Galois theory allows one to reduce certain problems in field theory, especially those related to field extensions, to problems in group theory. For questions about field theory and not Galois theory, use the (field-theory) tag instead. For questions about abstractions of Galois theory, use ...

learn more… | top users | synonyms

6
votes
1answer
120 views

Are there relations among Frobenii?

Let $G=\text{Gal}(\overline{\mathbf Q}/\mathbf Q)$, and for each prime $p$, choose an embedding $\overline{\mathbf Q} \hookrightarrow \overline{\mathbf Q_p}$. Let $\sigma_p$ be a choice of Frobenius ...
1
vote
2answers
78 views

Counterexamples (I assume) on field extensions

Concerning finite field extensions, with $E_1E_2$ the compositum: I proved $$[E_1E_2:K]=[E_1E_2:E_2] \cdot [E_2:K] \leq [E_1:K] \cdot [E_2:K]$$ Is it true that $[E_1E_2:K]$ has to divide $ [E_1:K] ...
0
votes
0answers
33 views

Galois representation associated to a number field

I think I'm missing something completely trivial. I want to know how to compute the Galois representation associated to an extension of $p$-adic fields. Let $p$ and $q$ be odd prime numbers. Fix ...
2
votes
1answer
38 views

Fixed Field of automorphisms (of $k(x)$ with $k$ a field) Induced by $I(x)=x$, $\varphi_1(x) = \frac{1}{1-x}$, $\varphi_2 (x)=\frac{x-1}{x}$?

Since $I(x)=x$, $\varphi_1(x)=\frac{1}{1-x}$, $\varphi_2 (x)=\frac{x-1}{x}$ form a group of order 3 the group is cyclic so it is generated by $\varphi_1$ then I have to find the fixed field of ...
4
votes
2answers
96 views

Galois group - extend homomorphism to automorphism

Let $K \subset L$ be a finite Galois extension, $M$ a field with $K \subset M \subset L$ and $G := \text{Aut}(L/K)$. I want to show that if $\sigma \, \colon M \longrightarrow L$ is a ...
0
votes
1answer
38 views

Galois theory in a different setting

Suppose that instead of wanting to express the roots of a polynomial equation with arithmetic operations and radicals we instead wanted expressed it with arithmetic operations and $\sin(x)$ ? What ...
3
votes
1answer
39 views

Let $F/\mathbb{Q}$ be a degree 4 extension, NOT Galois. Prove that the Galois closure of $F$ has Galois group either $S_4, A_4$ or $D_8$.

The question is as the title states. So if $F=\mathbb{Q}(\alpha)$ for some alpha that satisfies a degree 4 polynomial $p(x)$, then we are looking for the splitting field of $p(x)$? I'm not sure what ...
0
votes
1answer
19 views

Elements in one Galois extension that act trivially on another Galois extension

I am trying to pin down my lack of understanding in the following case, I know it must be something easy I am missing, so any pointers would be really appreciated. I have a (finite) Galois field ...
3
votes
2answers
69 views

Cubic polynomial - radical expression of roots

Let $f=X^3+X^2-2X-1$ be a polynomial with the three roots $x_1,x_2,x_3$ with $x_1=2\text{cos}(\frac{2 \pi}{7})$. We define $z:=(x_1-x_2)(x_1-x_3)(x_2-x_3)$. I want to find a radical expression for ...
1
vote
1answer
46 views

Galois Group of a cubic polynomial

Let $P=X^3+a_1X^2+a_2X+a_3 \in \mathbb{Q}[X]$ be irreducible, $x_1,x_2,x_3$ the roots of $P$ and $L:=\mathbb{Q}[x_1,x_2,x_3]$. The galois group of $P$ is isomorph to $S_3$. Now we define ...
1
vote
2answers
43 views

$\cos{nx}=p(\cos{x})$, general formula for $p$?

Does a general, closed formula for the $p$ polynom exist? $\cos{nx}=p(\cos{x})$
1
vote
1answer
29 views

Prove for $a,b \in \mathbb{F}_{p^n}$, if $p(x) = x^3 + ax +b$ is irreducible, then $-4a^3 - 27b^2$ is a square in $\mathbb{F}_{p^n}$.

