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

Lemma 10.103.9. 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 prime 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 10.15.2.) Then $\dim (R/xR) = \dim (R) - 1$ by Lemma 10.60.13. The module $M/xM$ is Cohen-Macaulay over $R/xR$ by Proposition 10.103.4 and Lemma 10.103.6. The support of $M/xM$ is $\mathop{\mathrm{Spec}}(R/xR)$ by Lemma 10.40.9. 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 10.72.8. By Lemma 10.103.7 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. $\square$

Comments (9)

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 is in the dimension case (although just by using that is not zero) and in the induction step, more precisely in the application of Proposition 10.103.4 (of course the proposition implies that is CM as well). What would happen if you did this proof with a CM module whose support isn't is that you would at some point in the induction end up with a module whose support is and the induction step would not work.

The final chain of primes is maximal because the original chain of primes in is maximal and because is exactly the collection of prime ideals in containing . 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 . This is the kind of thing the lemma is trying to rule out. The construction guarantees isn't in (supposing that's minimal), but one could still have , in which case would extend that final chain. I think this is the point where the argument breaks down if you try to run it with and , which is maximal CM but not supported on all of .

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 , which gives [0AAF]. (The case of minimal primes is handled via [0BUS], and that really uses that the support is all of .) 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 and 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 10.72.9 which is in some sense a generalization of this lemma.

Comment #6600 by WhatJiaranEatsTonight on

"Then any maximal chain of ideals". It should be "Then any maximal chain of primes".

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  • 6 comment(s) on Section 10.103: Cohen-Macaulay modules

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