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 ...

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Showing this field extension is not normal

This question is part of a homework assignment. We are asked if the field extension $\mathbb{Q(\sqrt[4]{-7})}/\mathbb{Q}$ is normal. Here is what I have so far: The obvious thing to do seems to be ...
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0answers
18 views

Uncountable field in Galois extension

Let $G$ be a finite cyclic group of order $m$. Then we can view $G$ as $Gal(T/K)$ for some Galois extension $T/K$. Is it possible to choose $K$ uncountable?
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0answers
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Given two dot products with the same vector in a prime finite field of 2 (Galois Field), how can one figure out future dot products?

I've stumbled upon an interesting "rule" derivation for the value of a dot product in $\mathbb{R}^{n}$ like this: Given an arbitrary vector $\vec a \in \mathbb{R}^{n}$ and the values of two dot ...
4
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1answer
37 views

Converse of Galois Theorem.

If $F\subset K\subset E$ field extension such that $K\subset E$ and $F\subset K$ is both finite and Galois extensions then $F\subset E$ is Galois extension. My intuition says that this is false but I ...
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0answers
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Let $f = x^4 + ax^2 + bx + c \in K[x]$ be an irreducible and separable polynomial and $L/K$ be its splitting field extension.

Let $disc(f) = D = \Delta^2$ be denoted as the discriminant of $f$ and assume that $\Delta \notin K.$ Let $g \in K[x]$ be the resolvent cubic of $f$ and assume there exists only one root which is $d ...
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0answers
24 views

question on Galois group

Let $f(x)\in F[x]$, let $E/F$ be a splitting field, and let $G=\text{Gal}(E/F)$ be the Galois group. 1) If $f(x)$ is irreducible, then $G$ acts transitively on the set of all roots of $f(x)$ (if ...
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1answer
22 views

$F/E$ a finite Galois extension, the integral closure of $E[X]$ in $F(X)$

If $F/E$ is a finite Galois extension, then one can show that $F(X)/E(X)$ is also a finite Galois extension of the same degree (a basis for $F/E$ is also a basis for $F(X)/E(X)$). Since $E[X]$ is a ...
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0answers
48 views

Splitting field of $1 + x^3 + x^6 + x^9$

I'm studying for a qualifying exam and wanted to know if my reasoning on this problem was correct: Let $L$ be the splitting field of $f(x) = 1 + x^3 + x^6 + x^9$ over $\mathbb{Q}$. Find the Galois ...
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1answer
51 views

On Galois closure

I'm working on this problem in Hungerford: For $\sigma \in Aut_F \bar{F}$, show that every finite extension of $K$, the fixed field of $\sigma$, is cyclic. For a finite extension $L$ of $K$, let $M$ ...
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1answer
18 views

degree of extension of rationals adjoin infinitely many roots

Consider a radical extension of $\mathbb{Q}$ in which infinitely many numbers are adjoined. Each of these numbers is the $n^{th}$ root of some integer, where $n$ can vary. Can any information be ...
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0answers
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Normal basis in every subgroup

Let $K$ be finite Galois extension of $k$, $G=Gal(K/k)$ and $k$ infinite. Prove: Exists $\beta\in K$ such that for every $N<G$, we have $\{\sigma(\beta):\ \sigma\in N\}$ is a basis for ...
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0answers
11 views

Is the Galois condition is necessary for the definition of cyclic extension?

Hungerford says that $F/K$ is cyclic, if $F/K$ is algebraic Galois and $Aut_K F$ is cyclic. Is it necessary that $F/K$ is Galois? Is there an example of extension which is cyclic, but not Galois?
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0answers
16 views

Cyclotomic polynomial identity [duplicate]

STATEMENT Show that $$\Phi_n(X)=\prod_{d\mid n}(X^{n/d}-1)^{\mu(d)}$$ where $\Phi_n(X)$ is the n-th cyclotomic polynomial. QUESTION Could anyone offer any help on how to show this result.
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0answers
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Galois group of $E/K$ & Galois group of the extension $E$ over Fixed field?

I've proved this result for myself, but I have doubt in my proof whether it is true : Let $E/K$ be a field extension and $G(E/K)$ its Galois group. Suppose $E^{G(E/K)}$ is its Fixed field, i.e. ...
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0answers
19 views

Example of degree 4 polynomial with Galois group $S_{4}$

I know the condition in terms of discriminant and resolvent cubic when Galois group of a degree $4$ polynomial over $\mathbb{Q}$ has Galois group $S_{4}$. But I am looking for an explicit example of ...
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0answers
12 views

Construct degree $n$ field extension with no intermediate field

I want to construct degree $n$ field extension with no intermediate field for each $n$. I know for any finite group $G$ there is a Galois extension $K/F$ so that $Gal(K/F)$ is $G$. So my idea was to ...
2
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1answer
27 views

Fixed fields,cyclic groups and Galois theory

Let $G=S \times T$ where $S$,$T$ are both finite cyclic groups. Question 1: Is it true that there exists a Galois finite extension $L/F$ such that $Gal(L/F) \cong S \times T$? (I can't recall if ...
3
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1answer
38 views

