Question

If finite product of maximal ideals of ring $R$ is zero, then $R$ is Noetherian$\iff R$ is Artinian

score 1k · Answer 1 · 2026-05-10 14:49:23

1k

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If finite product of maximal ideals of ring $R$ is zero, then $R$ is Noetherian$\iff R$ is Artinian

Published on 10 May 2026 - 14:49

score 534 · Answer 2 · 2026-05-10 17:16:48

534

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In an integrally closed, Noetherian, local, integral domain of dimension $1$, the maximal ideal $P$ is eventually principal

Published on 10 May 2026 - 17:16

score 70 · Answer 3 · 2026-04-13 17:48:02

70

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Show that any non-zero prime ideal of $R$ is invertible.

Published on 13 Apr 2026 - 17:48

score 1k · Answer 4 · 2026-04-10 19:01:30

1k

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How to show the following ring is not Noetherian

Published on 10 Apr 2026 - 19:01

score 65 · Answer 5 · 2026-05-11 02:50:34

65

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If $\mathfrak{p}$ is a prime such that $M_\mathfrak{p} \neq 0$, then $\mathfrak{p}$ contains an associated prime of $M$

Published on 11 May 2026 - 2:50

score 34 · Answer 6 · 2026-04-10 17:46:14

34

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A nice way to phrase this theorem about ideals in Noetherian rings?

Published on 10 Apr 2026 - 17:46

score 73 · Answer 7 · 2026-04-17 00:48:50

73

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Grillet's "Abstract Algebra", p. 148, ex. 3: Noetherian subring of a ring

Published on 17 Apr 2026 - 0:48

score 516 · Answer 8 · 2026-04-14 18:40:42

516

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Noetherian ring has an associated prime ideal

Published on 14 Apr 2026 - 18:40

score 238 · Answer 9 · 2026-04-16 21:42:37

238

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Is Axiom of Dependence Choice necessary for two definitions of Noetherian to be equivalent?

Published on 16 Apr 2026 - 21:42

score 213 · Answer 10 · 2026-05-10 18:27:58

213

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In a commutative local Noetherian ring $R$ with maximal ideal $J$, if $J$ is not nilpotent then $R$ is an integral domain.

Published on 10 May 2026 - 18:27

score 904 · Answer 11 · 2026-05-11 01:47:38

904

Views

Hartshorne Exercise III 3.2: $X$ is affine iff every component is affine

Published on 11 May 2026 - 1:47

score 198 · Answer 12 · 2026-04-16 14:25:05

198

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If $R$ is a right-Noetherian ring and the Jacobson radical satisfies the right Artin-Rees property, then $\bigcap^{\infty}_{n=1}\text{Jac}(R)^n = 0$

Published on 16 Apr 2026 - 14:25

score 251 · Answer 13 · 2026-04-17 14:44:06

251

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Ring of Laurent Polynomials is Noetherian Implies Polynomial Ring is Noetherian?

Published on 17 Apr 2026 - 14:44

score 144 · Answer 14 · 2026-05-11 05:03:54

144

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Non-Noetherian Domain Which is Locally Noetherian

Published on 11 May 2026 - 5:03

score 209 · Answer 15 · 2026-04-13 14:07:33

209

Views

Locally Noetherian Domain With Finitely Many Prime Ideals

Published on 13 Apr 2026 - 14:07

List Question

If finite product of maximal ideals of ring $R$ is zero, then $R$ is Noetherian$\iff R$ is Artinian

In an integrally closed, Noetherian, local, integral domain of dimension $1$, the maximal ideal $P$ is eventually principal

Show that any non-zero prime ideal of $R$ is invertible.

How to show the following ring is not Noetherian

If $\mathfrak{p}$ is a prime such that $M_\mathfrak{p} \neq 0$, then $\mathfrak{p}$ contains an associated prime of $M$

A nice way to phrase this theorem about ideals in Noetherian rings?

Grillet's "Abstract Algebra", p. 148, ex. 3: Noetherian subring of a ring

Noetherian ring has an associated prime ideal

Is Axiom of Dependence Choice necessary for two definitions of Noetherian to be equivalent?

In a commutative local Noetherian ring $R$ with maximal ideal $J$, if $J$ is not nilpotent then $R$ is an integral domain.

Hartshorne Exercise III 3.2: $X$ is affine iff every component is affine

If $R$ is a right-Noetherian ring and the Jacobson radical satisfies the right Artin-Rees property, then $\bigcap^{\infty}_{n=1}\text{Jac}(R)^n = 0$

Ring of Laurent Polynomials is Noetherian Implies Polynomial Ring is Noetherian?

Non-Noetherian Domain Which is Locally Noetherian

Locally Noetherian Domain With Finitely Many Prime Ideals

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