## Tag `0323`

## 10.154. The Cohen structure theorem

Here is a fundamental notion in commutative algebra.

Definition 10.154.1. Let $(R, \mathfrak m)$ be a local ring. We say $R$ is a

complete local ringif the canonical map $$ R \longrightarrow \mathop{\rm lim}\nolimits_n R/\mathfrak m^n $$ to the completion of $R$ with respect to $\mathfrak m$ is an isomorphism^{1}.Note that an Artinian local ring $R$ is a complete local ring because $\mathfrak m_R^n = 0$ for some $n > 0$. In this section we mostly focus on Noetherian complete local rings.

Lemma 10.154.2. Let $R$ be a Noetherian complete local ring. Any quotient of $R$ is also a Noetherian complete local ring. Given a finite ring map $R \to S$, then $S$ is a product of Noetherian complete local rings.

Proof.The ring $S$ is Noetherian by Lemma 10.30.1. As an $R$-module $S$ is complete by Lemma 10.96.1. Hence $S$ is the product of the completions at its maximal ideals by Lemma 10.96.8. $\square$Lemma 10.154.3. Let $(R, \mathfrak m)$ be a complete local ring. If $\mathfrak m$ is a finitely generated ideal then $R$ is Noetherian.

Proof.See Lemma 10.96.5. $\square$Definition 10.154.4. Let $(R, \mathfrak m)$ be a complete local ring. A subring $\Lambda \subset R$ is called a

coefficient ringif the following conditions hold:

- $\Lambda$ is a complete local ring with maximal ideal $\Lambda \cap \mathfrak m$,
- the residue field of $\Lambda$ maps isomorphically to the residue field of $R$, and
- $\Lambda \cap \mathfrak m = p\Lambda$, where $p$ is the characteristic of the residue field of $R$.

Let us make some remarks on this definition. We split the discussion into the following cases:

- The local ring $R$ contains a field. This happens if either $\mathbf{Q} \subset R$, or $pR = 0$ where $p$ is the characteristic of $R/\mathfrak m$. In this case a coefficient ring $\Lambda$ is a field contained in $R$ which maps isomorphically to $R/\mathfrak m$.
- The characteristic of $R/\mathfrak m$ is $p > 0$ but no power of $p$ is zero in $R$. In this case $\Lambda$ is a complete discrete valuation ring with uniformizer $p$ and residue field $R/\mathfrak m$.
- The characteristic of $R/\mathfrak m$ is $p > 0$, and for some $n > 1$ we have $p^{n - 1} \not = 0$, $p^n = 0$ in $R$. In this case $\Lambda$ is an Artinian local ring whose maximal ideal is generated by $p$ and which has residue field $R/\mathfrak m$.
The complete discrete valuation rings with uniformizer $p$ above play a special role and we baptize them as follows.

Definition 10.154.5. A

Cohen ringis a complete discrete valuation ring with uniformizer $p$ a prime number.Lemma 10.154.6. Let $p$ be a prime number. Let $k$ be a field of characteristic $p$. There exists a Cohen ring $\Lambda$ with $\Lambda/p\Lambda \cong k$.

Proof.First note that the $p$-adic integers $\mathbf{Z}_p$ form a Cohen ring for $\mathbf{F}_p$. Let $k$ be an arbitrary field of characteristic $p$. Let $\mathbf{Z}_p \to R$ be a flat local ring map such that $\mathfrak m_R = pR$ and $R/pR = k$, see Lemma 10.153.1. Then clearly $R$ is a discrete valuation ring. Hence its completion is a Cohen ring for $k$. $\square$Lemma 10.154.7. Let $p > 0$ be a prime. Let $\Lambda$ be a Cohen ring with residue field of characteristic $p$. For every $n \geq 1$ the ring map $$ \mathbf{Z}/p^n\mathbf{Z} \to \Lambda/p^n\Lambda $$ is formally smooth.

