# Coefficient rings and the Cohen structure theorem

*Adapted and self-checked by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026, from the Stacks Project Authors' proof in AI Integrated Stacks Project and its credited AI editorial companion. This combined lesson is distributed under GNU FDL 1.2 or later; its original material retains its CC0 dedication.*

Completion turns compatible approximations into actual elements. A coefficient ring goes further: it lifts the entire residue field coherently, so that elements can be expressed as power series in generators of the maximal ideal. In mixed characteristic, the coefficients must retain the prime integer rather than kill it. Cohen structure supplies exactly this distinction.

Read Completion, §§1–4 and Solution 7.1, first, along with Formally smooth, unramified and étale ring maps, Theorem 3.1, on the split conormal criterion. We also use the polynomial/direct-sum and filtered-colimit flatness proofs in Tor and flat modules, Proposition 3.1, the composition proof in Proposition 3.2, the localization proof in Theorem 3.3, and the ideal criterion in Theorem 2.1. The field-lifting and residue-extension constructions needed here are proved below, including their infinite cases. No Noetherian hypothesis is imposed on the coefficient-ring existence theorem.

This lesson incorporates the mathematical proof from AI Integrated Stacks Project, pinned at `565b10e987aba5969b21145a0833f42d69f96790`, algebra chapter, Cohen section, together with its source-linked editorial companion, supplementary proof 65. The exposition adds a direct prime-field lifting argument, states all intermediate maps, and proves the downstream regular complete-local specialization. Reuse and adaptations retain GNU FDL 1.2. Source and change notices appear at the end.

## 1. Lifting maps from the prime field

Formal smoothness means that maps can be lifted across every square-zero ideal. The residue field need not be finite, perfect, or finitely generated.

**Lemma 1.1.** Every field \(k\) of characteristic zero is formally smooth over \(\mathbb Q\). Every field \(k\) of characteristic \(p>0\) is formally smooth over \(\mathbb F_p\).

**Proof in characteristic zero.** Choose a transcendence basis \(B\), so \(k\) is algebraic separable over \(\mathbb Q(B)\). For a map \(k\to A/J\), \(J^2=0\), lift the images of the basis elements arbitrarily. Every nonzero denominator has invertible image modulo \(J\), hence is a unit in \(A\): lift its inverse and invert the resulting \(1+j\) by \(1-j\). Thus the basis lifts define a map \(\mathbb Q(B)\to A\).

Extend across each finite separable subextension by a finite tower of simple extensions, using the separable minimal polynomial at each step. A lift of the root is corrected by \(-f/f'\); its derivative is a unit and the square-zero Taylor identity proves existence. If two roots have the same residue, their difference times a unit is zero, proving uniqueness. The unique lifts of any two finite subextensions agree in their finite compositum, so give one compatible lift on their union, which is \(k\). This handles an arbitrary transcendence basis and arbitrary algebraic extension.

**Proof in characteristic \(p\).** A **\(p\)-basis** is a subset \(B\subset k\) such that the monomials \(\prod b^{e_b}\), with finite support and \(0\le e_b<p\), form a basis of \(k\) over \(k^p\). It exists by Zorn's lemma: a chain of \(p\)-independent sets has \(p\)-independent union, because each relation is finite. The subalgebra generated over \(k^p\) by a \(p\)-independent set is a field, since each finite subalgebra is a finite-dimensional domain. If an element \(a\) were outside the field generated by a maximal such set, its minimal polynomial there would be \(T^p-a^p\), irreducible of degree \(p\). To check irreducibility, in an algebraic closure all roots coincide with \(a\); an irreducible factor is purely inseparable and has degree one or \(p\). Degree one would put \(a\) in the field. Adjoining \(a\) would therefore preserve \(p\)-independence, a contradiction. This proves the basis assertion and the presentation

\[
k=k^p[T_b:b\in B]/(T_b^p-b^p:b\in B).
\tag{1}
\]

Now let \(\psi:k\to A/J\) be an \(\mathbb F_p\)-map. For \(c\in k^p\), write uniquely \(c=a^p\) and choose any lift \(\widetilde{\psi(a)}\in A\). Define

\[
\theta(c)=\widetilde{\psi(a)}^{\,p}.
\tag{2}
\]

This is independent of the lift, since \(j^p=0\) for \(j\in J\), and the Frobenius identities make it a ring map lifting \(\psi|_{k^p}\). Choose arbitrary lifts of \(\psi(b)\) for the basis elements. Their \(p\)-th powers are exactly \(\theta(b^p)\), so (1) defines a map \(k\to A\) reducing to \(\psi\). Both arguments also prove the nilpotent-ideal lifting property by iteration through successive square-zero quotients. \(\square\)

The positive-characteristic argument does not require a separating transcendence basis, which may fail to exist for an infinitely generated extension. A perfect field has empty \(p\)-basis; (2) then supplies the lift directly.

