Lemma 15.94.1. Let $K$ be a field. If $B$ is weakly étale over $K$, then

1. $B$ is reduced,

2. $B$ is integral over $K$,

3. any finitely generated $K$-subalgebra of $B$ is a finite product of finite separable extensions of $K$,

4. $B$ is a field if and only if $B$ does not have nontrivial idempotents and in this case it is a separable algebraic extension of $K$,

5. any sub or quotient $K$-algebra of $B$ is weakly étale over $K$,

6. if $B'$ is weakly étale over $K$, then $B \otimes _ K B'$ is weakly étale over $K$.

Proof. Part (1) follows from Lemma 15.93.8 but of course it follows from part (3) as well. Part (3) follows from Lemma 15.93.16 and the fact that étale $K$-algebras are finite products of finite separable extensions of $K$, see Algebra, Lemma 10.141.4. Part (3) implies (2). Part (4) follows from (3) as a product of fields is a field if and only if it has no nontrivial idempotents.

If $S \subset B$ is a subalgebra, then it is the filtered colimit of its finitely generated subalgebras which are all étale over $K$ by the above and hence $S$ is weakly étale over $K$ by Lemma 15.93.16. If $B \to Q$ is a quotient algebra, then $Q$ is the filtered colimit of $K$-algebra quotients of finite products $\prod _{i \in I} L_ i$ of finite separable extensions $L_ i/K$. Such a quotient is of the form $\prod _{i \in J} L_ i$ for some subset $J \subset I$ and hence the result holds for quotients by the same reasoning.

The statement on tensor products follows in a similar manner or by combining Lemmas 15.93.7 and 15.93.9. $\square$

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