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My goal is to show that a reduced Noetherian ring of dimension one is Cohen-Macaulay.

My question is on why the reduced condition is imposed. Let $ \mathfrak{p} \in \operatorname{Spec}(R) $. If $ \dim(R_{\mathfrak{p}}) = 0 $ we're done. If $\dim(R_{\mathfrak{p}}) = 1$ we just need to show that $\text{depth}(R_{\mathfrak{p}}) \ne 0 $. If $\text{depth}(R_{\mathfrak{p}}) = 0$ then $\mathfrak{p}R_{\mathfrak{p}}$ consists of zero-divisors, so $\mathfrak{p} R_{\mathfrak{p}} $ is a minimal associated prime, which is a contradiction since $\dim(R_{\mathfrak{p}}) = 1 $. Where did I need to assume that $R$ was reduced (i.e. contains no nonzero nilpotent elements)?

user26857
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Kevin Sheng
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    Let $R=k[x,y]/(x^2,xy)$. Then $R$ has dimension one, but not CM. – Mohan Dec 02 '16 at 18:12
  • I am a bit new to these concepts, could you explain where the nilpotent elements mess up my argument? I see that the maximal ideal $(x,y)$ is not a minimal associated prime which ruins my argument above. However, I am not sure how I can generalize this a reduced ring. – Kevin Sheng Dec 02 '16 at 18:16
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    What makes you say $\mathfrak pR_\mathfrak{p}$ is a minimal associated prime? – Bernard Dec 02 '16 at 19:15
  • I was wrong, it turns out that is the crucial piece I was missing. – Kevin Sheng Dec 02 '16 at 19:20
  • Actually the missing piece for proving the claim is the following: http://math.stackexchange.com/questions/1323929/in-a-reduced-ring-the-set-of-zero-divisors-equals-the-union-of-minimal-prime-ide. This shows that $pR_p$ can't be minimal provided $\dim R_p=1$. – user26857 Dec 02 '16 at 20:04

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