1
vote
1answer
22 views

Irreducibility in Galois/non Galois Extensions

Let $k$ be a field and $\alpha$ algebraic over $k$. Let $K$ be the Galois closure of $k(\alpha)$ (obtained by adding all conjugates of $\alpha$). If $f(x) \in k[x]$ is irreducible over $k[\alpha]$ is ...
2
votes
1answer
40 views

A Fixed Field of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$

I was working on a review problem from Dummit and Foote and came across the following issue. It is clear that the Galois group of the splitting field for the polynomial $(x^2-2)(x^2-3)(x^2-5)$ has ...
0
votes
0answers
22 views

Galois group of a splitting field

$f=x^4-5x^2+6 \in \operatorname Q[x]$ $f=(x^2-2)(x^2-3)$ $\operatorname F=\operatorname Q(\sqrt[2]{2},\sqrt[2]{3})$ $f$ is irreducible for Eisenstein's criterion, $\operatorname{char} \operatorname ...
1
vote
1answer
24 views

Galois group of the splitting field of $ x^6 - 5$

$f = x^6 - 5$ $\in \operatorname Q [x]$ I want to find a splitting field $\operatorname F$ of $f$ over $ \operatorname Q$ $\sqrt[6]{5}$ is a real root of $f$ $u$ is the 6-th root of unity then $ ...
1
vote
2answers
34 views

$K \le B\le F$, If $F$ Galois over $K$ $\Rightarrow$ $F$ a Galois over $B$

Let $F$ a Galois over $K$, and let $B$ be a subfield of $F$ such that : $K \le B\le F$ $\Rightarrow$ $F$ a Galois over $B$ PROOF: $F$ is a splitting field of $f \in K[x]$ separable over $K$. ...
-1
votes
0answers
27 views

Schanuel's conjecture and field extensions.

Doing a little bit of reading over the summer break before going into my masters year of my maths degree and i have been looking at Schanuel's conjecture which states that; Given any $n$ complex ...
1
vote
1answer
69 views

Solvability of polynomials over fields of characteristic zero

1) Let $K$ be a field, $\operatorname{char}(K)= 0$, and $f ∈ K [x]$ with $\deg(f)\le4$. Then $f$ is solvable by radicals. Proof: $\operatorname{Gal} (F/K) \cong S_4$ then $\operatorname{Gal}(F/K)$ ...
1
vote
1answer
51 views

How to show this is the minimal polynomial

I'm trying to the following problem. But I can't show some irreducibility of the polynomials. Put $L=\mathbb{C}(X,Y,Z)$, $\omega=\frac{-1+\sqrt{-3}}{2}$. Define two automorphism $\sigma, \tau$ of ...
2
votes
1answer
87 views

Radical Solvable quintic polynomial.

I am trying to solve the following exercise: Let $k$ be a field of characteristic 0, and let $f(x)\in k[x]$ be a polynomial of degree 5 with splitting field $E/k$. Prove that $f(x)$ is ...
3
votes
1answer
72 views

What is the difference between field theory and Galois theory

I am about to finish the book Galois theory by Harold Edwards. I am planning to study Galois theory at a more advanced level or field theory. I am unable to decide because I don't know the difference ...
2
votes
1answer
44 views

$(u+1)(u+2)\cdots(u+p)=u^p-u$

Let $E$ be a field with characteristic $p>0$. How should I prove that $$(u+1)(u+2)\cdots(u+p)=u^p-u$$ I can verify this equality for small $p=2,3,5$. Is there a way to prove this result in general? ...
3
votes
0answers
36 views

Proving that a certain maximal subfield has countably infinite index

Let $K$ be a maximal subfield of $\mathbb C$ which does not contain $\sqrt{2}$. I've shown that $\mathbb C$ is an algebraic extension of $K$, and that the Galois group of any finite extension of $K$ ...
2
votes
1answer
28 views

If $L_1/K$ and $L_2/K$ are not Galois (solvable), then $L_1L_2/K$ is not Galois (solvable)

This is part of an exam preparation: Prove/contradict: If $L_1/K$ and $L_2/K$ are not Galois, then $L_1L_2/K$ is not Galois. If $L_1/K$ and $L_2/K$ are not solvable Galois extensions, ...
1
vote
1answer
21 views

Normalizer of a subgroup of a Galois group

I wanted to check whether my solution for this problem was correct. Let $k \subseteq L \subseteq K$ be a finite extension of fields, with $K/k$ Galois $H$ the normalizer of $Aut(K/L)$ in $Aut(K/k)$. ...
0
votes
2answers
50 views

Find the galois group of the polynomial when a root is given

If $\alpha$ is a root of a polynomial $f(x)=x^3 +x^2-4x+1$ then show that $2 - 2\alpha - \alpha^2$ is also a root of $f(x)$. Use this fact to compute the Galois group of the splitting field of $f(x)$ ...
3
votes
2answers
136 views

Is the order of $\operatorname{Gal}(\bar{\Bbb Q}/\Bbb Q)$ known?

