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If $R$ is Noetherian local, then $R$ is CM iff $depth(R)=dim(R)$. If $R$ is only Noetherian, then $R$ is CM iff $R_P$ is CM for all prime ideals.

I want to ask, if $R$ is local, does the two definitions coincide?

The same question is true for regular ring since we have the formula $global.dim(R)=\sup(global.dim(R_P))$. Same for Gorenstein ring. But do we have a similar formula for depth?

T C
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  • I have to admit that I don't get this question. If $(R,m)$ is local and CM (with the second definition), then, in particular, $R_m=R$ is CM. – user26857 Jun 30 '18 at 17:29
  • Depth (in the sense used above) is defined only for local rings. So, in order to globalise the definition of CM one has to show that localisations of CM local rings are again CM, and then your second definition is the only reasonable choice to define CM for not necessarily local (noetherian) rings. – Fred Rohrer Jun 30 '18 at 19:47
  • @LongLêThiên When invert invertible elements like those from $R\setminus m$ you get noting new, that is, remain inside $R$. – user26857 Jul 02 '18 at 06:22

1 Answers1

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Here is a somewhat ad hoc proof of the fact that if $R$ is a local CM ring, then $R_P$ is CM for all prime ideals of $R$. Is this what you wanted?

We first prove that if $P$ is a prime ideal associated to $R$, the height $P=0$. If not, then $\dim R>0$ and the proof is by induction on the depth of $R=\dim R$. So, we have a non-zero element $a\in R$ such that $Pa=0$. Fix a non-zero divisor $x\in R$. Then, we may assume $a$ is not a multiple of $x$, since $\cap (x^n)=0$. If $a=x^nb, a\not\in (x^{n+1})$, then $Pb=0$, since $x\not\in P$. So, $b\not\in (x)$. Then, in $S=R/xR$, $a\neq 0$ and $PSa=0$, so height of $PS$ must be zero by induction ($PS$ is not a prime, but easy to see that we can find a prime ideal containing $PS$ doing what we want). Since $x\not\in P$, height of $P$ itself must be zero. This proves my assertion.

Again, we use the same induction. Let $P$ be any prime. If height of $P=0$, clearly $R_P$ is CM. If the height is positive, from the previous paragraph, we must have a non-zero divisor $x\in P$. Then go modulo $x$ to apply induction and thus the result. Notice that $x$ is a non-zero divisor in $R_P$ too.

Mohan
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  • "Here is a somewhat ad hoc proof" of what? (Do you want to show that $R_P$ is CM provided that $R$ is a local CM ring? I'm not sure if this is what the OP asked.) – user26857 Jul 02 '18 at 06:18
  • @user26857 I have made an edit. Am not sure, after what you say, that I am answering the right question. – Mohan Jul 02 '18 at 14:17