Proof.If $n = 1$, this follows from Proposition 10.152.9. For general $n$ we argue by induction on $n$. Namely, if $\mathbf{Z}/p^n\mathbf{Z} \to \Lambda/p^n\Lambda$ is formally smooth, then we can apply Lemma 10.136.12 to the ring map $\mathbf{Z}/p^{n + 1}\mathbf{Z} \to \Lambda/p^{n + 1}\Lambda$ and the ideal $I = (p^n) \subset \mathbf{Z}/p^{n + 1}\mathbf{Z}$. $\square$Theorem 10.154.8 (Cohen structure theorem). Let $(R, \mathfrak m)$ be a complete local ring.

- $R$ has a coefficient ring (see Definition 10.154.4),
- if $\mathfrak m$ is a finitely generated ideal, then $R$ is isomorphic to a quotient $$ \Lambda[[x_1, \ldots, x_n]]/I $$ where $\Lambda$ is either a field or a Cohen ring.

Proof.Let us prove a coefficient ring exists. First we prove this in case the characteristic of the residue field $\kappa$ is zero. Namely, in this case we will prove by induction on $n > 0$ that there exists a section $$ \varphi_n : \kappa \longrightarrow R/\mathfrak m^n $$ to the canonical map $R/\mathfrak m^n \to \kappa = R/\mathfrak m$. This is trivial for $n = 1$. If $n > 1$, let $\varphi_{n - 1}$ be given. The field extension $\mathbf{Q} \subset \kappa$ is formally smooth by Proposition 10.152.9. Hence we can find the dotted arrow in the following diagram $$ \xymatrix{ R/\mathfrak m^{n - 1} & R/\mathfrak m^n \ar[l] \\ \kappa \ar[u]^{\varphi_{n - 1}} \ar@{..>}[ru] & \mathbf{Q} \ar[l] \ar[u] } $$ This proves the induction step. Putting these maps together $$ \mathop{\rm lim}\nolimits_n~\varphi_n : \kappa \longrightarrow R = \mathop{\rm lim}\nolimits_n~R/\mathfrak m^n $$ gives a map whose image is the desired coefficient ring.Next, we prove the existence of a coefficient ring in the case where the characteristic of the residue field $\kappa$ is $p > 0$. Namely, choose a Cohen ring $\Lambda$ with $\kappa = \Lambda/p\Lambda$, see Lemma 10.154.6. In this case we will prove by induction on $n > 0$ that there exists a map $$ \varphi_n : \Lambda/p^n\Lambda \longrightarrow R/\mathfrak m^n $$ whose composition with the reduction map $R/\mathfrak m^n \to \kappa$ produces the given isomorphism $\Lambda/p\Lambda = \kappa$. This is trivial for $n = 1$. If $n > 1$, let $\varphi_{n - 1}$ be given. The ring map $\mathbf{Z}/p^n\mathbf{Z} \to \Lambda/p^n\Lambda$ is formally smooth by Lemma 10.154.7. Hence we can find the dotted arrow in the following diagram $$ \xymatrix{ R/\mathfrak m^{n - 1} & R/\mathfrak m^n \ar[l] \\ \Lambda/p^n\Lambda \ar[u]^{\varphi_{n - 1}} \ar@{..>}[ru] & \mathbf{Z}/p^n\mathbf{Z} \ar[l] \ar[u] } $$ This proves the induction step. Putting these maps together $$ \mathop{\rm lim}\nolimits_n~\varphi_n : \Lambda = \mathop{\rm lim}\nolimits_n~\Lambda/p^n\Lambda \longrightarrow R = \mathop{\rm lim}\nolimits_n~R/\mathfrak m^n $$ gives a map whose image is the desired coefficient ring.

The final statement of the theorem follows readily. Namely, if $y_1, \ldots, y_n$ are generators of the ideal $\mathfrak m$, then we can use the map $\Lambda \to R$ just constructed to get a map $$ \Lambda[[x_1, \ldots, x_n]] \longrightarrow R, \quad x_i \longmapsto y_i. $$ Since both sides are $(x_1, \ldots, x_n)$-adically complete this map is surjective by Lemma 10.95.1 as it is surjective modulo $(x_1, \ldots, x_n)$ by construction. $\square$

Remark 10.154.9. If $k$ is a field then the power series ring $k[[X_1, \ldots, X_d]]$ is a Noetherian complete local regular ring of dimension $d$. If $\Lambda$ is a Cohen ring then $\Lambda[[X_1, \ldots, X_d]]$ is a complete local Noetherian regular ring of dimension $d + 1$. Hence the Cohen structure theorem implies that any Noetherian complete local ring is a quotient of a regular local ring. In particular we see that a Noetherian complete local ring is universally catenary, see Lemma 10.104.8 and Lemma 10.105.3.

