# Reference request: compact objects in R-Mod are precisely the finitely-presented modules?

Let $R$ be a ring. According to this MO question, the modules $M \in R\text{-Mod}$ such that $\text{Hom}(M, -)$ preserves all filtered colimits (the compact objects) are precisely the finitely-presented modules. Where can I find a proof of this?

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Possibly in Adamek and Rosicky, Locally presentable and accessible categories. – Math Gems Feb 21 '12 at 0:15
@Math Gems: great! In fact Theorem 3.12 gives the general version of this result for categories of algebras. I would prefer not to have to absorb the previous portions of the book to understand this result so I would still appreciate references to a more self-contained proof... – Qiaochu Yuan Feb 21 '12 at 0:38
By the way: If $X$ is a quasicompact quasiseparated scheme, then I can show that the compact objects of $\mathrm{Qcoh}(X)$ are precisely the quasi-coherent sheaves which are locally of finite presentation. – Martin Brandenburg Mar 2 '12 at 8:47
Note that, if we require $\mathrm{Hom}(M,{-})$ to commute with all colimits, then $M$ is necessarily the zero module. See tiny object on the nLab. – Ingo Blechschmidt Feb 18 '15 at 17:54

This is easy enough to prove directly:

For the direction finitely presented $\implies$ compact, one can easily prove it by hand, and there is also the following somewhat slicker argument:

For any module $M$ and any filtered direct limit $N = \varinjlim_i N_i$, there is a natural transformation $\varinjlim_i Hom(M,N_i) \to Hom(M,N).$ This is certainly an isomorphism when $M = R$ (because it then reduces to the original equality $\varinjlim_i N_i = N$), and hence when $M = R^n$ for some $n \geq 0$, because both functors $\varinjlim_i Hom(\text{--},N_i)$ and $Hom(\text{--},N)$ commute with finite direct sums.

Since filtered direct limits are exact, both functors are also right exact.

Now if $M$ is finitely presented, choose a presentation $R^m \to R^n \to M \to 0,$ and apply both functors to it. By what has already been said together with the five lemma, we see that $\varinjlim_i Hom(M,N_i) \to Hom(N,M)$ is an isomorphism.

For the other directon, suppose that $M$ is compact. Like any module, $M$ can be written as a filtered direct limit of finitely presented modules $M_i$. Thus $Hom(M,M) = Hom(M,\varinjlim_i M_i) = \varinjlim_i Hom(M,M_i)$ (the last equality following by compactness). In particular, the identity $M \to M$ can be factored through the map $M_i \to M$ for some sufficiently large $i$. Thus the map $M_i \to M$ can be split, and so $M$ is a direct summand of the finitely presented module $M_i$. Consequently, $M$ is itself finitely presented.

A related fact, mentioned in the excerpt quoted by Math Gems:

If we restrict to filtered direct limits with injective transition maps (diret unions in the language of the excerpt quoted by Math Gems), then we similarly find, using the fact that any module can be written as the filtered union of its finitely generated submodules, that finitely generated modules are precisely those $M$ for which $Hom(M,\text{--})$ commutes with filtered direct limits having injective transition maps.

Added in response to the OP's questions below:

If $A \to B$ is a morphism of rings with $1$, then $M \mapsto Hom_A(B,M)$ (Homs of left $A$-modules, and $Hom_A(B,M)$ being endowed with the left $B$-module structure coming from the right $B$-module structure on $B$) is a right adjoint to the forgetful functor from left $B$-modules to left $A$-modules. This shows that formation of colimits commutes with this forgetful functor.

Alternatively, one can check by hand that the formula for filtered direct limits in $R$-mods is the same as in abelian groups. (Just use the fact that direct sums, kernels, and cokernels are the same.)

Also, any finitely generated $R$-module $M$ is of the form $R^n/N$ for some submodule $N$. We may write $N = \varinjlim_i N_i$ as the direct limit of its f.g. submodules. Then $M = \varinjlim_i R^n/N_i$ is the filtered direct limit of f.p. modules. Since any module is the direct limit of its f.g. submodules, we see that it is also the filtered direct limit of f.p. modules.

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Thanks! I don't know this material well so I need to ask a few naive follow-up questions. 1) I'm not sure why "this is certainly an isomorphism when $M = R$": that is, I'm not sure why the forgetful functor from $R\text{-Mod} \to \text{Ab}$ preserves colimits. Does it have a right adjoint? 2) I see why any module can be written as the filtered union of its finitely generated submodules, but I don't see why any module can be written as a filtered colimit of finitely presented modules. Can you expand on this? – Qiaochu Yuan Feb 21 '12 at 4:48
Thanks! The situation is clearer to me now. – Qiaochu Yuan Feb 21 '12 at 5:59

This is going to be a bit of a ramble, but hopefully relevant to the present question -- probably everything in here is in Adams and Rosicki, though.

