Question

Element of a splitting field that can be construct using ruler and compas

score 96 · Answer 1 · 2026-04-12 17:32:13

96

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Element of a splitting field that can be construct using ruler and compas

Published on 12 Apr 2026 - 17:32

score 78 · Answer 2 · 2026-04-11 18:17:08

78

Views

Galois automorphisms

Published on 11 Apr 2026 - 18:17

score 86 · Answer 3 · 2026-04-12 07:35:24

86

Views

ordered field $\implies$ $0$ cannot be written as the sum of two squares

Published on 12 Apr 2026 - 7:35

score 68 · Answer 4 · 2026-04-13 23:43:59

68

Views

Why is $0\cdot x$ defined in fields even though $F\setminus\{0\}$ is the set for multiplication

Published on 13 Apr 2026 - 23:43

score 85 · Answer 5 · 2026-04-13 08:58:23

85

Views

Equivalent statement on Markov and Gibbs random fields

Published on 13 Apr 2026 - 8:58

score 624 · Answer 6 · 2026-04-13 21:47:15

624

Views

Real and complex embeddings of a Galois extension

Published on 13 Apr 2026 - 21:47

score 53 · Answer 7 · 2026-04-16 03:19:51

53

Views

$\operatorname{Gal}(A/B)=\operatorname{Gal}(AC/BC)$

Published on 16 Apr 2026 - 3:19

score 64 · Answer 8 · 2026-04-16 19:05:21

64

Views

Field homomorphism in both directions

Published on 16 Apr 2026 - 19:05

score 1.2k · Answer 9 · 2026-04-10 16:57:44

1.2k

Views

Definition of an F-embedding

Published on 10 Apr 2026 - 16:57

score 359 · Answer 10 · 2026-04-15 17:44:18

359

Views

Show $xy-2$ is irreducible in $\mathbb{R}[x,y]$

Published on 15 Apr 2026 - 17:44

score 34 · Answer 11 · 2026-04-10 13:48:28

34

Views

In which field does this equation lies

Published on 10 Apr 2026 - 13:48

score 903 · Answer 12 · 2026-04-14 16:56:25

903

Views

If $\alpha \in E$ is algebraic over a $F$ and $E = F(\alpha)$, then $E$ is an algebraic extension.

Published on 14 Apr 2026 - 16:56

score 269 · Answer 13 · 2026-04-13 11:47:18

269

Views

Intersection of ramified Galois extension and maximal unramified abelian extension

Published on 13 Apr 2026 - 11:47

score 412 · Answer 14 · 2026-04-17 17:38:33

412

Views

Let $\beta$ be a zero of $f(x) =x^5+2x+ 4$ . Show that none of $\sqrt{2}$,$ \sqrt[3]{2}$ ,$ \sqrt[4]{2}$ belongs to $Q(\beta)$.

Published on 17 Apr 2026 - 17:38

score 116 · Answer 15 · 2026-04-11 05:02:41

116

Views

How to find a "good" extension of $\mathbb{R}$?

Published on 11 Apr 2026 - 5:02

List Question

Element of a splitting field that can be construct using ruler and compas

Galois automorphisms

ordered field $\implies$ $0$ cannot be written as the sum of two squares

Why is $0\cdot x$ defined in fields even though $F\setminus\{0\}$ is the set for multiplication

Equivalent statement on Markov and Gibbs random fields

Real and complex embeddings of a Galois extension

$\operatorname{Gal}(A/B)=\operatorname{Gal}(AC/BC)$

Field homomorphism in both directions

Definition of an F-embedding

Show $xy-2$ is irreducible in $\mathbb{R}[x,y]$

In which field does this equation lies

If $\alpha \in E$ is algebraic over a $F$ and $E = F(\alpha)$, then $E$ is an algebraic extension.

Intersection of ramified Galois extension and maximal unramified abelian extension

Let $\beta$ be a zero of $f(x) =x^5+2x+ 4$ . Show that none of $\sqrt{2}$,$ \sqrt[3]{2}$ ,$ \sqrt[4]{2}$ belongs to $Q(\beta)$.

How to find a "good" extension of $\mathbb{R}$?

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