The tag has no wiki summary.

learn more… | top users | synonyms

17
votes
0answers
669 views

Could unramified Galois groups satisfy a version of property tau?

This is an experiment: there is a question I want to mention in an article I'm writing, and I am not sure it's a sensible question, so I will ask it here first, in the hopes that if it's insensible ...
4
votes
2answers
600 views

Solving z^n=a+ib using only radicals of positive real numbers

Let $a+bi\in\mathbf{C}$ be a complex number with $a,b\in\mathbf{R}$. Then it is easy to find exact solutions of $z^2=a+ib$. For example let $z=u+iv$. Then $$ u^2+v^2=z\overline{z}=\sqrt{a^2+b^2} $$ ...
2
votes
2answers
336 views

Place stabilizers for the absolute Galois Group

Fix an algebraic closure, $\overline{\mathbb{Q}}$ for the rationals and consider the set, $B_p$, of all places of $\overline{\mathbb{Q}}$ over a fixed (possibly infinite) prime, $p$, of $\mathbb{Q}$. ...
2
votes
1answer
231 views

A cubic polynomial which contains a linear factor with irreducible residual quadratic form

Let $f(x)\in\mathbb{Z}[x_{1},\dots,x_{n}]$ be a cubic homogeneous polynomial, which factors as $f(x)=g(x)h(x)$ over $\mathbb{C}$ with $\mathrm{deg}(g)=1$ and $h$ irreducible over $\mathbb{C}$. Assume ...
1
vote
3answers
726 views

A polynomial whose galois group is D_8 [closed]

I need to construct such a polynomial, and more generally: given a group G, how can it be realized as a galois group?
2
votes
0answers
94 views

Explicit defining equations for the Leopoldt locus

Let $F$ be a number field, which we assume for simplicity to be Galois and totally real. Set $\mathcal{O}_p=\mathcal{O}_F\otimes_{\mathbf{Z}}\mathbf{Z}_p$. The norm map on $\mathcal{O}_F$ extends ...
0
votes
1answer
192 views

Is there a subfield $F$ of $\mathbb{R}$ such that $\mathbb{R}$ is a finite algebraic extension of $F$. [duplicate]

Possible Duplicate: Examples of algebraic closures of finite index The question is in the title. I can prove that if such field $F$ exist then the extension $\mathbb{R}/F$ cannot be of ...
42
votes
3answers
2k views

Forcing as a new chapter of Galois Theory

There is a (very) long essay by Grothendieck with the ominous title La Longue Marche à travers la théorie de Galois (The Long March through Galois Theory). As usual, Grothendieck knew what he was ...
3
votes
0answers
311 views

Galois groups and braid groups [closed]

Braid group can be viewed as a symmetry group with a "one more dimension to pass through". Is there any "Galois theory", where the braid groups plays analoguos role as a symmetry groups in a native ...
15
votes
2answers
751 views

$2$-categorical structure in Grothendieck's Galois Theory

Grothendieck's Galois Theory, as developed in SGA I, V.4, or very gently in Lenstra's notes, establishes an equivalence between profinite groups and Galois categories. We can put this into the ...
4
votes
2answers
941 views

non-continuous inverse Galois problem

Let $G=Gal(\bar{\mathbf{Q}}/\mathbf{Q})$ be the absolute Galois group over $\mathbf{Q}$. Q1: Is it possible to find a (necessarily non-closed) normal subgroup $K\leq G$ such that $G/K$ is free of ...
7
votes
1answer
150 views

Rationality conditions for determining Galois groups

Let $F$ be a field and $h \in F[x]$ be an irreducible, degree $n$ monic polynomial. Let $G$ denote the Galois group of $h$. It is well known that $G \subset A_n $ if and only if the discriminant of ...
2
votes
1answer
620 views