The problem is as the title states. We know that in this case determinant $D = -4a^3 -27b^2$, and also I know that if $G$ is the Galois group of $x^3 + ax + b$, then $$G \subset A_n \, \iff \sqrt{D} ...
2
votes
1answer
42 views

Charactristic polynomial of a F-linear transformation with respect to Galois group

Let $K$ be a Galois extension of $F$, and let $a \in K$. Let $L_a : K \to K$ be the $F$-linear transformation defined by $L_a(b)=ab$. Show that the characteristic polynomial of $L_a$ is $\prod_{\sigma ...
12
votes
1answer
124 views

When is a number in $\mathbb{Q}(\sqrt{a_1},\ldots,\sqrt{a_n})$?

Given an algebraic number $\alpha$ with minimal polynomial $P(x)$ of degree $2^n$, how can I decide if there are integers $a_1,\ldots,a_n$ such that ...
6
votes
1answer
75 views

Is the extension Galois if $\mathrm{Aut}(K)$ acts transitively on the non-ramified prime ideals?

Let $K/\mathbb Q$ be a finite extension such that $\mathrm{Aut}(K)$ acts transitively on the prime ideals that are not ramified above the same prime $p\in\mathbb N$. Is $K$ Galois? Thanks in advance. ...
18
votes
3answers
307 views

Why there is much interest in the study of $\operatorname{Gal}\left(\overline{\mathbb Q}/\mathbb Q\right)$?

Let's start for a simple quote from wikipedia: "No direct description is known for the absolute Galois group of the rational numbers. In this case, it follows from Belyi's theorem that the ...
3
votes
2answers
48 views

Splitting field of cyclotomic polynomials over $\mathbb{F}_2$.

Let $\Phi_5$ be the 5th cyclotomic polynomial and $\Phi_7$ the 7th. These polynomials are defined like this: $$ \Phi_n(X) = \prod_{\zeta\in\mathbb{C}^\ast:\ \text{order}(\zeta)=n} ...
0
votes
1answer
32 views

Char(F)=0, K/F is abelian and F contains a primitive root of unity

$\mathrm{Char}(F)=0, K/F$ is abelian . Let $n$ be a positive integer such that $f^n=1$ for every $f$ in $G(K/F)$ and $F$ contains a primitive $n^{th}$ root of unity. Prove there exist $x_1, x_2, …, ...
5
votes
2answers
75 views

Subgroup of class group

Let $K/\mathbb Q$ be a finite Galois extension with Galois group $G$, ring of integers $\mathcal O_K$ and $\mathcal Cl(K)$ its ideals class group. I want to show that $\mathcal Cl(K)^G$ is generated ...
0
votes
0answers
22 views

M/F normal, determine whether there is an automorphism $\sigma\in Gal(M/F)$ with $\sigma(a)=a'$ and $\sigma(b)=b'$

Let $M$ be a normal extension of $F$. Suppose that $a, a'$ are roots of $min(F,a)$ and that $b,b'$ are roots of $min(F,b)$,and that $min(F,a)\neq min(F,b)$. Determine whether or not there is an ...
31
votes
1answer
597 views

Is my field algebraically closed?

For a field $L$, let $\widetilde L$ be the splitting field of all irreducible polynomials over $L$ having prime-power degree. Question: Do we have $\widetilde{\mathbf Q}=\overline{\mathbf Q}$? ...
3
votes
1answer
45 views

Automorphisms of a field extension permute roots of irreducible factors.

Would someone mind confirming (or refuting) the following... Proposition. Let $K$ be a field and let $f \in K[X]$. Suppose $g \in K[X]$ is an irreducible factor of $f$. Let $L$ be a splitting field ...
2
votes
1answer
77 views

$L=K(\alpha)$ with $\alpha^p-\alpha=a\in K$

Let $K$ be a field of characteristic $p$. $L=K(\alpha)$ with $\alpha^p-\alpha=a\in K$, an extension of order $p$. Show that there does not exist $\beta \in L$ such that $\alpha^{p-1}=\beta ^p-\beta$. ...
2
votes
1answer
73 views

Having trouble considering a finite field $\mathbb{F}_{p^n}$ as a vector space over $\mathbb{F}_p$.

