Question

$R/\langle p^k\rangle$ is an associator (i.e. if $\langle a\rangle = \langle b\rangle,$ then $a$ and $b$ are associates) when $R$ is a PID.

score 116 · Answer 1 · 2016-06-17 15:26:03

116

Views

$R/\langle p^k\rangle$ is an associator (i.e. if $\langle a\rangle = \langle b\rangle,$ then $a$ and $b$ are associates) when $R$ is a PID.

Published on 17 Jun 2016 - 15:26

score 1.5k · Answer 2 · 2016-06-21 06:35:03

1.5k

Views

ED,PID and UFD and the relation between them

Published on 21 Jun 2016 - 6:35

score 567 · Answer 3 · 2016-06-21 11:52:24

567

Views

Prove that any finitely generated submodule of $R^+$ (the field of quotients) is free of rank $1$

Published on 21 Jun 2016 - 11:52

score 1.5k · Answer 4 · 2016-06-30 09:16:04

I wonder if $K$ is a field then $K[X]$ is a PID then the ideals of it have formula that is $<f(X)>$ when $f(X)$ is a polynomial with coefficent in $K$

score 500 · Answer 5 · 2016-07-03 06:47:19

500

Views

Generator of intersection of ideals in a PID via adjunction?

Published on 03 Jul 2016 - 6:47

score 262 · Answer 6 · 2016-07-03 11:43:09

262

Views

Is it true that all modules over $\mathbb{Q}$ are of form $\mathbb{Q}^n$ where $n\geq 0$

Published on 03 Jul 2016 - 11:43

score 2k · Answer 7 · 2013-04-14 02:59:56

2k

Views

Show that if $D$ is a domain but not a field then $D[x]$ is not a principal ideal domain.

Published on 14 Apr 2013 - 2:59

score 5k · Answer 8 · 2013-05-11 22:25:39

5k

Views

radical of sum of two ideals

Published on 11 May 2013 - 22:25

score 668 · Answer 9 · 2026-02-23 09:42:19

668

Views

Smith Normal Form for PIDs

Published on 23 Feb 2026 - 9:42

score 1k · Answer 10 · 2013-05-25 14:03:51

1k

Views

Why define vector spaces over fields instead of a PID?

Published on 25 May 2013 - 14:03

score 2k · Answer 11 · 2013-06-02 17:10:05

2k

Views

Every principal ideal domain satisfies ACCP.

Published on 02 Jun 2013 - 17:10

score 14.1k · Answer 12 · 2013-06-09 01:25:46

14.1k

Views

Why any field is a principal ideal domain?

Published on 09 Jun 2013 - 1:25

score 558 · Answer 13 · 2013-06-09 18:41:25

558

Views

Find a non free submodule of a free module over R which is not PID

Published on 09 Jun 2013 - 18:41

score 84 · Answer 14 · 2013-07-12 05:50:15

84

Views

Is quantum torus a principal ideal domain?

Published on 12 Jul 2013 - 5:50

score 105 · Answer 15 · 2013-07-30 03:02:00

105

Views

Ideals of nested PID's

Published on 30 Jul 2013 - 3:02

List Question

$R/\langle p^k\rangle$ is an associator (i.e. if $\langle a\rangle = \langle b\rangle,$ then $a$ and $b$ are associates) when $R$ is a PID.

ED,PID and UFD and the relation between them

Prove that any finitely generated submodule of $R^+$ (the field of quotients) is free of rank $1$

I wonder if $K$ is a field then $K[X]$ is a PID then the ideals of it have formula that is $<f(X)>$ when $f(X)$ is a polynomial with coefficent in $K$

Generator of intersection of ideals in a PID via adjunction?

Is it true that all modules over $\mathbb{Q}$ are of form $\mathbb{Q}^n$ where $n\geq 0$

Show that if $D$ is a domain but not a field then $D[x]$ is not a principal ideal domain.

radical of sum of two ideals

Smith Normal Form for PIDs

Why define vector spaces over fields instead of a PID?

Every principal ideal domain satisfies ACCP.

Why any field is a principal ideal domain?

Find a non free submodule of a free module over R which is not PID

Is quantum torus a principal ideal domain?

Ideals of nested PID's

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