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Questions tagged [galois-theory]

Galois theory, named after Évariste Galois, provides a connection between field theory and group theory. Using Galois theory, certain problems in field theory can be reduced to group theory, which is, in some sense, simpler and better understood.

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Maximal abelian subgroups of absolute Galois group

For both local and global fields, we have a good handle on the abelianization of the absolute Galois group of $K$. Essentially this allows us to "understand" all maps from $G_K$ to abelian ...
curious math guy's user avatar
2 votes
0 answers
150 views

Infinite tamely ramified $p$-extensions of $\mathbb{Q}$ contain infinite unramified subextensions?

Let $p$ be a prime. By a $ p $-extension we mean a Galois extension whose Galois group is a $ p $-group. Let $L$ be an infinite tamely ramified $p$-extension of $\mathbb{Q}$, i.e. all primes ramified ...
stupid boy's user avatar
4 votes
1 answer
243 views

Existence of intermediate field extensions for tamely ramified p-adic extensions

Let $p$ be a prime, and let $K/\mathbb{Q}_p$ be a tamely ramified finite extension of degree $n$. Let $q$ be a prime factor of $n$ with $q\neq p$. Must there exist an intermediate extension $L$ (...
Ralph Morrison's user avatar
3 votes
1 answer
384 views

Using the Lehmer quintic to solve $11$-degree equations and higher?

(This is a natural continuation of a previous post.) I. Quintic method Given the Lehmer quintic, $$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + ...
Tito Piezas III's user avatar
3 votes
1 answer
341 views

A similar relationship between the generic cubic and the Lehmer quintic?

I. Comparison It doesn't seem to be well-known that the generic cubic (prominent in this MO post) for $C_3 = A_3$, $$x^3-nx^2+(n-3)x+1 = 0$$ has the nice property that its roots $a,b,c$, if in correct ...
Tito Piezas III's user avatar
7 votes
2 answers
439 views

A method to generate solvable equations of degrees $p = 7, 13, 19, 31, 37,\dots$ using only cubics

I've always wondered if the DeMoivre method to generate an algebraic number $x_p$, $$x_p = u_1^{1/p}+u_2^{1/p}$$ of degree $p$ using only quadratic roots $u_i$ could be generalized using cubic roots $...
Tito Piezas III's user avatar
3 votes
1 answer
250 views

On the refined minimal ramification problem for $p$-groups

Let $p$ be a prime. The minimal ramification problem is to ask whether or not every finite $p$-group $G$ can be realized as the Galois group of a tamely ramified extension of $\mathbb{Q}$ with exactly ...
stupid boy's user avatar
2 votes
2 answers
381 views

On V. Arnold's trinities regarding PSL(2,5), PSL(2,7), and PSL(2,11)?

Given the Ramanujan theta function, $$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}$$ Let $q = e^{2\pi i \tau}$ and assume $\tau = \sqrt{-d}.$ Then the following functions for levels $...
Tito Piezas III's user avatar
5 votes
1 answer
273 views

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

Let $G_{\mathbb{Q}_p}$ denote the absolute Galois group of the $p$-adic field $\mathbb{Q}_{p}$. Also, their structure as abstract groups is completely known. It is well known that this group embeds ...
kindasorta's user avatar
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6 votes
0 answers
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Using the Rogers-Selberg identities to solve certain septics?

Given the Ramanujan theta function $f(a,b)$ and the Rogers-Selberg identities, \begin{align} U_1 &= \frac{f(-q,-q^6)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2+2n}} {(q^2;q^2)_n\,(-q;q)_{2n+1}} =...
Tito Piezas III's user avatar
-2 votes
1 answer
192 views

Finite normal extensions

Suppose that $K$ is a finite field extension of $F$. Is the following equivalent to the extension being normal? If $L$ is an extension of $K$ and $\sigma:K\to L$ fixes $F$, then $\sigma(K) = K$. I ...
Yoav Len's user avatar
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6 votes
1 answer
533 views

Construction of a symmetric polynomial in the roots that acts like the discriminant

The discriminant $\Delta(P)$ of a monic polynomial $P(x)=x^n + a_{n-1} x^{n-1} + \dotsb + a_0$ of degree $n$, when expanded (using elementary symmetric polynomials), is a symmetric polynomial of ...
rgvalenciaalbornoz's user avatar
3 votes
3 answers
368 views

Solving solvable septics using only cubics?

