Lemma 15.93.16. Let $B$ be an algebra over a field $K$. The following are equivalent

$B \otimes _ K B \to B$ is flat,

$K \to B$ is weakly étale, and

$B$ is a filtered colimit of étale $K$-algebras.

Moreover, every finitely generated $K$-subalgebra of $B$ is étale over $K$.

**Proof.**
Parts (1) and (2) are equivalent because every $K$-algebra is flat over $K$. Part (3) implies (1) and (2) by Lemma 15.93.14

Assume (1) and (2) hold. We will prove (3) and the finite statement of the lemma. A field is absolutely flat ring, hence $B$ is a absolutely flat ring by Lemma 15.93.8. Hence $B$ is reduced and every local ring is a field, see Lemma 15.93.5.

Let $\mathfrak q \subset B$ be a prime. The ring map $B \to B_\mathfrak q$ is weakly étale, hence $B_\mathfrak q$ is weakly étale over $K$ (Lemma 15.93.9). Thus $B_\mathfrak q$ is a separable algebraic extension of $K$ by Lemma 15.93.15.

Let $K \subset A \subset B$ be a finitely generated $K$-sub algebra. We will show that $A$ is étale over $K$ which will finish the proof of the lemma. Then every minimal prime $\mathfrak p \subset A$ is the image of a prime $\mathfrak q$ of $B$, see Algebra, Lemma 10.29.5. Thus $\kappa (\mathfrak p)$ as a subfield of $B_\mathfrak q = \kappa (\mathfrak q)$ is separable algebraic over $K$. Hence every generic point of $\mathop{\mathrm{Spec}}(A)$ is closed (Algebra, Lemma 10.34.9). Thus $\dim (A) = 0$. Then $A$ is the product of its local rings, e.g., by Algebra, Proposition 10.59.6. Moreover, since $A$ is reduced, all local rings are equal to their residue fields wich are finite separable over $K$. This means that $A$ is étale over $K$ by Algebra, Lemma 10.141.4 and finishes the proof.
$\square$

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