Lemma 10.154.10. Let $(R, \mathfrak m)$ be a Noetherian complete local ring. Assume $R$ is regular.

- If $R$ contains either $\mathbf{F}_p$ or $\mathbf{Q}$, then $R$ is isomorphic to a power series ring over its residue field.
- If $k$ is a field and $k \to R$ is a ring map inducing an isomorphism $k \to R/\mathfrak m$, then $R$ is isomorphic as a $k$-algebra to a power series ring over $k$.

Proof.In case (1), by the Cohen structure theorem (Theorem 10.154.8) there exists a coefficient ring which must be a field mapping isomorphically to the residue field. Thus it suffices to prove (2). In case (2) we pick $f_1, \ldots, f_d \in \mathfrak m$ which map to a basis of $\mathfrak m/\mathfrak m^2$ and we consider the continuous $k$-algebra map $k[[x_1, \ldots, x_d]] \to R$ sending $x_i$ to $f_i$. As both source and target are $(x_1, \ldots, x_d)$-adically complete, this map is surjective by Lemma 10.95.1. On the other hand, it has to be injective because otherwise the dimension of $R$ would be $< d$ by Lemma 10.59.12. $\square$Lemma 10.154.11. Let $(R, \mathfrak m)$ be a Noetherian complete local domain. Then there exists a $R_0 \subset R$ with the following properties

- $R_0$ is a regular complete local ring,
- $R_0 \subset R$ is finite and induces an isomorphism on residue fields,
- $R_0$ is either isomorphic to $k[[X_1, \ldots, X_d]]$ where $k$ is a field or $\Lambda[[X_1, \ldots, X_d]]$ where $\Lambda$ is a Cohen ring.

Proof.Let $\Lambda$ be a coefficient ring of $R$. Since $R$ is a domain we see that either $\Lambda$ is a field or $\Lambda$ is a Cohen ring.Case I: $\Lambda = k$ is a field. Let $d = \dim(R)$. Choose $x_1, \ldots, x_d \in \mathfrak m$ which generate an ideal of definition $I \subset R$. (See Section 10.59.) By Lemma 10.95.9 we see that $R$ is $I$-adically complete as well. Consider the map $R_0 = k[[X_1, \ldots, X_d]] \to R$ which maps $X_i$ to $x_i$. Note that $R_0$ is complete with respect to the ideal $I_0 = (X_1, \ldots, X_d)$, and that $R/I_0R \cong R/IR$ is finite over $k = R_0/I_0$ (because $\dim(R/I) = 0$, see Section 10.59.) Hence we conclude that $R_0 \to R$ is finite by Lemma 10.95.12. Since $\dim(R) = \dim(R_0)$ this implies that $R_0 \to R$ is injective (see Lemma 10.111.3), and the lemma is proved.

Case II: $\Lambda$ is a Cohen ring. Let $d + 1 = \dim(R)$. Let $p > 0$ be the characteristic of the residue field $k$. As $R$ is a domain we see that $p$ is a nonzerodivisor in $R$. Hence $\dim(R/pR) = d$, see Lemma 10.59.12. Choose $x_1, \ldots, x_d \in R$ which generate an ideal of definition in $R/pR$. Then $I = (p, x_1, \ldots, x_d)$ is an ideal of definition of $R$. By Lemma 10.95.9 we see that $R$ is $I$-adically complete as well. Consider the map $R_0 = \Lambda[[X_1, \ldots, X_d]] \to R$ which maps $X_i$ to $x_i$. Note that $R_0$ is complete with respect to the ideal $I_0 = (p, X_1, \ldots, X_d)$, and that $R/I_0R \cong R/IR$ is finite over $k = R_0/I_0$ (because $\dim(R/I) = 0$, see Section 10.59.) Hence we conclude that $R_0 \to R$ is finite by Lemma 10.95.12. Since $\dim(R) = \dim(R_0)$ this implies that $R_0 \to R$ is injective (see Lemma 10.111.3), and the lemma is proved. $\square$

