## 15.105 Weakly étale algebras over fields

If $K$ is a field, then an algebra $B$ is weakly étale over $K$ if and only if it is a filtered colimit of étale $K$-algebras. This is Lemma 15.104.16.

Lemma 15.105.1. 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 15.104.8 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. $\square$

Lemma 15.105.2. Let $K$ be a field. Let $A$ be a $K$-algebra. There exists a maximal weakly étale $K$-subalgebra $B_{max} \subset A$.

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 $B_{max}$). $\square$

Lemma 15.105.3. 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 15.105.2. 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 15.105.1.

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). $\square$

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

$B_{max}(A/K) \otimes _ K L \to B_{max}((A \otimes _ K L)/L)$

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

$B_{max}(A/K) \otimes _ K L' \to B_{max}((A \otimes _ K L)/L) \otimes _ L L' \to B_{max}((A \otimes _ K L')/L')$

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

$B_{max}((A \otimes _ K L)/L) \to B_{max}((A_{red} \otimes _ K L)/L)$

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

$B_{max}(A/K) = A \cap B_{max}(Q/K)$

and

$B_{max}((A \otimes _ K L)/L) = (A \otimes _ K L) \cap B_{max}((Q \otimes _ K L)/L)$

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

$K' \otimes _ K L \to \prod \nolimits _{\sigma \in \mathop{\mathrm{Hom}}\nolimits _ K(K', L)} L,\quad \alpha \otimes \beta \mapsto (\sigma (\alpha )\beta )_\sigma$

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

$A \otimes _ K L = A \otimes _{K'} (K' \otimes _ K L) = \prod \nolimits _{\sigma \in \mathop{\mathrm{Hom}}\nolimits _ K(K', L)} A \otimes _{K', \sigma } L$

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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