Question

Chain homotopy equivalence after tensoring

score 144 · Answer 1 · 2020-07-02 17:08:08

144

Views

Chain homotopy equivalence after tensoring

Published on 02 Jul 2020 - 17:08

score 392 · Answer 2 · 2020-07-04 14:36:34

392

Views

Every finitely generated free Module over a commutative Ring is flat

Published on 04 Jul 2020 - 14:36

score 40 · Answer 3 · 2020-07-05 15:43:45

40

Views

Are free Artinian modules finitely generated?

Published on 05 Jul 2020 - 15:43

score 301 · Answer 4 · 2020-07-07 16:53:42

301

Views

Are projective modules over $\mathbb{Z}[x_1,...,x_m]$ free?

Published on 07 Jul 2020 - 16:53

score 587 · Answer 5 · 2020-07-09 03:21:23

587

Views

Why is $\mathbb{Z}_{6}$ a free $\mathbb{Z}_{6}$ module?

Published on 09 Jul 2020 - 3:21

score 37 · Answer 6 · 2026-03-25 12:49:46

37

Views

If $f_x$ is injective whenever it is surjective on $M$, then is $f_x$ injective whenever it is surjective on $\bigoplus_{i\in \Bbb N}M$?

Published on 25 Mar 2026 - 12:49

score 40 · Answer 7 · 2020-07-15 06:56:27

40

Views

An Ideal Related to a Polynomial Ring: Theorem 3.4 of Chapter XXI of Lang's "Algebra" (Revised Third Edition)

Published on 15 Jul 2020 - 6:56

score 48 · Answer 8 · 2020-07-26 23:54:28

48

Views

Prove that if $P\oplus R^m\cong R^\infty $ , then $P\cong R^\infty$

Published on 26 Jul 2020 - 23:54

score 41 · Answer 9 · 2020-07-29 06:05:25

41

Views

A right $R$ module $M$ has finitely many submodules iff right $R$ module $M^n$ has finitely many submodules for any positive integer $n$.

Published on 29 Jul 2020 - 6:05

score 465 · Answer 10 · 2020-07-30 02:39:41

465

Views

Clarification in a proof that free modules are flat

Published on 30 Jul 2020 - 2:39

score 169 · Answer 11 · 2016-07-12 23:21:22

169

Views

$(R^{ \oplus A_1 })^{ \oplus A_2 } \cong R^{ \oplus (A_1 \times A_2)}$ - not a categorical proof

Published on 12 Jul 2016 - 23:21

score 257 · Answer 12 · 2016-07-25 18:28:40

257

Views

Property of free modules

Published on 25 Jul 2016 - 18:28

score 859 · Answer 13 · 2016-07-27 01:12:34

859

Views

free modules being isomorphic with different dimension

Published on 27 Jul 2016 - 1:12

score 423 · Answer 14 · 2026-03-22 07:25:13

423

Views

Verma flag (or standard filtration) on a direct summand.

Published on 22 Mar 2026 - 7:25

score 211 · Answer 15 · 2016-08-30 23:09:29

211

Views

$R^n \cong R^m$ if and only if $n=m$ without assuming the result for vector spaces

Published on 30 Aug 2016 - 23:09

List Question

Chain homotopy equivalence after tensoring

Every finitely generated free Module over a commutative Ring is flat

Are free Artinian modules finitely generated?

Are projective modules over $\mathbb{Z}[x_1,...,x_m]$ free?

Why is $\mathbb{Z}_{6}$ a free $\mathbb{Z}_{6}$ module?

If $f_x$ is injective whenever it is surjective on $M$, then is $f_x$ injective whenever it is surjective on $\bigoplus_{i\in \Bbb N}M$?

An Ideal Related to a Polynomial Ring: Theorem 3.4 of Chapter XXI of Lang's "Algebra" (Revised Third Edition)

Prove that if $P\oplus R^m\cong R^\infty $ , then $P\cong R^\infty$

A right $R$ module $M$ has finitely many submodules iff right $R$ module $M^n$ has finitely many submodules for any positive integer $n$.

Clarification in a proof that free modules are flat

$(R^{ \oplus A_1 })^{ \oplus A_2 } \cong R^{ \oplus (A_1 \times A_2)}$ - not a categorical proof

Property of free modules

free modules being isomorphic with different dimension

Verma flag (or standard filtration) on a direct summand.

$R^n \cong R^m$ if and only if $n=m$ without assuming the result for vector spaces

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