2
$\begingroup$

Let $A$ and $B$ be étale $k$-algebras and consider the functor (from the category of étale algebras to the category of finite sets) defined by $F=\hom_k(\_,k_s)$, where $k_s$ is a fixed separable closure of $k$.

If $f:A\to B$ is a morphism which induces a bijection $F(f):F(B)\to F(A)$, I want to prove that $f$ is an isomorphism.

I know that if $A=\prod_i L_i$ is a (finite) product of (finite) separable field extensions, then $\hom_k(A,k_s)$ is in bijection with $\coprod_i \hom_k(L_i,k_s)$ but I don't know how that helps.

(If that helps anyone to search this in the future, this is axiom G6 in order to show that the opposite category of finite étale algebras is Galois.)

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

It seems you know the structure theorem for finite étale $k$-algebras, so you should see that it suffices to prove the claim for the case where $A$ and $B$ are finite separable field extensions of $k$. This is essentially just Galois theory.

Indeed, if $A$ is a finite separable field extension of $k$, then the number of $k$-algebra homomorphisms $A \to k_s$ is equal to the dimension of $A$ as a $k$-vector space. So if $f : A \to B$ is a $k$-algebra homomorphism of finite separable field extensions of $k$ such that $F (f)$ is a bijection, then the dimensions of $A$ and $B$ (as $k$-vector spaces) are equal; but $f : A \to B$ is an $k$-linear map with trivial kernel, so it must be an isomorphism.

$\endgroup$
10
  • $\begingroup$ Hi @ZhenLin, by the structure theorem you mean that every finite étale algebra is a finite product of finite separable field extensions? If so, yes, I know it. I thought of basically the same things that you said but I just don't understand why $f$ has trivial kernel in the case of separable field extension and why it suffices to consider this case. I mean, for a general étale algebra, it is not true that $|\hom_k(A,k_s)|$ is the dimension of $A$ as a $k$-vector space. $\endgroup$
    – Gabriel
    Jul 8, 2020 at 14:53
  • $\begingroup$ It is a general fact that ring homomorphisms between fields are injective. For the case of a general finite étale algebra the reduction you need is that each factor in the domain maps to some factor of the codomain. $\endgroup$
    – Zhen Lin
    Jul 8, 2020 at 15:12
  • $\begingroup$ I'm sorry if this is too simple but it is still not clear to me. Why do each factor in the domain maps to some factor of the codomain? It isn't possible that two factors of the domain maps to the same factor in the codomain? It is also not possible to some factor to map to 0? $\endgroup$
    – Gabriel
    Jul 8, 2020 at 15:48
  • $\begingroup$ Have you tried to see what happens when the codomain only has one factor? Once you understand that case, you can handle the case where there are many factors by splitting into components. This fact about homomorphisms is true for finite products of field extensions in general, by the way. $\endgroup$
    – Zhen Lin
    Jul 8, 2020 at 16:07
  • 1
    $\begingroup$ $F$ may be defined as mapping into finite sets but it actually maps into sets with a Galois action. (Just compose each automorphism of $k_s$ with the homomorphisms into $k_s$!) The factors correspond to the orbits. Since $F (f)$ is an equivariant map it must map orbits to orbits; since $F (f)$ is an equivariant bijection the orbits are also in one-to-one correspondence, so the factors are also one-to-one. $\endgroup$
    – Zhen Lin
    Jul 12, 2020 at 14:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .