Question

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

score 312 · Answer 1 · 2026-04-17 07:18:03

312

Views

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

Published on 17 Apr 2026 - 7:18

score 52 · Answer 2 · 2026-04-17 02:51:17

52

Views

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

Published on 17 Apr 2026 - 2:51

score 190 · Answer 3 · 2026-04-13 19:22:18

190

Views

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

Published on 13 Apr 2026 - 19:22

score 87 · Answer 4 · 2026-04-14 16:32:59

87

Views

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

Published on 14 Apr 2026 - 16:32

score 73 · Answer 5 · 2026-04-14 09:51:24

73

Views

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

Published on 14 Apr 2026 - 9:51

score 89 · Answer 6 · 2026-04-11 08:12:46

89

Views

Cycles in an acyclic complex of projectives are projectives.

Published on 11 Apr 2026 - 8:12

score 78 · Answer 7 · 2026-04-14 15:49:30

78

Views

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

Published on 14 Apr 2026 - 15:49

score 29 · Answer 8 · 2026-04-10 13:56:07

29

Views

Understanding an example of a finitely generated projective module which is not free.

Published on 10 Apr 2026 - 13:56

score 85 · Answer 9 · 2026-04-12 20:58:40

85

Views

Projective modules restricted to smooth curves

Published on 12 Apr 2026 - 20:58

score 617 · Answer 10 · 2026-05-10 17:17:46

617

Views

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

Published on 10 May 2026 - 17:17

score 111 · Answer 11 · 2026-04-12 10:26:32

111

Views

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

Published on 12 Apr 2026 - 10:26

score 308 · Answer 12 · 2026-04-11 22:45:45

308

Views

Fractional ideal of a Dedekind domain A is a projective module

Published on 11 Apr 2026 - 22:45

score 236 · Answer 13 · 2026-04-15 16:48:43

236

Views

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

Published on 15 Apr 2026 - 16:48

score 832 · Answer 14 · 2026-04-14 09:54:35

832

Views

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

Published on 14 Apr 2026 - 9:54

score 35 · Answer 15 · 2026-04-17 02:29:34

35

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 17 Apr 2026 - 2:29

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 $R \ncong \bar{x}R \oplus \bar{x}R$

Trending Questions

Popular # Hahtags

Popular Questions