Lemma 15.94.3. Let $K$ be a field. For a $K$-algebra $A$ denote $B_{max}(A)$ the maximal weakly étale $K$-subalgebra of $A$ as in Lemma 15.94.2. Then

1. any $K$-algebra map $A' \to A$ induces a $K$-algebra map $B_{max}(A') \to B_{max}(A)$,

2. if $A' \subset A$, then $B_{max}(A') = B_{max}(A) \cap A'$,

3. if $A = \mathop{\mathrm{colim}}\nolimits A_ i$ is a filtered colimit, then $B_{max}(A) = \mathop{\mathrm{colim}}\nolimits B_{max}(A_ i)$,

4. the map $B_{max}(A) \to B_{max}(A_{red})$ is an isomorphism,

5. $B_{max}(A_1 \times \ldots \times A_ n) = B_{max}(A_1) \times \ldots \times B_{max}(A_ n)$,

6. if $A$ has no nontrivial idempotents, then $B_{max}(A)$ is a field and a separable algebraic extension of $K$,

Proof. Proof of (1). This is true because the image of $B_{max}(A') \to A$ is weakly étale over $K$ by Lemma 15.94.1.

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 15.94.1 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 15.93.14. 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 10.126.3 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 15.94.1, 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 15.93.16). By Algebra, Lemma 10.136.17 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 15.94.1).

Proof of (5). Omitted.

Proof of (6). Follows from Lemma 15.94.1 part (4). $\square$

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