Definition 10.104.1. A Noetherian local ring $R$ is called Cohen-Macaulay if it is Cohen-Macaulay as a module over itself.
10.104 Cohen-Macaulay rings
Most of the results of this section are special cases of the results in Section 10.103.
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 10.104.2.slogan 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 10.103.4. $\square$
Lemma 10.104.3. 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 10.103.9. $\square$
Lemma 10.104.4. 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 10.104.3. (Also, this is a special case of Lemma 10.103.10.) $\square$
Lemma 10.104.5. 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 10.103.11. $\square$
Definition 10.104.6. A Noetherian ring $R$ is called Cohen-Macaulay if all its local rings are Cohen-Macaulay.
Lemma 10.104.7. Suppose $R$ is a Noetherian Cohen-Macaulay ring. Any polynomial algebra over $R$ is Cohen-Macaulay.
Proof. Special case of Lemma 10.103.13. $\square$
Lemma 10.104.8. 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. This is a special case of Lemma 10.72.6. $\square$
Lemma 10.104.9. 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 10.104.8. $\square$
Lemma 10.104.10. 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 10.104.2 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 10.15.2. $\square$
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