3
$\begingroup$

Szamuely's book Galois groups and fundamental groups formulates several variants of the main theorem of Galois theory. This is the usual formulation (dual isomorphism of posets between intermediate fields and subgroups). Then there is also Grothendieck's version (dual equivalence of categories between finite étale algebras and actions of the Galois group).

Question: Does Grothendieck's version imply the usual formulation of the main theorem of Galois theory? If yes, why?

This question seems to be a very natural one - but as far as I can see it is not discussed in the book (am I missing something?). There are also two versions of the classification of covering spaces: firstly, as a bijection between subgroups and covers; secondly, as an equivalence of categories between covers of $X$ and $\pi_1(X,x)$-sets. Same question: does the equivalence of categories imply the bijection?

$\endgroup$

2 Answers 2

2
$\begingroup$

Basically it comes down the treatment of non-connected covers / non-transitive actions. Classical Galois theory – whether applied to field extensions or to covering spaces – focuses on connected covers and transitive group actions. Every transitive left action of a (discrete) group $G$ is isomorphic to the set of left cosets of some subgroup of $G$ (with the obvious left $G$-action), which is how subgroups enter the picture.

One way to make this precise is to define connectedness abstractly.

Definition. A connected object in a category $\mathcal{C}$ with finitary (resp. infinitary) coproducts is an object $X$ in $\mathcal{C}$ such that $\mathcal{C} (X, -) : \mathcal{C} \to \textbf{Set}$ preserves finitary (resp. infinitary) coproducts.

Remark. According to this definition, an initial object is never connected.

Example. A $G$-set is connected in the sense above if and only if it is transitive.

Example. A ring $A$ is connected in $\textbf{CRing}^\textrm{op}$ considered as a category with finitary (!!!) products if and only if $\operatorname{Spec} A$ is a connected topological space. (In this context, $\emptyset$ is not connected.) Equivalently, $A$ is connected in $\textbf{CRing}^\textrm{op}$ if and only if $A$ has exactly two idempotent elements, namely $0$ and $1$.

Example. A finite étale algebra over a field $k$ is connected in $\textbf{FÉt}_k{}^\textrm{op}$ if and only if it is a finite separable field extension of $k$.

So we can extract the objects studied in classical Galois theory, at least. To get the actual posets is a bit more difficult. This should not be surprising: after all, in the context of field extensions, this amounts to choosing an algebraic closure and embedding all the field extensions into that algebraic closure. But it can be done: this is what the fibre functor is for.

Let $\mathcal{C}$ be a category and let $U : \mathcal{C} \to \textbf{Set}$ be a functor. We may form the following category $\textbf{El} (U)$:

  • An object is a pair $(X, x)$ where $X$ is an object in $\mathcal{C}$ and $x \in U (X)$.

  • A morphism $(X, x) \to (Y, y)$ is a morphism $f : X \to Y$ in $\mathcal{C}$ such that $U (f) (x) = y$.

  • Composition and identities are inherited from $\mathcal{C}$.

Incidentally, $U : \mathcal{C} \to \textbf{Set}$ is representable if and only if $\textbf{El} (U)$ has an initial object.

In the case where $\mathcal{C}$ is the category of connected $G$-sets and $U$ is the forgetful functor, $\textbf{El} (U)$ is a preorder category, which can be canonically identified with the poset of open subgroups of $G$: just send $(X, x)$ to the stabiliser subgroup of $x$.

In the case where $\mathcal{C}$ is the opposite of the category of finite separable field extensions of $k$ and $U$ is the functor sending $K$ to the set of $k$-embeddings $\iota : K \to \bar{k}$, where $\bar{k}$ is a chosen algebraic closure of $k$, $\textbf{El} (U)$ is a preorder category, which can be canonically identified with the opposite of the poset of finite subextensions of $\bar{k}$: just send $(K, \iota)$ to the image of $\iota : K \to \bar{k}$.

Since Grothendieck's formulation asserts that the opposite of the category of finite étale $k$-algebras is equivalent to the category of finite $\textrm{Gal} (k)$-sets as categories equipped with fibre functors, restricting to the subcategory of connected objects and applying the construction above recovers the classical antitone isomorphism of posets.

$\endgroup$
1
$\begingroup$

Yes, it does imply it. That terminology may be different than what I studied, though.

A finite Galois extension $k\rightarrow L$ gives $L$ the structure of a finite $k$-algebra. The separable condition implies it is an étale algebra. As a $k$-algebra, $L$ has finite spectrum (the points are to be seen as roots of polynomials with coefficients in $k$), and is a finite $G$-set, with $G$ being the Galois group.

Here, the equivalence maps subextensions (which are morphisms in the category of $k$-algebras) into quotients of those $G$-sets (which are the spectrums, aka, the roots of the polynomial one gets by extending the field, which are permuted by the corresponding Galois groups).

Sorry, it's a messy explanation, since it's been a long time I studied this. I only know of this source: http://matematicas.unex.es/~navarro/acb.pdf

It's Juan Antonio Navarro's book on Commutative Algebra. Galois Theory is developed throughout the appendices on Finite algebras, Separable algebras, Galois theory, etc.

$\endgroup$
2
  • $\begingroup$ I don't really understand your argument. What is it that makes you conclude an isomorphism of posets from an equivalence of categories? $\endgroup$ Commented Feb 18, 2022 at 12:34
  • $\begingroup$ Inclusions are the same as morphisms. I can't really give much more detail, it's been ages seen I read that, and only very superficially. $\endgroup$
    – Compacto
    Commented Feb 18, 2022 at 13:14

You must log in to answer this question.

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