# Tagged Questions

**2**

votes

**1**answer

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

**2**answers

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

**1**answer

340 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

**1**answer

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

**2**answers

349 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

**1**answer

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

**2**answers

837 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

**4**answers

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

**2**answers

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 ...