Is the order of $\operatorname{Gal}(\bar{\Bbb Q}/\Bbb Q)$ known? And if so is there a description on how the order can be found? My initial thoughts is that because $\bar{\Bbb Q}$ is countable and ...
2
votes
0answers
35 views

Characterizing quadratic number fields that are subfields of cyclic quartic number fields [duplicate]

Given a quadratic number field $F = \mathbb{Q}(\sqrt{d})$, is there a way to determine whether or not $F \subset K$ for some quartic numberfield $K$ with $\operatorname{Gal}(K/\mathbb{Q}) \cong ...
1
vote
0answers
27 views

Splitting field and intermediate fields

Find splitting field $K$ of polynomial $x^3-7$ over $\mathbb{Q} (\sqrt{3} )$ and find $(K: \mathbb{Q} )$ $G(K/ \mathbb{Q} )$ Find intermediate fields between $\mathbb{Q}$ and $K$. So the ...
1
vote
1answer
49 views

Show that a sequence of fields exists

I do not have a clue how to solve the following problem: Let $K\subseteq L$ be Galois extension of degree $p^n$, where $p$ is prime and $n$ is natural. Show that there exists a sequence of subfields ...
1
vote
0answers
32 views

Galois extension of two fields

I am preparing myself to the exam in algebra and I have found the following exercise, which is quite hard for me. Do you have any suggestions for solving it? For fields $K\subseteq L,M\subseteq \bar ...
-1
votes
1answer
37 views

Prove that $u$ is algebraic over $\mathbb{Q}$ [closed]

Let $u$ be a root of the following equation: $$x^{3}+{\displaystyle \dfrac{1+i}{\sqrt{2}}x^{2}+\dfrac{-1+i\sqrt{3}}{2}x+1=0}$$ Prove that $u$ is algebraic over $\mathbb{Q}$ . Thanks in ...
2
votes
1answer
60 views

If $|\operatorname{Aut}_KF|=3$, must we have cube roots of unity?

Let $K$ be a field of zero characteristic. Let $F$ be a finite dimensional extension field of $K$ such that $|\operatorname{Aut}_K F|=3$. Must the equation $x^2+x+1=0$ have a root in $F$ ? Thank you
5
votes
1answer
35 views

Does the data of Galois group, ramified places, and inertia groups, determine a Galois number field?

Suppose I tell you that $K/\mathbb{Q}$ is a finite Galois extension, and I specify the Galois group $G$, and suppose further that I give you a finite list $S$ of places of $\mathbb{Q}$ and for each ...
-1
votes
0answers
23 views

Transitivity of norm and trace for intermediate fields of finite separable extensions

Let $K/F$ be a finite separable extension of fields. Show that $N_{K/F} = N_{E/F}\circ N_{K/E}$, and $Tr_{K/F} = Tr_{E/F}\circ Tr_{K/E}$ for any intermediate field K/E/F.
0
votes
0answers
22 views

Show that $K_1K_2=K_1(K_2)$ is abelian. [duplicate]

Let $L/F$ be a field extension and $L/K_i/F$ with $K_i/F$ abelian. Show that $K_1K_2=K_1(K_2)$ is abelian.
2
votes
2answers
29 views

Splitting field of $f$ as smallest field extension containing all BUT ONE zero of $f$

I'm just working with splitting fields and I have to prove something which I don't understand. Let $L$ be a splitting field of the polynomial $f$ over $K$ and $f = \prod_{i=1}^n(X-\alpha_i)$. ...
2
votes
1answer
34 views

Separability of a polynomial

I have a non zero polynomial $f\in F[X]$ where $F$ is a field. Let $L$ be a field extension of $F$ so that $f$ splits completely in $L[X]$, so $f(X)=c\prod_{i=1}^n (X-a_i)$ with $c,a_i\in L$. If ...
1
vote
0answers
21 views

Did I Do This Galois Theory Problem Right? Subfields of $\mathbb{Q}(\zeta_{12})$.

