Question

A free $A$-module $M$ which is also finitely-generated.

score 43 · Answer 1 · 2021-01-07 18:15:52

43

Views

A free $A$-module $M$ which is also finitely-generated.

Published on 07 Jan 2021 - 18:15

score 56 · Answer 2 · 2021-01-09 22:28:28

56

Views

$\mathbb{Z}^{\oplus \mathbb{N}} = \mathbb{Z}^{\oplus \mathbb{N}} \oplus \mathbb{Z}^{\oplus \mathbb{N}}$.

Published on 09 Jan 2021 - 22:28

score 495 · Answer 3 · 2021-01-12 22:04:47

495

Views

How to prove that there is an isomorphism between a free module and a direct sum of copies of $R$?

Published on 12 Jan 2021 - 22:04

score 55 · Answer 4 · 2021-01-13 11:39:49

55

Views

Prove that for every set there is a free module on that set

Published on 13 Jan 2021 - 11:39

score 21 · Answer 5 · 2026-02-23 11:29:31

21

Views

Taking quotients over isometric groups and determining the cokernel of transformations

Published on 23 Feb 2026 - 11:29

score 309 · Answer 6 · 2021-02-01 09:18:38

309

Views

If $F$ is free, then the functor $M \mapsto \operatorname{Hom}_A(F,M)$ is exact.

Published on 01 Feb 2021 - 9:18

score 162 · Answer 7 · 2021-02-06 16:10:14

162

Views

Showing that an ideal is not a free $R-$module.

Published on 06 Feb 2021 - 16:10

score 342 · Answer 8 · 2021-02-07 23:28:06

342

Views

What are the implications of a surjection between free modules?

Published on 07 Feb 2021 - 23:28

score 114 · Answer 9 · 2021-02-09 14:08:36

114

Views

Why I should show that no element of $R$ divides both $2$ and $1 + \sqrt{-5}$?

Published on 09 Feb 2021 - 14:08

score 75 · Answer 10 · 2021-02-15 05:31:34

75

Views

Why every module $P$ is the quotient of a free module implies the existence of this exact sequence specifically?

Published on 15 Feb 2021 - 5:31

score 27 · Answer 11 · 2026-03-28 08:10:31

27

Views

Understanding an example of a finitely generated projective module which is not free.

Published on 28 Mar 2026 - 8:10

score 368 · Answer 12 · 2021-02-28 16:28:09

368

Views

Tensor product of two spaces one of them is free module

Published on 28 Feb 2021 - 16:28

score 49 · Answer 13 · 2021-03-02 03:14:58

49

Views

understanding the contradiction of $I$ is not free.

Published on 02 Mar 2021 - 3:14

score 108 · Answer 14 · 2021-03-03 01:28:09

108

Views

$V\cdot\operatorname{Rad}(R)\subseteq\operatorname{Rad}(V)$ where equality holds if $V$ is free or projective.

Published on 03 Mar 2021 - 1:28

score 125 · Answer 15 · 2021-03-10 14:09:50

125

Views

maps forming a basis for integer function as $\mathbb{Z}$-module

Published on 10 Mar 2021 - 14:09

List Question

A free $A$-module $M$ which is also finitely-generated.

$\mathbb{Z}^{\oplus \mathbb{N}} = \mathbb{Z}^{\oplus \mathbb{N}} \oplus \mathbb{Z}^{\oplus \mathbb{N}}$.

How to prove that there is an isomorphism between a free module and a direct sum of copies of $R$?

Prove that for every set there is a free module on that set

Taking quotients over isometric groups and determining the cokernel of transformations

If $F$ is free, then the functor $M \mapsto \operatorname{Hom}_A(F,M)$ is exact.

Showing that an ideal is not a free $R-$module.

What are the implications of a surjection between free modules?

Why I should show that no element of $R$ divides both $2$ and $1 + \sqrt{-5}$?

Why every module $P$ is the quotient of a free module implies the existence of this exact sequence specifically?

Understanding an example of a finitely generated projective module which is not free.

Tensor product of two spaces one of them is free module

understanding the contradiction of $I$ is not free.

$V\cdot\operatorname{Rad}(R)\subseteq\operatorname{Rad}(V)$ where equality holds if $V$ is free or projective.

maps forming a basis for integer function as $\mathbb{Z}$-module

Trending Questions

Popular # Hahtags

Popular Questions