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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Complex conjugation in the Galois group of a polynomial

Suppose $P$ is an irreducible polynomial in $\mathbb Q[X]$, with exactly two non-real roots. Then we know these roots must be complex conjugates. Why must complex conjugation be an element of ...
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Galois group of a cubic over the function field $\mathbb Q(y)$

Given the polynomial $x^{3} - 3yx + 2 = 0$ with coefficients in $\mathbb Q(y)$ where $y$ is an indeterminate, which has discriminant $108(y^3 - 1)$, what is the Galois group of this polynomial? ...
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1answer
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Galois group of $X^4 + 4X^2 + 2$ over $\mathbb Q$.

I'd like to calculate the Galois group of the polynomial $f = X^4 + 4X^2 + 2$ over $\mathbb Q$. My thoughts so far: By Eisenstein, $f$ is irreducible over $\mathbb Q$. So $\mathrm{Gal}(f)$ must be ...
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3answers
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Is there a quick way to compute the degree of the splitting field of $x^3+x+1$ over $\mathbb{Q}$?

Is there a way to find the degree of the splitting field of $x^3+x+1$ over $\mathbb{Q}$? Just analyzing the roots shows that the polynomial is separable, so I suppose the splitting field would be a ...
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1answer
288 views

Calculating $\prod (\omega^j - \omega^k)$ where $\omega^n=1$.

Let $1, \omega, \dots, \omega^{n-1}$ be the roots of the equation $z^n-1=0$, so that the roots form a regular $n$-gon in the complex plane. I would like to calculate $$ \prod_{j \ne k} (\omega^j - ...
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Examples of Galois connections?

On TWF week 201, J. Baez explains the basics of Galois theory, and say at the end : But here's the big secret: this has NOTHING TO DO WITH FIELDS! It works for ANY sort of mathematical gadget! If ...
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Original works of great mathematician Évariste Galois

Through this question I wanted to know the original works of Galois. When I was reading Galois theory ( since from last month ) , I have been seeing one common line in every book, whose essence ...
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1answer
318 views

Derivations and separability of field extensions

This is written on page 4 of James E. Humphreys' Linear Algebraic Groups: A derivation $\delta: E \rightarrow L$ ($E$ a field, $L$ an extension field of $E$), is a map which satisfies $\delta(x+y) ...
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1answer
419 views

Galois group of a non-separable polynomial

In my notes, I am given the definition of the Galois group of a polynomial only in the case when the polynomial is separable (if $f$ is a separable polynomial over $K$ with splitting field $L$, then ...
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731 views

Algebraic conjugates

Suppose $L/K$ is an algebraic field extension. Take $\alpha_1 \in L$. Then $\alpha_1$ has minimal polynomial $f(x)$ over $K$. Let $\alpha_2, ... \alpha_k$ be the other roots of $f$ in $L$. The ...
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The implications of the insolvability of certain polynomials

So I've just been over a bit of Galois theory, and I'm trying to understand what the implications are for a polynomial's Galois group to not be solvable. My book says this means that there is no ...
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Basic example of extensions of residue fields.

Can anyone think of a simple example of the following: $B/A$ is an integral extension of DVRs with quotient fields $L$ and $K$ and residue fields $\bar{L}$ and $\bar{K}$, $L/K$ is finite dimensional ...
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1answer
407 views

Product in GF(16)

i need some help with a product in GF(16), where it is seen as an extension of GF(4)={0, 1, x, x+1} (where $x^2 = x + 1$ ) with the irreducible polynom $f(y) = y^2 + y + x$ So the elements in the ...
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1answer
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Splitting field implies Galois extension

I hope this isn't too elementary of a question, but I'm not sure I understand Artin's proof that if $K/F$ is a finite extension, then $K/F$ Galois is equivalent to $K$ being a splitting field over $F$ ...
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“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: The Galois group of $\pi$ is $\mathbb{Z}$. In what sense/framework is ...
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1answer
174 views

Counting endomorphisms of $\mathbf Q(\zeta _{n})$

If $\zeta= \zeta_{n}$, how does one count the homomorphisms $f:\mathbf{Q}(\zeta)\rightarrow \mathbf{Q}(\zeta)$?
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1answer
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For $n\ge 3, x_{1},…,x_{n} \in \mathbf{Q}^{\ast}$, $[\mathbf{Q}(\sqrt{x_{1}},…\sqrt{x_{n}}) : \mathbf{Q}] < 2^{n}$

For $n\ge 3, x_{1},...,x_{n} \in \mathbf{Q}^{\ast}$ and $[\mathbf{Q}(\sqrt{x_{1}},...\sqrt{x_{n}}) : \mathbf{Q}] < 2^{n}$ how can we conclude that there are non empty $I \subset \{1,...,n\}$ with ...
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Can a regular heptagon be constructed using a compass, straightedge, and angle trisector?

Euclid has a magical compass with which he can trisect any angle. Together with a regular compass and a straightedge, can he construct a regular heptagon?
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How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?

