TITLE: "Sums-compact" objects = f.g. objects in categories of modules?
QUESTION [29 upvotes]: Hello,
Let us call an object of an additive category sumpact (contraction of "sums" and "compact") if taking $Hom$ from it (considered as functor from the category to $Ab$) commutes with coproducts. Note that to be sumpact is weaker than to be compact (which means that $Hom$ from you commutes with filtered colimits).
Let us take, for our additive category, the category of left modules over some ring. It is known that compact objects in this category are exactly the finitely presented objects. What about sumpact objects?
It is clear that every finitely generated module is sumpact. When I try to prove the converse, I get into some pathological things.
Say, if a module has an increasing $\mathbb{N}$-sequence of submodules whose union is the whole module, and such that the union of every finite subsequence is not the whole module, then it is clear that this module is not a sumpact object (by considering the morphism from it to the direct sum of the quotients by members of our sequence). But it seems not clear (perhaps not true) that every non finitely generated module has such a sequence.
Also, when I check in the internet, it seems people put some condition: the ring is assumed to be perfect. Then indeed sumpact = f.g.
So my question is: for a general ring it is not true that sumpact implies f.g.? Can you give an example? Can you give an example when the ring is commutative? Can you indicate what perfect means and why then everything is OK?
Thank you

REPLY [12 votes]: If it's considered bad form to resurrect year-old threads, then please slap my wrist (gently, please; I'm new here!)
A fairly simple explicit example of a "sumpact" module that is not f.g. is as follows.
Let $R$ be the ring of functions from an uncountable set $X$ to, say, a field $k$. Let $M$ be the ideal of functions with countable support.
Then it's very easy to show that $M$ isn't f.g., and fairly easy to show that it is "sumpact", using no set theory beyond the fact that a countable union of countable sets is countable.
Edit to add details requested in comments:
To show that $M$ is "sumpact", suppose that $\alpha:M\to\bigoplus_{i\in I}N_i$ is a homomorphism that doesn't factor through a finite subsum. I.e., for infinitely many $i$ the composition $\pi_i\alpha:M\to\bigoplus_{i\in I}N_i\to N_i$ of $\alpha$ with projection onto the summand $N_i$ is non-zero. Replacing $I$ with a countable collection of such $i$ we can assume that $I$ is countable and that $\pi_i\alpha$ is non-zero for all $i\in I$.
For each $i\in I$ choose $f_i\in M$ so that $\pi_i\alpha(f_i)\neq0$. Then the union of the supports $\text{supp}(f_i)$ is countable, so there is some $f\in M$ with $\text{supp}(f)=\bigcup_{i\in I}\text{supp}(f_i)$.
But then the ideal generated by $f$ contains every $f_i$, and so $\pi_i\alpha(f)\neq0$ for every $i$, contradicting the fact that $\alpha(f)\in\bigoplus_{i\in I}N_i$.