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I am reading Theorem 27.5 here which states the following:

Let $f : X \to Y$ be a finite morphism of Noetherian schemes with $Y$ non-singular. Then $X$ is Cohen-Macaulay iff $f_\ast\mathcal{O}_X$ is locally free.

Now implicit in the proof is that if $X$ is Cohen Macaulay, then for any $y \in Y$, $(f_\ast \mathcal{O}_X)_y$ is Cohen-Macaulay. Why is this so? The definition of a Cohen-Macaulay scheme is that $\mathcal{O}_{X,x}$ be Cohen-Macaulay at every $x \in X$ which is not the same as saying that $(f_\ast \mathcal{O}_X)_y $ be Cohen-Macaulay.

user26857
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1 Answers1

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A ring $R$ is CM iff $R_{\mathfrak m}$ is CM for every maximal ideal $\mathfrak m$ of $R$, in which case $R_{\mathfrak p}$ is CM for every prime ideal $\mathfrak p$ of $R$. This is, for instance, in Eisenbud (Proposition 18.8). Choose an affine neighbourhood $U$ of $y$ and observe that $\mathcal O_X(f^{-1}(U))$ is CM because its localizations all are. But this means that $(f_\ast\mathcal O_X)(U)$ is CM, hence the localization at $y$ will also be CM.

Jesko Hüttenhain
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  • Thanks for your answer. There is a potential technicality in that $y \in Y$ may not be a closed point. Is this remedied by your first sentence? Thanks. –  Jan 13 '14 at 12:13
  • It is indeed. You can reformulate the first statement to "$R$ is CM $\Leftrightarrow$ All localizations at maximal ideals of $R$ are CM $\Leftrightarrow$ All localizations at prime ideals of $R$ are CM". Hence, even if $y$ corresponds only to a point and not a closed point, localization at $y$ will be CM. – Jesko Hüttenhain Jan 13 '14 at 12:18
  • Do you have an idea about the converse, namely if $(f_\ast \mathcal{O}X)_y$ is CM for all $y \in Y$ then $\mathcal{O}{X,x}$ is CM for all $x \in X$? –  Jan 13 '14 at 12:25
  • Well sure, the converse also holds. It is essentially the same argument. You might need that finite morphisms are affine. – Jesko Hüttenhain Jan 14 '14 at 10:01