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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Questions about $\mathrm{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$
Consider the extension $\mathbb{Q} \subset\mathbb{Q} (\sqrt{2}, \sqrt{3})$. How many elements are there in $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})?$ Describe all elements in ...
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1answer
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Why $F(\alpha_1)∩F(\alpha_2)=F$ is false
$\def\Q{\mathbb Q}$Let $\alpha_1$ and $\alpha_2$ be conjugates over a field $F$ such that $\alpha_2 \not\in F(\alpha_1)$. Is true that $F(\alpha_2)\cap F(\alpha_2)=F$.
Here is my attempt:
Is false. ...
3
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1answer
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Question about notation in a theorem about Galois theory from Lang's Algebra (chapter 6 §1, corollary 1.16)
I have a question about the notation in an assertion in Lang's Algebra, chapter 6 §1, corollary 1.16:
Let $K/k$ be finite Galois with group $G$, and assume that $G$ can be written as a direct ...
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1answer
42 views
Galois Theory Basic Algebra I Jacobson
I'm having troubles with this problem. I saw it in Jacobson's "Basic Algebra I".
I hope you can help me.
-Show that $E=\mathbb{Q}(\sqrt{2}, \sqrt{3}, u)$ where $u^2=(9-5\sqrt{3})(2-\sqrt{2})$ is ...
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3answers
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Automorphism of $\mathbb Q (\sqrt[5]{2}, \zeta_5)$
Show that if $\sigma \in \mathrm{Aut}(\mathbb Q (\sqrt[5]{2}, \zeta_5)/\mathbb Q)$, then $\sigma (\zeta_5) = \zeta^i_5$ for some $i=1,2,3,4$ and $\sigma (\sqrt[5]{2}) = \zeta^j_5 \sqrt[5]{2}$ for ...
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0answers
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Irreducible polynomial roots and representations for Galois field elements in normal basis
I am trying to understand some topics in the literature and came across the following problem. Say I have a field $GF(2^4)$ defined by the irreducible polynomial $r(z) = z^4 + z + 1$, and I want to ...
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1answer
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Uniformly solvable families of polynomials
It is a famous theorem that there is no "quintic formula", i.e. there is no formula which expresses the roots of a quintic polynomial $x^5+a_4x^4+\cdots+a_0$ in terms of $a_4,\ldots,a_0$ and rational ...
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3answers
36 views
automorphisms and field extension $E$ of $\mathbb{Q}$.
I want a hint. That is all I ask for. The question I am asked to prove is as follows:
Let $E$ be an extension field of $\mathbb{Q}$. Show that any automorphisn of $E$ acts as the identity on ...
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0answers
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Non-isomorphic simple extensions of the same degree of a field of positive characteristic
Let $K$ be a field of positive characterstic. I need to show that there exists $a$ and $b$ of the same degree over $K$ but $K(a)$ and $K(b)$ are not isomorphic.
I thought of an example where they are ...
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vote
1answer
21 views
Test for polynomial reducibility with binary coefficients
I'm learning about Galois Fields, in particular $GF(2^8)$, as they are applied to things like the AES algorithm and Reed-Solomon codes. Each of these rely on an irreducible 8th degree polynomial with ...
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2answers
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Find a monic polynomial over $\mathbb{Q}$ whose Galois group is $V_4$
I need to find a polynomial over $\mathbb{Q}$ whose Galois group is $V_4=\langle(12)(34),(13)(24)\rangle\subset S_4$. I can find examples of such polynomials, but I wonder if one can construct such a ...
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2answers
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GgT, (polynomial) division and finite fields…
Exercise: Let $f,g \in \mathbb{Z}_2[x]$ be the polynomials $f = x^6 + x^5 + x^4 + 1$ and $g = x^5 + x^4 + x^3 + 1$. Has the diophantic equation $f u + g u = x^4 + 1$ solutions $u,v \in \mathbb{Z}[x]$? ...
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2answers
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Galois Extensions and $n^{\text{th}}$ Roots
I've been studying for my prelims lately, and this problem has me stuck:
(a) Let $K$ be a field with no abelian Galois extensions. Suppose that $n$ is a positive integer and either $char(K)=0$ or ...
5
votes
1answer
42 views
finding fixed field of automorphism
Let $F$ be a field and let $g:F(x) \to F(x)$ be the automorphism which maps $x$ to $x+1$.
I need to find the fixed field of this automorphism.
So far I know $g$ fixes $F$. I want to use Galois ...
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votes
3answers
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Automorphism of $K$ extending to $K[x_1,\dots,x_n]$.
I think it is well known that if $K$ is field, and $\sigma \in Aut(K)$,
then the extended map $\sigma: K[x,y] \rightarrow K[x,y]$, given by
$\sum a_{ij}x^iy^j \mapsto \sum \sigma(a_{ij})x^iy^j$ is ...
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0answers
37 views
Show that an field extension is algebraic (normal).
Let $A/K$ be a field extension, I wanted to proof:
$A/K$ is normal iff for every irreducible polynomial $P \in K[x]$ which has a root in $A$, the field extension $A$ contains a splitting field for ...
