Question

Graded version of Quillen and Suslin's theorem (graded projectives of graded-free and finite rank modules are graded-free)

score 234 · Answer 1 · 2026-03-26 16:10:11

234

Views

Graded version of Quillen and Suslin's theorem (graded projectives of graded-free and finite rank modules are graded-free)

Published on 26 Mar 2026 - 16:10

score 168 · Answer 2 · 2021-03-19 08:19:09

168

Views

In a PID if the rank (finite) of a free module and its submodule are the same, then are they necessary equal?

Published on 19 Mar 2021 - 8:19

score 29 · Answer 3 · 2021-03-23 12:46:21

29

Views

Linear dependence relations in $R$-modules (of polynomials)

Published on 23 Mar 2021 - 12:46

score 56 · Answer 4 · 2021-04-03 08:18:05

56

Views

Linearly dependent sets of elements in finitely generated module over an integral domain

Published on 03 Apr 2021 - 8:18

score 133 · Answer 5 · 2021-04-07 12:58:16

133

Views

On cokernel via localization of morphism of modules

Published on 07 Apr 2021 - 12:58

score 30 · Answer 6 · 2021-04-08 12:43:19

30

Views

Is the map $h(m)=(rm_1,m_2)$ a homothety of $M=M_1\bigoplus M_2$?

Published on 08 Apr 2021 - 12:43

score 111 · Answer 7 · 2021-04-09 00:44:55

111

Views

Does the following isomorphism hold true for modules: $(M\cap N)/(aM\cap N)\cong (0:_Ma)\cap N ?$

Published on 09 Apr 2021 - 0:44

score 232 · Answer 8 · 2026-03-26 23:09:50

232

Views

On formal sums, free modules and free abelian groups.

Published on 26 Mar 2026 - 23:09

score 149 · Answer 9 · 2021-04-26 18:15:53

149

Views

Endomorphisms of modules of finite rank over integral domain have stabilizing kernels

Published on 26 Apr 2021 - 18:15

score 567 · Answer 10 · 2021-04-30 15:22:30

567

Views

If $R$-Module M has a submodule $N$ such that $N/M$ is free, then there exist a submodule $L$ such that $M = N \oplus L$

Published on 30 Apr 2021 - 15:22

score 66 · Answer 11 · 2021-05-02 02:31:14

66

Views

finite presentation $\mathcal{O}_X$ module over an integral scheme

Published on 02 May 2021 - 2:31

score 47 · Answer 12 · 2021-05-03 17:36:59

47

Views

$\Bbb{Q}/\Bbb{Z}$ as a $\Bbb{Z}$-module has no free non-trivial submodules

Published on 03 May 2021 - 17:36

score 59 · Answer 13 · 2021-05-05 16:19:36

59

Views

Proving wrong that $R$ is a field

Published on 05 May 2021 - 16:19

score 73 · Answer 14 · 2026-03-26 22:52:26

73

Views

Neukirch prop(2.10) Why $adM$ implies $M$ is free?

Published on 26 Mar 2026 - 22:52

score 66 · Answer 15 · 2021-05-15 09:14:04

66

Views

Are the following modules free and/or isomorphic?

Published on 15 May 2021 - 9:14

List Question

Graded version of Quillen and Suslin's theorem (graded projectives of graded-free and finite rank modules are graded-free)

In a PID if the rank (finite) of a free module and its submodule are the same, then are they necessary equal?

Linear dependence relations in $R$-modules (of polynomials)

Linearly dependent sets of elements in finitely generated module over an integral domain

On cokernel via localization of morphism of modules

Is the map $h(m)=(rm_1,m_2)$ a homothety of $M=M_1\bigoplus M_2$?

Does the following isomorphism hold true for modules: $(M\cap N)/(aM\cap N)\cong (0:_Ma)\cap N ?$

On formal sums, free modules and free abelian groups.

Endomorphisms of modules of finite rank over integral domain have stabilizing kernels

If $R$-Module M has a submodule $N$ such that $N/M$ is free, then there exist a submodule $L$ such that $M = N \oplus L$

finite presentation $\mathcal{O}_X$ module over an integral scheme

$\Bbb{Q}/\Bbb{Z}$ as a $\Bbb{Z}$-module has no free non-trivial submodules

Proving wrong that $R$ is a field

Neukirch prop(2.10) Why $adM$ implies $M$ is free?

Are the following modules free and/or isomorphic?

Trending Questions

Popular # Hahtags

Popular Questions