Question

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.

score 487 · Answer 1 · 2026-03-25 13:34:22

487

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 Mar 2026 - 13:34

score 48 · Answer 2 · 2019-08-26 11:26:28

48

Views

Is $I:=\{a\in \text{End}_D(V):\text{dim}_D(aV)<\infty\}$ the only proper ideal of $\text{End}_D(V)$?

Published on 26 Aug 2019 - 11:26

score 100 · Answer 3 · 2026-03-25 18:47:21

100

Views

Dual of a bimodule

Published on 25 Mar 2026 - 18:47

score 145 · Answer 4 · 2026-03-25 13:34:27

145

Views

Basis of a free algebra: linear independence of tensor powers

Published on 25 Mar 2026 - 13:34

score 175 · Answer 5 · 2019-08-27 10:20:19

175

Views

Support of $\mathbb{C}[x]/(x^2-1)$ as $\mathbb C[x]$-module

Published on 27 Aug 2019 - 10:20

score 328 · Answer 6 · 2019-08-27 10:45:54

328

Views

$\operatorname{Supp}(\mathbb{Q}/\mathbb{Z})$ as a $\mathbb{Z}$-module

Published on 27 Aug 2019 - 10:45

score 58 · Answer 7 · 2019-08-27 20:37:42

58

Views

Is this an equivalent way to define the $k$-th exterior power of an $R$-module?

Published on 27 Aug 2019 - 20:37

score 173 · Answer 8 · 2026-03-31 23:49:39

173

Views

Left ideal in a ring $R$ is a subgroup

Published on 31 Mar 2026 - 23:49

score 903 · Answer 9 · 2026-03-25 06:21:01

903

Views

Grading on the tensor product of graded modules over a non-graded ring

Published on 25 Mar 2026 - 6:21

score 65 · Answer 10 · 2019-08-29 06:34:23

65

Views

$M_p\cong N_p$ implies $M_q\cong N_q$ for $p\subset q$?

Published on 29 Aug 2019 - 6:34

score 182 · Answer 11 · 2026-03-26 04:31:29

182

Views

Show that $\mathbb{Q}$ is finitely generated $\mathbb{Q}[x_1,x_2,\dots]$-module

Published on 26 Mar 2026 - 4:31

score 128 · Answer 12 · 2019-08-29 17:37:23

128

Views

If $I\subset J(R)$, ann$(M/IM)\subset p$, then $p$ is maximal?

Published on 29 Aug 2019 - 17:37

score 264 · Answer 13 · 2026-03-25 14:23:38

264

Views

Prove that the kernel of a homomorphism between Modules Satisfies the Universal Property

Published on 25 Mar 2026 - 14:23

score 45 · Answer 14 · 2019-08-30 12:14:08

45

Views

Correspondance between modules over locations and modules over rings whose scalar multiplications form isomorphisms

Published on 30 Aug 2019 - 12:14

score 1k · Answer 15 · 2019-08-30 15:43:57

1k

Views

Turning an Abelian Group into a Vector Space

Published on 30 Aug 2019 - 15:43

List Question

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.

Is $I:=\{a\in \text{End}_D(V):\text{dim}_D(aV)<\infty\}$ the only proper ideal of $\text{End}_D(V)$?

Dual of a bimodule

Basis of a free algebra: linear independence of tensor powers

Support of $\mathbb{C}[x]/(x^2-1)$ as $\mathbb C[x]$-module

$\operatorname{Supp}(\mathbb{Q}/\mathbb{Z})$ as a $\mathbb{Z}$-module

Is this an equivalent way to define the $k$-th exterior power of an $R$-module?

Left ideal in a ring $R$ is a subgroup

Grading on the tensor product of graded modules over a non-graded ring

$M_p\cong N_p$ implies $M_q\cong N_q$ for $p\subset q$?

Show that $\mathbb{Q}$ is finitely generated $\mathbb{Q}[x_1,x_2,\dots]$-module

If $I\subset J(R)$, ann$(M/IM)\subset p$, then $p$ is maximal?

Prove that the kernel of a homomorphism between Modules Satisfies the Universal Property

Correspondance between modules over locations and modules over rings whose scalar multiplications form isomorphisms

Turning an Abelian Group into a Vector Space

Trending Questions

Popular # Hahtags

Popular Questions