Question

$\mathbb{R} ^ \mathbb{R}$ is a commutative ring with identity that is neither noetherian nor artinian.

score 165 · Answer 1 · 2026-05-10 16:05:04

165

Views

$\mathbb{R} ^ \mathbb{R}$ is a commutative ring with identity that is neither noetherian nor artinian.

Published on 10 May 2026 - 16:05

score 67 · Answer 2 · 2026-04-14 09:59:39

67

Views

Noether Equation on Artin and Milgram Galois Theory book

Published on 14 Apr 2026 - 9:59

score 122 · Answer 3 · 2026-04-16 00:21:52

122

Views

If a ring is noetherian, then so is its subring

Published on 16 Apr 2026 - 0:21

score 191 · Answer 4 · 2026-04-16 03:20:59

191

Views

Exercise VIII.3.9 in Hungerford's Algebra

Published on 16 Apr 2026 - 3:20

score 1.3k · Answer 5 · 2026-04-16 06:25:25

1.3k

Views

$K[X^2,X^3]\subset K[X]$ is a Noetherian domain and all its prime ideals are maximal

Published on 16 Apr 2026 - 6:25

score 51 · Answer 6 · 2026-05-03 10:25:48

51

Views

Existence of ideal in Cohen-Macaulay ring, going modulo which still gives Cohen-Macaulay

Published on 03 May 2026 - 10:25

score 383 · Answer 7 · 2026-05-10 16:12:08

383

Views

Ideal in polynomial ring which contains no non-zero prime ideal

Published on 10 May 2026 - 16:12

score 303 · Answer 8 · 2026-04-08 21:47:07

303

Views

Is every commutative ring a limit of noetherian rings?

Published on 08 Apr 2026 - 21:47

score 545 · Answer 9 · 2026-04-10 10:30:38

545

Views

Is it true every ideal of $\mathbb{Z}[x]$ is finitely generated?

Published on 10 Apr 2026 - 10:30

score 958 · Answer 10 · 2026-04-13 12:32:30

958

Views

Why does every component have to appear in a decomposition of a Noetherian space into irreducible components?

Published on 13 Apr 2026 - 12:32

score 136 · Answer 11 · 2026-05-10 14:49:36

136

Views

The finite generation of $M$, to conclude $M$ is Noetherian (when $R$ is Noetherian).

Published on 10 May 2026 - 14:49

score 96 · Answer 12 · 2026-05-11 02:50:05

96

Views

$R_{\mathfrak{p}}$ is integrally closed in its total quotient ring, then $R$ is integrally closed in its total quotient ring.

Published on 11 May 2026 - 2:50

score 230 · Answer 13 · 2026-05-10 14:48:37

230

Views

Consequence of epimorphism from Noetherian $R$-module

Published on 10 May 2026 - 14:48

score 245 · Answer 14 · 2026-05-10 18:30:39

245

Views

A Noetherian domain $R$ is a UFD iff $I_{fg}=\{h\in R \mid hg\in(f)\}$ is principal for all $f,g\in R$.

Published on 10 May 2026 - 18:30

score 115 · Answer 15 · 2026-04-15 17:30:19

115

Views

Let R be noetherian. If ideal $I$ is maximal with respect to the property that $R/I$ is not of finite length, prove $I$ is prime.

Published on 15 Apr 2026 - 17:30

List Question

$\mathbb{R} ^ \mathbb{R}$ is a commutative ring with identity that is neither noetherian nor artinian.

Noether Equation on Artin and Milgram Galois Theory book

If a ring is noetherian, then so is its subring

Exercise VIII.3.9 in Hungerford's Algebra

$K[X^2,X^3]\subset K[X]$ is a Noetherian domain and all its prime ideals are maximal

Existence of ideal in Cohen-Macaulay ring, going modulo which still gives Cohen-Macaulay

Ideal in polynomial ring which contains no non-zero prime ideal

Is every commutative ring a limit of noetherian rings?

Is it true every ideal of $\mathbb{Z}[x]$ is finitely generated?

Why does every component have to appear in a decomposition of a Noetherian space into irreducible components?

The finite generation of $M$, to conclude $M$ is Noetherian (when $R$ is Noetherian).

$R_{\mathfrak{p}}$ is integrally closed in its total quotient ring, then $R$ is integrally closed in its total quotient ring.

Consequence of epimorphism from Noetherian $R$-module

A Noetherian domain $R$ is a UFD iff $I_{fg}=\{h\in R \mid hg\in(f)\}$ is principal for all $f,g\in R$.

Let R be noetherian. If ideal $I$ is maximal with respect to the property that $R/I$ is not of finite length, prove $I$ is prime.

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