The Stacks project

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.

Comments (0)

There are also:

  • 4 comment(s) on Section 37.22: Cohen-Macaulay morphisms

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BUU. Beware of the difference between the letter 'O' and the digit '0'.