Lemma 37.22.8. Let $f : X \to S$ be a morphism of schemes which is flat and locally of finite presentation. Let $x \in X$ with image $s \in S$. Set $d = \dim _ x(X_ s)$. The following are equivalent

1. $f$ is Cohen-Macaulay at $x$,

2. there exists an open neighbourhood $U \subset X$ of $x$ and a locally quasi-finite morphism $U \to \mathbf{A}^ d_ S$ over $S$ which is flat at $x$,

3. there exists an open neighbourhood $U \subset X$ of $x$ and a locally quasi-finite flat morphism $U \to \mathbf{A}^ d_ S$ over $S$,

4. for any $S$-morphism $g : U \to \mathbf{A}^ d_ S$ of an open neighbourhood $U \subset X$ of $x$ we have: $g$ is quasi-finite at $x$ $\Rightarrow$ $g$ is flat at $x$.

Proof. Openness of flatness shows (2) and (3) are equivalent, see Theorem 37.15.1.

Choose affine open $U = \mathop{\mathrm{Spec}}(A) \subset X$ with $x \in U$ and $V = \mathop{\mathrm{Spec}}(R) \subset S$ with $f(U) \subset V$. Then $R \to A$ is a flat ring map of finite presentation. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $x$. After replacing $A$ by a principal localization we may assume there exists a quasi-finite map $R[x_1, \ldots , x_ d] \to A$, see Algebra, Lemma 10.125.2. Thus there exists at least one pair $(U, g)$ consisting of an open neighbourhood $U \subset X$ of $x$ and a locally1 quasi-finite morphism $g : U \to \mathbf{A}^ d_ S$.

Claim: Given $R \to A$ flat and of finite presentation, a prime $\mathfrak p \subset A$ and $\varphi : R[x_1, \ldots , x_ d] \to A$ quasi-finite at $\mathfrak p$ we have: $\mathop{\mathrm{Spec}}(\varphi )$ is flat at $\mathfrak p$ if and only if $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(R)$ is Cohen-Macaulay at $\mathfrak p$. Namely, by Theorem 37.16.2 flatness may be checked on fibres. The same is true for being Cohen-Macaulay (as $A$ is already assumed flat over $R$). Thus the claim follows from Algebra, Lemma 10.130.1.

The claim shows that (1) is equivalent to (4) and combined with the fact that we have constructed a suitable $(U, g)$ in the second paragraph, the claim also shows that (1) is equivalent to (2). $\square$

[1] If $S$ is quasi-separated, then $g$ will be quasi-finite.

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