Definition 10.102.1. Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. We say $M$ is *Cohen-Macaulay* if $\dim (\text{Supp}(M)) = \text{depth}(M)$.

## 10.102 Cohen-Macaulay modules

Here we show that Cohen-Macaulay modules have good properties. We postpone using Ext groups to establish the connection with duality and so on.

A first goal will be to establish Proposition 10.102.4. We do this by a (perhaps nonstandard) sequence of elementary lemmas involving almost none of the earlier results on depth. Let us introduce some notation.

Let $R$ be a local Noetherian ring. Let $M$ be a Cohen-Macaulay module, and let $f_1, \ldots , f_ d$ be an $M$-regular sequence with $d = \dim (\text{Supp}(M))$. We say that $g \in \mathfrak m$ is *good with respect to $(M, f_1, \ldots , f_ d)$* if for all $i = 0, 1, \ldots , d-1$ we have $\dim (\text{Supp}(M) \cap V(g, f_1, \ldots , f_ i)) = d - i - 1$. This is equivalent to the condition that $\dim (\text{Supp}(M/(f_1, \ldots , f_ i)M) \cap V(g)) = d - i - 1$ for $i = 0, 1, \ldots , d - 1$.

Lemma 10.102.2. Notation and assumptions as above. If $g$ is good with respect to $(M, f_1, \ldots , f_ d)$, then (a) $g$ is a nonzerodivisor on $M$, and (b) $M/gM$ is Cohen-Macaulay with maximal regular sequence $f_1, \ldots , f_{d - 1}$.

**Proof.**
We prove the lemma by induction on $d$. If $d = 0$, then $M$ is finite and there is no case to which the lemma applies. If $d = 1$, then we have to show that $g : M \to M$ is injective. The kernel $K$ has support $\{ \mathfrak m\} $ because by assumption $\dim \text{Supp}(M) \cap V(g) = 0$. Hence $K$ has finite length. Hence $f_1 : K \to K$ injective implies the length of the image is the length of $K$, and hence $f_1 K = K$, which by Nakayama's Lemma 10.19.1 implies $K = 0$. Also, $\dim \text{Supp}(M/gM) = 0$ and so $M/gM$ is Cohen-Macaulay of depth $0$.

Assume $d > 1$. Observe that $g$ is good for $(M/f_1M, f_2, \ldots , f_ d)$, as is easily seen from the definition. By induction, we have that (a) $g$ is a nonzerodivisor on $M/f_1M$ and (b) $M/(g, f_1)M$ is Cohen-Macaulay with maximal regular sequence $f_2, \ldots , f_{d - 1}$. By Lemma 10.67.4 we see that $g, f_1$ is an $M$-regular sequence. Hence $g$ is a nonzerodivisor on $M$ and $f_1, \ldots , f_{d - 1}$ is an $M/gM$-regular sequence. $\square$

Lemma 10.102.3. Let $R$ be a Noetherian local ring. Let $M$ be a Cohen-Macaulay module over $R$. Suppose $g \in \mathfrak m$ is such that $\dim (\text{Supp}(M) \cap V(g)) = \dim (\text{Supp}(M)) - 1$. Then (a) $g$ is a nonzerodivisor on $M$, and (b) $M/gM$ is Cohen-Macaulay of depth one less.

