Question

If $R$ is a PID and $M$ is finitely generated $R$-module, why is $M=\operatorname{coker}[R^m\to R^n]$

score 55 · Answer 1 · 2019-09-11 19:12:48

55

Views

If $R$ is a PID and $M$ is finitely generated $R$-module, why is $M=\operatorname{coker}[R^m\to R^n]$

Published on 11 Sep 2019 - 19:12

score 147 · Answer 2 · 2019-09-12 01:25:33

147

Views

Non-perfect ideal $I$ such that $\mathrm{projdim}(R/I)<\infty$

Published on 12 Sep 2019 - 1:25

score 681 · Answer 3 · 2019-09-14 06:03:10

681

Views

Quotient of annihilator module

Published on 14 Sep 2019 - 6:03

score 320 · Answer 4 · 2019-09-20 18:51:24

320

Views

A ring is Noetherian if its prime ideals are ﬁnitely generated: simplified proof

Published on 20 Sep 2019 - 18:51

score 287 · Answer 5 · 2019-09-28 11:13:00

287

Views

Can $A[x]$ be isomorphic to $A$ if $A$ is noetherian?

Published on 28 Sep 2019 - 11:13

score 163 · Answer 6 · 2019-09-28 20:25:22

163

Views

Localization and dual commute

Published on 28 Sep 2019 - 20:25

score 390 · Answer 7 · 2026-03-25 10:04:43

390

Views

Is an epimorphic endomorphism of a noetherian commutative ring necessarily an isomorphism?

Published on 25 Mar 2026 - 10:04

score 186 · Answer 8 · 2019-10-01 12:42:20

186

Views

Does the Cantor-Schröder-Bernstein Theorem hold in the category opposite to the category of noetherian commutative rings?

Published on 01 Oct 2019 - 12:42

score 709 · Answer 9 · 2014-03-12 20:16:32

709

Views

A non-zero and non-invertible element in a noetherian integral domain has a decomposition into irreducible elements

Published on 12 Mar 2014 - 20:16

score 546 · Answer 10 · 2014-03-14 02:43:18

546

Views

Noetherian Ring and Homomorphic Image

Published on 14 Mar 2014 - 2:43

score 3.3k · Answer 11 · 2026-03-25 12:27:25

3.3k

Views

In a noetherian integral domain every non invertible element is a product of irreducible elements

Published on 25 Mar 2026 - 12:27

score 267 · Answer 12 · 2014-03-19 18:01:19

267

Views

Ring isomorphisms maintain noetherian property?

Published on 19 Mar 2014 - 18:01

score 2.4k · Answer 13 · 2014-03-21 08:45:17

2.4k

Views

Finitely generated k-algebra is Noetherian

Published on 21 Mar 2014 - 8:45

score 823 · Answer 14 · 2014-03-23 15:41:09

823

Views

Primary ideals in Noetherian rings

Published on 23 Mar 2014 - 15:41

score 587 · Answer 15 · 2014-03-31 15:29:52

587

Views

Support of $\operatorname{Hom}(R/I, M)$

Published on 31 Mar 2014 - 15:29

List Question

If $R$ is a PID and $M$ is finitely generated $R$-module, why is $M=\operatorname{coker}[R^m\to R^n]$

Non-perfect ideal $I$ such that $\mathrm{projdim}(R/I)<\infty$

Quotient of annihilator module

A ring is Noetherian if its prime ideals are ﬁnitely generated: simplified proof

Can $A[x]$ be isomorphic to $A$ if $A$ is noetherian?

Localization and dual commute

Is an epimorphic endomorphism of a noetherian commutative ring necessarily an isomorphism?

Does the Cantor-Schröder-Bernstein Theorem hold in the category opposite to the category of noetherian commutative rings?

A non-zero and non-invertible element in a noetherian integral domain has a decomposition into irreducible elements

Noetherian Ring and Homomorphic Image

In a noetherian integral domain every non invertible element is a product of irreducible elements

Ring isomorphisms maintain noetherian property?

Finitely generated k-algebra is Noetherian

Primary ideals in Noetherian rings

Support of $\operatorname{Hom}(R/I, M)$

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