
In a local Cohen-Macaulay ring, any maximal chain of prime ideals has length equal to the dimension.

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. $\square$

Comment #2213 by David Savitt on

I don't follow the current version of this argument -- why is the chain in the last sentence necessarily maximal? It seems like something is missing because I think you should have to use that the support of M is really all of R and not just that M is maximal CM.

Comment #2214 by on

The proof seems fine to me. We are using that the support of $M$ is $\Spec(R)$ in the dimension $0$ case (although just by using that $M$ is not zero) and in the induction step, more precisely in the application of Proposition 10.102.4 (of course the proposition implies that $M/xM$ is CM as well). What would happen if you did this proof with a CM module whose support isn't $\Spec(R)$ is that you would at some point in the induction end up with a module whose support is $\{\mathfrak m\}$ and the induction step would not work.

The final chain of primes is maximal because the original chain of primes in $R$ is maximal and because $\Spec(R/xR)$ is exactly the collection of prime ideals in $R$ containing $x$. This kind of argument is used over and over again the proofs dealing with dimension theory in local Noetherian rings.

Comment #2215 by David Savitt on

The situation I'm worried about in the last sentence is that (a priori) we might have some other primes $\mathfrak{q}_0 \subset \mathfrak{q}'_0 \subset \mathfrak{p}_1$. This is the kind of thing the lemma is trying to rule out. The construction guarantees $x$ isn't in $\mathfrak{q}_0$ (supposing that's minimal), but one could still have $x \in \mathfrak{q}'_0$, in which case $\mathfrak{q}'_0/xR$ would extend that final chain. I think this is the point where the argument breaks down if you try to run it with $R=k[[x,y,z]]/(xy,xz)$ and $M = R/(x)$, which is maximal CM but not supported on all of $R$.

Comment #2216 by David Savitt on

BTW, I think this argument does at least show that every prime is contained in a chain of length $\dim(R)$, which gives [0AAF]. (The case of minimal primes is handled via [0BUS], and that really uses that the support is all of $R$.) Then if you prove that localization preserves CM (by one of the other standard arguments not invoking [0AAE]) you can combine these to check the catenary property.

Comment #2218 by on

Ah, OK, I see what you mean: clearly there are no primes in between the primes $\mathfrak p_i/(x)$ and $\mathfrak p_{i + 1}/(x)$ but there could be one at the beginning. This is indeed a flaw in the proof, thanks very much! Will fix this soon, hopefully later today.

Comment #2219 by on

One solution: use Lemma 10.71.9 which is in some sense a generalization of this lemma.

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• 6 comment(s) on Section 10.102: Cohen-Macaulay modules

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