3
votes
2answers
36 views
Solving equations by radicals
Let $\zeta$ be a complex number, $\zeta\neq 1, \zeta^3=1$. Then the expression
$$(x_1+\zeta x_2+\zeta^2x_3)^3$$
takes only two distinct values when we permute $x_j$'s. $\bf{Why\ this?}$
Hence it ...
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1answer
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Galois group of $x^8+2$
This is what I did to find the Galois group for $x^2+8$:
Splitting field: $K = \Bbb{Q}(\zeta_8, \zeta_{16}2^{1/8})$
Since $Gal(\Bbb{Q}(\zeta_8)/\Bbb{Q}) \cong Aut( \langle\zeta_8\rangle) \cong ...
1
vote
1answer
91 views
Calculating The Galois Group of the Splitting Field of $f=x^3-3$
If we let $f=x^3-3$ the let L be the splitting field for this polynomial I am trying to find $\Gamma(L:\mathbb{Q})$ and all intermediate field extensions.
Now as this is a splitting field and finite ...
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votes
1answer
31 views
Understanding how the Galois group acts on the character of a group representation
I'm trying to understand a proof sketch found here: http://mathoverflow.net/questions/10635/why-are-the-characters-of-the-symmetric-group-integer-valued
If $g$ is an element of order $m$ in a ...
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0answers
54 views
Galois representations and normal bases
I am not very familiar with the theory of Galois representations, but I do know a bit about both Galois theory and representation theory. Recently I learned about the notion of a normal basis for a ...
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Galois group of $x^8+2$.
For the splitting field of $x^8+2$, we have $\Bbb{Q}(\zeta_8, 2^{1/8})$, right? The minimal polynomial of $2^{1/8}$ over $\Bbb{Q}(\zeta_8)$ has degree 4. However, since there are no real roots of ...
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0answers
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Finding fixed fields and orders of elements.
The questions asks me to find a subgroup of order 8 and another of order 2 in the Galois group of $X^8-2$ over $\Bbb{Q}$ (to show that G is the internal semidirect product of them). It then asks ...
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Is simple extension $\mathbb{Q}(\cos\pi/9):\mathbb{Q}$ algebraic and normal?
I am trying to prove this question:
Q: show that the simple extension $\mathbb{Q}(\cos\pi/9):\mathbb{Q}$ is an algebraic extension and also normal extension?
Thank you
4
votes
1answer
46 views
A Galois Theory Question
Fix a prime $p$ and consider the equation $X^p-X-t^{-1}$ over $\mathbb F_p((t))$, the field of formal Laurent series over $\mathbb F_p$. What is the Galois group of this equation?
After fumbling ...
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1answer
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Finding the Galois group of $X^8-2$ over Q.
If K is the splitting field of $X^8-2$ over $\Bbb{Q}$, I want to find the galois group.
We know that $K=\Bbb{Q}(2^{1/8}, \zeta_8)$.
So first I want to look at $Gal(\Bbb{Q}(\zeta_8)/\Bbb{Q})$ and ...
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vote
1answer
83 views
Galois Group of $f(x)=x^4 - 10 x^2 + 1$
I am trying to calculate the Galois group of $f(x)=x^4 - 10 x^2 + 1\in\mathbb Q[x]$ over $\mathbb Q$.
In my notes it says that the four roots are $\pm\sqrt 2\pm\sqrt 3$. So the splitting field of ...
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votes
2answers
65 views
Show $\zeta_p \notin \mathbb{Q}(\zeta_p + \zeta_p^{-1})$
I asked a question here: [Writing a fixed field as a simple extension of $\mathbb{Q}$ ], but realised I couldn't justify why the given quadratic was irreducible.
Thus: Is there a way of showing ...
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1answer
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Writing a fixed field as a simple extension of $\mathbb{Q}$
I have $G = Gal(\mathbb{Q}(w) : \mathbb{Q})$ where $w = e^{2\pi i/p}$ for $p$ prime.
I have that there exists a (unique) element in $G$ of order 2, say $\phi$.
I'm trying to express the fixed field ...
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1answer
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The Fundamental Theorem of Galois Theory.
Let E/F be a finite Galois extension with Galois group G. If H is a subgroup of G,
let F(H) be the fixed field of H,and if K is an intermediate field,let G(K) be Gal(E/K),
the fixing group of ...
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1answer
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question on Galois theory
Can anybody help me with the following question ?
I start with a number field $F/\mathbb{Q}$ which is abelian (that is, a Galois extension of abelian Galois group). I know by the Kronecker-Weber ...
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votes
2answers
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If $f(x)$ is an irreducible polynomial of degree n, then the cardinality of its Galois group is divisible by $n$.
If $f(x)$ is an irreducible polynomial of degree $n$, then the cardinality of its Galois group is divisible by $n$.
I know I need to use the Tower Theorem, but I can't figure out how to get from ...
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vote
1answer
40 views
Compute the Automorphisms of tower of fields
Suppose we have the following tower of fields: $\mathbb Q \subset \mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})$. Compute Aut$(\mathbb Q(\sqrt{2})/\mathbb Q)$, Aut$(\mathbb Q(\sqrt[4]{2})/\mathbb ...