**Proof.**
Choose a $M$-regular sequence $f_1, \ldots , f_ d$ with $d = \dim (\text{Supp}(M))$. If $g$ is good with respect to $(M, f_1, \ldots , f_ d)$ we win by Lemma 10.102.2. In particular the lemma holds if $d = 1$. (The case $d = 0$ does not occur.) Assume $d > 1$. Choose an element $h \in R$ such that (i) $h$ is good with respect to $(M, f_1, \ldots , f_ d)$, and (ii) $\dim (\text{Supp}(M) \cap V(h, g)) = d - 2$. To see $h$ exists, let $\{ \mathfrak q_ j\} $ be the (finite) set of minimal primes of the closed sets $\text{Supp}(M)$, $\text{Supp}(M)\cap V(f_1, \ldots , f_ i)$, $i = 1, \ldots , d - 1$, and $\text{Supp}(M) \cap V(g)$. None of these $\mathfrak q_ j$ is equal to $\mathfrak m$ and hence we may find $h \in \mathfrak m$, $h \not\in \mathfrak q_ j$ by Lemma 10.14.2. It is clear that $h$ satisfies (i) and (ii). From Lemma 10.102.2 we conclude that $M/hM$ is Cohen-Macaulay. By (ii) we see that the pair $(M/hM, g)$ satisfies the induction hypothesis. Hence $M/(h, g)M$ is Cohen-Macaulay and $g : M/hM \to M/hM$ is injective. By Lemma 10.67.4 we see that $g : M \to M$ and $h : M/gM \to M/gM$ are injective. Combined with the fact that $M/(g, h)M$ is Cohen-Macaulay this finishes the proof.
$\square$

Proposition 10.102.4. Let $R$ be a Noetherian local ring, with maximal ideal $\mathfrak m$. Let $M$ be a Cohen-Macaulay module over $R$ whose support has dimension $d$. Suppose that $g_1, \ldots , g_ c$ are elements of $\mathfrak m$ such that $\dim (\text{Supp}(M/(g_1, \ldots , g_ c)M)) = d - c$. Then $g_1, \ldots , g_ c$ is an $M$-regular sequence, and can be extended to a maximal $M$-regular sequence.

**Proof.**
Let $Z = \text{Supp}(M) \subset \mathop{\mathrm{Spec}}(R)$. By Lemma 10.59.12 in the chain $Z \supset Z \cap V(g_1) \supset \ldots \supset Z \cap V(g_1, \ldots , g_ c)$ each step decreases the dimension at most by $1$. Hence by assumption each step decreases the dimension by exactly $1$ each time. Thus we may successively apply Lemma 10.102.3 to the modules $M/(g_1, \ldots , g_ i)$ and the element $g_{i + 1}$.

To extend $g_1, \ldots , g_ c$ by one element if $c < d$ we simply choose an element $g_{c + 1} \in \mathfrak m$ which is not in any of the finitely many minimal primes of $Z \cap V(g_1, \ldots , g_ c)$, using Lemma 10.14.2. $\square$

Having proved Proposition 10.102.4 we continue the development of standard theory.

Lemma 10.102.5. Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Let $M$ be a finite $R$-module. Let $x \in \mathfrak m$ be a nonzerodivisor on $M$. Then $M$ is Cohen-Macaulay if and only if $M/xM$ is Cohen-Macaulay.

**Proof.**
By Lemma 10.71.7 we have $\text{depth}(M/xM) = \text{depth}(M)-1$. By Lemma 10.62.10 we have $\dim (\text{Supp}(M/xM)) = \dim (\text{Supp}(M)) - 1$.
$\square$

Lemma 10.102.6. Let $R \to S$ be a surjective homomorphism of Noetherian local rings. Let $N$ be a finite $S$-module. Then $N$ is Cohen-Macaulay as an $S$-module if and only if $N$ is Cohen-Macaulay as an $R$-module.

**Proof.**
Omitted.
$\square$

Lemma 10.102.7. Let $R$ be a Noetherian local ring. Let $M$ be a finite Cohen-Macaulay $R$-module. If $\mathfrak p \in \text{Ass}(M)$, then $\dim (R/\mathfrak p) = \dim (\text{Supp}(M))$ and $\mathfrak p$ is a minimal prime in the support of $M$. In particular, $M$ has no embedded associated primes.

**Proof.**
By Lemma 10.71.9 we have $\text{depth}(M) \leq \dim (R/\mathfrak p)$. Of course $\dim (R/\mathfrak p) \leq \dim (\text{Supp}(M))$ as $\mathfrak p \in \text{Supp}(M)$ (Lemma 10.62.2). Thus we have equality in both inequalities as $M$ is Cohen-Macaulay. Then $\mathfrak p$ must be minimal in $\text{Supp}(M)$ otherwise we would have $\dim (R/\mathfrak p) < \dim (\text{Supp}(M))$. Finally, minimal primes in the support of $M$ are equal to the minimal elements of $\text{Ass}(M)$ (Proposition 10.62.6) hence $M$ has no embedded associated primes (Definition 10.66.1).
$\square$

Definition 10.102.8. Let $R$ be a Noetherian local ring. A finite module $M$ over $R$ is called a *maximal Cohen-Macaulay* module if $\text{depth}(M) = \dim (R)$.

