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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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 ...
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$\{1,\sqrt [3]{7},{\sqrt [3]{7 }}^{2}\}$ 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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1answer
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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
32 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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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
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1answer
83 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 ...
4
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1answer
108 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
37 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
30 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
41 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 ...
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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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1answer
136 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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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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1answer
33 views

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
59 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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29 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
55 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
30 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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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
48 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
47 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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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
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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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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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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
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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
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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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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 ...
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1answer
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Galois extensions questions

I'm working on answering this question but I'm unsure about alot the way I'm going about with the answers. The question is: Let L be a subfield of $\mathbb{C}$. a) Show that $\mathbb{Q}\subseteq ...
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Map from Galois extension to its third tensor power

Suppose $L/K$ is a finite Galois extension of fields, with group $G$. Then $G$ acts on $\mathrm{Spec}(L)$, which has the structure of a $G$-torsor over $\mathrm{Spec}(K)$. At the level of fields, this ...
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1answer
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Kummer-Dedekind's factorisation theorem

For a number field extension $K$ of $\mathbb{Q}$ one can factorise almost all prime ideals $(p)$ in the extension $K$, except finitely many, easily by factorising minimal polynomials in finite ...
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1answer
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Prove that $a$ is a $p$th power in $k$ if only if it is in $K$

Let $k\subset K$ be an extension having degree $[K:k]=n$ coprime to $p$. Prove that $a$ is a $p$th power in $k$ if only if it is in $K$ This is a problem in Galois theory - Miles Ried. I'm learning ...
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1answer
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Separable field extensions are Frobenius algebras

Wikipedia says that if $L/K$ is finite extension then $L/K$ is separable if and only if $L$ is a separable $K$-algebra. I am interested in the "only if" direction, which is outlined in the article. ...
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1answer
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Sppliting field and Galois theory and its Automorphism group

I'm studying elementary Galois theory and came across these two questions: If $L = Gal(x^n-1, \mathbb{Q})$ then $Aut_{\mathbb{Q}} L$ is abelian. This question is followed by If $ L = ...
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Embedding of a field in a cyclic extension

Show that $K=\mathbb{Q}(\sqrt {a})$ for $a\in \mathbb{Z}$, $a<0$ can not be embedded in a cyclic extension whose degree over $\mathbb{Q}$ divisible by 4. I have tried for order exactly ...
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What is the overall idea of Galois theory?

I am a third year undergraduate, doing a course on Field and Galois theory. Now, while I seem to understand most of the concepts locally, I do not seem to get the 'Whole picture' of what is happening ...
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Norm of an element in cyclotomic extension (Exercises VI.19 Lang's Algebra)

Let $\zeta$ be a primitive $n^{\rm{th}}$ root of unity. Let $K=\mathbb{Q}(\zeta)$. If $n=p^r (r\geq 1)$ is a prime power, show that $N_{K/F}(1-\zeta)=p$ If $n$ is divisible by at least two distinct ...
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1answer
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Equivalent conditions of a Galois extension (Exercise VI.4 in Lang's Algebra)

let $k$ be a field of characteristic $\neq 2$. Let $c\in k, c\notin k^2$. Let $F=k(\sqrt{c})$ . Let $\alpha=a+b\sqrt{c}$ with $a,b\in k$ not both $a,b=0$. Let $E=F(\sqrt{\alpha})$. Prove that the ...
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3answers
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How is $Gal(\mathbb Q(\sqrt{2+\sqrt{2}})/\mathbb Q)\cong \mathbb Z/4\mathbb Z$?

$\def\Gal{\operatorname{Gal}}$ I was working on homework, and the problem starts off by saying that I previously showed (I can't find where, though) that with $\def\Q{{\mathbb Q}}\def\Z{{\mathbb ...
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1answer
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The galois group of the polynomial $x^9+x^3+1$

What is the galois group of the polynomial $x^9+x^3+1$? Moreover give the bijection between subgroups and intermediate fields. Progress I think the order of the group is 108. But, there are many ...
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Normal basis theorem

Let $K$ be a finite Galois extension of, say, $\mathbb{Q}$. Then is known(and called normal basis theorem) that if i view $K$ as a representation of $Gal(K/\mathbb{Q})$ over $\mathbb{Q}$ it is ...
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1answer
25 views

Prove that if $f\in R[X]$ , then $\displaystyle\prod _{\sigma \in G}f^{\sigma}\in \mathbb{Z}[X].$

Let $K$ be an algebraic number field and $R$ be the ring of algebraic integers of $K.$ Denote by $h^{\sigma}$ the polynomial obtained from $h\in K[X]$ after applying to its coefficients the ...
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Showing an algebraic element of $\bar{\mathbb{Q}}$ is in $\bar{\mathbb{Q}}$ [duplicate]

This question really has me stumped. We define $\bar{\mathbb{Q}}$ to be the set of elements in $\mathbb{C}$ that are algebraic over $\mathbb{Q}$. I have shown that this is a field. Now, I'm ...
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29 views

how much certain mathematician has been cited in recent articles

I want to know how much Galois work has been cited in recent articles. So I'm looking for a tool that can help me do that, more simply is there a tool to know how much certain mathematician has been ...
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2answers
27 views

Prove that the number of automorphisms in $\mathbb Q[\alpha]$ equals $1$ $(|Aut\mathbb Q[\alpha]|)=1$

Please, help me to understand this problem: Let $\alpha=\sqrt[3]{2}$ be a root of the polynomial $x^3-2$. a) Prove that the number of automorphisms in $\mathbb Q[\alpha]$ equals $1$ $(|Aut\mathbb ...