Questions tagged [galois-extensions]
For questions about Galois extensions fields. We say $L/K$ is a Galois extension iff the subfield of $L$ that is fixed by automorphisms of $L$ which fix K is exactly $K$.
429
questions
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What is the Galois group of the polynomial $f(x)=x^3-3$ over $\mathbb{Q}$?
What is the Galois group of the polynomial $f(x)=x^3-3$ over $\mathbb{Q}$ ?
$f(x)=0 \ $ gives $x= 3^{1/3},~ \zeta_3 3^{1/3},~ \zeta_3^2 3^{1/3}$ over $\mathbb{C}$.
Thus the permutation of these $3$ ...
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1answer
47 views
Determine $\operatorname{Aut}_{\mathbb{Q}}L$ for $L = \mathbb{Q}[u]$, where $u$ is a root of $x^3 - 3x^2 + 3$
Now, I broke this down into two cases:
Case 1 - the other roots (let's call them $u_2$ and $u_3$) of the polynomial (let's call it $h(x)$) are in $\mathbb{Q}[u]$. In this case, $L = \operatorname{Gal}(...
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votes
1answer
40 views
Let $f(x) \in \mathbb{Q}[x]$ be an irreducible polynomial of degree $4$ with exactly $2$ real roots.
Show that the Galois group of $f$ over $\mathbb{Q}$ is either $S_4$ or the dihedral group of order $8$.
So I'm studying from quals remotely, in a state hundreds of miles away from my university that ...
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votes
1answer
102 views
Let $K$ be a Galois extension of $\mathbb{Q}$.
Let $K$ be a Galois extension of $\mathbb{Q}$ whose Galois group is isomorphic to $S_5$. Prove that $K$ is the splitting field of some polynomial of degree $5$ over $\mathbb{Q}$.
Since $K$ is a finite ...
4
votes
2answers
101 views
Determine the Galois group of $x^3 + 3x^2 - 1$ over $\mathbb{Q}$
So the only possible roots in $\mathbb{Q}$ are $1$ or $-1$ and neither are roots. So all I rely know is that the group is isomorphic to $S_3$ or $A_3$ and the polynomial has no rational roots.
Also ...
5
votes
2answers
67 views
For a complex number $\alpha $ which is algebraic over $\Bbb Q$, determining whether $\bar{\alpha}\in \Bbb Q(\alpha)$ or not
Let $\alpha =3^{1/3}+3^{5/4}i$, which is clearly algebraic over $\Bbb Q$. How can we determine whether $\Bbb Q(\alpha)$ contains $\bar{\alpha}$ or not?
This would be certainly true if $\Bbb Q(\alpha)$ ...
3
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1answer
44 views
Why is $\text{Gal}(K/\mathbb{Q}) \cong G_{\mathbb{Q}}/{\{\sigma \in G_{\mathbb{Q}}: \ \sigma|_K=id_K \}}$?
Here, in page $1$, the absolute Galois group is defined by $$G_{\mathbb{Q}}:=\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})=\{\sigma: \bar{\mathbb{Q} }\to \bar{\mathbb{Q}}, \ \text{field automorphism} \}$$ ...
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1answer
26 views
Cyclic Field Extensions of sum of radicals
Given a field $K$ of characteristic 0, which contains a primitive root of order 3,
I would like to show that the extension $K(\sqrt2+3^{\frac{1}{3}})/K$ is cyclic.
My attempt was to look at the "...
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0answers
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Lemma about certain extension of a Galois automorphism
I would like to know if the lemma below is correct. In such a case, a reference would be also useful. Thanks
Lemma. Let $L / K$ be a finite Galois extension and let $g \in G(L / K)$ (the Galois group ...
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1answer
23 views
Non Abelian Normal Field Extension with Abelian Subextensions
It is known that a subextion $L/F/K$ of an abelian (Galois) field extension $L/K$ is also abelian. The converse is not true: even when assuming that $L/K$ is Galois and $L/F$ and $F/K$ are abelian, $L/...
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3answers
57 views
What is the intuition behind mapping of elements from $GF(2^8)$ to $GF(((2^2)^2)^2)$?
I'm finding it very difficult to understand the concept of mapping elements from the extension field $GF(2^8)$, to $(GF(2)^2)^2)^2 $. I realize that the field that the elements of the field, $GF(2^8)$,...
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2answers
25 views
Abelian extensions $E|F$ and $F|K$ $\implies E|K$? [closed]
Let $E,F,K$ fields such that $K\subset F \subset E$
If $F|K$ and $E|F$ are abelian extensions, then $E|K$ is an abelian extension?
I can't find a counter example
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1answer
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Proof of : If $K$ is a finite extension of $F$, then $o(G(K,F)) \leq [K:F]$
$G(K,F)$ is the group of all automorphisms of $K$ that leave $F$ fixed. $F$ is a field of characteristic zero. $[K:F]$ is the dimension of $K$ as a vector space over $F$.