The Galois group and relations among the roots of a polynomial

Let $f(x) \in \mathbb{Z}[x]$ be a monic irreducible polynomial of degree $n$, and let $\alpha_1, \alpha_2, ... , \alpha_n \in \overline{\mathbb{Q}}$ be the $n$ distinct roots of $f(x)$. Following ...
2
votes
1answer
528 views

Solving polynomial equations in radicals provided all roots are rational

This question is related to this question of Joseph O'Rourke and this question of mine. Question. Let $f$ be a polynomial with integer coefficients. Suppose that all roots of $f$ are rational. ...
3
votes
1answer
345 views

Proof of a Simple Converse in Algebraic Number Theory

If $L/K$ is a Galois extension, then any prime $\mathfrak{p}$ of $K$ splits into a product ${\mathfrak P}_1^e\cdots {\mathfrak P}_g^e$ of primes in $L$, and the exponents on the primes are equal since ...
5
votes
3answers
508 views

Octic family with Galois group of order 1344?

Does the octic, $\tag{1} x^8+3x^7-15x^6-29x^5+79x^4+61x^3+29x+16 = nx^2$ for any constant n have Galois group of order 1344? Its discriminant D is a perfect square, $D = ...
16
votes
1answer
838 views

Connes-Kreimer Hopf algebra and cosmic Galois group

Hi, I'm interested in the relation between the two following constructions motivated by renormalization: Connes-Kreimer gives an interpretation of the renormalization procedure in the framework of ...
23
votes
5answers
2k views

Solubility of the quintic?

Over the p-adics, every Galois group is solvable. Does this imply that the quintic (and higher-order polynomials for that matter) can be solved by radicals over $\mathbb{Q}_p$? EDIT: The original ...
4
votes
2answers
337 views

Subject to some conditions, is it possible to conclude a subfield of an abelian extension generated by a unit is a cyclic extension

My research is mostly in the area of modular categories. In the course of my research I came across a constraining set of number theoretic conditions that I'd like to exploit. It has been pointed out ...
11
votes
2answers
1k views

Automorphisms of $\mathbb{C}$

Is it true that $G_{\mathbb{Q}}$, the absolute Galois group of $\mathbb{Q}$, is a subgroup of $Aut(\mathbb{C})$ ? Or a simpler question: can any automorphism of $\overline{\mathbb{Q}}$ be extended to ...
1
vote
0answers
166 views

$f(x_1,x_2,x_3,\ldots,x_n)$ Maximum how many different results can have with all permutation of inputs?

$\alpha _n=e^{2 \pi i/n}$ $$f(x_1,x_2,x_3,\ldots,x_n)=(x_1+\alpha _n x_2+ \alpha _n ^2 x_3+\cdots+\alpha _n ^{n-1} x_n)^n$$ Maximum how many different results can have with all permutation of ...
4
votes
1answer
1k views

Example of an algebraic number of degree 4 that is not constructible

The number $b:=\frac{\sqrt{2a}+\sqrt{4\sqrt{a^2-3}-2a}}{2}$ with $a:=\frac{\sqrt[3]{18+2\cdot\sqrt{65}}}{2}+\frac{2}{\sqrt[3]{18+2\cdot\sqrt{65}}}$ is a root of the irreducible polynomial ...
6
votes
1answer
677 views

For which fields is the inverse Galois problem known?

The inverse Galois problem is known for (or in Jarden's and Fried's terminology, the following fields are universally admissible) function fields over henselian fields (like $\mathbb{Q}_p(x)$); ...
6
votes
1answer
576 views

Can roots of any polynomial be expressed using Eulerian function?

I encountered an interesting function which is called "Eulerian" by the Wolfram's MathWorld: $$\phi(q)=\prod_{k=1}^{\infty} (1-q^{k})$$ It is interesting because it is claimed that roots of any ...
3
votes
1answer
339 views

What is the relation of the absolute Galois group and classical profinite groups?

