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

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    $\begingroup$ The usual definition of compact object in an abelian category, such as a module category, is that the covariant representable functor preserves filtered colimits, not just coproducts. In this case, compact is equivalent to finitely presented. $\endgroup$ Mar 23, 2011 at 11:34
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    $\begingroup$ Yes, thank you. That version I know. I wondered what happens when the definition which I use in the text is used. $\endgroup$
    – Sasha
    Mar 23, 2011 at 11:53
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    $\begingroup$ 1+, good question. By the way, I have seen many texts on homological algebra which define a compact object to be an object $X$ such that $Hom(X,-)$ preserves coproducts. $\endgroup$ Mar 23, 2011 at 17:22
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    $\begingroup$ @G. Rodrigues: I hat not noticed that Theo's little only requires that $\mathrm{Hom}(X,{-})$ preserves finite coproducts. I think this is a typo, since in an additive category, finite coproducts are "the same" as finite products, so $\mathrm{Hom}(X,{-})$ commutes with finite coproducts, for any $X$. However, this also shows that if $\mathrm{Hom}(X,{-})$ commutes with filtered colimits, it commutes with arbitrary coproducts (by the argument I gave above). $\endgroup$
    – user2035
    Nov 20, 2011 at 19:09
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    $\begingroup$ The condition that Hom(X, -) preserves coproducts is usually called connectedness of X. (Think about the category of topological spaces.) There's some flexibility about whether you ask for preservation of all small coproducts or just the finite ones. See for instance ncatlab.org/nlab/show/connected+object $\endgroup$ Nov 22, 2011 at 18:41

7 Answers 7


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:


(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.


$\bullet$ 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 $T_0$.]

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    $\begingroup$ +1: This is a great answer. I've been wondering about this sumpact problem for a while, and I'm glad to see it settled. I hope Sasha shows up to accept this; it completely answers what he wanted. I'll see if I can figure out an example to answer (c). I suppose it's worth noting (from the OP's original question) that (a) and (b) are equivalent if $R$ is perfect, which is pretty cool $\endgroup$ Nov 22, 2011 at 14:04
  • $\begingroup$ An example for (c) is in the linked paper: Lemma 1.2 and the remark afterwards that one of the inclusions is strict. The key observation seems to be that if you take the union of a strictly increasing $\kappa$-chain of sumpact modules (a.k.a. dually slender), then this is also sumpact. An example is the union of $\kappa$ many cyclic submodules. If $\kappa$ is uncountable then this module clearly is not finitely generated. I looked in T.Y.Lam's Lectures on Modules and Rings for a simpler example, but it's not there. Still, the example above is not so bad, & any example will need set theory $\endgroup$ Nov 22, 2011 at 14:22
  • $\begingroup$ Dear @David: Thank you very much for your kind comments! You seem to have a lot to say about this. I hope you'll post an answer of your own in a not too distant future. I appreciate all the comments you've made so far in this thread. $\endgroup$ Nov 22, 2011 at 14:37
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    $\begingroup$ Yes, please do share your thoughts, David. I'm following this discussion and would be very interested in collecting information (including proofs), so that I can write something up for the nLab. (Of course, anyone can get involved with the nLab.) $\endgroup$
    – Todd Trimble
    Nov 22, 2011 at 21:23
  • $\begingroup$ Dear @Todd: I agree with you, especially with the "including proofs". I hope you will write something up for the nLab. If you do, thank you in advance for letting us know. $\endgroup$ Nov 23, 2011 at 6:04

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.

  • $\begingroup$ Thank you! Those are very interesting comments. I have just two questions: 1) I don't know really what is perfect; you claim left perfect implies left Noetherian? 2) Is the fact that sum of injectives is injective for Noetherian ring a hard fact? Thanks. $\endgroup$
    – Sasha
    Mar 23, 2011 at 17:15
  • $\begingroup$ Why does $g$ exist? $\endgroup$ Mar 23, 2011 at 17:28
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    $\begingroup$ @Sasha: the relation between perfect and noetherian rings is not that close, but for instance, perfect + noetherian is the same as artinian, so my argument really applies to many more rings than just perfect rings. The fact that noetherianity is equivalent to the fact that direct sums of injectives is injetive is a very classical result. $\endgroup$ Mar 23, 2011 at 20:41
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    $\begingroup$ As I said in my comment on the main question, the standard term for "sumpact" is "connected" (modulo the question of whether we're dealing with finite coproducts or all of them). So Fernando's result says: a quotient of a connected set is connected. That makes sense! And it's true much more generally than Fernando states it. Sketch proof follows in next comment... $\endgroup$ Nov 22, 2011 at 22:49
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    $\begingroup$ ...Take a category in which monos, epis and coproducts behave "well"; I'm too lazy to figure out the exact hypotheses, but something like "regular category" should do. Take a connected object X and an epi e: X -> Y. I claim Y is connected. For take f: Y -> A+B. WLOG, fe factors through the inclusion A -> A+B, which I'll assume is mono. So we get a commutative square with an epi down the left-hand edge and a mono down the right. In the usual fashion, we can fill it in with a diagonal, i.e. a map Y -> A. This gives the required factorization. (I hope this can be made respectable.) $\endgroup$ Nov 22, 2011 at 22:53

