Question

Same as some localization at $\mathbb{Q}[X]$?

score 88 · Answer 1 · 2016-09-27 06:40:26

88

Views

Same as some localization at $\mathbb{Q}[X]$?

Published on 27 Sep 2016 - 6:40

score 3k · Answer 2 · 2016-10-06 07:45:41

3k

Views

a submodule of a noetherian module is also noetherian?

Published on 06 Oct 2016 - 7:45

score 1.2k · Answer 3 · 2016-10-09 10:45:56

1.2k

Views

Local ring with finitely generated maximal ideal and $\cap_{n>0}\mathfrak{m}^n=\{0\}$ is Noetherian?

Published on 09 Oct 2016 - 10:45

score 2k · Answer 4 · 2026-03-25 11:06:20

2k

Views

For $k$ a field, $k[x]$ is Noetherian but not Artinian?

Published on 25 Mar 2026 - 11:06

score 220 · Answer 5 · 2026-03-25 11:01:24

220

Views

Is $\oplus_{\mathbb{N}}\mathbb{Z}_2$ Artinian but not Noetherian?

Published on 25 Mar 2026 - 11:01

score 373 · Answer 6 · 2016-10-13 16:00:41

373

Views

If the source of a morphism is a noetherian scheme, the morphism is quasi-compact.

Published on 13 Oct 2016 - 16:00

score 1.1k · Answer 7 · 2016-10-17 00:16:23

1.1k

Views

Is every finitely-generated $\mathbb{Z}$-module Noetherian?

Published on 17 Oct 2016 - 0:16

score 166 · Answer 8 · 2016-10-21 12:18:32

166

Views

Height of Product of Ideals

Published on 21 Oct 2016 - 12:18

score 438 · Answer 9 · 2026-03-25 11:00:08

438

Views

Structure theorem for commutative Noetherian rings

Published on 25 Mar 2026 - 11:00

score 74 · Answer 10 · 2016-10-29 22:58:42

74

Views

If $R_3$ is a finitely generated $R_2$-module and $R_2$ is a finitely generated free $R_1$-module, can we say that $R_3$ is a free $R_1$-module?

Published on 29 Oct 2016 - 22:58

score 32 · Answer 11 · 2016-10-30 17:16:24

32

Views

Why $P_i+N$ is stationnary when $P_i$ are submodules of $M$?

Published on 30 Oct 2016 - 17:16

score 610 · Answer 12 · 2016-10-30 19:13:43

610

Views

finitely many minimal prime ideals in $\mathbb{Z}[x]$?

Published on 30 Oct 2016 - 19:13

score 113 · Answer 13 · 2026-03-25 16:03:43

113

Views

$M$ be a finitely generated module over commutative unital ring $R$ , $N,P$ submodules , $P\subseteq N \subseteq M$ and $M\cong P$ , is $M\cong N$?

Published on 25 Mar 2026 - 16:03

score 200 · Answer 14 · 2016-11-01 03:12:00

200

Views

A unique decomposition of an element to irreducible elements, not all of them primes

Published on 01 Nov 2016 - 3:12

score 143 · Answer 15 · 2016-11-02 14:50:50

143

Views

Equivalence to noetherian space uses Axiom of Choice?

Published on 02 Nov 2016 - 14:50

List Question

Same as some localization at $\mathbb{Q}[X]$?

a submodule of a noetherian module is also noetherian?

Local ring with finitely generated maximal ideal and $\cap_{n>0}\mathfrak{m}^n=\{0\}$ is Noetherian?

For $k$ a field, $k[x]$ is Noetherian but not Artinian?

Is $\oplus_{\mathbb{N}}\mathbb{Z}_2$ Artinian but not Noetherian?

If the source of a morphism is a noetherian scheme, the morphism is quasi-compact.

Is every finitely-generated $\mathbb{Z}$-module Noetherian?

Height of Product of Ideals

Structure theorem for commutative Noetherian rings

If $R_3$ is a finitely generated $R_2$-module and $R_2$ is a finitely generated free $R_1$-module, can we say that $R_3$ is a free $R_1$-module?

Why $P_i+N$ is stationnary when $P_i$ are submodules of $M$?

finitely many minimal prime ideals in $\mathbb{Z}[x]$?

$M$ be a finitely generated module over commutative unital ring $R$ , $N,P$ submodules , $P\subseteq N \subseteq M$ and $M\cong P$ , is $M\cong N$?

A unique decomposition of an element to irreducible elements, not all of them primes

Equivalence to noetherian space uses Axiom of Choice?

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