Consider the absolute Galois group $G = \mathrm{Gal}(\overline{\mathbb{Q}}: \mathbb{Q})$ and $G_p = \mathrm{Gal}(\overline{\mathbb{Q}_p}: \mathbb{Q}_p)$. Abelian class field theory gives us for the ...
1
vote
4answers
674 views

Explicit element in free group which is killed by every solvable quotient

The free group on two generators $F_2=\langle x,y|\rangle$ is the fundamental group of $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$. Now, there are plenty of galois covers of this space whose ...
8
votes
1answer
901 views

Is every finite group a quotient of the Grothendieck-Teichmuller group?

The Grothendieck-Teichmuller conjecture asserts that the absolute Galois group $Gal(\mathbb{Q})$ is isomorphic to the Grothendieck-Teichmuller group. I was wondering, would this conjecture imply the ...
4
votes
1answer
830 views

minimal polynomials of trig functions of ($k \pi/p$) and divisibility of coefficients by p

Take an odd prime $p$ and put $x_0:=\sum\limits_{j=0}^{p-1}\left(a_{j}\sqrt{p}\cos\dfrac{j\pi}p+b_{j}\sin\dfrac{j\pi}p +c_{j}\tan\dfrac{j\pi}p\right)$, where the $a_{ij}$ are integers. If $f$ denotes ...
2
votes
0answers
632 views

Cubic polynomials with “nice” roots, which can be expressed by trig functions of rational angles

Consider the cubic polynomial $x^3-ax+b$ for $a,b\in\mathbb N$. It has three real roots which, by Cardano's formula, can of course be written in closed form using thirds of angles or cube roots of ...
5
votes
0answers
265 views

Cutting and pasting in Galois theory

I want to ask who was the first to use cut-paste construction in Galois theory. This question is motivated from the trend in contemporary Galois theory to use patching methods to construct Galois ...
1
vote
1answer
201 views

Is the other extreme of Hilbert Irreducibility true?

Let $K$ be a number field (or perhaps more generally a Hilbertian field). Let $X_K\rightarrow \mathbb{P}^1_K$ be a regular (i.e. without extension of scalars) $G$-Galois branched cover. Hilbert's ...
4
votes
1answer
429 views

Symplectic groups Sp_{2m}(2) as 2-transitive permutation (i.e. Galois) groups

Hello, I am looking for information about the symplectic groups $Sp_{2m}(2)$ as permutation group acting on quadratic forms. Consider the block matrices ...
5
votes
0answers
745 views

“The Galois group of $\pi$ is $\mathbb{Z}$.”

Last year, in a talk of Michel Waldschmidt's, I remember hearing a statement along the lines of the title of this question, that is, "The Galois group of $\pi$ is $\mathbb{Z}$.". In what ...
0
votes
1answer
195 views

The image of generator under an automorphism of a cyclic function field

I'm reading the proof of Lemma 4.1 [1] which says: "Let $F = K(x,y), y^q = f(x)$, where $q$ is a prime different from characteristic of $K$. Let $Z := Gal(F/K(x))$ and we have $Z < G < ...
0
votes
1answer
330 views

A question related to Hilbert's Irreducibility Theorem

My question is whether for every extension of number fields $L\subset K$, and for every $f_0(x),...,f_n(x)$ in $K[x]$, there is some $\alpha\in L$ such that ...
24
votes
2answers
1k views

Solving the cubic by “radicals” in characteristics 2 and 3

This question has no justification other than a bit of fun. We all know that the cubic is solvable by "radicals" ($\root2\of{}$ and $\root3\of{}$) in characteristics $\neq2,3$. The formula was ...
10
votes
3answers
937 views

Can we always find such an irreducible polynomial of degree n where degree(p(x)-x^n)<= n/2?