Herstein proves this theorem (...
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46 views
Prove that $\sqrt{3} + (x^4 + 2x + 1) \not \in \mathbb{Q} [x]/(x^4 + 2x + 1)$
I know that $\theta = x + (x^4 + 2x + 1)$ is a root of $g(x) = x^4 + 2x + 1$. And $\{1,\theta,\theta^2,\theta^3\}$ is a basis for $\mathbb{Q} [x]/(x^4 + 2x + 1)$ over $\mathbb{Q} $.
If $\sqrt{3} + (x^...
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0answers
22 views
All roots of a polynomial in ring $F_2+uF_2+u^2F_2$, where $u^3=0$
Let $R=F_2+uF_2+u^2F_2$, where $u^3=0$, be a finite commutative ring. So $R=\{0,1,u,v,uv,u^2,v^2,v^3\}$, where $v=1+u$, $v^2=1+u^2$, $v^3=1+u+u^2$, $uv=u+u^2$. It is well known that
$$x^7-1=(x+v^3)(x^...
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2answers
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How to express the field $\mathbb Q(\sqrt[3]{2},\sqrt[3]{3})$ as a simple extension of $\mathbb Q$?
Iāve found $\mathbb Q(\sqrt[3]{2},\sqrt[3]{3},\zeta_3):\mathbb Q$ and the galois group $G=(C_3\times C_3)\rtimes C_2$ which has order 18. Now $\mathbb Q(\sqrt[3]{2},\sqrt[3]{3})$ is a subfield where $[...
0
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1answer
44 views
How to express a field L in the form $M(\sqrt{d})$?
If L is the splitting field of $t^{13}-1$,
then obviously $L=\mathbb Q(\zeta_{13})$.
Now an intermediate field $M=\mathbb Q(\zeta_{13}^{12}+\zeta_{13})$. How to express L in the form $M(\sqrt{d})$, ...
2
votes
1answer
86 views
Galois Theory without the Primitive Element Theorem
I have seen, in a few answers on MSE and in uploaded material from some courses, a proof of the primitive element theorem (PET) using Galois theory. It usually goes like this:
Let $F$ be a field and $...
2
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1answer
36 views
Group cohomology of Galois group of finite extension of finite fields
Let $E/F$ be a finite extension of finite fields; hence, it is a cyclic Galois extension, so let the Galois group be $G$. Hilbert's Theorem 90 states that $H^1(G, E^{\times})=0$.
My question is: How ...
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2answers
138 views
What is the Galois group of $(x^3-2)(x^3-3)$ over $\mathbb{Q}$? [duplicate]
I know that there is an answer in splitting field of $(x^3-2)(x^3-3)$ over $\mathbb Q$. I know the splitting field is $L=Q(Ī¶3,\sqrt[3]2,\sqrt[3]3)$. But I'm quite confused with [L:Q]=18. How can I get ...
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0answers
22 views
How can I prove that a finite extension $\mathbb{Q}(\alpha):\mathbb{Q}$ is not radical? [duplicate]
I have a finite extension of the kind $$\mathbb{Q}(\alpha):\mathbb{Q}$$ where $\alpha$ is a root of $$f(t)=t^3-3t+1$$ (I know all roots of this are real). How can I prove this extension is NOT radical?...
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1answer
43 views
For Galois extension $L:K$, does $L = K(\alpha)$ imply $\{\sigma_1(\alpha), \dots, \sigma_n(\alpha)\}$ is a basis for $L$ over $K$?
For Galois extension $L:K$ with Galois group $\{\sigma_1, \dots, \sigma_n\}$, does $L = K(\alpha)$ imply $\{\sigma_1(\alpha), \dots, \sigma_n(\alpha)\}$ is a basis for $L$ over $K$?
The proof that I'...
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votes
1answer
32 views
Galois correspondence of subgroups of $D_4$ with subfields of $\mathbb Q (\sqrt[4]{2},i)$
The Galois group of $\mathbb Q (\sqrt[4]{2},i)$ over $\mathbb Q$ is the Dihedral group $D_4$ = {$id, \sigma, \sigma^2, \sigma^3, \tau, \sigma\tau, \sigma^2\tau, \sigma^3\tau $}
Denoting $\sqrt[4]{2}$ ...
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0answers
38 views
Verify fundamental theorem of Galois for splitting field of $X^4\pm2\in\mathbb {Q}[X]$
My attempt:
Let $f(X) = X^4 ā 2$. Let $K$ be the splitting field of $f(X)$ over $\mathbb Q$. Now we have factorization:
$$X^4 ā 2 = (X^2ā \sqrt2)(X^2+ \sqrt2) = (X ā \sqrt[4]{2})(X + \sqrt[4]{2})(X ā\...
