2
votes
1answer
157 views

Bimodule version of IBN

Hello all, Does anyone have an example in mind of a ring $R$ for which $R^n\cong R^m$ as $R,R$ bimodules for some positive integers $n\neq m$? I would be a little surprised if someone showed no such ...
3
votes
2answers
377 views

Double orthogonal complement of a finite module

Crossposted from math.stackexchange since I'm not getting any answer. Let $W$ be the finite $\mathbb{Z}$-module obtained from $\mathbb{Z}_q^n$ with addition componentwise where $\mathbb{Z}_q$ is the ...
1
vote
1answer
341 views

Unimodular column property

Hi, I know that if $R$ is a ring such that every projective $R$-module finitely generated is free then $R$ has the unimodular column property. I would like to know if there is a ring $R$ that doesn't ...
7
votes
1answer
661 views

Direct sum of injective modules over non-Noetherian rings

Hi. I know, by the Bass-Papp theorem, that if every direct sum of injective $R$-modules is injective then $R$ is Noetherian. I would like to know if there exists a direct sum of injective $R$-modules ...
2
votes
2answers
350 views

Related to fractional ideals

$K$ a field, $A\subset K$ a subring, $M\subset K$ an $A$-submodule. Define $$(A:_{K}M):= \lbrace s\in K|sM\subset A\rbrace$$ Then it is easy to see that $$M\subset A\Longleftrightarrow A\subset ...
0
votes
1answer
136 views

Projectively splitting module

Is there a name for such class of modules $M$ such that $M\rightarrow N\rightarrow 0$ splits for every $N$ ?
1
vote
2answers
839 views

Extension problem

As I understand, if $0\rightarrow A\rightarrow X\rightarrow B\rightarrow 0$ is a short exact sequence of abelian groups, $\mbox{Ext }_{\mathbb{Z}}^{1}(B,A)$ gives all the isomorphism classes of what ...
10
votes
4answers
1k views

Linearly independent subsets of a free module

Is it true that the cardinality of every maximal linearly independent subset of a finitely generated free module $A^{n}$ is equal to $n$ (not just at most $n$, but in fact $n$)? Here $A$ is a nonzero ...
1
vote
2answers
424 views

Understanding the modules of semiprimitive rings

As far as I understand, a semiprimitive ring can be fully 'explored' by its simple modules, in the sense that a semiprimitive ring is the subdirect product of its simple modules (for brevity, I'll use ...