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votes
3answers
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Is $[L : K] = 2$, $f \in K[x]$ irreducible, then $\operatorname{deg}(f) \le 2$ valid?
Is it true that in every field extension of degree two, every irreducible polynomials has a degree smaller than two? And if so, how can I proof this?
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2answers
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If $[L : K ] = n$, then for every irreducible polynomial $f$ is $\operatorname{deg}(f) \le n$.
If a field extension is finite with degree $n$, how can I proof that every irreducible polynomial has a degree smaller then $n$. I guess that this is valid, but I am not sure how to proof this?
2
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1answer
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A basic question on factorization
Is the following true? If not, can anyone add some reasonable assumptions to make it true?
Statement. Let $E/F$ be a field extension and let $\alpha \in E \setminus F$ be algebraic over $F$. If $f(x) ...
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vote
2answers
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Galois Group of an irreducible cubic
I need to prove that if the Galois group of an irreducible cubic over $\mathbb Q$ is $\{id, \sigma, \sigma^2\}$, then all the roots of the cubic are real.
How do I even start this?
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0answers
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A question regarding linear disjiontness and the degree of a field extension
Let $K / k$ and $L / k$ be field extensions with $K$ and $L$ linearly disjoint over $k$. Suppose that $K' / K$ is an extension of finite degree, and that all fields are characteristic zero.
Then is ...
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2answers
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existence of an automorphism of $k^a$ whose fixed field is $k$
Let $k$ be a field such that every finite extension is cyclic. Show that there is an automorphism of $k^a$ over $k$ whose fixed field is $k$. Here $k^a$ is the algebraic closure of $k$.
P.S. It's a ...
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1answer
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Calculate a galois group
I am trying to calculate the galois group $\operatorname{Gal}( \mathbb{Z}_q (\vartheta_p) : \mathbb{Z}_q) $, where $p$ and $q$ are different primes, $\mathbb{Z}_q$ $q$-adic ring, $\vartheta_p$ a ...
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1answer
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There could no multiplication on a three dimensional vector space defined similar to complex multiplication, but what an field extension of degree 3?
According to
Wikipedia
there could be no multiplication on a three dimensional vector space like the multiplication over the complex numbers. But what about field extensions of degree three, in these ...
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1answer
159 views
A question regarding the finiteness of the degree of a field extension
Let $x$ and $y$ be transcendental and $y$ be algebraic over $\mathbb{Q}(x)$. Let $F$ be an algebraic Galois extension of $\mathbb{Q}(x)$ of infinite degree and let $\mathbb{Q}^{al}$ be the algebraic ...
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votes
2answers
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If all embeddings of $E/F$ are automorphisms, then $E/F$ is normal
Let $E$ be a (possibly infinite) field extension of $F$, and let $\Omega$ be an algebraic closure of $E$. I'm trying to prove that if $\sigma(E)=E$ for all $F$-algebra embeddings $\sigma \colon E\to ...
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Field-theoretic description of fixed field of central subgroups?
Given a Galois extension $E/F$ with Galois group $G$, and a subextension $E/K$ with Galois group $H$, is there a "field-theoretic" characterization of when $H$ is central (i.e. $H\leq Z(G)$)? By ...
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2answers
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3 questions on field extensions
I am trying to figure out some things regarding field extensions and some questions have arisen on the way.
Let $a$ be a positive integer which doesn't have a rational $nth$ root:
Is the splitting ...
4
votes
3answers
115 views
Quadratic subfield of cyclotomic field
Let $p$ be prime and let $\zeta_p$ be a primitive $p$th root of unity. Consider the quadratic subfield of $\mathbb{Q}(\zeta_p)$. For instance, for $p=5$ we get the quadratic subfield to be ...
3
votes
1answer
76 views
Why don't I end up with the same splitting field?
I've understood that the splitting field of $x^4+2$ and the splitting field of $x^4-2$ over $\mathbb{Q}$ are both the field $\mathbb{Q}(\sqrt[4]{2} , i)$. With degree $8$ over $\mathbb{Q}$. This ...
2
votes
1answer
43 views
Splitting field for $x^n+a$
What is a splitting field $E$ for $f(x)=x^n+a$ over the field $K$ of characteristic zero?
If I put $g(x)=x^{2n}-a^2=(x^n-a)(x^n+a)$. The splitting field $F$ of $g(x)$ is $K(\sqrt[n]{a},\alpha )$ ...
3
votes
1answer
72 views
Galois group of irreducible quartic with real coefficients
Let $K$ be a subfield of the real numbers and $f\in K[x]$ be an irreducible quartic. If $f$ has exactly two real roots, show that the Galois group of $f$ is $S_{4}$ or $D_{4}$ (I'm using the ...
8
votes
1answer
99 views
Galois group of $X^5 - X^3 - 2X^2 - 2X - 1$ over $\mathbb{Q}$.
So far I have found that this polynomial is irreducible, and its discriminant is a square. Therefore the Galois group it a transitive subgroup of $A_5$.
I found out that the only transitive ...