Recall that an object $X$ in a category $\mathcal{C}$ is called compact (as Qiaochu states in his question) if $\hom(X, \cdot)$ commutes with filtered colimits: intuitively, a map from $X$ into an ascending union must land in one of the factors. Compactness is thus a useful version of "smallness," and it turns out to be very useful (e.g. in formulating adjoint functor theorems) that in many relevant categories, the class of all objects (which may be gigantic) is controlled by the much more reasonable category of "small objects."

One way of saying this is that $\mathcal{C}$ is cocomplete, and there is a set of compact objects such that every object in $\mathcal{C}$ can be obtained as a colimit of objects in this set. Let me assume that $\mathcal{C}$ is compactly generated in this sense. (This is a bit more restrictive than the notion of locally presentable, where we don't assume that the generating set consists of compact objects but rather of small objects -- those such that homming out of them commutes with sufficiently filtered colimits.)

For instance, in the present question, the category of $R$-modules is generated by the $R$-module $R$ under colimits, and $R$ is compact. As another example, if $\mathcal{D}$ is a test (small) category, then the category of presheaves on $\mathcal{D}$ is generated by the representable ones under colimits, and the representable ones are compact (in fact, by Yoneda's lemma, homming out of a representable presheaf preserves all colimits -- the representables are tiny).

Anyway, the question asks when we can identify precisely what the compact objects are in the category of $R$-modules. I think we can do this in general for compactly generated categories.

Let $\mathcal{C}$ be a compactly generated category and $S$ the given set of compact objects. Note that $S$ might not consist of all the compact objects: $S$ might just consist of the representables. It is easy to check that a finite colimit of compact objects is compact (this is a consequence of the fact that in sets, finite limits and filtered colimits commute). The claim is that all compact objects are finite colimits of objects in $S$.

In the case of $R$-modules, every module generated under finite colimits from $R$ is finitely presented. So the only compact objects are the finitely presented ones. The same argument works in the category of $R$-algebras (where the compact ones are the finitely presented algebras and a generator is $R[x]$) or in groups (where the compact objects are the finitely presented groups).

To see this, fix any object in $\mathcal{C}$. Since we are assuming that the category is generated by $S$ under colimits, we can form a canonical way of writing $X$ as a colimit of objects in $S$. Namely, we take the (small) category consisting of pairs $(s, f)$ where $s \in S$ and $f: s \to X$ is a morphism. There is a natural functor from this category to $\mathcal{C}$ sending $(s, f)$ to $s$. (This is a generalization of the "category of elements" construction, which writes any presheaf as a canonical colimit of representables.) The claim is that the colimit of this functor gives $X$ again. In fact, if we have maps out of $s$ for every $s \in S$ and $s \to X$, we can get a map out of $X$ just by writing $X$ as a colimit of objects in $S$, and vice versa.

Let's call this generalized category of elements $\mathcal{E}$. We have thus chosen a diagram $D: \mathcal{E} \to \mathcal{C}$, taking values in $S$, such that $X$ is a colimit of $D$. Unfortunately, this needn't be a very good colimit.

Fortunately, if we replace $S$ with the collection $S'$ of objects which are finite colimits of $S$ (rather a small skeleton of this), we can get something better. Let $S'$ be the set in question; we are going to see that it consists of all compact objects. So we can run the same "category of elements"-type construction for $S'$ and, to any $X$, write down $X$ canonically as a colimit of objects of $S'$.

But there's something else you get here: the category of elements $(s' \to X)$ for $s' \in S'$ is actually a filtered category. (This is pretty easy: if we have $s_1' \to X$ and $s_2' \to X$, we can map them both to $s_1' \sqcup s_2' \to X$. If we have two maps $s_1' \rightrightarrows s_2' \to X$, we can form the coequalizer, which is in $S'$ and which maps to $X$.)

So we've thus found that we can write any object of $\mathcal{C}$ as a filtered colimit of objects of $S'$. (This generalizes the observation in Matt E's answer that any module is a filtered colimit of finitely presented ones.) Let $X$ be compact. Since $\hom(X, -)$ commutes with filtered colimits, if take the identity morphism $X \to X$, it must factor through one of the objects in $S'$, which means that $X$ is a retract of an object in $S'$. But $S'$ is closed under retracts (which are finite colimits). So $X \in S'$.

