## Tag `032D`

Chapter 10: Commutative Algebra > Section 10.154: The Cohen structure theorem

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. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 42492–42503 (see updates for more information).

```
\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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