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Questions tagged [galois-extensions]

For questions about Galois extensions of fields. We say that an algebraic extension $L/K$ is a Galois extension iff the subfield of $L$ that is fixed by automorphisms of $L$ which fix K is exactly $K$.

1 vote
2 answers
90 views

I'm trying to find the Galois group of $\mathbb{Q}\left(\zeta_{6}\right)/\mathbb{Q}$, where $\zeta_k=e^\frac{2\pi i}{k}$. We know that $\left\{1,\zeta_{6},\zeta_{6}^2,\zeta_{6}^3,\zeta_{6}^4,\zeta_{6}^...
wizzwizz4's user avatar
  • 189
5 votes
1 answer
195 views

In Milne (Fields and Galois Theory), by definition constructible elements are in $\mathbb{R}$. However, it is also argued (in remark 3.27) that given an algebraic constructible number, its Galois ...
Melanka's user avatar
  • 185
2 votes
0 answers
74 views

It seems to me that the beginning of Galois theory (Galois's work) is still not clear to me. To prove the quintic or higher degree polynomial is not solvable in radicals, Evertise Galois considered ...
Learner's user avatar
  • 584
1 vote
0 answers
110 views

How to find the degree and the ramification index of the splitting field of a certain polynomial over $\mathbb{Q}_p$? For instance, $f(x)=3+9x+3x^4+x^6$ over $\mathbb{Q}_3$? If $\gamma$ is one of the ...
8k14's user avatar
  • 311
0 votes
1 answer
50 views

Let $E/\mathbb{Q}$ be an elliptic curve with complex multiplication by $\mathbb{Q}(i)$, e.g. any curve of the form $E:y^2 = x^3 + Ax$ with $A \in \mathbb{Q}^\times$. I would like to compute generators ...
Oisin Robinson's user avatar
0 votes
1 answer
91 views

Let $K \subset F \subset E$ be a tower of extensions with $E/K$ being a Galois extension. How can I show that $g(F)=F \text{ for all }g\in {\rm Gal}(E/K) \iff F/K$ is normal? I tried using that $${\rm ...
Gabriela Martins's user avatar
0 votes
1 answer
83 views

Let $L/K$ is a finite field extension and assume it a Galois extension. I want to prove the following simplest form of the Fundamental theorem of Galois theory: $(a)$ For any subgroup $H$ of $\...
Learner's user avatar
  • 584
1 vote
0 answers
40 views

I am asking whether the following statement is true: given a prime $p$ a finite group $G$ of order coprime to $p$ an integer $g$ greater than 1 an algebraically closed field $K$ of characteristic $p$...
Maria Mazieri's user avatar
1 vote
1 answer
94 views

I believe the following problem comes from the UCLA algebra qualifying exams. Let $K / F$ be a finite Galois extension of fields and $\alpha \in K \setminus F$. Let $F \subseteq E \subseteq K$ such ...
zork zork's user avatar
  • 333
1 vote
0 answers
34 views

I'm currently studying étale cohomology and its application to the Weil Conjectures, following the Lecture Notes of Milne. In Chapter 27, he defines and reviews the properties of the Frobenius ...
Compacto's user avatar
  • 2,220
6 votes
2 answers
238 views

Let $p$ be an odd prime, and let $n_1,\dots,n_m\in\mathbb F_p$ be the roots of $x^{\frac{p-1}{2}}-1\in\mathbb F_p[x]$. Does $\zeta_p^{n_1}+\cdots+\zeta_p^{n_m}$ generate $\mathbb Q(\sqrt{p})$ if $p\...
Alann Rosas's user avatar
  • 6,892
2 votes
0 answers
50 views

Let $G$ be a finite group, $K$ be a field with characteristic zero, and $C$ be an algebraic closure of $K$. The algebras $CG$ and $KG$ are semisimple. Let $\chi$ be a $C$-character of $G$ ...
khashayar's user avatar
  • 2,674
3 votes
2 answers
146 views

I have a question about an answer to this post concerning finding fixed fields of subgroups of the Galois group $K$ of $f(x)=x^4-2$ over $\mathbb Q$. Part of the accepted answer involves determining ...
algebra learner's user avatar
4 votes
2 answers
168 views

I’m looking at the quotient ring $$ R := (\mathbb Z/2^{n}\mathbb Z)[X]\big/\bigl(X^{2^m}+1\bigr), $$ for example let's focus on $n=16,m=4$. I understand $X^{16}+1$ is irreducible over $\mathbb Z/2^{16}...
Joseph Johnston's user avatar
5 votes
2 answers
224 views

Let $F$ be a field and $K$ be a finite extension of $F$. (Assume all these fields have characteristic zero). Take $\alpha\in K$. Let $n$ be the total number of conjugates of $\alpha$ over the base ...
Naveen Kumar's user avatar

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