Question

Prove that prototype of prime ideal in homomorphism $f : A \to B$ is prime ideal.

score 70 · Answer 1 · 2020-10-25 21:27:34

70

Views

Prove that prototype of prime ideal in homomorphism $f : A \to B$ is prime ideal.

Published on 25 Oct 2020 - 21:27

score 92 · Answer 2 · 2020-10-26 19:45:25

92

Views

Proving $\operatorname{Hom}_A(M,A)$ is free given that $M$ is a free $A$-module both of rank $n$.

Published on 26 Oct 2020 - 19:45

score 215 · Answer 3 · 2020-10-27 09:56:27

215

Views

A problem about homomorphism on polynomial rings

Published on 27 Oct 2020 - 9:56

score 376 · Answer 4 · 2020-10-28 09:47:02

376

Views

Extension of a maximal ideal

Published on 28 Oct 2020 - 9:47

score 156 · Answer 5 · 2020-10-29 23:54:55

156

Views

How do ring morphisms work when dealing with polynomial rings?

Published on 29 Oct 2020 - 23:54

score 60 · Answer 6 · 2020-11-04 00:49:43

60

Views

Proving that $G$-action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr)$ is a homomorphism.

Published on 04 Nov 2020 - 0:49

score 33 · Answer 7 · 2020-11-04 01:45:21

33

Views

Is the statement of this question correct?

Published on 04 Nov 2020 - 1:45

score 31 · Answer 8 · 2020-11-04 04:34:40

31

Views

Prove the following Lemma in the polynomial rings.

Published on 04 Nov 2020 - 4:34

score 82 · Answer 9 · 2020-11-05 08:06:59

82

Views

Show that $\phi:\mathbb{Q}(\sqrt{3}) \rightarrow M_{2}(\mathbb{Q})$ maps identity to identity.

Published on 05 Nov 2020 - 8:06

score 196 · Answer 10 · 2026-03-26 22:53:19

196

Views

prove that $\phi(a+b \sqrt{2})$=$ \begin{pmatrix}{} a & 2b \\ b & a \end{pmatrix}$ is an isomorphism

Published on 26 Mar 2026 - 22:53

score 55 · Answer 11 · 2020-11-08 19:48:14

55

Views

$f: F[x] \to F[x]$ is a isomorphism if $F$ is a domain?

Published on 08 Nov 2020 - 19:48

score 23 · Answer 12 · 2020-11-13 09:56:26

23

Views

A Ring homomorphism problem

Published on 13 Nov 2020 - 9:56

score 1.1k · Answer 13 · 2020-11-20 09:51:37

1.1k

Views

Given a ring homomorphism $f: \mathbb C \to \mathbb C$, prove it is either the identity mapping or complex conjugation

Published on 20 Nov 2020 - 9:51

score 121 · Answer 14 · 2020-11-29 04:33:09

121

Views

Let $f:A \to B$ be a surjective homomorphism between rings. If $B$ and ker$f$ has ascending chain condition, then $A$ has it too.

Published on 29 Nov 2020 - 4:33

score 83 · Answer 15 · 2020-11-29 10:46:48

83

Views

Show that the image of a finite ring is also finite, when the map is a surjective ring homomorphism

Published on 29 Nov 2020 - 10:46

List Question

Prove that prototype of prime ideal in homomorphism $f : A \to B$ is prime ideal.

Proving $\operatorname{Hom}_A(M,A)$ is free given that $M$ is a free $A$-module both of rank $n$.

A problem about homomorphism on polynomial rings

Extension of a maximal ideal

How do ring morphisms work when dealing with polynomial rings?

Proving that $G$-action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr)$ is a homomorphism.

Is the statement of this question correct?

Prove the following Lemma in the polynomial rings.

Show that $\phi:\mathbb{Q}(\sqrt{3}) \rightarrow M_{2}(\mathbb{Q})$ maps identity to identity.

prove that $\phi(a+b \sqrt{2})$=$ \begin{pmatrix}{} a & 2b \\ b & a \end{pmatrix}$ is an isomorphism

$f: F[x] \to F[x]$ is a isomorphism if $F$ is a domain?

A Ring homomorphism problem

Given a ring homomorphism $f: \mathbb C \to \mathbb C$, prove it is either the identity mapping or complex conjugation

Let $f:A \to B$ be a surjective homomorphism between rings. If $B$ and ker$f$ has ascending chain condition, then $A$ has it too.

Show that the image of a finite ring is also finite, when the map is a surjective ring homomorphism

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