24
votes
What are dessins d'enfants?
Given a compact Riemann surface $X$, every holomorphic function on $X$ is constant. This is obvious if you think about holomorphic functions as locally conformal mappings, that is, transformations of $...
- 371
24
votes
Absolute Galois group, number theory and the Axiom of Choice
In the absence of the axiom of choice, it is still possible to define the "usual" algebraic closure of $\mathbb{Q}$ because you can just explicitly enumerate all polynomials with integer ...
- 76.9k
23
votes
Accepted
Profinite groups as absolute Galois groups
This is a very good question which is a big open problem. There are a number of theorems, some of them easy and some very difficult, and also a number of conjectures, restricting the class of groups ...
18
votes
Accepted
Classify all the fields with abelian absolute Galois group
Geyer in Unendliche algebraische Zahlkörper, über denen jede Gleichung auflösbar von beschränkter Stufe ist, Satz 1.13 and the paragraph after that, gives a full characterization of which abelian ...
- 1,979
17
votes
Absolute Galois group, number theory and the Axiom of Choice
There would be no consequences, for two reasons:
As Timothy Chow points out, if we define $\overline{\mathbb Q}$ as the set of complex numbers that are roots of a nonzero polynomial with rational ...
- 131k
13
votes
What "should" be the absolute galois group of a field with one element
The Galois group of the maximal abelian extension of $\mathbb Q$ (or any number field) is given (class field theory) as the quotient of the idele class group by the connected component of the identity ...
- 30.1k
13
votes
Teichmuller groupoids in Grothendieck's esquisse d'un programme
There is another type of important morphism between the (orbifold) fundamental groups of the moduli spaces $M_{g,\nu}\rightarrow M_{g',\nu'}$ that is considered in Grothendieck's tower. You can see ...
- 937
12
votes
Consequences of Shafarevich conjecture
The Shafarevich conjecture belongs to the broader program of Inverse Galois theory, and in that context it is just another step in that particular approach to understanding $\mathrm{Gal}(\overline{\...
- 17.4k
12
votes
Frobenius elements in infinite extensions
[This answer is an elaboration on what KConrad mentions in his comments below, posted while I was first writing this.]
At finite level one has a perfectly good theory of Frobenius elements attached ...
- 1,895
11
votes
Accepted
Involutions in the absolute Galois group (and the Axiom of Choice)
No, you shouldn't need any choice for this, and it should still be true if you replace $\overline{\mathbb{Q}}$ with any other algebraic closure of $\mathbb{Q}$. Let $K$ be a field (which in our ...
- 150k
10
votes
Accepted
Galois group for 0-dimensional motives
Motives of $0$-dimensional varieties are usually called Artin motives. The different fiber functors (essentially) all give rise to automorphism group isomorphic to the absolute Galois. There is one ...
- 17.1k
9
votes
Involutions in the absolute Galois group (and the Axiom of Choice)
Let me spell out a completely explicit elementary proof that visibly makes no use of choice.
Lemma 1. Let $\sigma$ be an automorphism of order $2$ of a field $K$ of characteristic $\ne2$, and let $F$ ...
- 43.4k
8
votes
Langlands program vs Shimura-Taniyama-Weil conjecture
To expand on zeno's answer, a Langlands-type formulation of the modularity conjecture would be:
(Taniyama-Shimura) $L(E,s)=L(f,s)$
Here $L(E,s)$ is the Hasse-Weil L-function of an elliptic curve $...
- 17.4k
8
votes
Accepted
Weight filtration on certain Galois representations
No, life is not so easy I'm afraid. For instance, the group $\mathrm{Ext}^1_{G_{\mathbf{Q}}}(\mathbf{Z}_\ell, \mathbf{Z}_\ell(n)) = H^1(\mathbf{Q}, \mathbf{Z}_\ell(n))$ has positive rank for all odd ...
- 35.8k
8
votes
Accepted
Commutator subgroup of the absolute Galois group - a closed subgroup
No, the abstract commutator subgroup $[G_K,G_K]$ of the absolute Galois group $G_K$ of a number field $K$ is never closed:
Write $[G,G]$ for the commutator subgroup of $G$ as an abstract group,
and $c(...
