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

Question

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

Answer 1

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 $T_0$.]

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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 $1_C$

Answer 7

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