Lemma 10.130.6. 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

$\xymatrix{ S_{\mathfrak q} \ar[r] & (S_ K)_{\mathfrak q_ K} \\ k[x_1, \ldots , x_ d]_{\mathfrak p} \ar[r] \ar[u] & K[x_1, \ldots , x_ d]_{\mathfrak p_ K} \ar[u] }$

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. $\square$

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