After the satisfying resolution of my question on the Kondo-Brumer quintic, I decided to revisit my old post on septic equations. I. Solution by eta quotients The septic mentioned in that post may not ...
Tito Piezas III's user avatar
0 votes
1 answer
243 views

Ramifications in Galois closures of number fields

Let $K/\mathbb{Q}$ be a finite Galois extension, and $L/K$ an infinite Galois extension such that the Galois group $ \operatorname{Gal}(L/K) $ is isomorphic to a closed subgroup of $ \operatorname{GL}...
stupid boy's user avatar
5 votes
1 answer
289 views

Transforming the Kondo quintic $5T2$ into the Lehmer quintic $5T1$?

I. Kondo-Brumer quintic The deceptively simple solvable quintic, $$x^5 + (a - 3)x^4 + (-a + b + 3)x^3 + (a^2 - a - 1 - 2b)x^2 + b x + a=0$$ is quite important for imaginary quadratic fields. For ...
Tito Piezas III's user avatar
1 vote
1 answer
305 views

Quadratic extension of local field

Let $F$ be a nonarchimedean local field of characteristic zero, and $E$ an extension of $F$ with $[E:F]=2^n$ for some $n$. Is it always possible to find a quadratic extension $M$ of $F$ such that $F\...
Windi's user avatar
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2 votes
1 answer
182 views

Measure on the places of $\bar{\mathbb Q}$

Consider the set $S$ of all places of $\mathbb Q$ (i.e. the set of all absolute values up to equivalence). Then we can consider $S$ as a measure space with the counting measure $\mu$. Therefore $\mu(\{...
manifold's user avatar
  • 321
0 votes
0 answers
80 views

Projection map $\pi:\left(\mathcal{O}/n\mathcal{O}\right)^\times \to\left(\mathcal{O}/\gcd(n,m)\mathcal{O}\right)^{\times}$ of a CM elliptic curve

In this paper the author has mentioned in page $693$ under section $2.2$ that for an Elliptic curve $E/\mathbb{Q}$ with CM by an order $\mathcal{O}$ of an imaginary quadratic field $K$ there is a ...
Anish Ray's user avatar
  • 309
2 votes
1 answer
159 views

Decomposition of the Galois group of the $m$-th division field of an elliptic curve with CM into a direct product of Galois groups

Let $E/\mathbb{Q}$ be an elliptic curve with CM from an imaginary quadratic field $K$. Let $K(E[m])$ denote $m$-th division field (number field obtained by adjoining the coordinates of the $m$-torsion ...
Anish Ray's user avatar
  • 309
27 votes
1 answer
2k views

A simple proof of the fundamental theorem of Galois theory

Update. It's now on the arXiv. Some time ago I found my "own" proof of the fundamental theorem of Galois theory. You can find a pdf with the proof (link removed, see arXiv). It is quite ...
Martin Brandenburg's user avatar
2 votes
1 answer
307 views

A question about unramified quadratic extension of number field

Is there any condition over a number field $K$ for an unramified quadratic extension of $K$ to admit an embedding into an unramified cyclic extension of degree 4 of $K$?
ayoub-chess's user avatar
2 votes
0 answers
146 views

Can we say anything about the zeros and Galois group of the polynomial $(x^p-a)^{p^2}-p^{p^2+1}x+p^{p^2} a=0$?

Let $p$ be an odd prime number and $\mathbb Q_p$ be the $p$-adic number field. Let $K=\mathbb Q_p(a)$ be the extension by $a=p^{\frac{p^2+1}{p^3-1}}$. Consider the polynomial $f(x)=(x^p-a)^{p^2}-p^{p^...
MAS's user avatar
  • 930
3 votes
1 answer
213 views

Is there a theory of "elementary closed form solution" at the operator level for differential equations?

We begin by considering the usual general first order linear equation of the form $$ a_0 y' + a_1 y + a_2 = 0 $$ Where $a_i,y \in \mathbb{C} \rightarrow \mathbb{C}$. Now it's well known from everyone'...
Sidharth Ghoshal's user avatar
7 votes
1 answer
890 views

When is $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ a Galois extension of $\mathbb{Q}$?

When is $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ a Galois extension of $\mathbb{Q}$? I was motivated by the question that $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ is a Galois extension of $\mathbb{Q}$. Here is a rough ...
Explorer1234's user avatar
3 votes
0 answers
128 views

Galois cohomology with coefficients in the integers of the Lubin-Tate extension

Let $K$ be a $p$-adic local field, and $L$ the Lubin-Tate extension obtained from $K$ by attaching roots of some Lubin-Tate formal $\mathcal{O}_{K}$-module with $Gal(L/K) \simeq \mathcal{O}_{K}^{\...
Piotr Pstrągowski's user avatar
5 votes
1 answer
438 views

What are the jumps in the ramification filtration of the absolute Galois group of a local field?

