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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 $K$.

2 votes
0 answers
47 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
  • 574
1 vote
0 answers
102 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
47 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
90 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
81 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
  • 574
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
92 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
32 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
234 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
  • 6,872
2 votes
0 answers
49 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
  • 2,613
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
167 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
2 votes
0 answers
39 views

Suppose $G$ is a finite group, $F$ is a field of characteristic zero, and $L/F$ is a field extension. Let $\varphi$ be an irreducible representation of $G$ over $F$, with character $\chi$, and in $L$ ...
tys's user avatar
  • 512
1 vote
0 answers
57 views

While the goal is multiplying polynomials, we can immediately reduce the problem to evaluating polynomials, then to multiply the evaluations. If one has a polynomial with coefficients and basis ...
Joseph Johnston's user avatar

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