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A $1$-dimensional integral domain is always Cohen-Macaulay (C-M). I know this fact, but I do not know how can I reach at. Maybe one should use, somehow, the fact that $R$ is C-M if and only if each localization $R_P$ at the prime ideals is so. Any leading would be thanked!

user26857
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karparvar
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    every non-zero element in an integral domain is non-zerodivisor. so depth is not $0$; so is $1$ – user 1 Jun 22 '15 at 06:06

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By definition, a ring $R$ is CM iff $R_m$ is CM for every maximal ideal $m$ of $R$. Since $R$ is a one-dimensional integral domain, $R_m$ is a one-dimensional integral domain, so the problem reduces to the local case. Now observe that the height of the maximal ideal is one, and the grade is at least one.

user26857
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