"Sums-compact" objects = f.g. objects in categories of modules?

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 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 $T_0$.]

I've got no real answers, but I've been thinking of this problem for fun and I'd like to share with you some more or less obvious facts that I've found out.

In general, the epimorphic image of a sumpact object is sumpact.

I'll show now that for left modules over a left noetherian ring, any submodule $N\subset M$ of a sumpact module $M$ is again sumpact. Let $$f\colon N\longrightarrow \bigoplus_{i\in I} P_i$$ be any morphism. Take injective envelopes $P_i\subset E_i$. By noetherianity, a direct sums of injectives is injective, therefore we can form a commutative square,

$$\begin{array}{rcccl} &N&\stackrel{\text{inclusion}}\longrightarrow&M&\\\ f\hspace{-10pt}&\downarrow&&\downarrow&\hspace{-10pt}g\\\ &\bigoplus_{i\in I} P_i&\stackrel{\text{inclusion}}\longrightarrow&\bigoplus_{i\in I} E_i& \end{array}$$

Since $M$ is sumpact, $g$ factors through the inclusion of finitely many summands, therefore so does $f$, because the horizontal arrows are injective, hence we are done.

Now we can follow your argument to show that in left modules over a left noetherian ring, sumpact implies finitely generated (and hence compact in the classical sense). Notice that being perfect is stronger that being left noetherian.

Suppose $M$ is a sumpact module. If $M$ is not finitely generated, we can find a strictly increasing sequence of submodules

$$\cdots\subset M_n\subset M_{n+1}\subset\cdots\subset M,\quad n\geq 1.$$

Your argument shows that the submodule

$$N=\bigcup_{n=1}^\infty M_n\subset M$$

is not sumpact, therefore $M$ cannot be sumpact.

So far, this is all I'm able to say.

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 $\alpha(f)\in\bigoplus_{i\in I}N_i$.