinian local ring. Its cokernel is zero modulo the nilpotent maximal ideal, hence zero by iterating that nilpotence; this uses no finiteness of the module. If $K$ is the kernel, flatness of the target makes $K$ inject into the free module after residue tensoring. The residue basis is an isomorphism, so $K/\mathfrak m_iK=0$, and the same nilpotence gives $K=0$. Thus the basis assertion holds, compatibly for all $n$, because the same lifts were chosen in $S$.

Taking inverse limits identifies $C$, as an $R_i$-module, with the completion of $\bigoplus_\alpha R_i$. Lemma 6.2 makes it flat over every $R_i$. A finite relation with coefficients in the colimit $S$ has all its coefficients represented at a common stage. Over that $R_i$, flatness of $C$ trivializes it by the equational criterion, and the same factorization works over $S$. Hence $C$ is flat over $S$. The map is local and the residue ring is nonzero, so it is faithfully flat. Since $C$ is Noetherian, the descent argument above makes $S$ Noetherian. This proves all assertions without circularly using flatness of a non-Noetherian completion. In the ordinary case formula (9) already identifies $C$ with $\widehat R$; the strict completion need not equal it. $\square$

Theorem 6.4 (dimension). For every local ring, $\dim R=\dim R^h=\dim R^{sh}$, with infinite dimension allowed.

Proof. Faithful flatness and going down lift every finite chain of $R$ to a chain of either henselization, so its dimension is at least $\dim R$. Conversely take a strict finite chain $\mathfrak q_0\subsetneq\cdots\subsetneq\mathfrak q_n$ in $S$. Choose an element in each difference $\mathfrak q_j\setminus\mathfrak q_{j-1}$ and represent all these elements at a common local étale stage $R_i$. Its contracted primes form a strict chain of length $n$. Their contractions to $R$ are also strict: two comparable distinct primes with the same contraction would give comparable distinct primes in one residue-field fibre of the étale algebra, whereas that fibre is a finite product of fields by the formal étale lesson, Theorem 7.2. Localization retains this property. Thus $n\le\dim R$. Taking suprema over finite chains proves the reverse inequality, including the case of unbounded chain lengths. No Noetherian argument occurs. $\square$

Two arithmetic examples. The ring $\mathbb Z_p$ is henselian by Completion, Solution 7.1 and Theorem 5.1. In contrast, $\mathbb Z_{(p)}$ is not henselian. For odd $p$, use $T^2-(1+p^2)$: its roots $1,-1$ modulo $p$ are simple, but $p^2<1+p^2<(p+1)^2$, so it has no rational root. A rational square root of an integer is an integer, by coprime numerator and denominator. For $p=2$, use $T^2-T-4$, whose two roots modulo two are simple but whose discriminant $17$ is not a rational square.

Algebraic series. Over any field, the subring

A_k=\{a\in k[[x]]:a\text{ is algebraic over }k(x)\} \tag{11}

is henselian. Exercise 7.4 proves this without identifying it with any henselization. For $k=\mathbb Q$, Exercise 7.6 proves that it is not complete. Thus henselianity supplies certain algebraic solutions while completeness supplies every compatible sequence of finite-order coefficients.

7. Exercises

Exercise 7.1 (easy). If $\operatorname{char}k\ne2$, prove that $k[x]_{(x)}$ is not henselian by testing $T^2-(1+x)$.

Exercise 7.2 (easy). Prove henselianity of every proper quotient of a henselian local ring directly from the root definition. Explain why the zero quotient is excluded.

Exercise 7.3 (medium). Suppose every finite algebra over a local ring is a product of local rings. Starting with a monic polynomial and a simple residue root, construct the lifted root and justify the rank-one algebra isomorphism.

Exercise 7.4 (medium). Prove that (11) is a local henselian ring with residue field $k$ and maximal ideal $xA_k$.

Exercise 7.5 (medium). Determine the henselization and strict henselization of a field $k$. Explain why a residue embedding is essential for the strict universal property.

