## 10.165 Geometrically Cohen-Macaulay algebras

This section is a bit of a misnomer, since Cohen-Macaulay algebras are automatically geometrically Cohen-Macaulay. Namely, see Lemma 10.129.6 and Lemma 10.165.2 below.

Lemma 10.165.1. Let $k$ be a field and let $k \subset K$ and $k \subset L$ be two field extensions such that one of them is a field extension of finite type. Then $K \otimes _ k L$ is a Noetherian Cohen-Macaulay ring.

Proof. The ring $K \otimes _ k L$ is Noetherian by Lemma 10.30.8. Say $K$ is a finite extension of the purely transcendental extension $k(t_1, \ldots , t_ r)$. Then $k(t_1, \ldots , t_ r) \otimes _ k L \to K \otimes _ k L$ is a finite free ring map. By Lemma 10.111.9 it suffices to show that $k(t_1, \ldots , t_ r) \otimes _ k L$ is Cohen-Macaulay. This is clear because it is a localization of the polynomial ring $L[t_1, \ldots , t_ r]$. (See for example Lemma 10.103.7 for the fact that a polynomial ring is Cohen-Macaulay.) $\square$

Lemma 10.165.2. Let $k$ be a field. Let $S$ be a Noetherian $k$-algebra. Let $k \subset K$ be a finitely generated 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. By Lemma 10.30.8 the ring $S_ K$ is Noetherian. Hence $S_{\mathfrak q} \to (S_ K)_{\mathfrak q_ K}$ is a flat local homomorphism of Noetherian local rings. Note that the fibre

$(S_ K)_{\mathfrak q_ K} / \mathfrak q (S_ K)_{\mathfrak q_ K} \cong (\kappa (\mathfrak q) \otimes _ k K)_{\mathfrak q'}$

is the localization of the Cohen-Macaulay (Lemma 10.165.1) ring $\kappa (\mathfrak q) \otimes _ k K$ at a suitable prime ideal $\mathfrak q'$. Hence the lemma follows from Lemma 10.161.3. $\square$

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