$\{1,\sqrt [3]{7},{\sqrt [3]{7 }}^{2}\}$ is a basis for $Q[\sqrt {2},\sqrt {3},\sqrt [3]{7}]$ over $Q[\sqrt {2},\sqrt{3}]$

I am trying to prove that $\{ 1,\sqrt [ 3 ]{ 7 } ,{ \sqrt [ 3 ]{ 7 } }^{ 2 }\} \quad is\quad a\quad basis\quad for\quad Q[\sqrt { 2 } ,\sqrt { 3 } ,\sqrt [ 3 ]{ 7 } ]\quad over\quad Q[\sqrt { 2 } ...
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Orthogonal Group and Special Orthogonal Group Isomorphism [closed]

Let M be a subset of R2 that contains a basis for R2 and M':= M x {0} = {(x,y,0)∈ M}. Prove that O(M) is isomorphic to SO(M')
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1answer
24 views

Splitting field of $x^ {a^n}$ −1 in Z/aZ[x]

What is the splitting field of the polynomial $x^ {a^n}$ −1 in \ the ring Z/aZ[x] with n natural? I´m working in some kind of proof of the best known theorem that says it´s impossible there exist a ...
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1answer
25 views

Showing two field extensions (of Q) are isomorphic using primitive elements, and why every element is primitive

I'm taking a class on Galois Theory in another language and the prof is saying my answer on this is incorrect and I'm wondering why, particularly since sometimes there's a language barrier. Basically ...
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2answers
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Aut$(K/F)$ permutes roots of polynomial.

Let Aut$(K/F)$ is the set of all automorphism from $F$ to $K$, where $K$ is a galois extension of $F$. Let $f(x) \in F[x]$ and $\alpha$ be a root of the polynomial $f(x)$. I am able to prove that for ...
3
votes
1answer
61 views

Extending field homomorphisms to automorphisms

I have $L/K$ a finite field extension and an irreducible polynomial which has two roots in $L$, $\alpha$ and $\beta$. I'm trying to show there is an automorphism of $L$ that fixes $K$ and switches ...
2
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0answers
29 views

Subgroups and corresponding subfields of galois group of $x^6 + x^3 + 1$

I think I have found the correct Galois group of $f = x^6 + x^3 + 1$ over $\mathbb{Q}$ to be $C_6$, the splitting field being $\mathbb{Q}(w)$ where $w$ is a complex 9th root of unity. Now I am trying ...
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0answers
36 views

Solvability by radicals of Polynomials defined by a recurrence relation

I want to determine the smallest integer $m$ such that the polynomial $P_{n}(x)$, $n\geq m$, given by : $$\left \lbrace \begin{array}{l} P_{n+1}(x) = P_n(x) (x-n-1) + \prod\limits_{i = 0}^n x-i\\ ...
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0answers
22 views

Transcendental extensions

I have to solve an exercise on transcendental extensions: 1) Show that a field extension $F \subset K$ which has transcendence degree at least 2, cannot be simple. 2) Two purely transcendental ...
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1answer
30 views

Calculating fixed fields

I have to show that $\mathbb{Q}(\zeta)^{<\sigma>}$ $=$ $\mathbb{Q}(\zeta + \frac{1}{\zeta})$ $\sigma \colon L \to L$ is defined by $\sigma(\alpha) = \overline{\alpha}$, where ...
2
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2answers
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For $p$ prime, show that $\Phi_{p^n}(x) = 1 + x^{p^{n-1}} + x^{2p^{n-2}} + \dots +x^{(p-1)^{p^n-1}}$

Well I attempted to try this but I failed to solve it: So $$\Phi_{p^{n}}(x)= \frac{x^{p^{n}} - 1}{\Phi_{1}(x) \Phi_{p^{2}}(x) \dots \Phi_{p^{n-1}}(x)}$$ Now I'm just stuck here. I saw a result ...
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0answers
37 views

Natural density of solvable quintics

A recent question asked about the topological density of solvable monic quintics with rational coefficients in the space of all monic quintics with rational coefficients. Robert Israel gave a nice ...
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0answers
62 views

Automorphism(Galois groups) and galois theory

I've been stuck on two last parts for two different questions, can someone please help me with these. The first question is: Let $\sigma\in Aut(L/\mathbb{Q})$, where $L$ is some subfield of ...
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0answers
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Galois group is $S_{n}$

If $K/F$ is Galois extension with Galois group $S_{n}$ then show that $K$ is the splitting field of a degree $n$ polynomial irreducible over $F$. We know $K$ is splitting field of some separable ...
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1answer
43 views

Degree 4 extension of $\mathbb {Q}$ with no intermediate field

I am looking for a degree $4$ extension of $\mathbb {Q}$ with no intermediate field. I know such extension is not Galois (equivalently not normal). So I was trying to adjoin a root of an irreducible ...
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0answers
23 views

Krull's Topology

Let $f_i : Gal(L/K) \mapsto Gal(L_i/K)$ the restriction $\sigma \mapsto \sigma_i=\sigma |_{L_i}$ and $\mathfrak{L}=(L_i)$ the system of a finite intermediate Galois extensions of $K$. We know this ...
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About Galois Covering Theory

so I am studying somethings about Galois Covering and I am writing a beamer to present for my friends of the university. But I would like of somethings about the author of Covering Galois Theory to ...
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0answers
45 views

Order of conjugate map

Let $\sigma :L\rightarrow L$, s.t $\sigma (\alpha)=\bar{\alpha}$. I've been asked to show that $\sigma\in Aut(L/K)$(the set of all automorphisms for the field extension) has order 1 or 2. I'm ...
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1answer
24 views

Order of automorphism = Order of field extension property

I've read that if $L/K$ is a field extension and $|Aut(L/K)|=|L/K|$, then L/K is a galois extension. I was wondering whether the converse is true, i.e if $|Aut(L/K)|\neq |L/K|$, then can we just ...
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2answers
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Are most rational quintics unsolvable?

It is well-known that, as polynomials of degree exceeding 4, there exist quintics whose roots cannot be solved for by radicals (Abel-Ruffini theorem). So we can divide the set of rational quintics ...
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1answer
24 views

Is Q(4th root(2))/Q a galois extension

I'm having some difficulty with the definition for galois extensions. The definition as read from my notes is $L/K$ is galois if $L^{Aut(L/K)}= K$. Where $Aut(L/K)$ is definied to be the set of all ...
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1answer
44 views

Degree of splitting field less than n!

I've been asked to prove that if a function $f\in \mathbb{Q}$ has degree $n$, then the splitting field of $f$ has degree less than or equal to $n!$. That is $\mathbb{Q}(\alpha_1, ...
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0answers
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Galois Group of a rational polynomial

How could I do to calculate the Galois group of a polynomial with rational coefficients ? Have you some links or books where I could find an answer ?
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Let $M/N$ be a Galois Extention where $L_1$ and $L_2$ intermediate subfields in between.

If $L_1/N$ is Galois show that: $Aut(L_1L_2/L_2) \cong Aut(L_1/L_1 \cap L_2)$ I attempted to try this question and I'm stuck on something. Since $L_1/N$ is Galois, we have that it is the splitting ...
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3answers
84 views

Calculating automorphism groups

I've been given a field $L=\mathbb{Q}(\sqrt[4]{2})$ and I have to find $Aut(\frac{L}{Q})$. Now i know that $\sqrt[4]{2}$ has minimal polynomial $x^4 -2$ over $\mathbb{Q}$, hence $[L : \mathbb{Q}] = ...
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1answer
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The absolute Galois group of a finite field is strongly complete

Let $k$ be a finite field. I am trying to prove that the absolute Galois group of $k$, i.e., $G = \operatorname{Gal}(\bar{k} / k)$ where $\bar{k}$ is an algebraic closure of $k$, is strongly ...
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0answers
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Normal extensions problem in Lang

This is a problem in Lang's Algebra. $F$ is finite normal extension over $k$ and $f(x)$ is irreducible in $k[x]$. If $f(x)=g(x)h(x)k(x) \in F[x]$ where $g(x),h(x)$ are monic irreducible factors in ...
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4answers
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Why can we prove mathematically that a formula to solve an (n+5) order polynomial does not exist?

I understand that the quadratic equation can solve any second order polynomial. Furthermore, equations exist for polynomials up to fourth order. However, without a graduate level degree and a deep ...
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P. Morandi on p-closure of a field

I am stuck on a step of the proof of Lemma 18.4 of Patrick Morandi, Field and Galois Theory. Let $p$ be a prime number and let $F$ be a field with $\mbox{char}(F) \neq p$. Morandi defines the ...
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0answers
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Is there a direct proof that pi is not the root of an algebraic equation whose degree is a power of 2 [duplicate]

All known proofs that the circle cannot be squared are based on Lindemann's theorem that $\pi$ is not analgebraic number. But this seems to be a case of using an atomic bomb to kill a fly. What ...
2
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1answer
15 views

Find a positive integer $n$ such that there is a subfield of $\mathbb{Q}_n$ that is not a cyclotomic extension of $\mathbb{Q}$

I'm in trouble with this simple exercise. If we denote $\mathbb{Q}_n:=\mathbb{Q}(\omega)$, where $\omega$ is a primitive $n$th of unity in $\mathbb{C}$, can we find a positive integer $n$ such that ...
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1answer
23 views

Let $G/H$ be a Galois extension and let $N_1$ and $N_2$ be subfields between $G$ and $H$

Show that: $Gal(G/(N_1N_2)) = Gal(G/N_1) \cap Gal(G/N_2)$ Um, one direction seems pretty obvious by definition: I believe it's that $Gal(G/N_1) \cap Gal(G/N_2) \subseteq Gal(G/N_1N_2)$ Now I have ...
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0answers
36 views

Is there any irreducible polynomial in $\mathbb{Q}$ whose Galois group is S_4?

As simple as that: Can we find an irreducible polynomial in $\mathbb{Q}$ such that, if $K$ is its splitting field over $\mathbb{Q}$, $Gal(K|\mathbb{Q})\cong S_4$? I've thought a lot and I have not ...