Question

Anderson’s exercise 8.3 on “generating and cogenerating”

score 80 · Answer 1 · 2018-11-14 19:09:30

80

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Anderson’s exercise 8.3 on “generating and cogenerating”

Published on 14 Nov 2018 - 19:09

score 220 · Answer 2 · 2018-11-15 01:14:59

220

Views

Computing $\mathrm{Hom}_{\mathbb Z}(\mathbb Z_n,G)$ by left-exactness

Published on 15 Nov 2018 - 1:14

score 45 · Answer 3 · 2018-11-15 03:12:22

45

Views

Let $A$ be a ring and $I$ an infinite set. ¿Does $A^{(I)}\cong A^{(J)}$ implies $|I|=|J|$?

Published on 15 Nov 2018 - 3:12

score 139 · Answer 4 · 2018-11-15 10:59:06

139

Views

R with different Jacobson radical and nilradical (both non-zero)

Published on 15 Nov 2018 - 10:59

score 83 · Answer 5 · 2026-03-26 19:01:49

$M_1$, $M_2$ are submodules of a module $M$, then $M = M_1 + M_2$ and $M_1 \cap M_2 = 0$ implies M is isomorphic to $M_1 \oplus M_2$?

score 40 · Answer 6 · 2026-03-17 21:59:03

40

Views

Trying to understand more about polynomials in noncommuting variables.

Published on 17 Mar 2026 - 21:59

score 503 · Answer 7 · 2026-03-25 05:07:43

503

Views

Every $R$-module $M$ has a maximal linearly independent subset $A$. If submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$.

Published on 25 Mar 2026 - 5:07

score 954 · Answer 8 · 2018-11-16 09:09:09

954

Views

$Ae$ indecomposable iff $e$ is a primitive idempotent

Published on 16 Nov 2018 - 9:09

score 238 · Answer 9 · 2018-11-16 18:10:23

238

Views

Dual of quasi-isomorphism of chain complexes.

Published on 16 Nov 2018 - 18:10

score 146 · Answer 10 · 2026-03-25 09:28:45

146

Views

If P is a Projective R Module and Q is an Injective R module then $P \otimes_{R} Q$ is Injective

Published on 25 Mar 2026 - 9:28

score 209 · Answer 11 · 2026-03-25 05:09:44

209

Views

Clarification on what is and is not a free abelian group

Published on 25 Mar 2026 - 5:09

score 109 · Answer 12 · 2026-03-25 05:10:00

109

Views

Understanding proof of: A submodule of a free module of finite rank is also free of finite rank?

Published on 25 Mar 2026 - 5:10

score 375 · Answer 13 · 2018-11-17 19:14:51

375

Views

finite product of artinian rings is artinian

Published on 17 Nov 2018 - 19:14

score 130 · Answer 14 · 2026-03-25 14:25:22

130

Views

Why is the decomposition of $\operatorname{Tor}(M)$ into cyclic modules a "subset" of the decomposition of $M$?

Published on 25 Mar 2026 - 14:25

score 111 · Answer 15 · 2018-11-18 14:43:58

111

Views

$A \oplus B$ a direct sum of $R$-modules. Possible to have a submodule $X$ of $A \oplus B$ s.t. $X \cap A = 0 = X \cap B$?

Published on 18 Nov 2018 - 14:43

List Question

Anderson’s exercise 8.3 on “generating and cogenerating”

Computing $\mathrm{Hom}_{\mathbb Z}(\mathbb Z_n,G)$ by left-exactness

Let $A$ be a ring and $I$ an infinite set. ¿Does $A^{(I)}\cong A^{(J)}$ implies $|I|=|J|$?

R with different Jacobson radical and nilradical (both non-zero)

$M_1$, $M_2$ are submodules of a module $M$, then $M = M_1 + M_2$ and $M_1 \cap M_2 = 0$ implies M is isomorphic to $M_1 \oplus M_2$?

Trying to understand more about polynomials in noncommuting variables.

Every $R$-module $M$ has a maximal linearly independent subset $A$. If submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$.

$Ae$ indecomposable iff $e$ is a primitive idempotent

Dual of quasi-isomorphism of chain complexes.

If P is a Projective R Module and Q is an Injective R module then $P \otimes_{R} Q$ is Injective

Clarification on what is and is not a free abelian group

Understanding proof of: A submodule of a free module of finite rank is also free of finite rank?

finite product of artinian rings is artinian

Why is the decomposition of $\operatorname{Tor}(M)$ into cyclic modules a "subset" of the decomposition of $M$?

$A \oplus B$ a direct sum of $R$-modules. Possible to have a submodule $X$ of $A \oplus B$ s.t. $X \cap A = 0 = X \cap B$?

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