15 votes
Accepted

Is a generic genus $g \geq 7$ curve a solvable cover of $\mathbb{P}^1$?

The answer is negative by results of Zariski (weak version) and the stronger result by Guralnick and Neubauer in Theorem B of Monodromy groups of branched coverings: the generic case. They show that ...
Peter Mueller's user avatar
13 votes
Accepted

Distinct characters with the same character values, outer automorphisms and Galois conjugation

Take $G = S_3 \times S_4$ and consider the unique two-dimensional irreducible representation of $S_3$ and the unique two-dimensional irreducible representation of $S_4$. These have the same character ...
Will Sawin's user avatar
  • 135k
12 votes
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irreducibility of the polynomial $ x^4 +1 $

Among $-1$, $2$, and $-2$, there are $0$, $1$ or $3$ squares. This determines the irreducible factorization of $X^4+1$. If none is a square, it is irreducible; If only $-1=i^2$ is a square, the ...
YCor's user avatar
  • 60.1k
11 votes
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How often does algebraic-conjugacy imply conjugacy?

It is reasonably standard to call a finite group $G$ a rational group if all its complex irreducible characters are rational-valued ( equivalently, if $g \in G,$ then $g$ is conjugate within $G$ to ...
Geoff Robinson's user avatar
10 votes
Accepted

A question on algebraic independence

This is true, but it's actually a bit subtle, because the completion $\hat A \to \hat B$ of an injective local homomorphism $A \to B$ of local domains need not be injective. Given $f_1,\ldots,f_n \in ...
R. van Dobben de Bruyn's user avatar
10 votes
Accepted

Understanding absolute Galois group from its representations

The slogan "number theorists aim to understand $\operatorname{Gal}(\overline{\mathbb{Q}} / \mathbb{Q})$" is one that gets used a lot, but it's perhaps a tiny little bit misleading. ...
David Loeffler's user avatar
9 votes
Accepted

Galois action on algebraic K-theory of finite fields

$K(\overline{\mathbb{F}_p})$ is, after completion at any prime $\ell \neq p$, equivalent to $ku^\wedge_\ell$. It's true that the equivalence relies on a non-canonical embedding, but the conclusion ...
Achim Krause's user avatar
  • 8,459
9 votes
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Fields in which $ -1 $ can't be written as sum of two square elements

In the notation of Lam's Quadratic forms over fields, the Stufe (a German word) or level (its English translation) $s(F)$ of a field is the minimal $n$ such that $-1$ is the sum of $n$ squares. A ...
PseudoNeo's user avatar
  • 477
9 votes

A cyclic Galois extension over $ \mathbb{Q}(\omega)$

The answer is ``yes". Note first that in the question we may without loss restrict to cyclic extension of $2$-power degree, since the odd part of the order would split off as a direct factor, ...
Joachim König'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
  • 27.4k
8 votes
Accepted

Generalizing Klein's order 7 formula $y\left(y^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)y+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3 = j$ to order 13?

The smallest $m$ for which $\mathrm{PGL}(2,13)$ acts faithfully and transitively on a set of $13m$ elements is $m=6$, with point stabiliser a dihedral group of order $28$. I think you may be seeing a ...
Dave Benson's user avatar
  • 11.6k
8 votes
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Trying to understand the topology of the Weil group for the local Langlands conjecture

Note that in a topological group $G$, any subgroup $H$ containing an open subgroup $U$ is itself open: we can write $H$ as a disjoint union of $hU$ for a set of coset representatives for $H/U$, and ...
R. van Dobben de Bruyn's user avatar
7 votes
Accepted

Splitting fields of degree 4 irreducible polynomials containing a fixed quadratic extension

Yes, this can be done. Let $K$ be a $S_4$-quartic field, $C$ its cubic resolvent field, and $L$ the Galois closure of $K$. By Galois correspondence, $L$ contains a unique quadratic subfield $Q$ which ...
Stanley Yao Xiao's user avatar
7 votes
Accepted

Learning Inverse Galois Theory

I would like to suggest that a good place to start is with John Thompson's work on the subject. He initiated the modern approach using the notion of "rigid" tuples of conjugacy classes.
Dave Benson's user avatar
  • 11.6k
6 votes
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Good and bad reduction for twists of algebraic curves

Let $B$ be a Dedekind scheme with function field $K$. (Think of $B=\mathrm{Spec } \ \mathbb{Z}_{p}$ for simplicity, so that $K=\mathbb{Q}_p$.) Let $C$ be a smooth proper geometrically connected curve ...
Ariyan Javanpeykar's user avatar
6 votes

Does there exist a framework for determining if a power series is "differentially algebraic"

I had not seen this post before, but although the differential equation may be complicated, the result is easy: any modular form on $\Gamma_0(4)$ (but this is true much more generally), here $\theta(\...
Henri Cohen's user avatar
  • 11.5k
6 votes

