
## 10.103 Cohen-Macaulay rings

Most of the results of this section are special cases of the results in Section 10.102.

Definition 10.103.1. A Noetherian local ring $R$ is called Cohen-Macaulay if it is Cohen-Macaulay as a module over itself.

Note that this is equivalent to requiring the existence of a $R$-regular sequence $x_1, \ldots , x_ d$ of the maximal ideal such that $R/(x_1, \ldots , x_ d)$ has dimension $0$. We will usually just say “regular sequence” and not “$R$-regular sequence”.

Lemma 10.103.2. Let $R$ be a Noetherian local Cohen-Macaulay ring with maximal ideal $\mathfrak m$. Let $x_1, \ldots , x_ c \in \mathfrak m$ be elements. Then

$x_1, \ldots , x_ c \text{ is a regular sequence } \Leftrightarrow \dim (R/(x_1, \ldots , x_ c)) = \dim (R) - c$

If so $x_1, \ldots , x_ c$ can be extended to a regular sequence of length $\dim (R)$ and each quotient $R/(x_1, \ldots , x_ i)$ is a Cohen-Macaulay ring of dimension $\dim (R) - i$.

Proof. Special case of Proposition 10.102.4. $\square$

Lemma 10.103.3. Let $R$ be Noetherian local. Suppose $R$ is Cohen-Macaulay of dimension $d$. Any maximal chain of ideals $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ n$ has length $n = d$.

Proof. Special case of Lemma 10.102.9. $\square$

Lemma 10.103.4. Suppose $R$ is a Noetherian local Cohen-Macaulay ring of dimension $d$. For any prime $\mathfrak p \subset R$ we have

$\dim (R) = \dim (R_{\mathfrak p}) + \dim (R/\mathfrak p).$

Proof. Follows immediately from Lemma 10.103.3. (Also, this is a special case of Lemma 10.102.10.) $\square$

Lemma 10.103.5. Suppose $R$ is a Cohen-Macaulay local ring. For any prime $\mathfrak p \subset R$ the ring $R_{\mathfrak p}$ is Cohen-Macaulay as well.

Proof. Special case of Lemma 10.102.11. $\square$

Definition 10.103.6. A Noetherian ring $R$ is called Cohen-Macaulay if all its local rings are Cohen-Macaulay.

Lemma 10.103.7. Suppose $R$ is a Noetherian Cohen-Macaulay ring. Any polynomial algebra over $R$ is Cohen-Macaulay.

Proof. Special case of Lemma 10.102.13. $\square$

Lemma 10.103.8. Let $R$ be a Noetherian local Cohen-Macaulay ring of dimension $d$. Let $0 \to K \to R^{\oplus n} \to M \to 0$ be an exact sequence of $R$-modules. Then either $M = 0$, or $\text{depth}(K) > \text{depth}(M)$, or $\text{depth}(K) = \text{depth}(M) = d$.

Proof. If $d = 0$, then every nonzero $R$-module has depth $0$ and the lemma is true. Assume $d > 0$. Then $\text{depth}(K) > 0$ as $K$ is a submodule of a module of depth $> 0$. Hence the lemma holds if $\text{depth}(M) = 0$. Assume both $\text{depth}(M) > 0$ and $d > 0$. Then we choose $x \in \mathfrak m$ which is a nonzerodivisor on $M$ and on $R$. Then $x$ is a nonzerodivisor on $M$ and on $K$ and it follows by an easy diagram chase that $0 \to K/xK \to (R/xR)^{\oplus n} \to M/xM \to 0$ is exact. Using Lemmas 10.71.7 and 10.103.2 we find the result follows from the result for $K/xK$ over $R/xR$ which has smaller dimension. $\square$

Lemma 10.103.9. Let $R$ be a local Noetherian Cohen-Macaulay ring of dimension $d$. Let $M$ be a finite $R$ module of depth $e$. There exists an exact complex

$0 \to K \to F_{d-e-1} \to \ldots \to F_0 \to M \to 0$

with each $F_ i$ finite free and $K$ maximal Cohen-Macaulay.

Proof. Immediate from the definition and Lemma 10.103.8. $\square$

Lemma 10.103.10. Let $\varphi : A \to B$ be a map of local rings. Assume that $B$ is Noetherian and Cohen-Macaulay and that $\mathfrak m_ B = \sqrt{\varphi (\mathfrak m_ A) B}$. Then there exists a sequence of elements $f_1, \ldots , f_{\dim (B)}$ in $A$ such that $\varphi (f_1), \ldots , \varphi (f_{\dim (B)})$ is a regular sequence in $B$.

Proof. By induction on $\dim (B)$ it suffices to prove: If $\dim (B) \geq 1$, then we can find an element $f$ of $A$ which maps to a nonzerodivisor in $B$. By Lemma 10.103.2 it suffices to find $f \in A$ whose image in $B$ is not contained in any of the finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ r$ of $B$. By the assumption that $\mathfrak m_ B = \sqrt{\varphi (\mathfrak m_ A) B}$ we see that $\mathfrak m_ A \not\subset \varphi ^{-1}(\mathfrak q_ i)$. Hence we can find $f$ by Lemma 10.14.2. $\square$

Comment #244 by Olaf Schnuerer on

In Lemma 9.101.8 M (or K) might be the zero module. Then $depth(M)=\infty$, and $depth(K)=d$ if $n>0.$ My suggestion:

If $depth(M), then $depth(K)=depth(M)+1.$ If $depth(M)\geq d$, then $depth(K) \in \{d, \infty\}.$

Comment #246 by on

BY Definition 10.71.1 depth as the supremum of the lengths of regular sequences. By Definition 10.67.1 there are no regular sequences for a zero module (not even one of length zero). So the supremum is $-\infty$ and not $\infty$. Right?

However, I do agree the statement of the lemma is misleading if either $M$ or $K$ is zero. Perhaps the best solution is to explicitly make a list of cases in the statement.

Comment #247 by Olaf Schnuerer on

You are right, I thought depth of M was defined as infimum of i's such that Ext^i(k,M) is nonzero.

Comment #2217 by David Savitt on

BTW, what happened to the statement of [00NE]? From the above comments it sounds like it used to say that depth(K) = depth(M) + 1 in the first alternative [ rather than just depth(K) > depth(M) ], and I think that was used in [00NG] to see that the complex has length exactly d-e rather than at most d-e. Maybe one doesn't care, though, since the one tag referencing this result doesn't use that.

Comment #2224 by on

@#2217: No, Lemma 10.103.8 never used to say $\text{depth}(K) = \text{Depth}(M) + 1$. You can check by looking at the history of the lemma here: history. It would be easy to add this...

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