- This includes the condition that $\bigcap \mathfrak m^n = (0)$; in some texts this may be indicated by saying that $R$ is complete and separated. Warning: It can happen that the completion $\mathop{\rm lim}\nolimits_n R/\mathfrak m^n$ of a local ring is non-complete, see Examples, Lemma 100.6.1. This does not happen when $\mathfrak m$ is finitely generated, see Lemma 10.95.3 in which case the completion is Noetherian, see Lemma 10.96.5. ↑

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 42188–42524 (see updates for more information).

```
\section{The Cohen structure theorem}
\label{section-cohen-structure-theorem}
\noindent
Here is a fundamental notion in commutative algebra.
\begin{definition}
\label{definition-complete-local-ring}
Let $(R, \mathfrak m)$ be a local ring. We say $R$ is a
{\it complete local ring} if the canonical map
$$
R \longrightarrow \lim_n R/\mathfrak m^n
$$
to the completion of $R$ with respect to $\mathfrak m$ is an
isomorphism\footnote{This includes the condition
that $\bigcap \mathfrak m^n = (0)$; in some texts this may be indicated
by saying that $R$ is complete and separated. Warning: It can happen
that the completion $\lim_n R/\mathfrak m^n$ of a local ring is
non-complete, see
Examples, Lemma \ref{examples-lemma-noncomplete-completion}.
This does not happen when $\mathfrak m$ is finitely generated, see
Lemma \ref{lemma-hathat-finitely-generated} in which
case the completion is Noetherian, see
Lemma \ref{lemma-completion-Noetherian}.}.
\end{definition}
\noindent
Note that an Artinian local ring $R$ is a complete local ring
because $\mathfrak m_R^n = 0$ for some $n > 0$. In this section
we mostly focus on Noetherian complete local rings.
\begin{lemma}
\label{lemma-quotient-complete-local}
Let $R$ be a Noetherian complete local ring.
Any quotient of $R$ is also a Noetherian complete local ring.
Given a finite ring map $R \to S$, then $S$ is a product of
Noetherian complete local rings.
\end{lemma}
\begin{proof}
The ring $S$ is Noetherian by Lemma \ref{lemma-Noetherian-permanence}.
As an $R$-module $S$ is complete by Lemma \ref{lemma-completion-tensor}.
Hence $S$ is the product of the completions at its maximal ideals
by Lemma \ref{lemma-completion-finite-extension}.
\end{proof}
\begin{lemma}
\label{lemma-complete-local-ring-Noetherian}
Let $(R, \mathfrak m)$ be a complete local ring.
If $\mathfrak m$ is a finitely generated ideal then
$R$ is Noetherian.
\end{lemma}
\begin{proof}
See Lemma \ref{lemma-completion-Noetherian}.
\end{proof}
\begin{definition}
\label{definition-coefficient-ring}
Let $(R, \mathfrak m)$ be a complete local ring.
A subring $\Lambda \subset R$ is
called a {\it coefficient ring} if the following conditions hold:
\begin{enumerate}
\item $\Lambda$ is a complete local ring with maximal ideal
$\Lambda \cap \mathfrak m$,
\item the residue field of $\Lambda$ maps isomorphically to the
residue field of $R$, and
\item $\Lambda \cap \mathfrak m = p\Lambda$, where $p$ is the characteristic
of the residue field of $R$.
\end{enumerate}
\end{definition}
\noindent
Let us make some remarks on this definition. We split the discussion
into the following cases:
\begin{enumerate}
\item The local ring $R$ contains a field. This happens if
either $\mathbf{Q} \subset R$, or $pR = 0$ where $p$ is the