Let $\omega$ be a primitive $12$th root of unity. (i) What is $[ \mathbb{Q}(\omega) : \mathbb{Q}]?$ (ii) List the distinct conjugates of $\omega + \omega^{-1}$. (iii) What is $Aut(\mathbb{Q}(\omega ...
1
vote
2answers
38 views

Is my answer correct on this Galois Theory problem? Find the lattice of subfields of $\mathbb{Q}(\zeta_9)$

Problem: let $\zeta$ be a primitive $9$th root of unity, and $K = \mathbb{Q}(\zeta)$. Describe the lattice of subfields of $K$, give generators for each subfield and list its degree over ...
0
votes
0answers
30 views

Splitting fields and isomorphic

Please check these statements whether those are true Let $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ be extension fields of $\mathbb{Q}$, the rational numbers. Since they are not isomorphic as ...
4
votes
1answer
51 views

Find all subfields in extension $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[4]{2})$

I want to find all intermediate subfields of extension $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[4]{2})$. I guess that $\mathbb{Q}(\sqrt[4]{2})$ is not a splitting field, since we would have polynomial ...
4
votes
1answer
47 views

A Galois Extension over a field of characteristic p

Hi! I've been having a bit of trouble with this question, namely the very last part. It's from a past paper for a Galois Theory course, in which I have an exam on Monday. Basically, I can show every ...
3
votes
1answer
61 views

Help with computing Galois group of $x^4 - 3$.

Let $f(x) = x^4 - 3$. I believe $Gal(f(x)) = Gal(\mathbb{Q}(\sqrt[4]{3}, i)/\mathbb{Q})$, and then we have $$\sigma_1 = \begin{cases} \sqrt[4]{3} \rightarrow \zeta^n\sqrt[4]{3}, ...
0
votes
1answer
59 views

Why is $x^2-2ux+1$, where $u = \cos(\frac{2\pi}{n})$, irreducible in $\mathbb Q(u)$?

My textbook states that $x^2-2ux+1$, where $u = \cos(\frac{2\pi}{n})$ for some $n \in \mathbb N$ is clearly irreducible in $\mathbb Q(u)$. Is this obvious? I tried to write it as a product of ...
2
votes
1answer
39 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 ...
3
votes
1answer
45 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 ...
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. ...
0
votes
1answer
14 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 ...
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
29 views

Reference for Galois Theory of infinite field extensions.

I would like to ask what is in your opinion the best place to learn about Infinite Galois Theory that requires not much knowledge of topology. I am searching for a text that explains the notions ...
0
votes
0answers
15 views

Consider $n$ a product of distinct primes $p_j$, with each $p_j-1$ a power of 2, then the regular $n$-gon is constructible.

I am having trouble showing that if $n$ is a product of distinct primes $p_j$, with each $p_j-1$ a power of $2$, then the regular $n$-gon is constructible. Apparently it could be useful to use that ...
3
votes
1answer
53 views

Is there a nice topology on Aut(C)?

Let $G=\mathrm{Aut}(\mathbf C)$, the group of field automorphisms of the complex numbers. It is a very large group (see this MSE question and the nice answer by Andres). For instance, there even ...
1
vote
1answer
30 views

A doubt regarding splitting field.

$\Bbb Q(\omega)= \Bbb Q(\sqrt3,\iota)$ This is written in my text book that i am following. But I think this is a typo. Since $\Bbb Q(\sqrt3,\iota)$ is a larger field. in which $\Bbb Q(\omega)$ is ...
0
votes
1answer
60 views

The Galois Group of $x^4 - 5x^2 + 6$

As the title suggests, I'm asked to describe the Galois group of the polynomial $x^4 - 5x^2 + 6 \in \mathbb{Q}[x]$ over $\mathbb{Q}$. I am pretty certain I have 95% of the problem completed. I'm just ...
0
votes
1answer
41 views

Does $E$ a finite field and $F\subset E$ imply that $E$ is Galois over $F$?

Is this the case? I don't know whether to go fishing for a counterexample or to try to prove it.
1
vote
3answers
89 views

Field extensions - if $(E : F) = n$ then $(E(x) : F(x)) = n$

Well, pretty much everything is in the title - I'm looking for the proof of the following statement: if we have a field extension $F \subset E$ then the degree of the extension $F(x) \subset E(x)$ ...
1
vote
0answers
34 views

Topology in Infinite Galois Theory.

I am a final year undergraduate student in Mathematics. I have a good background in algebra up to Galois theory of finite extensions of fields. I have started trying to understand the Galois theory of ...
0
votes
0answers
30 views

$G_f^\theta$ is $A_4$ or $S_4$?

Let $f(x)\in \mathbb{Q} [x]$ irreducible polynomial of degree 4, $u \in \mathbb{C}$ a root of $f(x)$. Prove that there are not subfields $K$ such that $\mathbb{Q} \subset K \subset \mathbb{Q} (u)$ if ...
1
vote
2answers
71 views

Irreducible Polynomials over a Finite Field

I am motivated by this question, and want to find a solution to the following problem. Question: How many monic, irreducible polynomials of degree $n$ are there over the finite field ...