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

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    $\begingroup$ Resurrecting old threads is perfectly fine, and if you give a good answer, you may even earn a "Necromancer" badge. $\endgroup$ Apr 18, 2012 at 20:05
  • $\begingroup$ Can you please say something about the proof? I don't see how it works. $\endgroup$
    – HeinrichD
    Apr 7, 2017 at 9:04
  • $\begingroup$ @HeinrichD I've added a proof that $M$ is sumpact. I presume the fact that it isn't finitely generated is clear? $\endgroup$ Apr 7, 2017 at 9:28
  • $\begingroup$ Thank you. I wanted to construct $f$ directly from the $f_i$, but this did not work since $R$ has no infinite sums. But as you write, we only need to consider the support. $\endgroup$
    – HeinrichD
    Apr 7, 2017 at 12:23

A module $M$ such that $\mathrm{Hom}(M,-)$ preserves (infinite) direct sums is called dually slender. A ring is called steady if every dually slender module over it is finitely generated. A google research shows that in the last 15 years a lot of work has been devoted to the study of dually slender modules and steady rings, in particular by Jan Trlifaj and Jan Zemlicka. See also the comprehensive list of references in: Bashir, Kepka, Němec "Modules commuting (via Hom) with some colimits." (online).

An important characterization of dually slender modules is the following:

$M$ is dually slender iff for every chain of submodules $M_1 \subseteq M_2 \subseteq ...$ whose union is $M$, there is some $n$ with $M = M_n$.

You can find the proof as Lemma 1.1 in: Jan Zemlicka, "Class of dually slender modules" (online). In the introduction to Jan Zemlikca, "Steadiness of regular semiartinian rings with primitive factors artinian" (online) it is noted that three constructions of non-finitely generated dually slender modules are known - many references are given. An explicit example is finally given in Jan Zemlikca, "$\omega_1$-generated uniserial modules over chain rings" (online), Example 2.7:

Take the reverse of the natural order on the ordinal number $\omega_1$ and consider the lexicographic order on $\mathbb{Z}^{(\omega_1)}$. Pick a valuation domain $R$ whose value group is $\mathbb{Z}^{(\omega_1)}$. Then $R$ is not steady. In fact, this follows from a more general result (Corollary 2.6) which says that (a transfinite version of) the Krull dimension of steady chain rings is countable.


As I promised I would do some time ago (in a comment on Pierre-Yves Gaillard's answer), and with the hope this is useful, I have written an article in the nLab which ties together some of the results in the answers already given into a single narrative. Here and there I provide a few extra glosses and some background information.


From "Abelian CAtegories and its application to Rings and Modules " by Popescu N, par. 3.5 pag 88:

Popescu using the names "small" and "finitely presented" for the yours "sumcompact" and "compact" respectively.

Popescu call a object $X$ (of a Grothendick abelian category $\mathcal{C}$)
of "finite type" if for any direct union of subobjets $Y=\cup_{i\in I}Y_i$ the natural morphism $Colim_{i\in I} \mathcal{C}(X, Y_i)\to \mathcal{C}(X, Y))$ is a isomorphism, this is equivalent to:

for any directed union of subobjets $X=\cup_{i\in I}X_i$ there is a $i_o\in I$ such that $X=X_{i_0}$.

In a category of modules finitely presented is equivalent to the usal definition (there is a exact $0\to A\to X\to C\to 0$ with $A,\ B$ finitely generated), and finite type is equivalent to finitely generated.

From 5.4 of Popescu book a finitely generated module is small (sumcompact). And of course exist finitely generated modules that aren't finitely presented. then we have the implications:

finitely presented $\Rightarrow$ finitely generated $\Rightarrow$ small (suncompact) and *finitely .generated*$\not\Rightarrow$ finitely presented

Then cannot have that *small (sumcompact)*$\Rightarrow$ finitely presented.

