Lemma 15.105.4. Let $L/K$ be an extension of fields. Let $A$ be a $K$-algebra. Let $B \subset A$ be the maximal weakly étale $K$-subalgebra of $A$ as in Lemma 15.105.2. Then $B \otimes _ K L$ is the maximal weakly étale $L$-subalgebra of $A \otimes _ K L$.

**Proof.**
For an algebra $A$ over $K$ we write $B_{max}(A/K)$ for the maximal weakly étale $K$-subalgebra of $A$. Similarly we write $B_{max}(A'/L)$ for the maximal weakly étale $L$-subalgebra of $A'$ if $A'$ is an $L$-algebra. Since $B_{max}(A/K) \otimes _ K L$ is weakly étale over $L$ (Lemma 15.104.7) and since $B_{max}(A/K) \otimes _ K L \subset A \otimes _ K L$ we obtain a canonical injective map

The lemma states that this map is an isomorphism.

To prove the lemma for $L$ and our $K$-algebra $A$, it suffices to prove the lemma for any field extension $L'$ of $L$. Namely, we have the factorization

hence the composition cannot be surjective without $B_{max}(A/K) \otimes _ K L \to B_{max}((A \otimes _ K L)/L)$ being surjective. Thus we may assume $L$ is algebraically closed.

Reduction to finite type $K$-algebra. We may write $A$ is the filtered colimit of its finite type $K$-subalgebras. Using Lemma 15.105.3 we see that it suffices to prove the lemma for finite type $K$-algebras.

Assume $A$ is a finite type $K$-algebra. Since the kernel of $A \to A_{red}$ is nilpotent, the same is true for $A \otimes _ K L \to A_{red} \otimes _ K L$. Then

is injective because the kernel is nilpotent and the weakly étale $L$-algebra $B_{max}((A \otimes _ K L)/L)$ is reduced (Lemma 15.105.1). Since $B_{max}(A/K) = B_{max}(A_{red}/K)$ by Lemma 15.105.3 we conclude that it suffices to prove the lemma for $A_{red}$.

Assume $A$ is a reduced finite type $K$-algebra. Let $Q = Q(A)$ be the total quotient ring of $A$. Then $A \subset Q$ and $A \otimes _ K L \subset Q \otimes _ A L$ and hence

and

by Lemma 15.105.3. Since $-\otimes _ K L$ is an exact functor, it follows that if we prove the result for $Q$, then the result follows for $A$. Since $Q$ is a finite product of fields (Algebra, Lemmas 10.25.4, 10.25.1, 10.31.6, and 10.31.1) and since $B_{max}$ commutes with products (Lemma 15.105.3) it suffices to prove the lemma when $A$ is a field.

Assume $A$ is a field. We reduce to $A$ being finitely generated over $K$ by the argument in the third paragraph of the proof. (In fact the way we reduced to the case of a field produces a finitely generated field extension of $K$.)

Assume $A$ is a finitely generated field extension of $K$. Then $K' = B_{max}(A/K)$ is a field separable algebraic over $K$ by Lemma 15.105.3 part (6). Hence $K'$ is a finite separable field extension of $K$ and $A$ is geometrically irreducible over $K'$ by Algebra, Lemma 10.47.13. Since $L$ is algebraically closed and $K'/K$ finite separable we see that

is an isomorphism (Fields, Lemma 9.13.4). We conclude

Since $A$ is geometrically irreducible over $K'$ we see that $A \otimes _{K', \sigma } L$ has a unique minimal prime. Since $L$ is algebraically closed it follows that $B_{max}((A \otimes _{K', \sigma } L)/L) = L$ because this $L$-algebra is a field algebraic over $L$ by Lemma 15.105.3 part (6). It follows that the maximal weakly étale $K' \otimes _ K L$-subalgebra of $A \otimes _ K L$ is $K' \otimes _ K L$ because we can decompose these subalgebras into products as above. Hence the inclusion $K' \otimes _ K L \subset B_{max}((A \otimes _ K L)/L)$ is an equality: the ring map $K' \otimes _ K L \to B_{max}((A \otimes _ K L)/L)$ is weakly étale by Lemma 15.104.11. $\square$

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