characteristic of $R/\mathfrak m$. In this case a coefficient ring
$\Lambda$ is a field contained in $R$ which maps isomorphically to
$R/\mathfrak m$.
\item The characteristic of $R/\mathfrak m$ is $p > 0$ but no
power of $p$ is zero in $R$. In this case $\Lambda$ is a complete
discrete valuation ring with uniformizer $p$ and residue field $R/\mathfrak m$.
\item The characteristic of $R/\mathfrak m$ is $p > 0$, and for some
$n > 1$ we have $p^{n - 1} \not = 0$, $p^n = 0$ in $R$. In this case
$\Lambda$ is an Artinian local ring whose maximal ideal is
generated by $p$ and which has residue field $R/\mathfrak m$.
\end{enumerate}
The complete discrete valuation rings with uniformizer $p$
above play a special role and we baptize them as follows.
\begin{definition}
\label{definition-cohen-ring}
A {\it Cohen ring} is a complete discrete valuation ring with
uniformizer $p$ a prime number.
\end{definition}
\begin{lemma}
\label{lemma-cohen-rings-exist}
Let $p$ be a prime number.
Let $k$ be a field of characteristic $p$.
There exists a Cohen ring $\Lambda$ with $\Lambda/p\Lambda \cong k$.
\end{lemma}
\begin{proof}
First note that the $p$-adic integers $\mathbf{Z}_p$ form a Cohen ring
for $\mathbf{F}_p$. Let $k$ be an arbitrary field of characteristic $p$.
Let $\mathbf{Z}_p \to R$ be a flat local ring map such that
$\mathfrak m_R = pR$ and $R/pR = k$, see
Lemma \ref{lemma-flat-local-given-residue-field}.
Then clearly $R$ is a discrete valuation ring. Hence its
completion is a Cohen ring for $k$.
\end{proof}
\begin{lemma}
\label{lemma-cohen-ring-formally-smooth}
Let $p > 0$ be a prime.
Let $\Lambda$ be a Cohen ring with residue field of characteristic $p$.
For every $n \geq 1$ the ring map
$$
\mathbf{Z}/p^n\mathbf{Z} \to \Lambda/p^n\Lambda
$$
is formally smooth.
\end{lemma}
\begin{proof}
If $n = 1$, this follows from
Proposition \ref{proposition-characterize-separable-field-extensions}.
For general $n$ we argue by induction on $n$.
Namely, if $\mathbf{Z}/p^n\mathbf{Z} \to \Lambda/p^n\Lambda$ is
formally smooth, then we can apply Lemma \ref{lemma-lift-formal-smoothness}
to the ring map
$\mathbf{Z}/p^{n + 1}\mathbf{Z} \to \Lambda/p^{n + 1}\Lambda$
and the ideal $I = (p^n) \subset \mathbf{Z}/p^{n + 1}\mathbf{Z}$.
\end{proof}
\begin{theorem}[Cohen structure theorem]
\label{theorem-cohen-structure-theorem}
Let $(R, \mathfrak m)$ be a complete local ring.
\begin{enumerate}
\item $R$ has a coefficient ring (see
Definition \ref{definition-coefficient-ring}),
\item if $\mathfrak m$ is a finitely generated ideal, then
$R$ is isomorphic to a quotient
$$
\Lambda[[x_1, \ldots, x_n]]/I
$$
where $\Lambda$ is either a field or a Cohen ring.
\end{enumerate}
\end{theorem}
\begin{proof}
Let us prove a coefficient ring exists.
First we prove this in case the characteristic of the residue field $\kappa$
is zero. Namely, in this case we will prove by induction
on $n > 0$ that there exists a section
$$
\varphi_n : \kappa \longrightarrow R/\mathfrak m^n
$$
to the canonical map $R/\mathfrak m^n \to \kappa = R/\mathfrak m$.
This is trivial for $n = 1$. If $n > 1$, let $\varphi_{n - 1}$ be given.
The field extension $\mathbf{Q} \subset \kappa$ is formally smooth by
Proposition \ref{proposition-characterize-separable-field-extensions}.