## 2. Lifting formal smoothness across the base

**Lemma 2.1.** Let \(A\to B\) be flat, let \(I\subset A\) satisfy \(I^2=0\), and assume \(B/IB\) is formally smooth over \(A/I\). Then \(B\) is formally smooth over \(A\).

**Proof.** Take a polynomial presentation \(P=A[X_t:t\in T]\twoheadrightarrow B\) with kernel \(J\); the variable set may be arbitrary. Flatness of \(B\) makes

\[
J\cap IP=IJ.
\tag{3}
\]

Indeed tensor the exact sequence \(0\to J\to P\to B\to0\) with \(A/I\); its left injectivity follows from flatness of \(B\). The kernel of \(J/IJ\to P/IP\) is exactly the left side of (3) modulo \(IJ\).

Put \(E=J/J^2\), \(F=\Omega_{P/A}\otimes_P B=\bigoplus_t B\,dX_t\), and let \(d:E\to F\) be the conormal map. By (3), reduction modulo \(I\) is the conormal sequence for \(P/IP\to B/IB\). The split conormal criterion supplies a left inverse \(\bar r:F/IF\to E/IE\). Lift its value on each free basis vector to obtain \(r_0:F\to E\). For

\[
\Delta=1_E-r_0d
\]

we have \(\Delta(E)\subset IE\). Since it is \(B\)-linear and \(I^2B=0\), its square is zero. Therefore

\[
r=(1_E+\Delta)r_0,
\qquad rd=(1_E+\Delta)(1_E-\Delta)=1_E.
\tag{4}
\]

The conormal map has a left inverse. Theorem 3.1 of the formal-smoothness lesson constructs a section of \(P/J^2\to B\) from precisely such a left inverse, and proves the lifting property. It applies with arbitrary variable sets and without Noetherianity. \(\square\)

The direction of the splitting in (4) matters. It is a left inverse on the conormal module, rather than an assertion that projective differentials alone force smoothness.

## 3. Prescribing the residue field of a flat local extension

**Lemma 3.1.** For any local ring \((A,\mathfrak n,k_0)\) and any field extension \(k/k_0\), there is a flat local map \(A\to B\) with maximal ideal \(\mathfrak nB\) and specified residue field \(B/\mathfrak nB=k\).

**Proof.** First extend by one element \(\alpha\) of the desired residue field. If it is transcendental, use \(A[T]_{\mathfrak nA[T]}\). Its maximal ideal is generated by \(\mathfrak n\), its residue field is \(k_0(T)\), and polynomial extension and localization are flat. If \(\alpha\) is algebraic with monic minimal polynomial \(\bar f\), lift its coefficients and use \(A[T]/(f)\). It is free over \(A\), with basis \(1,T,\ldots,T^{d-1}\), and its quotient by \(\mathfrak n\) is \(k_0(\alpha)\). Every maximal ideal contracts to the maximal ideal of \(A\), by the integral maximal-ideal theorem in *Integral extensions*, Theorem 3.3. Thus there is only one maximal ideal, namely \(\mathfrak nB\). No separability is needed in this construction.

Well-order the set of elements of \(k\). At a successor step adjoin the next element by the preceding construction, using the currently attained residue field. At a limit step take the directed colimit of the preceding rings. This ring is local: each element outside the union of the maximal ideals was already a unit at some stage. Its maximal ideal is \(\mathfrak n\) times the colimit, and its residue field is the union of the preceding subfields. Every stage is flat over the original \(A\), by composition and, at limits, directed-colimit flatness. After the whole well-ordered set has been processed, the residue field is exactly \(k\). The recursion ranges over a set and uses no maximality argument over a proper class. \(\square\)

## 4. Cohen rings for all positive-characteristic fields

A **Cohen ring** is a complete discrete valuation ring whose uniformizer is a prime integer \(p\).

**Theorem 4.1.** For every field \(k\) of characteristic \(p>0\), there is a Cohen ring \(C\) with \(C/pC=k\). For each \(n\ge1\), the map \(\mathbb Z/p^n\mathbb Z\to C/p^nC\) is flat and formally smooth.