- 1,979
8
votes
Accepted
absolute Galois group of the function field of a curve over $\mathbb{F}_p^{alg}$
The theorem is true. It seems to have been proved independently by Florian Pop and by David Harbater. In Pop's paper, it is the corollary on p. 556.
MR1334484 (96k:14011)
Pop, Florian
Étale Galois ...
8
votes
Algebraically closed fields with only finite orbits
The answer is no. Let $K$ be an algebraically closed field, and let $k$ be the algebraic closure of the prime field contained in $K$. Pick a transcendence basis $B$ of $K/k$, which is nonempty if $k\...
- 26.3k
8
votes
Accepted
Galois action on automorphisms of a curve
$K$-defined would mean that the subgroup itself is defined over $K$, not the elements.
Formally, this means that there are equations over $K$ which are satisfied by the coefficients of polynomials ...
- 131k
7
votes
Accepted
Is co-restriction in Galois cohomology in fact the norm map via Kummer isomorphism?
The corestriction map on cohomology is indeed the norm in degree zero (see Tate's notes on Galois cohomology for example). By a dimension shifting argument, it then easily follows that the ...
- 4,665
7
votes
Interesting (combinatorial) actions of the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$
You should read the paper
I. Bauer, F. Catanese, F. Grunewald: Faithful actions of the absolute Galois group on connected components of moduli spaces, Invent. Math. 199, No. 3, 859-888 (2015). ZBL1318....
- 64.8k
7
votes
Accepted
Dessins d'enfants and absolute Galois group
For the sake of getting this off the unanswered stack...
Please refer to the following reference:
Guillot, P. An elementary approach to dessins d'enfants and the Grothendieck-Teichmüller group, ...
- 11.1k
7
votes
Langlands program vs Shimura-Taniyama-Weil conjecture
The Taniyama conjecture says that the L-series of an elliptic curve over Q is automorphic (more specifically, arises from a modular form). Langlands conjectures that every L-series arising from ...
- 641
7
votes
Is the absolute Galois group the same as the automorphism group?
This has already been answered in the comments, but perhaps you can see it more clearly like this.
Take the isomorphism $\mathrm{Gal}(\bar{\mathbb{Q}}|\mathbb{Q}) \cong \varprojlim(K|\mathbb{Q})$ as ...
- 17.4k
7
votes
Galois group for 0-dimensional motives
What follows is a hands-on explanation (not a complete proof!) of why the Tannakian group $$G_{\textrm{dR}}(AM(\mathbb{Q}))$$ of the category of Artin motives is a nontrivial inner form of $$G_{\...
- 71
5
votes
Accepted
induced isomorphism in continuous cohomology
Here is one recent result in this direction, taken from I. Efrat and J. Minac, Galois groups and cohomological functors, Trans. of the AMS, http://www.ams.org/journals/tran/0000-000-00/S0002-9947-2016-...
- 749
4
votes
Frobenius elements in infinite extensions
You can easily define it for an infinite extension $\bar{K}|K$, and an unramified maximal ideal $\mathfrak{p} \in O_K$. Take $\mathfrak{P} \in O_\bar{K}$ maximal over $\mathfrak{p}$. The Frobenius ...
- 17.4k
4
votes
Accepted
Another fix field of a certain galois group action
The right way to do this would be to give a short, efficient, abstract argument. I, however, will give a fairly messy example to show that $u^{1/p}\in\hat L$.
I’ll use $K=E=\Bbb F_p((u))$, and set $...
- 4,048
4
votes
Accepted
Fix field of a certain galois group action
This question is completely answered in
J. Ax – “Zeros of polynomials over local fields—The Galois action”, J. Algebra 15 (1970), p. 417–428.
You get the completion of the perfection of E.
- 6,876
4
votes
Profinite groups as absolute Galois groups
In order to study absolute Galois groups of fields one may also use another consequence of the aforementioned Rost-Voevodsky theorem: in fact, for any field $K$, the cohomology algebra $H^*(G_K,\...
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