Question

If $R/I$ and $R/J$ are Noetherian (resp. Artinian), then so is $R/(I \cap J)$

score 966 · Answer 1 · 2016-05-14 08:20:51

966

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If $R/I$ and $R/J$ are Noetherian (resp. Artinian), then so is $R/(I \cap J)$

Published on 14 May 2016 - 8:20

score 114 · Answer 2 · 2026-04-30 11:22:50

114

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$M$ is a $\Bbb Z/36 \Bbb Z$-module and $\bar 6 \cdot M = \{0\}$. Prove: $M$ is Noetherian $\iff$ it's Artinian

Published on 30 Apr 2026 - 11:22

score 108 · Answer 3 · 2026-04-10 02:00:14

108

Views

Quotient of free algebra - Noetherian/Artinian?

Published on 10 Apr 2026 - 2:00

score 2k · Answer 4 · 2026-05-07 19:22:34

2k

Views

Maximal ideal in a local artinian ring.

Published on 07 May 2026 - 19:22

score 1.9k · Answer 5 · 2026-03-25 17:21:37

1.9k

Views

Classification of local Artin (commutative) rings which are finite over an algebraically closed field

Published on 25 Mar 2026 - 17:21

score 82 · Answer 6 · 2026-03-25 17:18:22

82

Views

Is every finitely generated linearly compact module Artinian?

Published on 25 Mar 2026 - 17:18

score 223 · Answer 7 · 2026-03-25 17:19:16

223

Views

What is wrong in my proof for "every linearly compact module is Artinian"

Published on 25 Mar 2026 - 17:19

score 787 · Answer 8 · 2026-03-25 17:24:30

787

Views

Given an Artinian ring $(A, \mathfrak{m})$, show that $\mathfrak{m}$ is nilpotent.

Published on 25 Mar 2026 - 17:24

score 42 · Answer 9 · 2026-03-25 17:24:54

42

Views

Are these rings Artinian? Are the composition series correct?

Published on 25 Mar 2026 - 17:24

score 329 · Answer 10 · 2026-03-25 17:21:30

329

Views

Prove that the ring $\mathbb{Q}[x,y,z]\over (x^2,y^3,z^7)$ is Artin.

Published on 25 Mar 2026 - 17:21

score 1k · Answer 11 · 2026-03-25 17:23:32

1k

Views

Artinian ring without zero divisors is a field

Published on 25 Mar 2026 - 17:23

score 713 · Answer 12 · 2026-04-17 05:21:09

713

Views

Decomposition of the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}[i]$ into a product of fields

Published on 17 Apr 2026 - 5:21

score 271 · Answer 13 · 2026-03-25 17:19:48

271

Views

About right ideals of a $\mathbb{Q}$-subalgebra of $M_2(\mathbb{R})$, $A = \begin{bmatrix} \mathbb{Q} & \mathbb{R}\\ 0 & \mathbb{R} \end{bmatrix}$

Published on 25 Mar 2026 - 17:19

score 113 · Answer 14 · 2026-03-25 17:22:44

113

Views

A simple Artinian left quotient ring of a left Noetherian domain

Published on 25 Mar 2026 - 17:22

score 287 · Answer 15 · 2026-03-25 17:21:32

287

Views

The ring of quotients of the first Weyl algebra

Published on 25 Mar 2026 - 17:21

List Question

If $R/I$ and $R/J$ are Noetherian (resp. Artinian), then so is $R/(I \cap J)$

$M$ is a $\Bbb Z/36 \Bbb Z$-module and $\bar 6 \cdot M = \{0\}$. Prove: $M$ is Noetherian $\iff$ it's Artinian

Quotient of free algebra - Noetherian/Artinian?

Maximal ideal in a local artinian ring.

Classification of local Artin (commutative) rings which are finite over an algebraically closed field

Is every finitely generated linearly compact module Artinian?

What is wrong in my proof for "every linearly compact module is Artinian"

Given an Artinian ring $(A, \mathfrak{m})$, show that $\mathfrak{m}$ is nilpotent.

Are these rings Artinian? Are the composition series correct?

Prove that the ring $\mathbb{Q}[x,y,z]\over (x^2,y^3,z^7)$ is Artin.

Artinian ring without zero divisors is a field

Decomposition of the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}[i]$ into a product of fields

About right ideals of a $\mathbb{Q}$-subalgebra of $M_2(\mathbb{R})$, $A = \begin{bmatrix} \mathbb{Q} & \mathbb{R}\\ 0 & \mathbb{R} \end{bmatrix}$

A simple Artinian left quotient ring of a left Noetherian domain

The ring of quotients of the first Weyl algebra

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