Consider unitary polynomials of degree $n$ over $GF(2)$. That is, polynomials of the form $p(x) = \sum_{i=0}^n a_i x^i$ where $a_i\in GF(2)$ and $a_n=1$. Can we always find such an irreducible ...
1
vote
3answers
788 views

Constructive proof of algebraic elements forming a subfield

Let $F \leqslant E$ be a field extension. If $a, b \in E$ are algebraic over $F$ then $a+b$ and $ab$ are algebraic as well. There is an short proof of this using the extension $E(a,b)$: $[E(a,b):E]$ ...
15
votes
2answers
2k views

Galois theory for polynomials in several variables

I feel a bit ashamed to ask the following question here. What is (actually, is there) Galois theory for polynomials in $n$-variables for $n\geq2$? I am preparing a large audience talk on ...
1
vote
1answer
773 views

What does Gal(Q_p/Q) mean? [closed]

What does $\mathrm{Gal}(\overline{\mathbb{Q}}_{p}/\mathbb{Q})$ mean? ($p$ is a prime number.) If it is defined as $\mathrm{Aut}(\overline{\mathbb{Q}}_{p}/\mathbb{Q})$, then does it have any property ...
6
votes
0answers
602 views

Automorphisms of local fields

It is an amusing coincidence (at least it appears to be a coincidence to me) that any completion of the field $\mathbb{Q}$ has trivial automorphism group as an abstract field, i.e. when ignoring the ...
12
votes
2answers
2k views

Why should the anabelian geometry conjectures be true?

I had probed friends of mine about Grothendieck's motivation for making the anabelian geometry conjectures, and they gave me the following explanation: If $X$ is a hyperbolic curve over some field ...
6
votes
1answer
318 views

Galois groups at closed points from Galois group at generic point?

Consider the finite map $\mathbb{A}^1_\mathbb{Q}\rightarrow \mathbb{A}^1_\mathbb{Q}$ given by $z\mapsto z^5-z$. The fiber over generic point is the field extension $\mathbb{Q}(t)[z]/(z^5-z-t)$ over ...
41
votes
1answer
2k views

Which small finite simple groups are not yet known to be Galois groups over Q?

The subject line pretty much says it all. To expand just a little bit: 1) What is the smallest simple group that is not yet known to occur as a Galois group over $\mathbb{Q}$? (Variants: not known ...
-4
votes
2answers
966 views

what part of using vieta's formulas violates quintic non-solvability? [closed]

You can write the n roots of an n degree polynomial in terms of its n coefficients, i.e., "Vieta's" formulas. You can solve this system of nonlinear equations using Newton's method and the Jacobian. ...
5
votes
5answers
989 views

Galois theory and algorithms

Steven Weintraub's book {\em A Guide to Advanced Linear Algebra} includes the following remark: "Of course, there is no algorithm for factoring polynomials, as we know from Galois theory." I can't ...
3
votes
1answer
475 views

What is the general statement of Hilbert 90?

I know two generalizations of Hilbert 90, but I don't if there is a statement that contains both: The first statement Let $K$ be a field, then $H^1(Gal(K), GL_n(K^{sep}))=0$. The second statement ...
6
votes
1answer
708 views

Quadratic extension of quadratic extension

Let $K_1$ be a perfect field. Let $K_2/K_1$ and $K_3/K_2$ be quadratic extensions. Let $K_4/K_3$ be the Galois closure of $K_3$ over $K_1$. Is it true that either $K_3 = K_4$ or $K_4/K_3$ is quadratic ...
0
votes
1answer
450 views

Multiplication of matrices in GF(2) and R

$H$ is an $n \times n$ matrix with elements in $ \{ -1,1 \}$ $G$ is an $n \times k$ matrix with elements in $GF(2)$ and also upper triangular, invertable $m$ is an $k \times 1$ vector with elements ...
19
votes
0answers
1k views

more on “Transalgebraic Theories” (a 19th century yoga)?

Among the talks at occasion of the Galois Bicentennial, one is about "Transalgebraic Theories". Unfortunately I found only this article describing that fascinating idea as " an extremely powerful ...