This is Exercise 18.14 from Algebra, Isaacs. $p_{1}\ ,\ p_{2}\ ,\ ... p_{n}$ are different prime numbers. How to show that $$\mathbb{Q}[\sqrt{p_{1}}, \sqrt{p_{2}}, \ldots, \sqrt{p_{n}} ] = ...
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1answer
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The smallest Galois extension of $\mathbb{Q}(x^3)$ containing $\mathbb{Q}(x)$

What's the smallest Galois extension of $\mathbb{Q}(x^3)$ containing $\mathbb{Q}(x)$? For example, let E be the smallest Galois extension of $\mathbb{Q}(x^3)$ containing $\mathbb{Q}(x)$. Then, do i ...
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1answer
219 views

Galois group over $\mathbb Q$ is $\mathbb Z/2\mathbb Z$

For each integer $n \geq 2$ , find a polynomial of degree $n$ with non-rational roots, whose Galois group over $\mathbb{Q}$ is $\mathbb{Z}/2\mathbb{Z}$. Anybody can help me?
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1answer
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Maple: Galois Field with characteristic 2 $\mathrm{GF}(2^m)$ - how to convert polynomials into binary vectors and vice versa?

In GF($2^m$) each element is a polynomial also it is a binary message. I wonder how to make maple help me convert vectors of maple Bits into GF elements and back? Also it is not the question of ...
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Condition that $\mbox{Char}(K) $ doesn't divide $m$ in the definition of $m$th cyclotomic extension of $K$

This should hopefully be a very simple question: "Suppose $\mbox{char}K = 0$ or $p$, where $p\not| \ m $. The $m$th cyclotmic extension of $K$ is just the splitting field $L$ over $K$ of $X^m - 1$" ...
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2answers
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Constructing an explicit isomorphism between finite extensions of finite fields

Suppose $K$ is a finite field, $K = \mathbb F_{p^s}$. If we take an irreducible polynomial $f$ of degree $d$ over $K$, then the splitting field $L$ of $f$ is $K(\alpha)$ where $f$ is the minimal ...
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1answer
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Splitting fields of polynomials over finite fields

I can't follow a statement in my notes: "Let $K$ be a finite field, with $f \in K[X]$ an irreducible polynomial of degree $d$. Then any finite extension $L/K$ is normal, and so if $L$ contains one ...
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1answer
65 views

Notation question involving discriminant identity

Let $f \in K[X] $ be an irreducible, separable polynomial, and let $M/K$ be a splitting field for $f$. Let $\alpha \in M$ be a root of $f$. Then $D(f) = (-1)^{d(d-1)/2} N_{K/k} (f'(\alpha))$, ...
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3answers
527 views

Relationship between discriminant of a separable polynomial and its Galois group

Probably a straightforward question: Let $f \in K[X]$ be a polynomial with disctinct roots $ \alpha_1, ..., \alpha_d$ in a splitting field $L$. Set $\Delta = \Pi_{i <j} (\alpha_i - \alpha_j) $. ...
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1answer
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surjectivity of group homomorphisms

I don't know if the next thing is true, but I'm not able to find a counterexample: suppose you have a surjective group homomorphism of finite groups $f:G \rightarrow G'$ and normal subgroups $H ...
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Addition and multiplication in a Galois Field

I am attempting to generate QR codes on an extremely limited embedded platform. Everything in the specification seems fairly straightforward except for generating the error correction codewords. I ...
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1answer
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Splitting field of a separable polynomial is separable

Probably a stupid question, but.. Why is the splitting field of a separable polynomial necessarily separable? Thanks. Follow up question Show that if $F$ is a splitting field over $K$ for $P \in ...
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1answer
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Action of a finite automorphism group on a field

Let $G$ be a finite group of automorphisms acting on a field $L$, with fixed field $K$. My notes say "Let $ \alpha \in L $. Consider the set $ \{\sigma(\alpha)| \sigma \in G\}$ and suppose its ...
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1answer
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Prove that if $[L:K] = 4$ and $ \mbox{Aut}(L/K) \cong C_2 \times C_2 $ then $ L = K(\sqrt{a},\sqrt{b}) $

Prove that if $[L:K] = 4$ and $ \mbox{Aut}(L/K) \cong C_2 \times C_2 $ then $ L$ is of the form $K(\sqrt{a},\sqrt{b}) $. I know that the extension is Galois, and so I can use the Galois ...
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0answers
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Galois group of solvable quintic is subgroup of Fr20

Why is it true that any solvable quintic polynomial in has a Galois group that is a subgroup on the Frobenius group of order 20? Thanks in advance.
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0answers
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Can a subgroup $ \mbox{Gal}(L/K) $ fix $K$?

