Question

tensor product of non free modules which is free

score 758 · Answer 1 · 2019-07-01 22:22:52

758

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tensor product of non free modules which is free

Published on 01 Jul 2019 - 22:22

score 35 · Answer 2 · 2026-03-25 11:16:51

35

Views

Question about isomorphism of direct sums (Step needed to show uniqueness of Elementary Divisors Theorem)

Published on 25 Mar 2026 - 11:16

score 418 · Answer 3 · 2019-07-30 02:09:22

418

Views

Noncommutative rings, matrices and homomorphisms of free modules

Published on 30 Jul 2019 - 2:09

score 484 · Answer 4 · 2019-08-06 20:32:23

484

Views

Decomposition of an abelian group subject to certain relations

Published on 06 Aug 2019 - 20:32

score 36 · Answer 5 · 2026-03-21 21:00:27

36

Views

About finitely presented algebras and free product

Published on 21 Mar 2026 - 21:00

score 73 · Answer 6 · 2019-08-10 00:36:52

73

Views

Linear independent set of the right cardinality is a basis.

Published on 10 Aug 2019 - 0:36

score 512 · Answer 7 · 2019-08-11 14:22:54

512

Views

Step in Bourbaki Commutative algebra exercise

Published on 11 Aug 2019 - 14:22

score 93 · Answer 8 · 2019-08-11 22:30:25

93

Views

Expression for the sum of submodules of a free $\Bbb{Z}$-module

Published on 11 Aug 2019 - 22:30

score 171 · Answer 9 · 2019-08-12 23:04:57

171

Views

GAP — compute kernel of a matrix with coefficients in a finite ring

Published on 12 Aug 2019 - 23:04

score 527 · Answer 10 · 2019-08-14 03:25:00

527

Views

For $F$ a finitely generated free $R$-module, prove $F \cong \text{Hom}_R (F, R)$

Published on 14 Aug 2019 - 3:25

score 214 · Answer 11 · 2019-08-15 14:02:47

214

Views

An equivalent definition of division rings (in module theory)

Published on 15 Aug 2019 - 14:02

score 412 · Answer 12 · 2019-08-20 22:48:36

412

Views

Extension of linearly independent set to a basis of a free module over a quotient of a Dedekind domain

Published on 20 Aug 2019 - 22:48

score 486 · Answer 13 · 2019-08-25 06:11:59

486

Views

Clarification: If $R$ is a ring, then $R^n\cong R^m$ as left $R$-modules if and only if they are also isomorphic as right $R$-modules.

Published on 25 Aug 2019 - 6:11

score 144 · Answer 14 · 2019-08-26 18:09:06

144

Views

Basis of a free algebra: linear independence of tensor powers

Published on 26 Aug 2019 - 18:09

score 600 · Answer 15 · 2019-09-03 15:07:06

600

Views

Every finitely generated module of a PID is a direct sum of a free module and of a torsion module

Published on 03 Sep 2019 - 15:07

List Question

tensor product of non free modules which is free

Question about isomorphism of direct sums (Step needed to show uniqueness of Elementary Divisors Theorem)

Noncommutative rings, matrices and homomorphisms of free modules

Decomposition of an abelian group subject to certain relations

About finitely presented algebras and free product

Linear independent set of the right cardinality is a basis.

Step in Bourbaki Commutative algebra exercise

Expression for the sum of submodules of a free $\Bbb{Z}$-module

GAP — compute kernel of a matrix with coefficients in a finite ring

For $F$ a finitely generated free $R$-module, prove $F \cong \text{Hom}_R (F, R)$

An equivalent definition of division rings (in module theory)

Extension of linearly independent set to a basis of a free module over a quotient of a Dedekind domain

Clarification: If $R$ is a ring, then $R^n\cong R^m$ as left $R$-modules if and only if they are also isomorphic as right $R$-modules.

Basis of a free algebra: linear independence of tensor powers

Every finitely generated module of a PID is a direct sum of a free module and of a torsion module

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