Question

If $ F(a_1, \dots, a_n)$ is splitting field of a separable polynomial $f$, then $[F(a_1, \dots, a_i):F] \leq n(n-1) \cdots (n-i+1)$

score 68 · Answer 1 · 2026-03-26 14:30:28

68

Views

If $ F(a_1, \dots, a_n)$ is splitting field of a separable polynomial $f$, then $[F(a_1, \dots, a_i):F] \leq n(n-1) \cdots (n-i+1)$

Published on 26 Mar 2026 - 14:30

score 118 · Answer 2 · 2026-03-26 14:33:20

118

Views

Splitting Field as Subfield Generated by Roots

Published on 26 Mar 2026 - 14:33

score 45 · Answer 3 · 2026-03-29 12:05:26

45

Views

Understanding a proof that $[L : L^G] = |G|$ when $G$ acts faithfully on a field $L$

Published on 29 Mar 2026 - 12:05

score 63 · Answer 4 · 2020-03-23 11:57:51

63

Views

Show that $\sqrt{2} \notin \mathbb{Q}(a)$ for real numbers a.

Published on 23 Mar 2020 - 11:57

score 313 · Answer 5 · 2020-03-23 15:43:28

313

Views

Let$ K$ be a finite extension of $F$. If $L_1,L_2$ are subfields of $K$ containing $F$ and $L_1⊂ L_2$ then $K$ is simple extension of $F$.

Published on 23 Mar 2020 - 15:43

score 24 · Answer 6 · 2020-03-23 16:31:18

24

Views

Solving equation for a field extension

Published on 23 Mar 2020 - 16:31

score 33 · Answer 7 · 2020-03-23 17:47:35

33

Views

Intersection of Galois extensions with additional algebraic element.

Published on 23 Mar 2020 - 17:47

score 107 · Answer 8 · 2026-03-28 12:58:44

107

Views

Which of the following pairs is an extension of fields?

Published on 28 Mar 2026 - 12:58

score 462 · Answer 9 · 2020-03-24 07:13:41

462

Views

Galois Extension, Chinese Remainder Theorem, and Quotients of Polynomial Rings

Published on 24 Mar 2020 - 7:13

score 256 · Answer 10 · 2020-03-25 13:08:57

256

Views

There are infinitely many non-isomorphic cubic Galois extension of $\mathbb Q$

Published on 25 Mar 2020 - 13:08

score 90 · Answer 11 · 2020-03-25 14:54:37

90

Views

Name a field ($\ne \Bbb{R}$ or $\Bbb{C}$) which contains a root of $x^5+2x^3+4x^2+6$.

Published on 25 Mar 2020 - 14:54

score 103 · Answer 12 · 2020-03-27 02:04:30

103

Views

Prove that $F(a) = F(a + c)$ and $F(a) = F(ca)$. (Assume $c \ne 0$.)

Published on 27 Mar 2020 - 2:04

score 77 · Answer 13 · 2020-03-28 06:51:14

77

Views

Proves that $F(a^2) \subseteq F(a)$ and $F(a + b) \subseteq F(a,b)$

Published on 28 Mar 2020 - 6:51

score 65 · Answer 14 · 2020-03-28 20:31:05

65

Views

Prove if $a$ is a root of $x^2-2$ and $b$ is a root of $\displaystyle x^2-\frac{1}{2}$, then $\Bbb{Q}(a)\cong \Bbb{Q}(b)$

Published on 28 Mar 2020 - 20:31

score 85 · Answer 15 · 2026-03-28 12:59:21

85

Views

Galois group of $F/K$ where $F/K$ is galois with $[F:K]=8$ and where there is an intermediate field $E$ with $[E:K]=4$ such that $E/K$ is not galois

Published on 28 Mar 2026 - 12:59

List Question

If $ F(a_1, \dots, a_n)$ is splitting field of a separable polynomial $f$, then $[F(a_1, \dots, a_i):F] \leq n(n-1) \cdots (n-i+1)$

Splitting Field as Subfield Generated by Roots

Understanding a proof that $[L : L^G] = |G|$ when $G$ acts faithfully on a field $L$

Show that $\sqrt{2} \notin \mathbb{Q}(a)$ for real numbers a.

Let$ K$ be a finite extension of $F$. If $L_1,L_2$ are subfields of $K$ containing $F$ and $L_1⊂ L_2$ then $K$ is simple extension of $F$.

Solving equation for a field extension

Intersection of Galois extensions with additional algebraic element.

Which of the following pairs is an extension of fields?

Galois Extension, Chinese Remainder Theorem, and Quotients of Polynomial Rings

There are infinitely many non-isomorphic cubic Galois extension of $\mathbb Q$

Name a field ($\ne \Bbb{R}$ or $\Bbb{C}$) which contains a root of $x^5+2x^3+4x^2+6$.

Prove that $F(a) = F(a + c)$ and $F(a) = F(ca)$. (Assume $c \ne 0$.)

Proves that $F(a^2) \subseteq F(a)$ and $F(a + b) \subseteq F(a,b)$

Prove if $a$ is a root of $x^2-2$ and $b$ is a root of $\displaystyle x^2-\frac{1}{2}$, then $\Bbb{Q}(a)\cong \Bbb{Q}(b)$

Galois group of $F/K$ where $F/K$ is galois with $[F:K]=8$ and where there is an intermediate field $E$ with $[E:K]=4$ such that $E/K$ is not galois

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