Understanding absolute Galois group from its representations

$\newcommand{\Q}{\mathbb{Q}} \DeclareMathOperator{\Gal}{Gal}$ Here is an example of theorem that may be in the style that you are looking for, in the sense that the statement does not involve Galois ...
Aurel's user avatar
  • 4,878
5 votes

Splitting fields of degree 4 irreducible polynomials containing a fixed quadratic extension

This works even for any symmetric group and any quadratic number field, and can be done very explicitly. When $n$ is even, let $f(t,X) = X^{n-1}((n-1)X-n) + t$, and when $n$ is odd, let $f(t,X) = X^n-...
Joachim König's user avatar
5 votes
Accepted

On reducing degree-$12$ equations with Mathieu group $M_{11}$ to its degree-$11$ resolvent?

The answer is yes: $M_{11}$ in its action of degree $12$ has a subgroup of index $11$ (some people call it $M_{10}$) such that there is a $6$-element subset $A$ of the $12$ points which $M_{11}$ acts ...
Peter Mueller's user avatar
5 votes
Accepted

How do "Kummer closures" of fields look?

A solvable extension is iterated Kummer if and only if for all primes $\ell$ dividing its degree, the extension adjoining the $\ell$th roots of unity is iterative Kummer. Proof: For if, first adjoin ...
Will Sawin's user avatar
  • 135k
4 votes
Accepted

Norm of $2^{i}$-th primitive root

Let $K=\mathbb{Q}[\sqrt{-2}]$. Then $\frac{1}{2}\sqrt{-2}(1-i)$ is an eighth root of unity in $L$ whose norm is $\frac{1}{2}\sqrt{-2}(1-i)\frac{1}{2}\sqrt{-2}(1+i)=\frac{1}{4}(-2)(1-i^2)=-1\in K$. [I'...
Dave Benson's user avatar
  • 11.6k
4 votes

Automorphism of positive characteristic field

If $K$ is finite with order $q$, then we have the nice theorem that the $q$th power map generates ${\rm Gal}(L/K)$ when $L$ is an arbitrary finite extension of $K$, but there is no analogue of this ...
KConrad's user avatar
  • 49.5k
4 votes
Accepted

If degree $N$ polynomials always have a root, does there still exist an irreducible polynomial of degree $N+1$?

The answer to the first question is yes: For $n\ge5$ let $L$ be a Galois extension of $\mathbb Q$ with Galois group the alternating group $A_n$. Suppose that $L$ is contained in the composition $E$ of ...
Peter Mueller's user avatar
4 votes
Accepted

Statistics of action of Galois group of number field on primes over unramified rational primes

This is an elaboration of Chris Wuthrich's comment. Let $p$ be unramified (i.e. $p$ does not divide the discriminant of the Galois extension $K / \mathbb{Q}$), and let $\mathfrak{P}$ be a prime in $\...
GH from MO's user avatar
  • 98.2k
4 votes
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Finding a sextic analogue to the solvable octic $\frac{(x + 1)^6(x^2 + x + 7)}x = -k^3$ where $e^{(\pi/3)\sqrt{d}}\approx k^3+41.999999999999\dots$

Didn't think about eta quotients, but if the suggested analogies of ramification types are anything to go by, then $$j_2 = \frac{(x+1)^4(x^2+6x+25)}{x}$$ (which matches the prescribed discriminant ...
Joachim König's user avatar
4 votes

Does there exist a framework for determining if a power series is "differentially algebraic"

For the lazy reader I have wrapped Richard Stanley's very helpful comment into an answer. Richard Stanley's Enumerative Combinatorics Volume 2 on Page 282 reviews Jacobis Result that $$ y = \sum_{n=-\...
Sidharth Ghoshal's user avatar
4 votes
Accepted

Decomposition groups for the Galois module $\mu_8$

If I understand this, $E = \mathbf Q(\sqrt{7},\zeta_8) = F(\zeta_8)$. Since $\mathbf Q(\zeta_8)/\mathbf Q$ ramifies only at $2$ (I am ignoring infinite places), $E/F$ can ramify only at a place over $...
KConrad's user avatar
  • 49.5k
4 votes
Accepted

A subgroup of index 2 for a solvable finite group transitively acting on a finite set of cardinality $2m$ where $m$ is odd

At @MikhailBorovoi's request, I copy here two comments 1 2 from p-adic field extension of degree 2n without a subfield of degree 2? as an answer to Question 4. (The comment suggested that they answer ...
LSpice's user avatar
  • 11.3k
4 votes

Existence of a symmetric matrix satisfying certain irreducible conditions

Let $\mathbb{F}$ be a field of characteristic two, and let $K=\mathbb{F}(t)$, the field of rational functions over $\mathbb{F}$. Let $p(x)=x^2-t$. If $A$ is a symmetric matrix with entries in $K$ then ...
Dave Benson's user avatar
  • 11.6k

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