Galois theory

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Galois theory

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

651 posts

#1 h Apr 4, 2025, 3:04 PM

Y by

Prove that there does not exist a positive integer $n$ such that the $n$ th cyclotomic field over $\mathbb{Q}$ is an extension of the field $\mathbb{Q}(\sqrt[3]{5})$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3148 posts

yofro

#2 h Apr 4, 2025, 8:02 PM

Y by

Since $\mathbb{Q}(\zeta_n)$ is Abelian, then $\mathbb{Q}(\sqrt[3]{5})$ would have to correspond to a normal subgroup of the Galois group of $\mathbb{Q}(\zeta_n)$ , and hence it would have to be Galois. But $\mathbb{Q}(\sqrt[3]{5})$ is not Galois; the Galois closure of $\mathbb{Q}(\sqrt[3]{5})$ is $\mathbb{Q}(\sqrt[3]{5}, \omega)$ , where $\omega$ is a primitive cube root of unity.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

651 posts

#3 h Apr 5, 2025, 5:33 AM

Y by

yofro wrote:

Since $\mathbb{Q}(\zeta_n)$ is Abelian, then $\mathbb{Q}(\sqrt[3]{5})$ would have to correspond to a normal subgroup of the Galois group of $\mathbb{Q}(\zeta_n)$ , and hence it would have to be Galois. But $\mathbb{Q}(\sqrt[3]{5})$ is not Galois; the Galois closure of $\mathbb{Q}(\sqrt[3]{5})$ is $\mathbb{Q}(\sqrt[3]{5}, \omega)$ , where $\omega$ is a primitive cube root of unity.

can't understand your explanation

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3148 posts

yofro

#4 h Apr 8, 2025, 5:34 AM

Y by

ILOVEMYFAMILY wrote:

yofro wrote:

Since $\mathbb{Q}(\zeta_n)$ is Abelian, then $\mathbb{Q}(\sqrt[3]{5})$ would have to correspond to a normal subgroup of the Galois group of $\mathbb{Q}(\zeta_n)$ , and hence it would have to be Galois. But $\mathbb{Q}(\sqrt[3]{5})$ is not Galois; the Galois closure of $\mathbb{Q}(\sqrt[3]{5})$ is $\mathbb{Q}(\sqrt[3]{5}, \omega)$ , where $\omega$ is a primitive cube root of unity.

can't understand your explanation

It's phrased in the language of the fundamental theorem of Galois theory; there's a correspondence between intermediate fields and subgroups of the Galois group, and I'm utilizing this correspondence.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a