1
vote
1answer
17 views
Galois-property extends across different roots?
Suppose the extension $\mathbb{Q}(a)/\mathbb{Q}$ is Galois for some root $a$ of the polynomial $f(x)$. If $b$ is another root of $f(x)$, is $\mathbb{Q}(b)/\mathbb{Q}$ also Galois? I'm thinking it ...
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0answers
22 views
Char $p$ fields, extension/subextension
Let $K = \mathbb{F}_{p}$ (A prime field with char $= p$)
Let $C|K$ an algebraic closure of $K$
Then if $F$ is a field such that $K\subset F \subset C$, and $F \not= C \implies |F|<\infty$
Can ...
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0answers
21 views
Galois group of $x^p-q$ over $\mathbb{Q}$ [duplicate]
I am trying to compute the Galois group of the polynomial $x^p-q$ over $\mathbb{Q}$ where $p$ and $q$ are prime numbers (not necessarily distinct). I know that the polynomial's splitting field is $\...
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0answers
36 views
Criterion for completely splitting primes in a radical extension
Given $d \in \mathbb{Z}^+$ and $a \in \mathbb{Q}^*$, let $K = \mathbb{Q}(\zeta_d, a^{1/d})$, where $\zeta_d$ is a primitive $d$th root of unity.
I'm trying to prove that: "A prime number $p$ splits ...
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1answer
65 views
Let $X = \{ \sqrt{p} : p \text{ is prime} \}$, $Y \subseteq X$ and $\sqrt{p} \not\in Y$. Show that $[\mathbb{Q}(Y)(\sqrt{p}) : \mathbb{Q}(Y)] = 2$.
I am trying to solve Problem 22 from Chapter 5 of Patrick Morandi's Field and Galois Theory:
Let $K = \mathbb{Q}(X)$, where $X = \{ \sqrt{p} : p \text{ is prime} \}$. Show that $K$ is Galois ...
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1answer
96 views
Splitting of primes and other properties of $\mathbb{Q}[\omega]$ for $\omega=e^{2\pi i/m}$
Reading through Marcus I came to this exercise part of which already have answers in this same site (Splitting of primes in real cyclotomic field ) but no complete answer can be found and I'm having ...
2
votes
2answers
64 views
Description of the decomposition and the inertia group in terms of the product $\mathbb{Z}^*_{p^k}\times \mathbb{Z}^*_n $
Let $\omega^{\frac{2\pi}{m}}$, we fix a prime p and write $m=p^kn$ with $p\not| \, n$.
We know that the Galois group of $\mathbb{Q}[\omega]$ over $\mathbb{Q}$ is isomorphic to $\mathbb{Z}^*_m$ that ...
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0answers
53 views
Find Galois group of $\mathbb{Q}\sqrt{(2+\sqrt2)(3+\sqrt3)}$ [duplicate]
Find Galois group of $Q=\mathbb{Q}\sqrt{(2+\sqrt2)(3+\sqrt3)}$.
I know the minimal polynomial is $f(x)=x^8 -24x^6+144x^4-288x^2+144$. It's irreducible over $\mathbb{Q}$, hence, $[Q:\mathbb Q]=8$.
...
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0answers
14 views
Prime inert in each subfields never occurs
Suppose you are working with $K=\mathbb{Q}[\sqrt{m},\sqrt{n}]$ and $K_1=\mathbb{Q}[\sqrt{m}],K_2=\mathbb{Q}[\sqrt{n}], K_3=\mathbb{Q}[\sqrt{k}]$ where $m,n$ are squarefree integers and $k = mn/gcd(m,n)...
2
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1answer
35 views
Decomposition group does not depend on the prime
Suppose you are working with an abelian Galois group $G=G(L/K)$ of the Galois extension $L/K$. You know the Decomposition group is:
$D=D(Q,P) = \{ \sigma \in G : \sigma(Q) = Q \}$
where $Q$ (in $L$...
2
votes
1answer
71 views
An extension corresponding to a subgroup of Galois group
Let $G$ be the Galois group of $f(x)=x^6-2x^4+2x^2-2$ over $\mathbb{Q}$. Describe an extension corresponding to any of it's proper subgroups of maximal order (i.e. find generators of this extension).
...
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0answers
52 views
Relation between subgroup topology and Krull topology for an intermediate field of a Galois extension
Let $E/K$ be a Galois extension and let $F$ be an intermediate field such that $K\subseteq F\subseteq E$. Then $E/F$ is a Galois extension too and $H=\mbox{Gal}(E/F)$ is a closed subgroup of $G=\mbox{...
