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



"Sums-compact" objects = f.g. objects in categories of modules?
It seems to me the references in this [Mathematics - Stack Exchange answer](https://math.stackexchange.com/questions/82957/preservation-of-direct-sums-and-finite-generation/82958#82958) contain the requested information. 


**EDIT 1.** Here is an excerpt from Hyman Bass's book **Algebraic K-Theory**, W. A. Benjamin (1968), p. 54: 



> 
> Exercise. 
> 
> 
> (a) Show that a module $P$ is finitely generated if and only if the union of a totally ordered family of proper submodules of $P$ is a proper submodule.
> 
> 
> (b) Show that $\text{Hom}\_A(P,\bullet)$ preserves coproducts if and only if the union of every (countable) chain of proper submodules is a proper submodule.
> 
> 
> (c) Show that the conditions in (a) and (b) are not equivalent. (Examples are not easy to find.)
> 
> 
> 


**EDIT 2.** Here is a solution to Exercise (a) above. Let $R$ be an associative ring with $1$, and $A$ an $R$-module. If $A$ is finitely generated, then the union of a totally ordered set of proper submodules is clearly a proper submodule. Let's prove the converse: 


Assume that $A$ is not finitely generated. Let $Z$ be the set of those submodules $B$ of $A$ such that $A/B$ is is not finitely generated. The poset $Z$ is nonempty and has no maximal element. By Zorn's Lemma, there is a nonempty totally ordered subset $T$ of $Z$ which has no upper bound. Letting $U$ be the union of $T$, we see that $A/U$ is finitely generated. There is thus a finitely generated submodule $F$ of $A$ which generates $A$ modulo $U$. Then the $B+F$, where $B$ runs over $T$, form a totally ordered set of proper submodules whose union is $A$. QED 


I'd be most grateful to whoever would post a solution to the other exercises in Bass's list. (I haven't been able to do them.) The following references might help, but I haven't been able to find them online: 


* R. Rentschler, Sur les modules M tels que $\text{Hom}(M,-)$ commute avec les sommes directes, C. R. Acad. Sci. Paris Sér. A-B 268 (1969), 930-933. [Update: see Edit 3 below.]
* P.C. Eklof, K.R. Goodearl and J. Trlifaj, Dually slender modules and steady rings, Forum Math. 9 (1997), 61-74.


This paper is [available online](http://www.karlin.mff.cuni.cz/katedry/ka/preprint/alg05_17.pdf), but I don't understand it: 


* Jan Zemlicka, Classes of dually slender modules, Proc. Algebra Symposium Cluj 2005, 129-137.


**EDIT 3.** 


$\bullet$ Rentschler's paper 


R. Rentschler, Sur les modules M tels que $\text{Hom}(M,-)$ commute avec les sommes directes, C. R. Acad. Sci. Paris Sér. A-B 268 (1969), 930-933


is available [here](http://www.iecn.u-nancy.fr/~gaillapy/MO/rentschler.pdf) in one click, and [there](http://gallica.bnf.fr/ark:/12148/bpt6k480296q/f936.image.langEN) in a few clicks. [I'm also giving the second option because it's a trick worth knowing.] Thanks to Stéphanie Jourdan for having found this link! 


$\bullet$ Exercise (b) in Bass's list is in fact the easiest. [Sorry for not having realized that earlier.] Here is a solution. --- Let $R$ be an associative ring with $1$, let $A$ be an $R$-module, and let "map" mean "$R$-linear map". 


If $A\_0\subset A\_1\subset\cdots$ is a sequence of proper submodules of $A$ whose union is $A$, then the natural map from $A$ to the direct product of the $A/A\_n$ induces a map from $A$ to the direct sum of the $A/A\_n$ whose components are all nonzero. 


Conversely, let $f$ be a map from $A$ to a direct sum $\oplus\_{i\in I}B\_i$ of $R$-modules such that the set $S$ of those $i$ in $I$ satisfying $f\_i\neq0$ [obvious notation] is is infinite. By choosing a countable subset of $S$ we get a map $g$ from $A$ to a direct sum $\oplus\_{n\in \mathbb N}C\_n$ of $R$-modules such that $g\_n\neq0$ for all $n$. It is easy to check that the 
$$
A\_n:=\bigcap\_{k > n}\ \ker(g\_k), 
$$ 
form an increasing sequence of proper submodules of $A$ whose union is $A$.


**EDIT 4.** [Version of Nov. 26, 2011, UTC.] The following result is implicit in Rentschler's paper, and solves Bass's Exercise (c): 



> 
> **Theorem.** Let $T$ be a nonempty ordered set $ ( \* ) $ with no maximum. Then there is a domain $A$ which has the following property. If $P$ denotes the poset of proper sub-$A$-modules of the field of fractions of $A$, then there is an increasing $ ( \* ) $ map $f:T\to P$ such that $f(T)$ is cofinal in $P$. 
> 
> 
> 


$ ( \* ) $ Since I'm using references written in French while writing in English (or at least trying to), I adhere strictly to linguistic conventions. In particular: 


ordered set = ensemble totalement ordonné, 


poset = ensemble ordonné, 


increasing = strictement croisssant. 


**Proof.** Let $T\_0$ be the ordered set opposite to $T$, let $\mathbb Z^{(T\_0)}$ be the free $\mathbb Z$-module over $T\_0$ equipped with the lexicographic order. Then $\mathbb Z^{(T\_0)}$ is an abelian ordered group (groupe abélien totalement ordonné). By Example 6 in Section V.3.4 of Bourbaki's **Algèbre commutative**, there is a field $K$ and a surjective valuation $$
v:K\to\mathbb Z^{(T\_0)}\cup \{ \infty \}.
$$ 
Say that a subset $F$ of $\mathbb Z^{(T\_0)}$ is a **final segment** if 
$$F\ni x < y\in\mathbb Z^{(T\_0)}
$$ 
implies $y\in F$. Attach to each such $F$ the subset 
$$
S(F):=v^{-1}(F)\cup \{ 0 \}
$$ 
of $K$. Then $A:=S(F\_0)$, where $F\_0$ is the set of nonnegative elements of $\mathbb Z^{(T\_0)}$, is a subring of $K$. Moreover, by Proposition 7 in Section V.3.5 of the book quoted above, $F\mapsto S(F)$ is an increasing bijection from the final segments of $\mathbb Z^{(T\_0)}$ to the sub-$A$-modules of $K$. 


Write $e\_{t\_0}$ for the basis element of $\mathbb Z^{(T\_0)}$ corresponding to $t\_0\in T\_0$. Then the intervals 
$$
I\_{t\_0}:=[-e\_{t\_0},\infty)
$$ 
are cofinal in the set of all proper final segments of $\mathbb Z^{(T\_0)}$, and we have $I\_{t\_0}\subset I\_{u\_0}$ if and only if $t\le u$. [We denote an element $t$ of $T$ by $t\_0$ when we view it as an element of $T\_0$.]