In other words, a maximal Cohen-Macaulay module over a Noetherian local ring is a finite module with the largest possible depth over that ring. Equivalently, a maximal Cohen-Macaulay module over a Noetherian local ring $R$ is a Cohen-Macaulay module of dimension equal to the dimension of the ring. In particular, if $M$ is a Cohen-Macaulay $R$-module with $\mathop{\mathrm{Spec}}(R) = \text{Supp}(M)$, then $M$ is maximal Cohen-Macaulay. Thus the following two lemmas are on maximal Cohen-Macaulay modules.

Lemma 10.102.9. Let $R$ be a Noetherian local ring. Assume there exists a Cohen-Macaulay module $M$ with $\mathop{\mathrm{Spec}}(R) = \text{Supp}(M)$. Then any maximal chain of ideals $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ n$ has length $n = \dim (R)$.

**Proof.**
We will prove this by induction on $\dim (R)$. If $\dim (R) = 0$, then the statement is clear. Assume $\dim (R) > 0$. Then $n > 0$. Choose an element $x \in \mathfrak p_1$, with $x$ not in any of the minimal primes of $R$, and in particular $x \not\in \mathfrak p_0$. (See Lemma 10.14.2.) Then $\dim (R/xR) = \dim (R) - 1$ by Lemma 10.59.12. The module $M/xM$ is Cohen-Macaulay over $R/xR$ by Proposition 10.102.4 and Lemma 10.102.6. The support of $M/xM$ is $\mathop{\mathrm{Spec}}(R/xR)$ by Lemma 10.39.9. After replacing $x$ by $x^ n$ for some $n$, we may assume that $\mathfrak p_1$ is an associated prime of $M/xM$, see Lemma 10.71.8. By Lemma 10.102.7 we conclude that $\mathfrak p_1/(x)$ is a minimal prime of $R/xR$. It follows that the chain $\mathfrak p_1/(x) \subset \ldots \subset \mathfrak p_ n/(x)$ is a maximal chain of primes in $R/xR$. By induction we find that this chain has length $\dim (R/xR) = \dim (R) - 1$ as desired.
$\square$

Lemma 10.102.10. Suppose $R$ is a Noetherian local ring. Assume there exists a Cohen-Macaulay module $M$ with $\mathop{\mathrm{Spec}}(R) = \text{Supp}(M)$. Then for a prime $\mathfrak p \subset R$ we have

**Proof.**
Follows immediately from Lemma 10.102.9.
$\square$

Lemma 10.102.11. Suppose $R$ is a Noetherian local ring. Let $M$ be a Cohen-Macaulay module over $R$. For any prime $\mathfrak p \subset R$ the module $M_{\mathfrak p}$ is Cohen-Macaulay over $R_\mathfrak p$.

**Proof.**
Choose a maximal chain of primes $\mathfrak p = \mathfrak p_ c \subset \mathfrak p_{c - 1} \subset \ldots \subset \mathfrak p_1 \subset \mathfrak m$. If we prove the result for $M_{\mathfrak p_1}$ over $R_{\mathfrak p_1}$, then the lemma will follow by induction on $c$. Thus we may assume that there is no prime strictly between $\mathfrak p$ and $\mathfrak m$.

