2
$\begingroup$

Here is my question:

Does there exist an infinite commutative ring $R$ with identity with an indecomposable injective (unitary) $R$-module $M$ of larger cardinality than $R$ with the additional property that $M$ has a minimum submodule (that is, a nonzero $R$-submodule $N$ such that $N\leq K$ for every nonzero $R$-submodule $K$ of $M$)? This question is equivalent to the question obtained by dropping the adjective "indecomposable injective."

This is a bit out of my research area, but I would really like to know the answer for a paper I'm working on. Any references would be much appreciated.

Thanks!

Improve this question
$\endgroup$

2 Answers 2

Reset to default
4
$\begingroup$

Let $k$ be a field and $V$ a $k$-vector space, and let $R=k\oplus V$, where $V$ is a square zero ideal.

Then the $k$-linear dual $R^\ast=k^\ast\oplus V^\ast$ is an injective $R$-module with unique minimal submodule $k^\ast\oplus0$, but in general $R^\ast$ has larger cardinality than $R$. For example, if $k$ and $\dim_k(V)$ are countable, then $R$ is countable, but $R^\ast$ has the cardinality of the continuum.

Improve this answer
$\endgroup$
1
$\begingroup$

No if $R$ is a field, because in that case the only module with a minimal submodule is the free module of rank $1$ which has the same cardinality as $R$. Muahahaha!

More seriously, I think Theorem 3.3 of this pdf file shows that what you want happens if and only if $R$ is not Noetherian. (I only took 5 seconds to check this so you'd better double check.)

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ You have a frightening laugh, Count Dracula. $\endgroup$
    – Jason Starr
    Commented Jun 23, 2015 at 15:19
  • $\begingroup$ Thanks for taking the time to respond, Dracula (I too appreciated the laugh, as it tipped me off to the fact that I wasn't careful enough in the wording of the question). What I meant to ask was this: does there exist an infinite commutative ring $R$ with the property that there is a large indecomposable injective module $M$ over $R$ with a minimum submodule. Yes, any such ring must be non-Noetherian; this follows from early (and famous) results of Eben Matlis. I don't think the paper you referenced answers the question; my guess is that "most" non-Noetherian rings don't have such a module $\endgroup$
    – Greg Oman
    Commented Jun 23, 2015 at 21:51
  • 1
    $\begingroup$ There do exist nonnoetherian rings which do not have a big injective of the type sought. Any infinite power of a finite ring is such an example. If $|R|=n < \omega\leq\kappa$, then any module $M$ over the nonnoetherian ring $S:=R^{\kappa}$ that has a least nonzero submodule must be annihilated by an ultrafilter ideal of $S$. Hence $M$ may be thought of as a module over an ultrapower of $R$. Since $R$ is finite, it is isomorphic to any of its ultrapowers. Thus, $M$ may be considered to be a subdirectly irreducible $R$-module, forcing $|M|$ to divide $|R|=n$. $\endgroup$
    – Keith Kearnes
    Commented Jul 7, 2015 at 20:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .