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An object $M$ of an abelian category is called of finite type iff for every directed set of subobjects $M_i$ of $M$ whose sum is $M$ there exists some $i$ with $M = M_i$. Is the direct sum $M \oplus N$ of two objects $M,N$ of finite type again of finite type?

So let $P_i \subseteq M \oplus N$ be a directed set of subobjects whose sum is $M \oplus N$. If $M_i$ denotes the projection of $P_i$ to $M$ and $N_i$ the one to $N$, then it is easy to see that there is some $i$ with $M_i = M$ and $N = N_i$. But does not show yet $P_i = M \oplus N$!

More general: If $0 \to M' \to M \to M'' \to 0$ is an exact sequence, where $M',M''$ are of finite type, does this imply that $M$ is of finite type?

If there are counterexamples: Is it at least true in a Grothendieck-category? What are other reasonable definitions for "finitely generated" which generalize the ones for modules over rings or quasi-coherent modules over nice schemes?

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I think that the preferred notion in general is finite presentation, which reduces to finite generation in the Noetherian case. It has a very nice categorical definition, namely that the contravariant hom functor defined by $A$, that is, $Hom(A,-)$ commutes with filtered colimits. – Harry Gindi Nov 15 '10 at 19:45
@Martin: You seem to have outlined quite a bad way of attacking the problem---your approach seems to not even work in the category of modules over a ring. What happens if you try to generalise the standard proof for the category of modules over a ring, rather than an idea that doesn't work in this setting? – Kevin Buzzard Nov 15 '10 at 19:52
@Kevin: I know the approach does not work. But also for modules I don't know an element-free proof. – Martin Brandenburg Nov 15 '10 at 19:53
@Martin: :-(. Then perhaps it is false in general? Ouch. I thought about it a bit and I can't see an element-free proof either. – Kevin Buzzard Nov 15 '10 at 19:55
Can one cheat and use some sort of Freyd-Mitchell embedding theorem? If not then I give up :-) – Kevin Buzzard Nov 15 '10 at 20:00
up vote 3 down vote accepted

At least if we have a Grothendieck category everything seems OK: Suppose $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0$ is exact with $M'$ and $M''$ of finite type. Assume $\{M_i\}$ is a directed collection of subobjects of $M$ such that $\sum_i M_i=M$. We then have $M'=M'\bigcap\sum_i M_i=\sum_i M'\bigcap M_i$ and hence $M'=M'\bigcap M_{i_0}$ for some $i_0$. Throwing away all indices which are not $\geq i_0$ we may assume $M'\subseteq M_i$ for all $i$. We then get that $M''=\sum_i M_i/M'$ and hence $M''=M_{i_1}/M'$ for some $i_1$

Addendum: Stealing some ideas from Sándor's reply we can get the statement without extra axioms. Note that finite generation is formulated in terms of $\sum_iM_i=M$ which is the same as $\mathrm{lim}M_i\to M$ being surjective (as the sum is image of the limit). Now, with notations as before we put $M''_i$ to be the image of $M_i$ in $M''$. As $\mathrm{lim}M_i\to M$ is surjective we get that so is $\mathrm{lim}M_i''\to M''$ and hence $M''_i=M''$ for some $i$ and after throwing away we can assume this is always true. This means that we get an exact sequence $0\to M'_i\to M'\to M/M_i\to0$ and as $\mathrm{lim}M/M_i=0$ (by right exactness of directed colimits) we get that $\mathrm{lim}M'_i\to M'$ is surjective (again by right exactness) and hence that $M'_i=M'$ for some $i$ but then $M_i=M$ as $M''_i=M''$, which means that $M=M_{i}$.

