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Questions tagged [projective-module]

For questions related to projective modules, their structures, and properties.

1 vote
2 answers
84 views

Let $K$ be a number field, that is, a finite extension of ${\mathbb Q}$. Let $R={\mathcal O}_K$ be the Dedekind domain of algebraic integers in $K$. It is well-known that an ideal $I\subset R$ is a ...
Three aggies's user avatar
  • 5,566
1 vote
1 answer
74 views

Let $R$ be a commutative ring with unity and let $R \operatorname{-Mod}$ be the category of $R$-modules. Question: How can we show that for an $R$-module $M$, the following are equivalent? The ...
Elia Immanuel Auer's user avatar
2 votes
1 answer
52 views

For a commutative (not necessarily Noetherian) ring $R$, define the following two quantities $$fpd(R):=\sup\{\operatorname{pd}_R(M) | M \text{ is a finitely presented }R\text{-module of finite ...
uno's user avatar
  • 1,901
1 vote
0 answers
117 views

Let $R$ be an associative ring, $P$ is a finitely generated projective right $R$-module. Denote $E=\operatorname{End}_R(P,P)^{\operatorname{op}}$ the opposite of the ring of endomorhisms, and $P^*=\...
Alex's user avatar
  • 6,826
1 vote
1 answer
69 views

I'm familiar with the proof that finitely-generated projective modules over a local Noetherian ring are free, but I'm a little stuck on directly showing a small variation of that result: Let $(A, \...
zork zork's user avatar
  • 333
-2 votes
1 answer
78 views

The question is: Show that if we have a diagram in which the row is exact, $P$ is projective, and $h\circ f = 0$, then there exists a $k:P → M$ such that $f = g\circ k$. I know that $\text{Im}g=\ker ...
bunnie's user avatar
  • 23
2 votes
1 answer
77 views

in Exercise 8.10 from Module Theory: An Approach to Linear Algebra by T. S. Blyth, the author defined a terminology as follows: If $M$ and $N$ are $R$-modules, an $R$-linear map $f:M\to N$ factor ...
Bernard Pan's user avatar
  • 3,175
1 vote
1 answer
84 views

We know the following fact: Let $A$ and $B$ are finitely self-injective $k$-algebras whose semi-simple quotients are separable, $Z$ is a finitely generated $A$-$B$-bimodule, where $k$ is a field. ...
Zhenxian Chen's user avatar
0 votes
0 answers
40 views

We encountered the following problem related to tensoring and lifting morphisms: Let $\cdots \to A_n \to \cdots \to A_0 \to A \to 0$ be a projective resolution of $A$. Tensoring this complex with a ...
ChaosLord's user avatar
0 votes
0 answers
86 views

Let $R$ be a (possibly noncommutative) ring. Recall that an ideal $I \subseteq R$ is called idempotent if $I^2=I$, meaning that every element of $I$ can be written as a finite sum of products of ...
Mourad Khattari's user avatar
3 votes
1 answer
69 views

A ring $R$ is called a left SF-ring (resp. right SF-ring) if every simple left (resp. right) $R$-module is flat. It is well-known that if $R$ is a von Neumann regular ring, then $R$ is both a left and ...
Mourad Khattari's user avatar
1 vote
1 answer
166 views

This question is from the 14th Yau-Contest. Let $R$ be a commutative ring with identity. Let $S=\prod_{\mathbb{N}} R$ the product of countably many copies of $R$, and $I=\oplus_{\mathbb{N}} R \subset ...
ANT's user avatar
  • 57
0 votes
1 answer
108 views

Let $G$ be any fixed group (possibly infinite). Suppose there is a projective $\mathbb{Z}G$-module $P$ and a $\mathbb{Z}G$-module $F$ which is free over $\mathbb{Z}$. Then I want to show that the ...
Milind's user avatar
  • 73
3 votes
2 answers
150 views

I'm new to homological algebra. I was studying projective modules from Brown's book and was trying to work out some examples. I read here that $\mathbb{Q}$ is not a projective module over $\mathbb{Z}$,...
Jenna Fargo's user avatar
0 votes
1 answer
55 views

For $C_{2}$ (cyclic group of order $2$) there are two irreducible representations. Let $E_{0}$ be the trivial one and $E$ the sign one. They form the basis of $R_{\mathbb{C}}(C_{2})$. Do I understand ...
Matthew Willow's user avatar

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