Hence we can find the dotted arrow
in the following diagram
$$
\xymatrix{
R/\mathfrak m^{n - 1} &
R/\mathfrak m^n \ar[l] \\
\kappa \ar[u]^{\varphi_{n - 1}} \ar@{..>}[ru] & \mathbf{Q} \ar[l] \ar[u]
}
$$
This proves the induction step. Putting these maps together
$$
\lim_n\ \varphi_n : \kappa \longrightarrow
R = \lim_n\ R/\mathfrak m^n
$$
gives a map whose image is the desired coefficient ring.
\medskip\noindent
Next, we prove the existence of a coefficient ring in the case
where the characteristic of the residue field $\kappa$ is $p > 0$.
Namely, choose a Cohen ring $\Lambda$ with $\kappa = \Lambda/p\Lambda$,
see Lemma \ref{lemma-cohen-rings-exist}. In this case we will prove by
induction on $n > 0$ that there exists a map
$$
\varphi_n :
\Lambda/p^n\Lambda
\longrightarrow
R/\mathfrak m^n
$$
whose composition with the reduction map $R/\mathfrak m^n \to \kappa$
produces the given isomorphism $\Lambda/p\Lambda = \kappa$. This is trivial
for $n = 1$. If $n > 1$, let $\varphi_{n - 1}$ be given.
The ring map $\mathbf{Z}/p^n\mathbf{Z} \to \Lambda/p^n\Lambda$
is formally smooth by Lemma \ref{lemma-cohen-ring-formally-smooth}.
Hence we can find the dotted arrow
in the following diagram
$$
\xymatrix{
R/\mathfrak m^{n - 1} &
R/\mathfrak m^n \ar[l] \\
\Lambda/p^n\Lambda \ar[u]^{\varphi_{n - 1}} \ar@{..>}[ru] &
\mathbf{Z}/p^n\mathbf{Z} \ar[l] \ar[u]
}
$$
This proves the induction step. Putting these maps together
$$
\lim_n\ \varphi_n :
\Lambda = \lim_n\ \Lambda/p^n\Lambda
\longrightarrow
R = \lim_n\ R/\mathfrak m^n
$$
gives a map whose image is the desired coefficient ring.
\medskip\noindent
The final statement of the theorem follows readily. Namely, if
$y_1, \ldots, y_n$ are generators of the ideal $\mathfrak m$,
then we can use the map $\Lambda \to R$ just constructed
to get a map
$$
\Lambda[[x_1, \ldots, x_n]] \longrightarrow R,
\quad x_i \longmapsto y_i.
$$
Since both sides are $(x_1, \ldots, x_n)$-adically complete
this map is surjective by Lemma \ref{lemma-completion-generalities}
as it is surjective modulo $(x_1, \ldots, x_n)$ by
construction.
\end{proof}
\begin{remark}
\label{remark-Noetherian-complete-local-ring-universally-catenary}
If $k$ is a field then the power series ring $k[[X_1, \ldots, X_d]]$
is a Noetherian complete local regular ring of dimension $d$.
If $\Lambda$ is a Cohen ring then $\Lambda[[X_1, \ldots, X_d]]$
is a complete local Noetherian regular ring of dimension $d + 1$.
Hence the Cohen structure theorem implies that any Noetherian
complete local ring is a quotient of a regular local ring.
In particular we see that a Noetherian complete local ring is
universally catenary, see Lemma \ref{lemma-CM-ring-catenary}
and Lemma \ref{lemma-regular-ring-CM}.
\end{remark}
\begin{lemma}
\label{lemma-regular-complete-containing-coefficient-field}
Let $(R, \mathfrak m)$ be a Noetherian complete local ring.
Assume $R$ is regular.
\begin{enumerate}
\item If $R$ contains either $\mathbf{F}_p$ or $\mathbf{Q}$, then $R$
is isomorphic to a power series ring over its residue field.
\item If $k$ is a field and $k \to R$ is a ring map inducing
an isomorphism $k \to R/\mathfrak m$, then $R$ is isomorphic
as a $k$-algebra to a power series ring over $k$.
\end{enumerate}
\end{lemma}
\begin{proof}
In case (1), by the Cohen structure theorem