Exercise 7.6 (hard). For $A_{\mathbb Q}$ in (11), show that its maximal-ideal completion is $\mathbb Q[[x]]$ and that the completion map is not surjective. Deduce that $\mathbb Q[x]_{(x)}^h$ is henselian but not complete, using its universal map to $A_{\mathbb Q}$ and (9).

8. Solutions

Solution 7.1. Reduction gives $T^2-1$; both roots are simple since two is nonzero. A root in the local polynomial ring would give a square root of $1+x$ in $k(x)$. Write a rational square as $a(x)^2/b(x)^2$ with coprime polynomials. Its exponent of the irreducible factor $x+1$ is even, whereas that exponent in $1+x$ is one. This is impossible. In particular the local ring fails the root condition.

Solution 7.2. For $I\subsetneq R$, locality gives $I\subseteq\mathfrak m$, residue field $R/\mathfrak m$, and maximal ideal $\mathfrak m/I$. Lift a monic polynomial over $R/I$ coefficient by coefficient to a monic polynomial over $R$. The simple residue root has exactly the same equation and derivative condition. Henselianity of $R$ supplies a root, whose image is the required root in $R/I$. The zero ring has no maximal ideal, so it is not a local ring under the convention of this lesson.

Solution 7.3. Let $D=R[T]/(f)$. Its special fibre splits as the Artinian factors attached to the powers of the distinct monic irreducible divisors of $\bar f$. The simple root gives the factor $\kappa[T]/(T-\bar a)=\kappa$. The assumed local product decomposition of $D$ reduces to this special-fibre decomposition: each factor's quotient is local, and its unique maximal ideal lies over $\mathfrak m$. Thus some factor $D_1$ has reduction $\kappa$.

It is a direct summand of the finite free $R$-module $D$, hence finite projective and therefore finite free over the local ring. Its rank is the dimension of its reduction, namely one. Nakayama says that its unit generates it. In any chosen free basis, the coefficient of that unit is outside $\mathfrak m$, hence invertible. The unit is therefore itself a basis, and the structural map $R\to D_1$ is an algebra isomorphism. Project the class of $T$ to $D_1$ and use its inverse to obtain $a\in R$. Then $f(a)=0$ and $\bar a$ is its residue.

Solution 7.4. Finitely many elements algebraic over $k(x)$ lie in a finite field extension of $k(x)$: adjoining them successively gives a tower of finite extensions. Their sums and products are algebraic, so $A_k$ is a ring. A series with nonzero constant term has an inverse in $k[[x]]$; its inverse is algebraic as an element of the same finite extension. Thus these elements are units in $A_k$. A series with zero constant term is $xb$ with $b\in k[[x]]$, and $b=a/x$ is algebraic, so belongs to $A_k$. Therefore $A_k$ is local with maximal ideal $xA_k$, and constant-term evaluation gives residue field $k$.

Let $f\in A_k[T]$ be monic with a simple root in $k$. Its coefficients lie in a finite extension $L$ of $k(x)$. The complete local ring $k[[x]]$ supplies a root $a$ with that residue, by the Hensel theorem of Completion. The equation makes $a$ algebraic over $L$, so the tower $k(x)\subseteq L\subseteq L(a)$ is finite. Thus $a$ belongs to $A_k$, proving henselianity in every characteristic.

Solution 7.5. A field is henselian: a residue root is already an element of the ring and the residue polynomial is the original polynomial. The ordinary universal property therefore gives $k^h=k$. The chosen $k^{\mathrm{sep}}$ is a filtered union of finite separable extensions of $k$, each étale by the étale field theorem. It is strictly henselian, and its maximal ideal is zero. For a henselian local $A$ with a specified embedding $k^{\mathrm{sep}}\to\kappa_A$, Theorem 1.2 uniquely lifts that embedding on every finite separable subextension. The maps agree on their overlaps by uniqueness and yield a unique $k$-algebra map $k^{\mathrm{sep}}\to A$. Its inverse image of $\mathfrak m_A$ is zero, so it is local. This is the strict universal property, hence $k^{sh}=k^{\mathrm{sep}}$.

