Lemma 10.130.7. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $K/k$ be a field extension, and set $S_ K = K \otimes _ k S$. Let $\mathfrak q \subset S$ be a prime of $S$. Let $\mathfrak q_ K \subset S_ K$ be a prime of $S_ K$ lying over $\mathfrak q$. Then $S_{\mathfrak q}$ is Cohen-Macaulay if and only if $(S_ K)_{\mathfrak q_ K}$ is Cohen-Macaulay.

**Proof.**
During the course of the proof we may (finitely many times) replace $S$ by $S_ g$ for any $g \in S$, $g \not\in \mathfrak q$. Hence using Lemma 10.115.5 we may assume that $\dim (S) = \dim _{\mathfrak q}(S/k) =: d$ and find a finite injective map $k[x_1, \ldots , x_ d] \to S$. Note that this also induces a finite injective map $K[x_1, \ldots , x_ d] \to S_ K$ by base change. By Lemma 10.116.6 we have $\dim _{\mathfrak q_ K}(S_ K/K) = d$. Set $\mathfrak p = k[x_1, \ldots , x_ d] \cap \mathfrak q$ and $\mathfrak p_ K = K[x_1, \ldots , x_ d] \cap \mathfrak q_ K$. Consider the following commutative diagram of Noetherian local rings

By Lemma 10.130.1 we have to show that the left vertical arrow is flat if and only if the right vertical arrow is flat. Because the bottom arrow is flat this equivalence holds by Lemma 10.100.1. $\square$

## 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Your email address will not be published.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.

## Comments (0)