The Stacks project

Proof. Part (1) follows from Lemma 15.104.8 but of course it follows from part (3) as well. Part (3) follows from Lemma 15.104.16 and the fact that étale $K$-algebras are finite products of finite separable extensions of $K$, see Algebra, Lemma 10.143.4. 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 15.104.16. 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 15.104.7 and 15.104.9. $\square$

Proof. Let $B_1, B_2 \subset A$ be weakly étale $K$-subalgebras. Then $B_1 \otimes _ K B_2$ is weakly étale over $K$ and so is the image of $B_1 \otimes _ K B_2 \to A$ (Lemma 15.105.1). Thus the collection $\mathcal{B}$ of weakly étale $K$-subalgebras $B \subset A$ is directed and the colimit $B_{max} = \mathop{\mathrm{colim}}\nolimits _{B \in \mathcal{B}} B$ is a weakly étale $K$-algebra by Lemma 15.104.14. Hence the image of $B_{max} \to A$ is weakly étale over $K$ (previous lemma cited). It follows that this image is in $\mathcal{B}$ and hence $\mathcal{B}$ has a maximal element (and the image is the same as $B_{max}$). $\square$

Proof. Proof of (1). This is true because the image of $B_{max}(A') \to A$ is weakly étale over $K$ by Lemma 15.105.1.

Proof of (2). By (1) we have $B_{max}(A') \subset B_{max}(A)$. Conversely, $B_{max}(A) \cap A'$ is a weakly étale $K$-algebra by Lemma 15.105.1 and hence contained in $B_{max}(A')$.

Proof of (3). By (1) there is a map $\mathop{\mathrm{colim}}\nolimits B_{max}(A_ i) \to A$ which is injective because the system is filtered and $B_{max}(A_ i) \subset A_ i$. The colimit $\mathop{\mathrm{colim}}\nolimits B_{max}(A_ i)$ is weakly étale over $K$ by Lemma 15.104.14. Hence we get an injective map $\mathop{\mathrm{colim}}\nolimits B_{max}(A_ i) \to B_{max}(A)$. Suppose that $a \in B_{max}(A)$. Then $a$ generates a finitely presented $K$-subalgebra $B \subset B_{max}(A)$. By Algebra, Lemma 10.127.3 there is an $i$ and a $K$-algebra map $f : B \to A_ i$ lifting the given map $B \to A$. Since $B$ is weakly étale by Lemma 15.105.1, we see that $f(B) \subset B_{max}(A_ i)$ and we conclude that $a$ is in the image of $\mathop{\mathrm{colim}}\nolimits B_{max}(A_ i) \to B_{max}(A)$.

Proof of (4). Write $B_{max}(A_{red}) = \mathop{\mathrm{colim}}\nolimits B_ i$ as a filtered colimit of étale $K$-algebras (Lemma 15.104.16). By Algebra, Lemma 10.138.17 for each $i$ there is a $K$-algebra map $f_ i : B_ i \to A$ lifting the given map $B_ i \to A_{red}$. It follows that the canonical map $B_{max}(A_{red}) \to B_{max}(A)$ is surjective. The kernel consists of nilpotent elements and hence is zero as $B_{max}(A_{red})$ is reduced (Lemma 15.105.1).

Proof of (5). Omitted.

Proof of (6). Follows from Lemma 15.105.1 part (4). $\square$

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. $\square$