It seems to me the references in this Mathematics - Stack Exchange answer 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, 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 in one click, and there 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 segements 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 $T_0$.]