Corollary 2

Tag 0D8I, permanent. Cite it as [Bourbaki, Tag 0D8I].

Commutative Algebra, chapter X, § 6. Extension des scalaires dans les algèbres régulières

Let $k$ be a field and $A$ a $k$ -algebra essentially of finite type. In order that $A$ should be absolutely regular, it is necessary and sufficient that the $A$ -module $\Omega_k(A)$ be projective and that for every minimal prime ideal $\mathfrak{q}$ of $A$ , the $k$ -algebra $A_{\mathfrak{q}}$ be separable.

Conversely, suppose that the $A$ -module $\Omega_k(A)$ is projective and that the $k$ -algebra $A_q$ is separable for every minimal prime ideal $q$ of $A$ . Let $p$ be a prime ideal of $A$ , and $q$ a minimal prime ideal of $A$ contained in $p$ . Since the $A_p$ -module $\Omega_k(A)_p$ is free (II, § 3, No. 2, Cor. 2 of Prop. 5), we have

[\kappa(p) \otimes_A \Omega_k(A) : \kappa(p)] = [\kappa(q) \otimes_A \Omega_k(A) : \kappa(q)] .

The $k$ -algebra $A_q$ is Artinian and separable, hence absolutely regular (No. 4, Example 2). Th. 1 implies

[\kappa(q) \otimes_A \Omega_k(A) : \kappa(q)] = \deg.\operatorname{tr}_k(\kappa(q)) .

It then follows from Th. 1 that the $k$ -algebra $A_p$ is absolutely regular, whence the corollary.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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