Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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
show 2 more comments

2 Answers

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 $M=M_{i}$.

share|improve this answer
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
add comment

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.