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28 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 $...
Compacto's user avatar
  • 584
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 ...
Timothy Chow's user avatar
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19 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 ...
Arno Fehm's user avatar
  • 2,051
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 ...
Will Sawin's user avatar
  • 148k
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 ...
Leila Schneps's user avatar
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 ...
Felipe Voloch's user avatar
13 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{\...
Myshkin's user avatar
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12 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 ...
David E Speyer's user avatar
10 votes
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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 ...
Matthias Wendt's user avatar
9 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 ...
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 $F$ ...
Emil Jeřábek's user avatar
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\...
Wojowu's user avatar
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8 votes
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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 ...
Will Sawin's user avatar
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8 votes
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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(...
Arno Fehm's user avatar
  • 2,051
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_{\...
user585094's user avatar
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....
Francesco Polizzi's user avatar
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 ...
R.P.'s user avatar
  • 4,746
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-...
user05811's user avatar
  • 749
4 votes

Finite-order automorphisms in the absolute Galois group of a $p$-adic field?

As Seewoo Lee points out, this is obstructed by the Artin–Schreier theorem. Suppose that such $\sigma$ exists. By Artin–Schreier, it has order 2, and $C^\sigma \subset C$ is real-closed, and $C = C^\...
Theo Johnson-Freyd's user avatar
4 votes

$G_{\mathbb Q}$ and primes of $\overline{\mathbb{Z}}$

There are infinitely many prime ideals $\mathfrak p$ in $\overline{\mathbf Z}$ that lie over $p$ since you can find an arbitrarily large (finite) number of prime ideals lying over $p$ in suitable ...
KConrad's user avatar
  • 50.6k
4 votes
Accepted

Shafarevich's conjecture on Galois groups over fields ramified at finitely many places

In the footnote on Page 11 of [Fernando Q. Gouvea: Deformations of Galois Representations], The reference is [I.R. Shafarevich, Algebraic number fields, Proceedings of the international Congress of ...
stupid boy's user avatar
3 votes
Accepted

Relation between $G_{\mathbb{Q}_p}$ for different primes

As noted in the remarks above, there are different (although conjugate) embeddings $G_{\mathbb{Q}_p}\rightarrow G_\mathbb{Q}$. Let us fix one of them and denote its image by $G_p$. If we fix prime ...
Arno Fehm's user avatar
  • 2,051
2 votes
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Tits algebra of the quasi-split semisimple algebraic groups

I found and read the original paper <Représentations linéaires irréductibles d'un groupe réductif sur un corps quelconque> by Jacques Tits. In Theorem 3.3 of this paper, it is shown that the ...
YJ Kim's user avatar
  • 321
2 votes
Accepted

maximal pro-l-quotients of absolute Galois groups

I assume your field has characteristic $p>0$. Then the maximal pro-$l$ quotient of the absolute Galois group is torsion-free. Indeed, by results of E. Becker, Euklidische Korper und euklidische ...
user05811's user avatar
  • 749
1 vote

Meaning of epimorphism from full Galois group to some group

If your map is continuous, its kernel is normal and closed, hence of the form $Gal(\overline{\mathbb{Q}}/F)$, where $F/\mathbb{Q}$ is finite Galois (I think you can find all the results you need and ...
GreginGre's user avatar
  • 1,766
1 vote

Recovering the Zariski topology from the Zariski topology over an extension

Let $A$ be a finitely generated $k$-algebra, and let $K$ be the algebraic closure of $k$. I explain how to recover $spec(A)$ from $spec(A\otimes K)$. Equivalently, I explain how to recover $spm(A)$ ...
anon's user avatar
  • 76
1 vote

Consequences of Shafarevich conjecture

One additional aspect may be worth mentioning: When trying to understand the structure of an absolute Galois group ${\rm Gal}(\overline{K}/K)$ of a field $K$, it is often helpful to keep in mind its ...
user05811's user avatar
  • 749

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