Let $k$ be a (complete) discretely valued field and $\ell$ a Galois extension of $k$, possibly infinite. The Galois group $\Gamma=\text{Gal}(\ell/k)$ of $\ell$ over $k$ admits a descreasing, $\mathbb ...
David Schwein's user avatar
5 votes
1 answer
511 views

Cycle type in Galois group from ramified primes

Let $P \in \mathbb Z[X]$ be monic, separable, of degree $d$, $K$ its splitting field over $\mathbb Q$ and $G$ the Galois group of $K$ over $\mathbb Q$. Now, let $p$ be a prime number unramified in $K$....
A. Bailleul's user avatar
  • 1,322
6 votes
0 answers
375 views

How to construct this non-geometric mod $p$ Galois representation?

Let $ K $ be a totally real cubic field. I am concerned with the following kind of Inverse Galois Problem: Is there a prime $ p>5 $ and a continuous representation $ \bar{\rho}:G_{K}\to {\rm GL}_{...
Nobody's user avatar
  • 863
1 vote
0 answers
135 views

Reference request: arithmetical implications of an ambient Galois extension

This is a cross-post of a question I asked on StackExchange. See there for further details. Let $L/K$ be a Galois extension of algebraic number fields of finite degree over $\mathbb{Q}$, with group $G$...
Matthé van der Lee's user avatar
6 votes
0 answers
220 views

Can we deduce that both $\alpha$ and $\beta$ are algebraic over F if $F[\alpha,\beta]=F(\alpha,\beta)$

So after reading David Cox. book on Galois Theory, it can be shown that “Given a field F, $F[\alpha]=F(\alpha)$ if and only if $\alpha$ is algebraic over F” (See Prop. 4.1.14). Later, it was mentioned ...
Explorer1234's user avatar
-2 votes
1 answer
504 views

In Galois theory, why solvable groups must have their quotient groups be Abelian? [closed]

The definition of solvable groups can be regarded as two constraints, one is that there must be a sequence of normal subgroups, and the other is that the quotient groups between these sequences are ...
Ray's user avatar
  • 1
0 votes
0 answers
120 views

Colimits in the category of suplattices

I want to compute coequalizers in the category $\mathcal{S}up$ of complete lattices and $\bigvee$-preserving maps. One way (I think?) is to use the dual equivalence $$ \mathcal{Sup} \leftrightarrows \...
Hans's user avatar
  • 147
8 votes
1 answer
355 views

The distribution of certain Galois groups

Let $f(x)$ be a polynomial of degree $d$ with integer coefficients. Let $G_p^+$ be the Galois group of the polynomial $f(x)-y$ over $\overline{\mathbb{F}}_p(y)$ and $G_p$ be the Galois group of the ...
Alexander Kalmynin's user avatar
6 votes
2 answers
329 views

Algebraic numbers which prescribed degree which does not belong to some fields

In my research it would be great if the following result is valid. In what follows, $\overline{\mathbb{Q}}$, $\overline{\mathbb{Q}}_n$ and $\overline{\mathbb{Q}}_{<n}$ denotes the set of algebraic ...
Jean's user avatar
  • 515
4 votes
0 answers
83 views

The cyclic analogue of the gonality of the superelliptic curve $s^n = t^m + 1$

For naturals $n$, $m > 1$ consider the superelliptic curve $C\!: s^n = t^m + 1$, for simplicity, over an algebraically closed field of zero characteristic or large characteristic $p \nmid n$, $m$. ...
Dimitri Koshelev's user avatar
3 votes
1 answer
393 views

Completion of infinite degree extension of perfectoid fields is perfectoid?

