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This is from CM Rings (Bruns & Herzog) p 64 2.1.20. Show that a one dimensional Noetherian local ring has a maximal Cohen Maucaulay module.

flag	asked Jan 30 2012 at 22:48 Andrei 35●5

Mahdi Majidi-Zolbanin · Accepted Answer · 2012-01-31 02:19:42Z UTC

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Take $R/\mathfrak{p}$, where $\mathfrak{p}$ is a minimal prime ideal of $R$.

answered Jan 30 2012 at 23:12

In this case R/p has dimension 1, but why it has depth 1? R is not necessarily CM, so R can have depth 0. – Andrei Jan 31 2012 at 9:37

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$R/p$ is a domain, so has depth 1, both as a ring itself and as a module over the original ring. – Graham Leuschke Jan 31 2012 at 11:43

sorry, that's obvious – Andrei Jan 31 2012 at 20:51

maximal Cohen-Macaulay module [closed]