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Stupid question: how did you interchange the sum and the intersection with $M'$? – Kevin Buzzard Nov 15 '10 at 20:17
That's exactly the Ab5 axiom. – Torsten Ekedahl Nov 15 '10 at 20:19
I am totally confused. Let $V$ be a 2-dimensional vector space. Let $M$ and $P1$ and $P2$ be distinct one-dimensional subspaces. Then $V=P1+P2$ but $M$ is not $(M\cap P1)+(M\cap P2)$. What am I missing? – Kevin Buzzard Nov 15 '10 at 20:23
Aah---I see what I'm missing---the word "directed" :-) Very nice! – Kevin Buzzard Nov 15 '10 at 20:28

EDITs: 1) edited to make it work for the general case of a short exact sequence 2) edited some steps following Torsten's comments below.

I believe the following definition is equivalent to Martin's: Let $M$ be an object of an abelian category. $M$ is of finite type if for any directed system of objects $\{P_i\}$ admitting maps $\{P_i\to M\}$ consistent with the directed system $\{P_i\}$ such that the induced $\lim P_i\to M$ is surjective, there exists a $j$ such that $P_i\to M$ is surjective for all $i\geq j$.

Proposition. Let $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0$ be exact with $M'$ and $M''$ of finite type. Then $M$ is of finite type.

Proof: Let $(M\to Q_i):={\rm coker} (P_i\to M)$ so one has exact sequences: $$ P_i \to M \to Q_i \to 0. $$ Then $$ \lim P_i \to M \text{ is surjective} \quad \Leftrightarrow \quad \lim Q_i =0. $$ Now let $(K'_i\to M'):=\ker (M'\to M\to Q_i)$ and $(Q_i\to C'_i):={\rm coker } (M'\to M\to Q_i)$. By construction we have surjective maps: $$ M \to Q_i \to C_i $$ that composed with $M'\to M$ is the zero map. Hence we obtain a surjective map $\gamma_i : M''\to C'_i$. Similarly, for the surjective map $M \to \lim Q_i \to \lim C_i$ composed with $M'\to M$ we obtain an induced map $\gamma : M\to \lim C_i$. Observe that $\gamma$ has to be the same as $\lim\gamma_i$. However, since $\lim Q_i =0$, it follows that $\gamma=0$ and hence $\lim \gamma_i=0$. Let $K_i'':=\ker\gamma_i$. Then it follows that $\lim K_i''\to M''$ is surjective and hence for some $j$, $K''_i\to M''$ is surjective for all $i\geq j$. However, that implies that $\gamma_i=0$ and hence $C_i=0$ for $i\geq j$.

This in turn implies that $M'\to Q_i$ is surjective for $i\geq j$. Since $\lim Q_i =0$, it then follows that $\lim K'_i\to M'$ is surjective and hence there exists a $j'\geq j$ such that $K'_i\to M'$ is surjective for all $i\geq j'$. It follows that (the surjective) $M'\to Q_i$ is the zero map. Therefore $Q_i=0$, and hence $P_i\to M$ is surjective for $i\geq j'$. Q.E.D.

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Your proposed sticky point is not one I think. However, without Ab5 directed colimits are not left exact so I don't think your conclusion that $\mathrm{lim} Q_i = 0$ implies $\mathrm{lim}K_i=M$ is justified (there seems to be problems with both injectivity and surjectivity of $\mathrm{lim}K_i\to M$). – Torsten Ekedahl Nov 15 '10 at 20:45
@Torsten: Thanks! Yeah, I thought this would be the issue, but I did not want to clutter it with proposed "sticky points". However, in some sense, this is an issue already where I indicated, as I wrote $\Leftrightarrow$ even though there only $\Rightarrow$ is needed, but if it were true there it would have worked for the $K_i$. – Sándor Kovács Nov 15 '10 at 20:53
1) It is true that direct limits are always right exact. 2) I don't see why $M'\to Q_i$ is surjective which seems to be needed to get the surjectivity of $\mathrm{lim} K'_i \to M'$. 3) Note that without Ab5 finite generation should be formulated as $\mathrm{lim}M_i\to M$ surjective implies $M_j=M$ for some $j$. – Torsten Ekedahl Nov 15 '10 at 21:23
@Torsten: I think I fixed it. – Sándor Kovács Nov 15 '10 at 23:55

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