Question

If $p, q$ are prime integers, then $\mathbb{Q}(\sqrt{p})$ is not isomorphic (as a field) to $\mathbb{Q}(\sqrt{q})$

score 113 · Answer 1 · 2018-11-26 06:28:31

113

Views

If $p, q$ are prime integers, then $\mathbb{Q}(\sqrt{p})$ is not isomorphic (as a field) to $\mathbb{Q}(\sqrt{q})$

Published on 26 Nov 2018 - 6:28

score 375 · Answer 2 · 2026-03-26 12:07:01

375

Views

Polynomials $P(x)\in k[x]$ satisfying condition $P(x^2)=P(-x)P(x)$

Published on 26 Mar 2026 - 12:07

score 71 · Answer 3 · 2018-11-27 14:22:03

71

Views

Fields such that all finite extensions are Galois

Published on 27 Nov 2018 - 14:22

score 64 · Answer 4 · 2018-11-28 12:55:55

64

Views

Centre of algebra under field extension

Published on 28 Nov 2018 - 12:55

score 154 · Answer 5 · 2018-11-29 19:57:34

154

Views

Find a factorisation in a cubic field.

Published on 29 Nov 2018 - 19:57

score 32 · Answer 6 · 2026-03-27 14:58:45

32

Views

Question about Galois group of a splitting field

Published on 27 Mar 2026 - 14:58

score 2.1k · Answer 7 · 2026-03-24 20:33:29

2.1k

Views

Field Extension over the Field of Rational Functions is Finite

Published on 24 Mar 2026 - 20:33

score 349 · Answer 8 · 2026-03-27 15:05:35

349

Views

Characterising the irreducible polynomials in positive characteristic whose roots generate the (cyclic) group of units of the splitting field

Published on 27 Mar 2026 - 15:05

score 258 · Answer 9 · 2018-11-30 23:10:57

258

Views

A particular field generated from a set

Published on 30 Nov 2018 - 23:10

score 59 · Answer 10 · 2018-11-30 23:40:37

59

Views

Why is the basis the same for $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $\mathbb{Q}(\sqrt{3},\sqrt{6})$?

Published on 30 Nov 2018 - 23:40

score 38 · Answer 11 · 2018-12-02 00:54:42

38

Views

Field Extensions - Algebraic Extensions

Published on 02 Dec 2018 - 0:54

score 617 · Answer 12 · 2026-03-26 06:06:47

617

Views

Compute the irreducible polynomials over Q for $a=\sqrt{2}+\sqrt{5}$ and $b=\sqrt[3]{2}+\sqrt{5}$

Published on 26 Mar 2026 - 6:06

score 2.3k · Answer 13 · 2018-12-02 17:53:57

2.3k

Views

Find a basis for $\mathbb{Q}(\sqrt{2}+\sqrt{5})$ over $\mathbb{Q}(\sqrt{5})$.

Published on 02 Dec 2018 - 17:53

score 225 · Answer 14 · 2018-12-02 23:40:12

225

Views

Splitting fields for $x^3-3$ and $x^5-1$

Published on 02 Dec 2018 - 23:40

score 902 · Answer 15 · 2026-03-27 14:59:14

902

Views

Describing the Galois group of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ without computing the extension degree (proof check please)

Published on 27 Mar 2026 - 14:59

List Question

If $p, q$ are prime integers, then $\mathbb{Q}(\sqrt{p})$ is not isomorphic (as a field) to $\mathbb{Q}(\sqrt{q})$

Polynomials $P(x)\in k[x]$ satisfying condition $P(x^2)=P(-x)P(x)$

Fields such that all finite extensions are Galois

Centre of algebra under field extension

Find a factorisation in a cubic field.

Question about Galois group of a splitting field

Field Extension over the Field of Rational Functions is Finite

Characterising the irreducible polynomials in positive characteristic whose roots generate the (cyclic) group of units of the splitting field

A particular field generated from a set

Why is the basis the same for $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $\mathbb{Q}(\sqrt{3},\sqrt{6})$?

Field Extensions - Algebraic Extensions

Compute the irreducible polynomials over Q for $a=\sqrt{2}+\sqrt{5}$ and $b=\sqrt[3]{2}+\sqrt{5}$

Find a basis for $\mathbb{Q}(\sqrt{2}+\sqrt{5})$ over $\mathbb{Q}(\sqrt{5})$.

Splitting fields for $x^3-3$ and $x^5-1$

Describing the Galois group of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ without computing the extension degree (proof check please)

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