If $M_\mathfrak p = 0$, then the lemma holds. Assume $M_\mathfrak p \not= 0$. We have $\dim (\text{Supp}(M_\mathfrak p)) \leq \dim (\text{Supp}(M)) - 1$ as a chain of primes in the support of $M_\mathfrak p$ is a chain a primes in the support of $M$ not including $\mathfrak m$. Thus it suffices to show that the depth of $M_\mathfrak p$ is at least the depth of $M$ minus $1$. We will prove by induction on the depth of $M$ that there exists an $M$-regular sequence $f_1, \ldots , f_{\text{depth}(M) - 1}$ in $\mathfrak p$. This will prove the lemma since localization at $\mathfrak p$ is exact. Since $\text{depth}(M) = \dim ((\text{Supp}(M)) \geq \dim (\text{Supp}(M_\mathfrak p)) + 1 \geq 1$ we see that the base case happens when the depth of $M$ is $1$ and this case is trivial. Assume the depth of $M$ is at least $2$.

Let $I \subset R$ be the annihilator of $M$ such that $\mathop{\mathrm{Spec}}(R/I) = V(I) = \text{Supp}(M)$ (Lemma 10.39.5). By Lemmas 10.102.6 and 10.102.9 every maximal chain of primes in $V(I)$ has length $\geq 2$. Hence none of the minimal primes of $V(I)$ are equal to $\mathfrak p$. Thus we can use Lemma 10.14.2 to find a $f_1 \in \mathfrak p$ which is not contained in any of the minimal primes of $V(I)$. Then $f_1$ is a nonzerodivisor on $M$ and $M/f_1M$ has depth exactly one less by Lemma 10.102.3. By induction we can extend to an $M$-regular sequence $f_1, \ldots , f_ r \in \mathfrak p$ with $r = \text{depth}(M) - 1$ as desired. $\square$

Definition 10.102.12. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. We say $M$ is *Cohen-Macaulay* if $M_\mathfrak p$ is a Cohen-Macaulay module over $R_\mathfrak p$ for all primes $\mathfrak p$ of $R$.

By Lemma 10.102.11 it suffices to check this in the maximal ideals of $R$.

Lemma 10.102.13. Let $R$ be a Noetherian ring. Let $M$ be a Cohen-Macaulay module over $R$. Then $M \otimes _ R R[x_1, \ldots , x_ n]$ is a Cohen-Macaulay module over $R[x_1, \ldots , x_ n]$.

**Proof.**
By induction on the number of variables it suffices to prove this for $M[x] = M \otimes _ R R[x]$ over $R[x]$. Let $\mathfrak m \subset R[x]$ be a maximal ideal, and let $\mathfrak p = R \cap \mathfrak m$. Let $f_1, \ldots , f_ d$ be a $M_\mathfrak p$-regular sequence in the maximal ideal of $R_{\mathfrak p}$ of length $d = \dim (\text{Supp}(M_{\mathfrak p}))$. Note that since $R[x]$ is flat over $R$ the localization $R[x]_{\mathfrak m}$ is flat over $R_{\mathfrak p}$. Hence, by Lemma 10.67.5, the sequence $f_1, \ldots , f_ d$ is a $M[x]_{\mathfrak m}$-regular sequence of length $d$ in $R[x]_{\mathfrak m}$. The quotient

has support equal to the primes lying over $\mathfrak p$ because $R_\mathfrak p \to R[x]_\mathfrak m$ is flat and the support of $M_{\mathfrak p}/(f_1, \ldots , f_ d)M_{\mathfrak p}$ is equal to $\{ \mathfrak p\} $ (details omitted; hint: follows from Lemmas 10.39.4 and 10.39.5). Hence the dimension is $1$. To finish the proof it suffices to find an $f \in \mathfrak m$ which is a nonzerodivisor on $Q$. Since $\mathfrak m$ is a maximal ideal, the field extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak m)$ is finite (Theorem 10.33.1). Hence we can find $f \in \mathfrak m$ which viewed as a polynomial in $x$ has leading coefficient not in $\mathfrak p$. Such an $f$ acts as a nonzerodivisor on

and hence acts as a nonzerodivisor on $Q$. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (6)

Comment #684 by Keenan Kidwell on

Comment #2211 by David Savitt on

Comment #2221 by Johan on

Comment #2700 by Tanya Kaushal Srivastava on

Comment #2842 by Johan on

Comment #2846 by Johan on