16
$\begingroup$

I've been learning about Grothendieck's Galois theory, and I just haven't been able to understand the fundamental theorem properly. Let's phrase the fundamental theorem in the case of fields:

Let $k$ be a field, and $k_s$ its separable closure. There is an anti-equivalence between the category of finite separable $k$-algebras and the category of finite sets equipped with a continuous Gal($k_s/k$)-action. Under this equivalence, separable field extensions of $k$ correspond to sets with a transitive action, and Galois extensions of $k$ correspond to finite quotients of Gal($k_s/k$).

Everywhere I read that the above theorem generalizes the fundamental theorem of Galois theory, but I don't quite understand this. For example, given a Galois extension, why do subfields correspond to subgroups of its Galois group (which is the finite quotient of Gal($k_s/k$))? All I can tell is that a subfield $K$ of a Galois extension $L$ corresponds to a surjective map of Gal$(k_s/k)$-sets $$\text{Hom}_k(L,k_s) \to \text{Hom}_k(K,k_s),$$ and the left-hand side can be identified with a group. How does that realise the right-hand side as a subgroup?

Moreover, how do we obtain that the normal subgroups of the Galois group of a finite extension correspond to normal subfields of that extension?

Many thanks.

$\endgroup$
  • 2
    $\begingroup$ i think you need the full explicit description of this equivalence to answer your question. One direction looks like this: given a Finite set with continuous action by Galois group, split this set into orbits, consider the stabilizer of an element in the orbits, this is an open subgroup of the absolute Galois group, and by Galois theory this gives you a finite extension. The direct product of these finite extensions gives you the desired algebra. The other direction is similar. So you will see that in the case of field extension this theorem is just the Fdamental theorem of Galois theory. $\endgroup$ – Saberization Jun 10 '18 at 19:58
  • $\begingroup$ I'm interested in the maths behind the claim "and by Galois theory this gives you a finite extension". Doesn't this use the fundamental theorem (infinite version)? I don't want to be assuming the fundamental theorem here. $\endgroup$ – user Jun 12 '18 at 10:49
4
+25
$\begingroup$

Let $L$ be a Galois extension of $k$ embedded in $k_s$. Then $L$ is the union of its finite Galois subextensions $L'$,and $\hom(L,k_s)$ is the projective limit of $\hom(L',k_s)$ along Galois subextensions $L'$. Grothendieck's theorem gives you a structure of group on $\hom(L',k_s)$ for each finite subextension and these are compatible with the limit, hence you get a structure of profinite group on $\hom(L,k_s)=Gal(L/k)$.

Now take a subextension $K$ of $L$, as you say there is a natural map $\hom(L,k_s)\to\hom(K,k_s)$. But now you say something wrong: $\hom(K,k_s)$ is not a subgroup, the subgroup is the inverse image in $\hom(L,k_s)=Gal(L/k)$ of the fixed embedding $K\subset L\subset k_s$ which is an element of $\hom(K,k_s)$. It is a subgroup since you can see it as a stabilizer of the action of $\hom(L,k_s)$ on $\hom(K,k_s)$.

In this construction, if $K$ is a normal subextension of $L$ then Grothendieck's theorem gives you a group structure on $\hom(K,k_s)$ compatible with the group structure on $\hom(L,k_s)$, hence the inverse image in $\hom(L,k_s)$ of the fixed embedding $K\subset L\subset k_s$ is a kernel of an homomorphism, and hence it is normal.

On the other hand, if $H\subset Gal(L/k)=\hom(L,k_s)$ is a closed subgroup, $\hom(L,k_s)/H$ has a natural structure of profinite set with continuos action of $Gal(k_s/k)$ induced by the projection $Gal(k_s/k)\to \hom(L,k_s)=Gal(L/k)$ (this projection comes with the construction, we are not using classical Galois theory). Then, since the projection $Gal(k_s/k)\to \hom(L,k_s)/H$ is surjective (because $Gal(k_s/k)\to \hom(L,k_s)$ is surjective) and hence the action is transitive, Grothendieck's theorem gives you a separable extension $K/k$ with a natural identification $Gal(L/k)/H=\hom(K,k_s)$, hence with a natural map $\hom(L,k_s)\to\hom(K,k_s)$ which in turn gives you an embedding $K\subset L$.

If moreover $H$ is normal, then $Gal(L/k)/H$ is a group and hence Grothendieck's theorem tells you that $K$ is Galois.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.