**Proof.** Solution 7.1 of *Completion* proves that \(\mathbb Z_p\) is a complete DVR with uniformizer \(p\). Apply Lemma 3.1 to \(\mathbb Z_p\) and \(k/\mathbb F_p\). We obtain a flat local ring \(B\), with maximal ideal \(pB\) and residue field \(k\). Put \(C=\varprojlim_n B/p^nB\). Because the completion ideal is principal, Proposition 2.2 of *Completion* gives

\[
C/p^nC=B/p^nB
\tag{5}
\]

and \(p\)-adic completeness. An element with nonzero residue is a unit by the convergent geometric-series inverse, so \(C\) is local with maximal ideal \(pC\).

Multiplication by \(p\) on \(B\) is injective by flatness over \(\mathbb Z_p\). If \(pc=0\) in \(C\), its coordinate modulo \(p^{n+1}\) satisfies \(pc_{n+1}=0\). Injectivity on \(B\) implies \(c_{n+1}\in p^nB/p^{n+1}B\), so \(c_n=0\). Every coordinate is zero and \(c=0\).

For a nonzero \(c\in C\), separatedness gives a largest \(v\) with \(c\in p^vC\). Write \(c=p^vu\). The factor \(u\) is a unit, since otherwise \(c\in p^{v+1}C\). Products of nonzero elements stay nonzero because \(p\) is a nonzerodivisor. Every nonzero ideal is generated by the smallest power of \(p\) occurring among its nonzero elements. Thus \(C\) is a DVR and, in particular, Noetherian. This directly verifies the DVR conclusion without assuming \(B\) was Noetherian.

Write \(A_n=\mathbb Z/p^n\mathbb Z\), \(C_n=C/p^nC\). Every ideal of \(A_n\) is \((p^i)\). Its tensor map into \(C_n\) is

\[
C_n/p^{n-i}C_n\longrightarrow C_n,
\qquad [c]\longmapsto p^ic
\quad(0\le i<n).
\tag{6}
\]

If the displayed image is zero, cancellation of \(p^i\) in \(C\) puts a lift of \(c\) in \(p^{n-i}C\); hence (6) is injective. The zero ideal causes no exception. The ideal criterion proves flatness. At \(n=1\), Lemma 1.1 gives formal smoothness. Inductively use the square-zero ideal \((p^n)\subset A_{n+1}\), whose quotient map is \(A_n\to C_n\); flatness and Lemma 2.1 give formal smoothness at \(n+1\). \(\square\)

For example, starting from \(k=\mathbb F_p(u)\), the flat local ring \(\mathbb Z_p[T]_{(p)}\), followed by its \(p\)-adic completion, constructs a Cohen ring with this imperfect residue field. The theorem also allows arbitrarily many generators and infinite inseparable towers in the residue field.

## 5. Coefficient subrings in every complete local ring

A **coefficient subring** of a complete local ring \((R,\mathfrak m,k)\) is a complete local subring \(D\subset R\), inducing the given residue field isomorphism, with

\[
D\cap\mathfrak m=pD,
\tag{7}
\]

where \(p=\operatorname{char}k\), and \(p=0\) means that \(D\) is a field. Throughout, completeness includes separatedness.

**Theorem 5.1 (coefficient-ring existence).** Every complete local ring has a coefficient subring. This assertion does not require its maximal ideal to be finitely generated.

**Proof.** If \(\operatorname{char}k=0\), every nonzero integer is a unit in \(R\), so \(R\) contains \(\mathbb Q\). Starting with \(k=R/\mathfrak m\), inductively lift a chosen map \(k\to R/\mathfrak m^{n-1}\) to \(R/\mathfrak m^n\) by Lemma 1.1. The kernel \(\mathfrak m^{n-1}/\mathfrak m^n\) is square-zero for \(n\ge2\). The chosen maps are compatible, so their inverse limit gives \(k\to R\) splitting the residue map. Its image is a field and therefore a coefficient subring.

If \(\operatorname{char}k=p>0\), choose a Cohen ring \(C\) from Theorem 4.1 with specified identification \(C/pC=k\). Construct maps

\[
\phi_n:C/p^nC\longrightarrow R/\mathfrak m^n
\tag{8}
\]

inductively. Start with the given residue identification. At stage \(n\ge2\), view the preceding map as the composite
\(C/p^nC\to C/p^{n-1}C\xrightarrow{\phi_{n-1}}R/\mathfrak m^{n-1}\).
Both target rings are \(\mathbb Z/p^n\mathbb Z\)-algebras because \(p\in\mathfrak m\), and the square-zero reduction permits a lift by Theorem 4.1. Thus \(\phi_n\) reduces exactly to this composite. Passing to the limit gives \(\phi:C\to R\) with the selected residue isomorphism.

It remains to prove that the image \(D\) really satisfies the coefficient conditions. The kernel is a proper ideal of the DVR \(C\), so is either zero or \((p^e)\) for some \(e\ge1\). Hence \(D\cong C\), or \(D\cong C/(p^e)\). In the latter case it is an Artinian local ring, complete for its nilpotent maximal ideal. Its residue kernel is exactly \(pD\), since \(\phi(c)\) has zero residue precisely when \(c\in pC\). This proves (7) and the residue isomorphism. If \(pR=0\), the kernel is \((p)\) and \(D\) is a coefficient field. If no power of \(p\) vanishes in \(R\), the kernel is zero. If \(p\) has exact nilpotence exponent \(e\), the kernel is \((p^e)\). The image in that last case is not a DVR. \(\square\)

## 6. The power-series presentation

**Theorem 6.1 (Cohen structure).** If \((R,\mathfrak m,k)\) is complete local and \(\mathfrak m=(y_1,\ldots,y_d)\) is finitely generated, then

\[
R\cong L[[X_1,\ldots,X_d]]/J,
\tag{9}
\]

where \(L\) is a field in equal characteristic and may be taken to be a Cohen ring in positive residue characteristic. The coefficient map to \(R\) need not be injective in the \(p\)-torsion case.

**Proof.** In equal characteristic use the coefficient field from Theorem 5.1. Otherwise use the Cohen ring \(C\) and its map \(\phi\), before taking its image. Define the map from the power-series ring by

\[
\Phi\left(\sum_{\alpha\in\mathbb N^d}a_\alpha X^\alpha\right)
=\lim_{N\to\infty}\sum_{|\alpha|<N}\phi(a_\alpha)y^\alpha.
\tag{10}
\]

The tails lie in \(\mathfrak m^N\); completeness and separatedness make this well-defined. Finite convolution modulo \(\mathfrak m^N\) proves that it is a ring map. Its source is complete for the variable ideal \((X_1,\ldots,X_d)\), and its target is complete for the image ideal \(\mathfrak m\). This statement concerns these respective ideals, even in mixed characteristic.

For surjectivity, first lift the residue of \(r\in R\) by a coefficient. If the remaining error lies in \(\mathfrak m^N\), express it as a finite sum \(\sum_{|\alpha|=N}b_\alpha y^\alpha\). Lift the residues of the \(b_\alpha\)'s to coefficients and add the corresponding homogeneous polynomial of degree \(N\). This leaves error in \(\mathfrak m^{N+1}\). The corrections form a power series whose image is \(r\), by separatedness. This gives (9) with \(J=\ker\Phi\). For \(d=0\), the maximal ideal is zero and the coefficient map itself is onto the field \(R\). \(\square\)

In particular this applies to every complete Noetherian local ring. It also implies that a complete local ring with finitely generated maximal ideal is Noetherian: the field and Cohen coefficient rings are Noetherian, and their finite-variable formal power-series rings are Noetherian by Solution 7.3 of *Noetherian and Artinian rings*, applied one variable at a time. Quotients are Noetherian.

**Corollary 6.2.** A complete regular local ring of equal characteristic, of dimension \(d\), is isomorphic to \(k[[X_1,\ldots,X_d]]\) for a coefficient field \(k\). In particular a complete equal-characteristic DVR is \(k[[T]]\), with no perfectness assumption on \(k\).

**Proof.** Lift a basis of \(\mathfrak m/\mathfrak m^2\). Nakayama gives \(d\) generators, so (10) gives a surjection \(k[[X_1,\ldots,X_d]]\to R\). The source is a regular local domain of dimension \(d\), by the polynomial-completion calculation and Theorem 4.1 of *Completion*. If its kernel contained a nonzero \(F\), the one-nonzerodivisor dimension theorem in *Dimension theory of Noetherian local rings*, Theorem 3.2, would make its quotient by \(F\) have dimension \(d-1\). The further quotient \(R\) would have dimension at most \(d-1\), a contradiction. The kernel is zero. A DVR is regular of dimension one by *Discrete valuation rings, normal rings and Serre's criterion*, Theorem 1.2. \(\square\)

## 7. Exercises and solutions

**Exercise 7.1 (easy).** For \(R=\mathbb Z/p^e\mathbb Z\), identify a Cohen ring mapping to it and its coefficient image. Which is the DVR?

