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 ...
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 ...
12
votes
Accepted
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 ...
11
votes
Accepted
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 ...
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 ...
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.
...
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 ...
9
votes
Accepted
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 ...
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, ...
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\...
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 ...
8
votes
Accepted
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 ...
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 ...
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.
6
votes
Accepted
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 ...
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(\...
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 ...
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-...
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 ...
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 ...
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'...
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 ...
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 ...
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 $\...
4
votes
Accepted
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 ...
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=-\...
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 $...
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 ...
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 ...
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