4
votes
2answers
102 views
Galois group of $x^6-2x^4+2x^2-2$ over $\mathbb{Q}$
Find Galois group of $x^6-2x^4+2x^2-2$ over $\mathbb{Q}$ and describe an extension corresponding to any of it's proper subgroups of maximal order.
I know that the roots are $$\sqrt{\frac{1}{3}\left(2 ...
2
votes
1answer
49 views
Intuition for equivalent definitions of Galois extension
Assume $K/F$ is a finite field extension. Then the following are equivalent:
$|\operatorname{Gal}(K/F)| = [K:F]$.
The fixed field of $\operatorname{Gal}(K/F)$ is $F$.
$K$ is the splitting field of ...
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0answers
42 views
Is $\operatorname{Gal}(\overline{\mathbb{F}_p} \,/\, \mathbb{F}_p)$ cyclic?
I know if $L/K$ is a extension of finite fields then $\text{Gal}(L/K)=\langle \phi \rangle$ where $\phi:L\longrightarrow L$ is the Frobenius automorphism.
How can I show that it is also true for the ...
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1answer
35 views
Degree of extension $\mathbb{C}/K$, where $K$ is maximal with the property $\sqrt{2} \notin K$
This question has been asked before but not really answered, but my query is a bit separate. To summarise the details: $K$ is a field maximal with respect to the property $\sqrt{2}\notin K$, any ...
1
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0answers
82 views
Why this polynomial reducible? (composite field)
In galois field of prime 2, in composite field $GF((({2}^2)^2)^2)$,
There are irreducible polynomials and reducible polynomials.
$GF(2^2):Q_1(x) = x^2+x+1,$
$GF((2^2)^2):Q_2(x) = x^2+x+\phi,$ $\...
0
votes
1answer
29 views
Galois Group of K/Q and Embeddings K --> C
If any, what is the relation between $\text{Gal}(K/\mathbb{Q})$ and the set of embeddings (say $E$) of $K \to \mathbb{C}$? I ask this for two reasons:
(1) The orders of $E$ and $\text{Gal}(K/\mathbb{...
2
votes
0answers
18 views
Divide any given angle into $n$ equal parts using paper folding
I have known that it is possible to trisect an arbitrary angle by Hisashi Abe's method. Of course, it is easy to divide an angle into two or four equal parts using paper folding.
My question is ...
5
votes
1answer
136 views
Galois Group of $x^{6}-2x^{3}-1$
I was trying to compute the normal closure of $\mathbb{Q}[\alpha]$, where $\alpha = \sqrt[3]{1+\sqrt{2}}$.
I had a reallyyyy hard time proving that $x^{6}-2x^{3}-1$ is irreducible. I proved that it ...
1
vote
1answer
114 views
Arithmetic in GF$(2^{32})$ using GF$(2^{16})$ and extensions
Ultimately, I'm looking to implement arithmetic in GF$(2^{32})$. I have a library that implements arithmetic in GF$(2^{16})$ using look-up tables for log and anti-log to implement multiplication, and ...
0
votes
1answer
25 views
For every $H \subset \operatorname{Gal}(E / \mathbb{Q})$ find the fixed field $E^H$
Let $E$ be the splitting field of $x^3 - 2$ over $\mathbb{Q}$.
I proved that $E = \mathbb{Q}(\sqrt{2},\omega)$. Where $\omega$ is a primitive root of unity.
And I also know that $G := \operatorname{...
2
votes
2answers
70 views
Can there exist a finite extension $K$ where $K$ is Galois over $Q(i)$ but K is not Galois over $Q?$
Can there exist a finite extension $K$ where $K$ is Galois over $\mathbb{Q}(i)$ but $K$ is not Galois over $\mathbb{Q}$?
I am trying to come up with a specific example to show it is possible. My ...
0
votes
1answer
92 views
Linearly independent vectors over a field and its subfield
Let $\mathrm{F}_q$ be a subfield of $\mathrm{F}_{q^m}$. $\mathrm{F}_{q^m}$ can be seen as an $m$-dimensional vector space over $\mathrm{F}_q$. Let $v_1,\ldots, v_k \in \mathrm{F}_{q^m}^n$ be linearly ...
1
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1answer
55 views
Exactness of Inflation restriction sequence, Galois Cohomology
I am trying to prove the following.
Let $K/k$ be a finite Galois extension, $G= G(K/k)$, $k \subset F \subset K$ with $K/k$ normal and $H=G(K/F)$. Then:
$ \rho : C^{2} (G,A) \rightarrow C^{2} (H,A) $ ...
1
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0answers
51 views
Subfields of splitting field of $x^{15}-1$ over $\mathbb Q$
I'd like to find the subfields of splitting field of $x^{15}-1$ over $\mathbb Q$
I solved the followings step.
The splitting field L of $x^{15}-1$ is same as the splitting field of $(x^3-1)(x^5-1)$. ...