This argument based on "categories of elements" seems to be pretty standard in category theory, when one wants to show that an object with a "good" property (e.g. compactness) can be obtained as a special colimit of certain special good objects.

For instance, the same approach is used to prove that the tiny objects in a presheaf category are precisely the retracts of representables. (A consequence: the presheaf category on a category $\mathcal{C}$ determines the idempotent completion up to equivalence.)

Another application is to Lazard's theorem that a flat $R$-module is a filtered colimit of free modules. Namely, given an $R$-module $M$, one can consider the category of finite free modules $F \to M$ together with morphisms over $M$. This expresses $M$ canonically as a colimit of finite free modules. When $M$ is flat, one can show (using the identity $\hom(P, M) \simeq \hom(P, R) \otimes M$ valid for a finitely presented module $P$) that this category turns out to be filtered.

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Thanks, Akhil! You've given me some good stuff to digest. – Qiaochu Yuan Feb 25 '12 at 6:44

This is a complement to Matt E's nice answer. Let us prove

Lemma A. A direct summand of a finitely presented $R$-module is finitely presented.

Consider the following condition $(*)$ on a finitely generated $R$-module $C$:

$(*)$ If $$0\to A\to B\to C\to0$$ is an exact sequence of $R$-modules, and if $B$ is finitely generated, then so is $A$.

Lemma B. Condition $(*)$ holds if and only if $C$ is finitely presented.

The implication "$(*)\implies C$ finitely presented'' is clear. It is also clear that a direct summand of an $R$-module satisfying $(*)$ satisfies $(*)$. So, we are left with proving that finitely presented $R$-modules satisfy $(*)$.

Let
$$0\to A\to B\to C\to0,\quad F\to G\to C\to0$$ be respectively an exact sequence and a finite presentation of $C$. There are morphisms making the diagram $$\begin{matrix} F&\to&G&\to&C&\to&0\\ \downarrow&&\downarrow&&||&&||\\ A&\to&B&\to&C&\to&0 \end{matrix}$$ commute. By the Snake Lemma, the vertical arrows have isomorphic cokernels. QED

Now let us rephrase Matt E's proof that finitely presented $\implies$ compact. (I'll use the setup of Categories and Sheaves by Kashiwara and Schapira.)

Let $M$ be a finitely presented $R$-module and $F:I\to\mathcal M$ a functor from a small filtrant category to the category of $R$-modules (category which depends on the underlying universe). We must show that the natural map $$\lim_{\longrightarrow}\text{Hom}(M,F)\to\text{Hom}\left(M,\lim_{\longrightarrow}F\right)$$ is bijective.

As $M$ is finitely presented, there is a finite category $J$ and a functor $G:J\to\mathcal M$ such that $G(j)$ is free and finitely generated for all $j$, and $$\lim_{\longrightarrow}G\simeq M.$$ Then there are natural isomorphisms $$\lim_{\longrightarrow}\text{Hom}(M,F)$$ $$\simeq\lim_{\substack{\longrightarrow\\ i}}\text{Hom}\left(\lim_{\substack{\longrightarrow\\ j}}G(j),F(i)\right)$$ $$\simeq\lim_{\substack{\longrightarrow\\ i}}\lim_{\substack{\longleftarrow\\ j}}\text{Hom}(G(j),F(i))$$ $$\simeq\lim_{\substack{\longleftarrow\\ j}}\lim_{\substack{\longrightarrow\\ i}}\text{Hom}(G(j),F(i))$$ $$\simeq\lim_{\substack{\longleftarrow\\ j}}\text{Hom}\left(G(j),\lim_{\substack{\longrightarrow\\ i}}F(i)\right)$$ $$\simeq \text{Hom}\left(\lim_{\substack{\longrightarrow\\ j}}G(j),\lim_{\substack{\longrightarrow\\ i}}F(i)\right)$$ $$\simeq \text{Hom}\left(M,\lim_{\longrightarrow}F\right).$$ The third isomorphism follows from the fact that, in the category of sets, small filtrant limits commute with finite projective limits. The fourth isomorphism results from the assumption that $G(j)$ is free and finitely generated. The other isomorphisms are obvious.

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According to Simion Breaz [1] this result is due to H. Lenzing [2]. Since the introduction of Breaz's paper may be of general interest, I have appended an excerpt below.

[2] H. Lenzing. Endlich prasentierbare Moduln, Arch. Math. (Basel) 20 (1969), 262–266.

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