For the dependence on the embedding, take $k=\mathbb R$ and its closure $\mathbb C$. The identity and complex conjugation are two distinct $\mathbb R$-maps $\mathbb C\to\mathbb C$; they induce different residue embeddings. Prescribing one selects one lift. Without that prescription uniqueness fails.

Solution 7.6. In $A=A_{\mathbb Q}$, the maximal ideal is $xA$. Every class modulo $x^n$ has a polynomial representative of degree below $n$. Conversely, a member of $A$ whose first $n$ coefficients vanish equals $x^nb$, and $b=a/x^n$ still belongs to $A$. Hence, compatibly in $n$,

A/x^nA=\mathbb Q[x]/(x^n),\qquad\widehat A=\mathbb Q[[x]]. \tag{12}

There are countably many polynomials over the countable field $\mathbb Q(x)$. Each nonzero polynomial has only finitely many roots in the field $\mathbb Q((x))$. Their union contains $A$, so $A$ is countable. But $\mathbb Q[[x]]$ is uncountable: its series with coefficients zero or one contain a copy of the set of binary sequences, which is uncountable by the diagonal argument. Thus (12) is not surjective on $A$.

Set $R=\mathbb Q[x]_{(x)}$. The inclusion into $A$ is local: rational functions regular at zero have algebraic expansions, and a nonzero value at zero makes them units. The universal property gives a local map $R^h\to A$. Suppose $R^h$ were complete. Formula (9), together with Completion, formula (12), would identify its canonical completion map with an isomorphism $R^h\to\mathbb Q[[x]]$, extending the polynomial inclusion. Since its maximal ideal is $xR^h$, the local map to $A$ preserves all powers of $x$. Its composite into $\mathbb Q[[x]]$ induces the identity on every truncated polynomial quotient: by (9), every such class in $R^h$ is represented by a polynomial. Consequently that composite is precisely the completion map. If the latter were surjective, every series would lie in the image of $A$, contradicting its countability. Therefore $R^h$ is henselian and not complete. No equality $R^h=A$ was needed.

References and proof scope

The Stacks project provides the henselian characterizations, permanence and universal constructions at the tags cited above. Timothy J. Ford, Commutative Algebra, version of 23 September 2026, Chapter 6, Section 4, treats the lifting of idempotents to integral algebras over complete rings, Hensel's Lemma and coefficient fields of complete local rings. The incorporated proofs in Section 6 retain the source credit and licence below.

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Proof dependencies. Theorems 6.3–6.4 prove the Noetherian equivalence and arbitrary-local dimension equality in (10). Maps between étale algebras are proved in the formal étale lesson, Proposition 7.4. Its Theorem 7.6 gives the exact written programme provider for principal standard étale neighborhoods and links its written algebraic Zariski proof in Morphisms of schemes, Theorem 1.1, Sections 1–2. The Cohen lesson, Theorem 6.1, supplies the complete finite-maximal-ideal Noetherian statement used here. All henselian criteria, permanence, universal properties, faithful-flatness and exercise arguments appear above.

Sources and adaptation history

The Noetherianity and dimension proofs in Section 6 incorporate the Stacks Project Authors’ henselization arguments from the pinned AI Integrated Stacks Project more-algebra chapter. The completed-free-module step is proved here directly with Artin–Rees, and flatness is established before Noetherian descent. This preserves the arbitrary-local-ring dimension statement and avoids assuming Noetherianity of the henselization. Incorporated source.

History: Stacks Project Authors, original construction; OpenAI GPT-6.1 Sol, mathematical exposition and completed verifications, October 2026.

Copyright and licence

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts and no Back-Cover Texts. A copy is supplied as GNU Free Documentation License 1.2.

The original course material remains available under its CC0 dedication. This combined edition, including the incorporated and adapted proof, is distributed under GNU FDL 1.2 or later. The source authors, incorporated source titles and mathematical adaptations are identified above. History: Stacks Project Authors, original source; the credited AI Integrated Stacks Project editorial contributors, where used; OpenAI GPT-6.1 Sol, course adaptation, October 2026.