As the title states, I'm having trouble considering a finite field $\mathbb{F}_{p^n}$ as a vector space V over $\mathbb{F}_p$. Clearly it is dimension $n$. How can we work with this vector space? ...
0
votes
0answers
25 views

E is a Galois extension of F

15.32 Let $E$ be a finite extension of a field $F$ with dimension $n$. Show that $|\operatorname{Gal}_F(E)| = n$ if, and only if, $E$ is a Galois extension of $F$. Please some help to find the ...
0
votes
0answers
15 views

Field extension, primitive root of unity solvable by radicals

I'm stuck with this exercise. I'd like to show, that for a field $K$ with $\mathrm{char}(K)=0$ and $\omega_n$ being a primitive n-th root of unity $K(\omega_n)/K$ is solvable. I thought I could do ...
0
votes
1answer
25 views

Galois group of $K(\sqrt[p]a)$ over $K$

If $p$ is a prime number and $a$ is an element of the field $K$, what can be said about the Galois group of $K(\sqrt[p]a)$ over $K$? (That is, $\sqrt[p]a$ is an element whose $p$-th power is $a$) I ...
3
votes
1answer
50 views

A question about Galois extension

Let $n\geq 2$. Let $K$ be a field which contains $n$ distinct n-th roots of unity. Let $L=K(\sqrt[n]{a})$ with $a\in K$ and $[L:K]=n$. Show: there exists a Galois extension $K\hookrightarrow M$ such ...
1
vote
0answers
28 views

How can i give this isomorphism between a Galois group and $S_3$?

I have shown that if $x$ is the nontrivial cubic root of the unity and that if $y$ is the real cubic root of $2$, then $Q(x,y)$ is a Galois extension whose Galois group has order $6$. I know that the ...
6
votes
1answer
69 views

Galois group of $X^5-n$

In the following situation I want to find the Galois group of a specific polynomial: Let $n > 1$ be a square-free integer, $f =X^5 - n \in \mathbb{Q}[X]$, $x:=\sqrt[5]{n}$ and $\zeta=\text{e}^{2 ...
3
votes
1answer
29 views

Minimal polynomial in Galois extension

Let $K \subset L$ be a Galois extension and $x \in L$ such that $L=K[x]$. $H \leq \text{Aut}(L|K)$ is a subgroup of the galois-group. I want to show that the minimal polynomial of $x$ over ...
3
votes
0answers
130 views

If $f(x)\in F[x]$ is solvable by radicals then its Galois group $Gal(E/F)$ is solvable.

In all the books that I've checked the use of roots of unity (as hypothesis) is very crucial to prove that if $f(x)\in F[x]$ is solvable by radicals then its Galois group $Gal(E/F)$ is solvable, but ...
1
vote
1answer
23 views

A Galois extension of two roots.

Let $x$ be one of the nontrivial cubic roots of the unity and let $y$ be the cubic root of $2$. Prove that $\mathbb{Q}(x,y)$ is a Galois extension and then, show that is isomorphic to one of the ...
5
votes
2answers
91 views

Galois group of irreducible Quartic polynomial over $\mathbb{Q}$

Actual Question is : What are all possible galois groups of an irreducible Quartic polynomial over $\mathbb{Q}$ As polynomial is irreducible, Galois group is transitive subgroup of $S_4$. I ...
0
votes
1answer
39 views

Problem with a Galois extension [duplicate]

How can i prove that Q(a,b) is a Galois extension and that its Galois group is of order 6, if a is a root of the cubic that is not 1 and b is the cubic root of 2? Can you give me a little hint for ...
3
votes
3answers
72 views

$\mathbb{Q}(\sqrt[3]{2}, \zeta_{9})$ Galois group

How do I calculate the degree of $\mathbb{Q}(\sqrt[3]{2}, \zeta_{9})$ over $\mathbb{Q}$. Should it be 18, as $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$, and $[\mathbb{Q}(\zeta_{9}):\mathbb{Q}] = 6$? ...
1
vote
1answer
32 views

Primitive Element and Minimal Polynomial

Let $K = \mathbb{Q}(\sqrt{2},\sqrt{5},\sqrt{7})$: I know that $K/F$, with $F =\mathbb{Q}$, is Galois and has degree 8. By the primitive element theorem there is a a $\theta$ such that $K = F(\theta)$, ...
0
votes
1answer
13 views