$EDIT$ I get a mistake simply gived a answer to another question (sorry, mistake), I find the answer as exercises in "Rings os Quotients" B. Strenstrom, Springer Verlag 1975, pag 134 n.13.

I try to do a proof (I hope):

THEOREM) for a abject $C$ in a Grothendieck category (we think simply to a module category) the following are equivalent:

1) For any sequence of subobject like $C_1\subset C_2\subset\ldots C$ we have $C=C_m$ for some $m$.

2) For any sequence of subobject like $M_1\subset M_2\subset\ldots M$ with union $M$ we have that $(C, M)=\cup_n(C, M_n)$ (naturally).

3) The functor $(C, -)$ commute by denumerable coproducts.

4) The functor $(C, -)$ commute by coproducts.

5) The functor $(C, -)$ commute by directed unions (i.e. $C$ is $f.g.$).

PROOF. $(1\Rightarrow 2):$ we have to proof that any $f: C\to M $ has image in some $M_m$, if we put $C_n:=f^{-1}(M_n)$ we done.

$(2\Rightarrow 3):$ of course $(C,-)$ commute by finite coproduts (they are biproducts), we have to prove that a $f: C\to M$, with $M=\coprod_n X_n$ as a factorization on a finite summands, let $M_n:=\coprod_{i\leq n} X_i$ we done.

$(3\Rightarrow 4):$ we have to prove that a $f: C\to M$, with $M=\coprod_{i\in I} X_i$ as a factorization on a finite summands, suppose the opposite: then we have an infinite denumerable set of indices $i_0, i_1\ldots \in I$ such that for any integer $n$ exist a $x_n\in C$ with $f(x_n)_{i_n}\neq 0$

then we consider $J:= I\setminus ${$i_0, i_1\ldots$} and the quotient map $\pi: \coprod_{i\in I}X_i\to (\coprod_{i\in I}X_i)/(\coprod_{j\in J}X_j)\cong\coprod_n X_{i_n} $ and the composition $\pi\circ f: C\to \coprod_n X_{i_n}$, this map isnt factorizable to a finite summands (absurd).

$(4\Rightarrow 5):$ Let $M=\cup_{i\in I} M_i$ where $I$ is a directed order. We can suppose $I$ cofinite i.e. for any $i\in I$ exist only finite $j$ such that $j\leq i$ (e.g. "Shape Theory" Sibe MArdiesic NH 1982 T.2 pag. 10). Then the natural map $\pi: M\to \coprod_{i\in I} M/M_i$ with $(\pi(x))_i=\pi_i(x)$, $\pi_i: M\to M/M_i$ natural, is well defined. We have to proof that any $f: C\to M$ has a image on some $M_j$, considering $\pi\circ f: C\to \coprod_i M/M_i$ then this map has a factorization on finite summands $M/M_{i_1},\ldots M/M_{i_N}$, if some $M_{i_n}$ is $M$ the assert is trivial

if no we can have a $j\in I$ strictly greater of any $i_1,\ldots i_N$, then $f\circ \pi_j: C\to M\to M/M_j $ is the $0$ map, then the image of $f$ is in $M_j$.

$(5\Rightarrow 2):$ Trivial

$(2\Rightarrow 1):$ Let $M:=M,\ M_n:=C_n $ and considerind $1_C$

  • $\begingroup$ Reading the OP's post, it seems he already knew all the implications and non-implications in your answer. What he wanted to know was whether sumpact implies finitely generated, not finitely presented. He already knew it couldn't imply finitely presented because sumpact is strictly weaker than compact (which is equivalent to finitely presented). Still, I hope you leave this answer up because it's nice to have the terminology from algebra all in one place. $\endgroup$ Nov 22, 2011 at 13:59
  • $\begingroup$ @David White. Thank you for notification. I Edit the answere and try to give a better answer. $\endgroup$ Nov 22, 2011 at 20:51

Group of rational numbers is sumpact. In older terminology sumpact for abelian categories is called small.

  • $\begingroup$ Can you give a reference for your second sentence? I spend a fair amount of time with abelian categories and the only definition of a small object I know is the one on nLab, i.e. $\kappa$-compact for some regular cardinal $\kappa$. This uses filtered colimits rather than coproducts. ncatlab.org/nlab/show/small+object $\endgroup$ Nov 22, 2011 at 13:45
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    $\begingroup$ See the answer by Fernando Muro for the proof that over noetherian rings, sumpact = f.g. $\endgroup$ Nov 22, 2011 at 15:17

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