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Let $R$ be a commutative Noetherian ring. $I=(x_1,...,x_n)$ is an ideal generated by $n$ elements such that $\operatorname{height}I=n$. If $R$ is Cohen-Macaulay, then every associated prime of $I$ is minimal over $I$.

This is the statement of unmixedness theorem from Eisenbud's commutative algebra book. His definition of Cohen-Macaulay ring is

A ring such that $\operatorname{depth}P=\operatorname{height}P$ for every maximal ideal $P$ of $R$ is called a Cohen-Macaulay ring.

However, in his proof of unmixedness theorem he uses a corollary which states:

Cohen-Macaulay rings are universally catenary. In a local Cohen-Macaulay ring every associated prime of $R$ is minimal.

We know that $R/I$ is Cohen-Macaulay, but since he never assumes in the unmixedness theorem that $R$ is local, how could one conclude that every associated prime of $I$ is minimal over $I$ in the theorem from his corollary?

user782932
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  • An aasociated prime of $I$ is an associated prime of $R/I$ and conversely. Your question is how he deduced the non-local case from the local case? – user26857 Mar 25 '22 at 13:38
  • @user26857 yes,I wonder if it is true in non-local case the associate primes are all minimal? – user782932 Mar 25 '22 at 15:36

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First of all, let me mention that the first claim is Corollary 18.14.

In Proposition 18.13 it is proved that $R/I$ is Cohen-Macaulay. Let $\mathfrak p\in\operatorname{Ass}_R(R/I)$. Then $\mathfrak pR_{\mathfrak p}\in\operatorname{Ass}_{R_{\mathfrak p}}(R_{\mathfrak p}/IR_{\mathfrak p})$. By Corollary 18.10 we have that $\mathfrak pR_{\mathfrak p}$ is minimal over $IR_{\mathfrak p}$, so $\mathfrak p$ is minimal over $I$.

user26857
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  • It seems to me by applying thse same argument of proposition 18.10 of the local case, we have for any Cohen-Macaulay ring, the associate primes are minimal. So, maybe the $R$ of 18.10 in the last sentence just refers to any Cohen-Macaulay ring. – user782932 Mar 25 '22 at 17:39
  • I mean the property assocaite primes are minimal holds for any CM ring. Is this right? – user782932 Mar 25 '22 at 17:49
  • Yes, it is right. – user26857 Mar 25 '22 at 20:48