Question

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

score 310 · Answer 1 · 2026-03-29 14:59:08

310

Views

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

Published on 29 Mar 2026 - 14:59

score 50 · Answer 2 · 2021-02-06 09:44:48

50

Views

rank$(R\otimes_R S)$ is the composition of $f^∗$ and rank$(P)$.

Published on 06 Feb 2021 - 9:44

score 187 · Answer 3 · 2021-02-08 21:40:11

187

Views

Finitistic dimension conjeture for $A^{op} $ implies the strong Nakayama conjecture for A

Published on 08 Feb 2021 - 21:40

score 84 · Answer 4 · 2021-02-09 21:52:25

84

Views

proving $ \mathbb{Z}/12 \not\cong \mathbb{Z}/6 \oplus \mathbb{Z}/2$ as $\mathbb{Z}/12$ modules

Published on 09 Feb 2021 - 21:52

score 70 · Answer 5 · 2021-02-12 17:05:06

70

Views

Show gl dim$R=\sup\{pd_RI\}$, where I runs through all right ideals of $R$, where $R$ is not Wedderburn

Published on 12 Feb 2021 - 17:05

score 86 · Answer 6 · 2021-02-12 19:21:32

86

Views

Cycles in an acyclic complex of projectives are projectives.

Published on 12 Feb 2021 - 19:21

score 76 · Answer 7 · 2026-03-29 15:03:44

76

Views

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

Published on 29 Mar 2026 - 15:03

score 27 · Answer 8 · 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 82 · Answer 9 · 2021-02-20 12:14:23

82

Views

Projective modules restricted to smooth curves

Published on 20 Feb 2021 - 12:14

score 614 · Answer 10 · 2026-03-25 05:07:38

614

Views

Show that if $M$ is projective then $\operatorname{Hom}_R(M,.)$ is an exact functor.

Published on 25 Mar 2026 - 5:07

score 109 · Answer 11 · 2026-03-29 14:59:10

109

Views

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

Published on 29 Mar 2026 - 14:59

score 307 · Answer 12 · 2026-03-27 06:56:49

307

Views

Fractional ideal of a Dedekind domain A is a projective module

Published on 27 Mar 2026 - 6:56

score 234 · Answer 13 · 2026-03-26 16:10:11

234

Views

Graded version of Quillen and Suslin's theorem (graded projectives of graded-free and finite rank modules are graded-free)

Published on 26 Mar 2026 - 16:10

score 830 · Answer 14 · 2026-03-27 21:35:34

830

Views

Show that $\mathbb{Q}/\mathbb{Z}$ is an injective module and that it is not a projective module

Published on 27 Mar 2026 - 21:35

score 32 · Answer 15 · 2021-04-13 13:41:49

32

Views

Let $K$ be a field and set $R= K[x]/(x^2)$ so that $R=K \dotplus K\bar{x}$ with $\bar{x}^2=0$. Show that $R \ncong \bar{x}R \oplus \bar{x}R$

Published on 13 Apr 2021 - 13:41

List Question

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

rank$(R\otimes_R S)$ is the composition of $f^∗$ and rank$(P)$.

Finitistic dimension conjeture for $A^{op} $ implies the strong Nakayama conjecture for A

proving $ \mathbb{Z}/12 \not\cong \mathbb{Z}/6 \oplus \mathbb{Z}/2$ as $\mathbb{Z}/12$ modules

Show gl dim$R=\sup\{pd_RI\}$, where I runs through all right ideals of $R$, where $R$ is not Wedderburn

Cycles in an acyclic complex of projectives are projectives.

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.

Projective modules restricted to smooth curves

Show that if $M$ is projective then $\operatorname{Hom}_R(M,.)$ is an exact functor.

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

Fractional ideal of a Dedekind domain A is a projective module

Graded version of Quillen and Suslin's theorem (graded projectives of graded-free and finite rank modules are graded-free)

Show that $\mathbb{Q}/\mathbb{Z}$ is an injective module and that it is not a projective module

Let $K$ be a field and set $R= K[x]/(x^2)$ so that $R=K \dotplus K\bar{x}$ with $\bar{x}^2=0$. Show that $R \ncong \bar{x}R \oplus \bar{x}R$

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