There are finite distinct restrictions to a subfield

Consider the field extension $L\subseteq K\subseteq \mathbb C$ where $K/L$ is finite. I must show that the set $\{\sigma_{|K}\,:\,\sigma\in\operatorname{Gal}( \mathbb C/L)\}$ is finite, but I have ...
5
votes
1answer
187 views

Galois extension with galois group $A_4$

Suppose $char F \neq 2$ and $K/F$ is a degree three galois extension with $Gal(K/F)\cong \mathbb{Z}/(3)$. Is there a bijection between extensions $N/F$ with galois group $A_4$ and the order four ...
5
votes
1answer
89 views

Galois Group of $x^{14}+x^7-1$ over $\mathbb{Q}$

So consider the polynomial $f(x)=x^{14}+x^7-1$ defined over $\mathbb{Q}$. We want to determine its Galois Group. So let's look for the splitting field, $L$ say, to give us an idea of the size of the ...
3
votes
1answer
44 views

Galois Group of $x^4+x-1$ over $\mathbb{F}_3$

Consider the finite field $\mathbb{F}_3$ and define the polynomial $f(x)=x^4+x-1$ over $\mathbb{F}_3$. I want to find its Galois Group. I observe that $f$ has no root over $\mathbb{F}_3$, so if it ...
1
vote
1answer
30 views

maximal abelian extension of exponent $q-1$ of $\mathbb F_q((t))$

I would like to find the maximal abelian extension of exponent $q-1$ of $K=\mathbb F_q((t))$ and find its Galois group. Due to Kummer theory this extension is $K(\sqrt[q-1]{K^*})$ and it's Galois ...
29
votes
1answer
412 views

Is Frobenius the only magical automorphism?

The Frobenius automorphism is special because the $p$-power map makes sense in any characteristic $p$ ring, which allows us to canonically extend the Galois-theoretic Frobenius to any such ring. I ...
0
votes
0answers
21 views

K/F is finite separable extension show that N_K/F = N_E/F * N_K/E and Tr_K/F = Tr_E/F * Tr_K/E for any intermediate field K/E/F.

K/F is finite separable extension show that $N_{K/F} = N_{E/F} N_{K/E}$ and $Tr_{K/F}$ = $Tr_{E/F}Tr_{K/E}$ for any intermediate field K/E/F If x is an element of K then $N_{K/F}$ is the norm ...
2
votes
0answers
23 views

MAGMA Commands for Galois Theory calculations

Perhaps this is walking over old ground (or the wrong place to ask this), but I'm looking to use MAGMA to perform certain calculations in Galois Thoery. The motivation of this question is to create an ...
0
votes
0answers
21 views

Prove if $F$ is an infinite field and $u,v$ are algebraic and separable over $F$ that there exists $x \in F$ such that $F(u,v) = F(u+xv)$.

Prove if $F$ is an infinite field and $u,v$ are algebraic and separable over $F$ that there exists $x \in F$ such that $F(u,v) = F(u+xv)$. Can someone give any hint? Thanks
3
votes
3answers
79 views

Show [F(a):F] = r

Suppose that K/F is Galois and a in K has precisely r distinct images under G(K/F). Show [F(a):F] = r I'll really appreciate if someone could give some hints? Thank you!
2
votes
2answers
49 views

When has a polynomial a non-real and a real root?

Let $g \in \mathbb{Q}[X]$ be an irreducible and separable polynomial which has a real and a complex root in $\mathbb{C}$. Show that in this situation $\text{Gal}(K|\mathbb{Q})$ is not abelian, where ...
1
vote
1answer
27 views

Automorphism group of a non-normal field extension

Consider finite field extensions $L>K>F$ such that $L/F$ is Galois, and $K/F$ is separable. I am particularly interested in the case $F=\mathbb{Q}$. By Galois theory, $K/F$ is normal iff ...
1
vote
0answers
22 views

Why G(P/F) and G(Q/F) abelian implies PQ abelian?

Suppose L/F is a field extension, then L/P/F with P/F abelian and L/Q/F with Q/F abelian implies PQ=P(Q) abelian. After trying to pick element from PQ, I still can't find the relation between PQ and ...