**Solution 7.1.** Use \(\mathbb Z_p\to\mathbb Z/p^e\mathbb Z\). Its kernel is \((p^e)\), and its image is all of \(R\), complete with maximal ideal \(pR\). The source \(\mathbb Z_p\) is a DVR; the image is Artinian, and for \(e=1\) it is a field. For \(e>1\), \(p\) is a nonzero nilpotent in the image, so that image cannot be a domain or DVR.

**Exercise 7.2 (medium).** Let \(R\) be a complete DVR with residue field \(k\) and characteristic \(p>0\), and choose a uniformizer \(\pi\). Prove that every element has a unique expansion \(\sum_{n\ge0}a_n\pi^n\) after a coefficient field has been fixed. Apply it to imperfect \(k\).

**Solution 7.2.** Theorem 5.1 supplies the field, without perfectness. Lift the residue of \(r\), subtract it and divide by \(\pi\); repeat. Completeness gives the expansion. If two expansions differ, their first nonzero differing coefficient, at degree \(n\), gives valuation exactly \(n\), so their difference cannot vanish. This proves existence and uniqueness, and the coefficient map identifies \(R\) with \(k[[T]]\). The argument is unchanged for \(k=\mathbb F_p(u)\) or any other imperfect field. It is the exact specialization used when a complete equal-characteristic DVR is replaced by a formal-series ring.

**Exercise 7.3 (hard).** For \(R=\mathbb Z_p[[T]]/(pT)\), explain why mixed characteristic does not guarantee that \(p\) is a nonzerodivisor in \(R\), and determine the kernel of its natural coefficient map.

**Solution 7.3.** The quotient is complete Noetherian local by the complete-quotient theorem, with maximal ideal \((p,T)\). It has characteristic zero: evaluating \(T=0\) gives a retraction to \(\mathbb Z_p\), so the coefficient map is injective. The class of \(T\) is nonzero, as reduction modulo \(p\) shows, but \(pT=0\). Thus \(p\) is a zero divisor in \(R\). No power of \(p\) vanishes, so Theorem 5.1 correctly allows a coefficient DVR embedded in a ring with other zero divisors. Its coefficient map has kernel zero. Confusing these two assertions would incorrectly exclude this complete local ring from Cohen structure.

## Attribution, changes and licence

The Stacks Project Authors supply the open Cohen proof and its flat-residue-field and formal-smoothness prerequisites. The version incorporated here is [AI Integrated Stacks Project, pinned source](https://github.com/KokunoYumeto/unofficial-stacks-project-ai-drafts/blob/565b10e987aba5969b21145a0833f42d69f96790/algebra.tex#L46360), especially `lemma-flat-local-given-residue-field`, `lemma-lift-formal-smoothness`, `lemma-cohen-rings-exist`, `lemma-cohen-ring-formally-smooth` and `theorem-cohen-structure-theorem`. The fork's [source-linked editorial proof 65](https://github.com/KokunoYumeto/unofficial-stacks-project-ai-drafts/blob/565b10e987aba5969b21145a0833f42d69f96790/ai-integrated/candidates/commons/stacks/errata/r58/proof-source/successor/notes/ALGEBRA_COHEN_NOTES_20260929.tex) is credited to OpenAI Codex, GPT-6 Astra, Ultra. It clarifies the compatible quotient maps, coefficient image, completion ideals and boundary cases.

GPT-6.1 Sol checked the incorporated arguments and rewrote their teaching order, supplied the prime-field \(p\)-basis lifting proof, made the transfinite residue extension explicit, used the direct coordinate proof of injectivity of \(p\), and integrated the regular-local consequence and three solved exercises. All main proofs and their required lifting support appear here or at the exact preceding course locators; readers need no external book to fill them.

This adapted lesson, including its added exposition, is licensed under the **GNU Free Documentation License, version 1.2 or later**, with no Invariant Sections, no Front-Cover Texts and no Back-Cover Texts. The [licence text](COPYING-GFDL-1.2.txt) and transparent Markdown source accompany the lesson. The separate original material retains its CC0 dedication; incorporated proofs retain the source licence. History: Stacks Project Authors, *Commutative Algebra*, as incorporated in AI Integrated Stacks Project at the pinned revision; OpenAI GPT-6 Astra, source-linked editorial companion, September 2026; OpenAI GPT-6.1 Sol, this course adaptation, October 2026.

## Copyright and licence

Copyright (C) 2005–2025 Johan de Jong. The incorporated source is *The Stacks Project*, as distributed in AI Integrated Stacks Project at revision `565b10e987aba5969b21145a0833f42d69f96790`.

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](COPYING-GFDL-1.2.txt).

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.