Is completion of infinite degree extension of perfectoid fields perfectoid ? It is known that finite extension of perfectoid fields is also perftoid from tilting correspondence, but what about ...
Duality's user avatar
  • 1,531
5 votes
2 answers
1k views

Absolute Galois group, number theory and the Axiom of Choice

Richard Taylor once explained his research (in number theory) in a very simple way: understanding the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. It is known that in ...
THC's user avatar
  • 4,547
2 votes
0 answers
148 views

Embeddings and images of number fields in $\mathbb{C}$ [closed]

Let $\mathbb{Q}(\alpha)$ be a number field, and suppose that $[\mathbb{Q}(\alpha) : \mathbb{Q} ] = m$. Then there are precisely $m$ different embeddings of $\mathbb{Q}(\alpha)$ into $\mathbb{C}$, ...
THC's user avatar
  • 4,547
4 votes
0 answers
245 views

Dessins d'enfants and the absolute Galois group

If I am not mistaken, Alexander Grothendieck introduced dessins d'enfants as they are known today, in order to better understand the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}...
THC's user avatar
  • 4,547
0 votes
0 answers
78 views

Local field such that the value group of $K^\text{perf}$ ( perfect closure of $K$) is $\bigcup_{n\geqq1}(1/p^n) \Bbb{Z}$

Let $K$ be a local field of positive characteristic. I'm looking for a $K$ which satisfies the following condition. Value group of $K^\text{perf}$ (perfect closure of $K$) is $\bigcup_{n\geqq1}(1/p^n)...
Duality's user avatar
  • 1,531
4 votes
1 answer
354 views

Can a general quintic be solved using inverse beta regularized function?

Tyma Gaidash has recently posted solutions to some quintics in terms of Inverse Beta Regularized function. He also found the closed form for the equation $\cos x=x$ using the same Inverse Beta ...
Anixx's user avatar
  • 10.1k
5 votes
1 answer
345 views

Is it possible to solve sextic equations using the Fox H function?

Although the Kampé de Fériet function can solve the sextic equation, the details about it are shrouded in the fog of more than a century ago. In contrast, we know more about the Fox H function, and we ...
Ember Edison's user avatar
5 votes
1 answer
286 views

Rationality of field embeddings

After my earlier question question turned out to have a negative answer (Thank you to all respondents!), here is a more modest one. Both a positive answer and a counterexample would help my work. If ...
Lauritz's user avatar
  • 135
7 votes
0 answers
149 views

Finding when a certain product in a cyclotomic field is equal to one

For the following, fix $m\geq 2$, and let $a_{0},\dots,a_{m-1}\in\mathbb{Z}$ be such that $\sum_{k=0}^{m-1}a_{k}=0$. I would like to find the exact conditions on the $a_{k}$ so that the following ...
Ian Montague's user avatar
6 votes
2 answers
280 views

Cancellation of irreducibility for Galois conjugates

Motivation: Take an algebraic number $\lambda$. In my research, I've stumbled upon the question in which cases the expression $\sum_{\sigma \in S} \sigma(\lambda)$, where $S$ is a subset of the field ...
Lauritz's user avatar
  • 135
9 votes
2 answers
698 views

Non-trivial automorphisms and descent

In this expository paper by Low it says: Roughly speaking, a topos in the sense of Grothendieck is the category of sheaves on a kind of generalised space whose “points” may have non-trivial ...
user481494's user avatar
5 votes
1 answer
417 views

On the solvability of the equation $ax^p+bx^{p-1}+cx+d=0$ by radicals

Joint with Qing-Hu Hou at Tianjin Univ., we seek for explicit criteria via coefficients for the solvability of an algebraic equation by radicals. In this direction, we formulate the following ...
Zhi-Wei Sun's user avatar
  • 15.6k
6 votes
2 answers
817 views

What is $\mathbb Q^{\mathrm{hypoab}}$?

$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Gal{Gal}\newcommand{\ab}{\mathrm{ab}}$Let $G(\mathbb Q) = \Gal(\overline{\mathbb Q} / \mathbb Q)$ be the absolute Galois group. It's well-known that ...
Tim Campion's user avatar
  • 63.9k
3 votes
1 answer
343 views

A criterion for the equation $ax^n+bx+c=0$ not solvable by radicals via $a,b,c$ and $n$

Galois revealed that an algebraic equation $f(x)=0$ with coefficients in a field $K$ of zero characteristic is solvable by radicals if and only if the Galois group of $f(x)$ over $K$ is solvable. ...
Zhi-Wei Sun's user avatar
  • 15.6k
6 votes
1 answer
466 views

Is there a conjectured dependence on $n$ in van der Waerden's conjecture?

Bhargava 2021 proves van der Waerden's conjecture about Galois groups of random integer polynomials: over all $x^n + a_{n-1} x^{n-1} + \cdots + a_0 = 0$ with $a_k \in \{-H, \ldots, H\}$, the number of ...
Geoffrey Irving's user avatar

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