4
$\begingroup$

I am thinking about the following excerpt of Patrick Morandi's "Field and Galois Theory" from chapter 1 section 2, page 21: enter image description here Here, $K$ is a field, and $\mathcal{F}(G)$ denotes the subfield $\{ x \in K : \sigma (x) = x\ \forall x \in G \}$ of $K$. $\text{Gal}(K/F)$ is the set of automorphisms of $K$ which fix $F$, which is also a subgroup of $\text{Aut}(K)$.

I understand the proof but I am looking for some insight in to the way it works. This could include a (categorical or representation-theoretical) generalization of the result, or context surrounding it, or possibly a reference where I could read about this generalized viewpoint.

To be completely specific about what I am asking for in terms of a categorical viewpoint, I was thinking something along the following lines: there is a contravariant adjoint correspondence between $\mathcal{F}$ and $\text{Gal}(K/-)$, so that for a group $G \subset \text{Gal}(K/F)$ and a field $L \subset K$ containing $F$, we have $L \subset \text{Fix}(G)$ if and only if $G \subset \text{Gal}(K/L)$. One can ask when this correspondence is a contravariant poset isomorphism. It shouldn't be too hard to show that $F = \mathcal{F} ( \text{Gal}(K/F))$ is a necessary and sufficient criterion. The proposition above therefore gives another such criterion.

In light of the above remark, consider the following situation: let $G$ be a group and let $\mathcal{C}$ be a category. View $G$ as a category $\mathcal{C}_G$ in the ordinary way, and identify $G$ with the unique object in $\mathcal{C}_G$. Consider a functor $\Phi: \mathcal{C}_G \rightarrow \mathcal{C}$ (an action of $G$ on $\mathcal{C}$), and set $\Phi (G) = K$ for $K \in \mathcal{C}$. We say that an element $g \in G$ fixes a subobject $L \stackrel{l}{\rightarrow} K$ if the following diagram commutes: $\require{AMScd}$ \begin{CD} K @>{\Phi(g)}>> K \\ @A{l}AA @A{l}AA\\ L @>{\text{Id}_L}>> L \end{CD} Let $F \stackrel{f}{\rightarrow} K$ be the sum of the subobjects of $K$ fixed by all $g \in G$. Let $\mathcal{G}$ be the category of subgroups of $G$ and $\mathcal{K}$ be the category of subobjects $l : L \rightarrow K$ of $K$ such that $f \leq l$.

We then have a pair of contravariant right adjoint functors $\mathcal{Fix} : \mathcal{G} \rightarrow \mathcal{K}$ and $\mathcal{Gal} : \mathcal{K} \rightarrow \mathcal{G}$, where $H \subset G$ is sent to the sum of subobjects $L$ in $\mathcal{K}$ fixed by $H$ and a subobject $L$ of $\mathcal{K}$ is sent to the set of elements of $G$ which fix $L$.

The question becomes, what properties applicable to the situation above can be required of $\mathcal{Fix}$ and $\mathcal{Gal}$ so that they form an inclusion reversing poset isomorphism? (Is $\mathcal{Fix} (\mathcal{Gal} (F)) = F$ enough? is there then a condition analogous to the proposition in question?).

$\endgroup$
  • 1
    $\begingroup$ Have you come across the book by Janeldize & Borceux, Galois Theories? They go from classical Galois theory to categorical Galois theory via several generalisations. $\endgroup$ – Mozibur Ullah Apr 2 '18 at 2:05
  • $\begingroup$ Thanks I’ll check that out $\endgroup$ – Dean Young Apr 2 '18 at 8:35
  • 1
    $\begingroup$ The other thing to look out for is Grothendiecks Galois categories which he uses in his own formulation of Galois theory; this is historically earlier than Borceuxs theory. $\endgroup$ – Mozibur Ullah Apr 2 '18 at 8:44
  • $\begingroup$ Thanks again. Are these distinct, or is Borceuxs possibly a development of Grothendieck's ideas? $\endgroup$ – Dean Young Apr 2 '18 at 14:47
  • 1
    $\begingroup$ You're welcome; they're distinct but related; for example, one of the early chapters of Borceuxs book is an exposition of Grothendiecks Galois theory, and they then go on to develop it further by generalising this; however, they don't use the technology of Galois categories and fibre functors; this technology (Galois categories) has inspired other developments, such as the Tannakian formalism, which approximates the category of representations of a group. But this I think is straying pretty far from your concerns. $\endgroup$ – Mozibur Ullah Apr 2 '18 at 15:09

Your Answer

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

Browse other questions tagged or ask your own question.