Question

When irreducibility of a polynomial implies its irreducibility over an extension field?

score 79 · Answer 1 · 2026-03-31 11:54:24

79

Views

When irreducibility of a polynomial implies its irreducibility over an extension field?

Published on 31 Mar 2026 - 11:54

score 209 · Answer 2 · 2016-10-23 18:37:42

209

Views

Degree of $\mathbb{Q}(\sqrt[4]{3},\sqrt[5]{3})$ over $\mathbb{Q}$

Published on 23 Oct 2016 - 18:37

score 427 · Answer 3 · 2016-10-24 23:14:21

427

Views

When two primitive elements of a simple field extension are conjugate?

Published on 24 Oct 2016 - 23:14

score 84 · Answer 4 · 2026-03-25 07:39:50

84

Views

Algebraic closures and monomorphisms

Published on 25 Mar 2026 - 7:39

score 169 · Answer 5 · 2026-03-27 02:38:03

169

Views

An exercise about field extensions and roots

Published on 27 Mar 2026 - 2:38

score 30 · Answer 6 · 2026-03-30 01:47:00

30

Views

$[K(\beta):K] \ \deg{f_i} = [K(\alpha): K] \ \deg{g_i}$

Published on 30 Mar 2026 - 1:47

score 133 · Answer 7 · 2016-10-28 16:16:01

133

Views

$E'=\mathbb{Q}(\sqrt2, \sqrt{1+i})$ is a normal closure of the degree-4 extension $\mathbb{Q}(\sqrt{1+i})/\mathbb{Q}$

Published on 28 Oct 2016 - 16:16

score 3.7k · Answer 8 · 2016-10-30 16:16:08

3.7k

Views

Field extension of degree two is normal

Published on 30 Oct 2016 - 16:16

score 239 · Answer 9 · 2016-11-05 09:45:42

Prove that there exists no algebraically closed field $F$ such that $\bar{\mathbb{Q}} \subsetneq F \subsetneq \overline{\mathbb{Q}(\pi)}$

score 159 · Answer 10 · 2016-11-06 13:00:20

159

Views

Is $K/F$ also Galois when $K/k$ is Galois and $k \subset F \subset K$?

Published on 06 Nov 2016 - 13:00

score 1.5k · Answer 11 · 2016-11-06 16:29:25

1.5k

Views

Show that if $F \subset D \subset E$ then D is a field

Published on 06 Nov 2016 - 16:29

score 130 · Answer 12 · 2016-11-07 01:39:35

130

Views

$N|K$ normal $\Rightarrow F \in K[X]$ factors into $F=Q_1\cdots Q_r$ with $\text{deg}(Q_1)=\dots=\text{deg}(Q_r)$

Published on 07 Nov 2016 - 1:39

score 196 · Answer 13 · 2026-04-03 17:02:26

196

Views

Constructing an explicit isomorphism between two fields

Published on 03 Apr 2026 - 17:02

score 337 · Answer 14 · 2026-03-26 17:32:59

337

Views

If a commutator subgroup is the identity, then G is solvable.

Published on 26 Mar 2026 - 17:32

score 104 · Answer 15 · 2016-11-10 15:02:23

104

Views

Irreducible in Base Field vs Irreducible in Quadratic Extension

Published on 10 Nov 2016 - 15:02

List Question

When irreducibility of a polynomial implies its irreducibility over an extension field?

Degree of $\mathbb{Q}(\sqrt[4]{3},\sqrt[5]{3})$ over $\mathbb{Q}$

When two primitive elements of a simple field extension are conjugate?

Algebraic closures and monomorphisms

An exercise about field extensions and roots

$[K(\beta):K] \ \deg{f_i} = [K(\alpha): K] \ \deg{g_i}$

$E'=\mathbb{Q}(\sqrt2, \sqrt{1+i})$ is a normal closure of the degree-4 extension $\mathbb{Q}(\sqrt{1+i})/\mathbb{Q}$

Field extension of degree two is normal

Prove that there exists no algebraically closed field $F$ such that $\bar{\mathbb{Q}} \subsetneq F \subsetneq \overline{\mathbb{Q}(\pi)}$

Is $K/F$ also Galois when $K/k$ is Galois and $k \subset F \subset K$?

Show that if $F \subset D \subset E$ then D is a field

$N|K$ normal $\Rightarrow F \in K[X]$ factors into $F=Q_1\cdots Q_r$ with $\text{deg}(Q_1)=\dots=\text{deg}(Q_r)$

Constructing an explicit isomorphism between two fields

If a commutator subgroup is the identity, then G is solvable.

Irreducible in Base Field vs Irreducible in Quadratic Extension

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