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