Question essentially in the title. I know that if $ K \subseteq L $ is a finite extension, then $L/K$ is Galois if and only if the fixed field of $ \mbox{Aut}(L/K)$ is $K$. But is it possible for ...
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3answers
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Recognizing when a tower of Galois extensions gives a Galois extension

For the easiest case, assume that $L/E$ is Galois and $E/K$ is Galois. Under what conditions can we conclude that $L/K$ is Galois? I guess the general case can be a bit tricky, but are there some ...
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2answers
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Number of embeddings $ \tilde{\theta} : K(\alpha) \to L $ extending $\theta : K \to L $

Proposition Let $ K(\alpha)/K $ be a finite simple extension, with $ f \in K[X] $ the minimal polynomial for $ \alpha$. Given a field extension $ \theta : K \to L $, the number of embeddings $ ...
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1answer
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How can one compute the Galois group of this unusual extension?

Been reviewing some Galois theory, and computing Galois groups has been relatively routine, but one has me stumped. Suppose $\omega$ is a primitive 37th root of unity. Let's set ...
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2answers
427 views

Are quartic minimal polynomials over $\mathbb{Q}$ always reducible over $\mathbb{F}_p$?

This situation arose while studying biquadratic extensions. Let $\mathbb{Q}(\alpha)$ is some biquadratic extension, with $m(x)$ the minimal polynomial of $\alpha$. Suppose that ...
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1answer
495 views

Finding some missing subfields of a splitting field of $x^4-7$

I was looking at the Galois group of the splitting field of $x^4-7$ over $\mathbb{Q}$. I found it to be $\mathbb{Q}(\sqrt[4]{7},i)$, and the Galois group to be the dihedral group of order $8$. Now ...
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Why does an irreducible polynomial split into irreducible factors of equal degree over a Galois extension?

I've been struggling to prove this fact over the past day or so. Suppose $f(x)\in F[X]$ is irreducible over a field $F$ with $\deg(f)=n$, and let $L$ be the splitting field of $f(x)$ over $F$ with ...
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1answer
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Inertia groups generate Galois Group

While reading a paper about the Kronecker-Weber Theorem, I noticed a theorem saying that for a Galois extension $K/\mathbb{Q}$, its Galois group is generated by $I_p$s, being the inertia groups of ...
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1answer
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$\mathbb{Q}(\sqrt{1-\sqrt{2}})$ is Galois over $\mathbb{Q}$

I am trying to show that $E = \mathbb{Q}\left(\sqrt{1-\sqrt{2}}\right)$ is galois over $\mathbb{Q}$. The extension has the minimal polynomial ...
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1answer
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Inclusion among subextensions of a cyclic field extension

I have some troubles with this problem: Let $E/F$ be a Galois extension with Galois group cyclic. Prove that two intermediate $B_1$ and $B_2$ satisfy $B_1\subseteq B_2$ if $[E:B_2]$ divides ...
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1answer
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Question about field norm from Dummit and Foote 14.2.17(d)

I'm trying to solve the last part of an exercise in Dummit and Foote. Let $K/F$ be any finite separable extension, and let $\alpha\in K$. Let $L$ be a Galois extension of $F$ containing $K$ and ...
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343 views

Normal extension of any degree

Could any one tell me how to show that for any positive integer n, there exists a normal extension of rational number field of degree n?
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1answer
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What is the Galois group of the splitting field of $X^8-3$ over $\mathbb{Q}$?

I've computed the splitting field of $x^8-3$ over $\mathbb{Q}$ to be $\mathbb{Q}(\sqrt[8]{3},\zeta_8)=\mathbb{Q}(\sqrt[8]{3},\sqrt{2},i)$, which is of degree 32 over $\mathbb{Q}$. The possible ...
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2answers
156 views

Chain of fields that are Galois over all subfields

Is there an example of fields $F_1$, $F_2$, and $F_3$ such that $\mathbb{Q}\subset F_1\subset F_2\subset F_3$ such that $[F_3:\mathbb{Q}]=8$ and each field is Galois over all its subfields but $F_2$ ...
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2answers
660 views

Galois extensions of degrees $p$ and $p^{n-1}$ given a Galois extension of $p^n$

Suppose $K$ is a Galois extension of a field $F$ of degree $p^n$ for a $p$ a prime. I want to see if there are Galois extensions of degrees $p$ and $p^{n-1}$ over $F$. If $G=\text{Gal}(K/F)$, then ...
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1answer
893 views

Fastest way to compute subfields of $\mathbb{Q}(\sqrt[8]{2},i)$ which are Galois over $\mathbb{Q}$?

I have the lattice of subfields of the splitting field $\mathbb{Q}(\sqrt[8]{2},i)$ over $x^8-2$, and the corresponding lattice of subgroups of the Galois group $G$ of the splitting field. I'm now ...
3
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1answer
222 views

Fixed field of Galois group of an infinite field $E$ is equal to $E$

I am trying to prove that if $E$ is an infinite field, then the fixed field of $Gal(E(x)/E)$ is $E$. The first part of the question was to find all automorphisms $$x\longmapsto ...