(Theorem \ref{theorem-cohen-structure-theorem})
there exists a coefficient ring which must be a field
mapping isomorphically to the residue field. Thus
it suffices to prove (2). In case (2) we pick
$f_1, \ldots, f_d \in \mathfrak m$ which
map to a basis of $\mathfrak m/\mathfrak m^2$ and we consider
the continuous $k$-algebra map $k[[x_1, \ldots, x_d]] \to R$
sending $x_i$ to $f_i$. As both source and target are
$(x_1, \ldots, x_d)$-adically complete, this map is surjective by
Lemma \ref{lemma-completion-generalities}. On the other hand, it
has to be injective because otherwise the dimension of
$R$ would be $< d$ by Lemma \ref{lemma-one-equation}.
\end{proof}
\begin{lemma}
\label{lemma-complete-local-Noetherian-domain-finite-over-regular}
Let $(R, \mathfrak m)$ be a Noetherian complete local domain.
Then there exists a $R_0 \subset R$ with the following properties
\begin{enumerate}
\item $R_0$ is a regular complete local ring,
\item $R_0 \subset R$ is finite and induces an isomorphism on
residue fields,
\item $R_0$ is either isomorphic to $k[[X_1, \ldots, X_d]]$ where $k$
is a field or $\Lambda[[X_1, \ldots, X_d]]$ where $\Lambda$ is a Cohen ring.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $\Lambda$ be a coefficient ring of $R$.
Since $R$ is a domain we see that either $\Lambda$ is a field
or $\Lambda$ is a Cohen ring.
\medskip\noindent
Case I: $\Lambda = k$ is a field. Let $d = \dim(R)$.
Choose $x_1, \ldots, x_d \in \mathfrak m$
which generate an ideal of definition $I \subset R$.
(See Section \ref{section-dimension}.)
By Lemma \ref{lemma-change-ideal-completion} we see that $R$
is $I$-adically complete as well.
Consider the map $R_0 = k[[X_1, \ldots, X_d]] \to R$
which maps $X_i$ to $x_i$.
Note that $R_0$ is complete with respect to the ideal
$I_0 = (X_1, \ldots, X_d)$,
and that $R/I_0R \cong R/IR$ is finite over $k = R_0/I_0$
(because $\dim(R/I) = 0$, see Section \ref{section-dimension}.)
Hence we conclude that $R_0 \to R$ is finite by
Lemma \ref{lemma-finite-over-complete-ring}.
Since $\dim(R) = \dim(R_0)$ this implies that
$R_0 \to R$ is injective (see Lemma \ref{lemma-integral-dim-up}),
and the lemma is proved.
\medskip\noindent
Case II: $\Lambda$ is a Cohen ring. Let $d + 1 = \dim(R)$.
Let $p > 0$ be the characteristic of the residue field $k$.
As $R$ is a domain we see that $p$ is a nonzerodivisor in $R$.
Hence $\dim(R/pR) = d$, see Lemma \ref{lemma-one-equation}.
Choose $x_1, \ldots, x_d \in R$
which generate an ideal of definition in $R/pR$.
Then $I = (p, x_1, \ldots, x_d)$ is an ideal of definition of $R$.
By Lemma \ref{lemma-change-ideal-completion} we see that $R$
is $I$-adically complete as well.
Consider the map $R_0 = \Lambda[[X_1, \ldots, X_d]] \to R$
which maps $X_i$ to $x_i$.
Note that $R_0$ is complete with respect to the ideal
$I_0 = (p, X_1, \ldots, X_d)$,
and that $R/I_0R \cong R/IR$ is finite over $k = R_0/I_0$
(because $\dim(R/I) = 0$, see Section \ref{section-dimension}.)
Hence we conclude that $R_0 \to R$ is finite by
Lemma \ref{lemma-finite-over-complete-ring}.
Since $\dim(R) = \dim(R_0)$ this implies that
$R_0 \to R$ is injective (see Lemma \ref{lemma-integral-dim-up}),
and the lemma is proved.
\end{proof}
```

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