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Hello!

I am trying to understand the structure of the smallest abelian subcategory of an abelian category that contains one object $X$ and all endomorphisms of that object (or rather containing a finite number of object with all morphisms between, but this seems the same to me by just taking the sum). Just taking the intersection over all subcategories containing my object and morphisms seems a bad idea, since sums, kernels etc are just determined up to isomorphism. I am not bothered about having "too many" isomorphic objects in my subcategory, though. I stumbled over the notion of subquotients. Since kernels, cokernels are subquotients and subquotients of subquotients are subquotients it seems to me as if I could describe the objects of my subcategory the following way: Just take all objects isomorphic to subquotients of finite direct sums of $X$. Does this make sense? Thank you!

Jonas

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I'm not sure I understand exactly what you want to do. You want the smallest subcategory containing one object $X$ and all endomorphisms of $X$. Okay, so just take the category consisting of $X$ and all of its endomorphisms. If you mean something else by subcategory, then what? If I'm not horribly mistaken, you need to distinguish between "abelian subcategory" and "sub-abelian category" because in the former the kernels, cokernels, etc. don't need to agree with those in the ambient category (or something like that). –  Qiaochu Yuan Mar 13 '12 at 15:44
    
Oh yeah, sorry I didn't mention that the subcategory should be abelian. I see the problem with the distinction between sub-abelian categories and abelian subcategories. Haven't thought about it before. I guess I would prefer the subcategories to abelian subcategories, but I guess an answer to either would help me to understand the task better. –  JvW Mar 13 '12 at 15:57
    
There was a similar terminological discussion here: mathoverflow.net/questions/86090 –  Tom Leinster Mar 13 '12 at 17:45
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If you have a fixed ambient category, there is no reason to worry about objects only being defined up to isomorphism. It is very reasonable to consider only those subcategories that are "closed under isomorphism": if $X$ is in your subcategory and $X \overset\sim\to Y$ in your ambient category, then demand that $Y$ is in your subcategory. You should also of course demand that the subcategory be full. –  Theo Johnson-Freyd Mar 14 '12 at 3:03

2 Answers 2

It is not true in general that the abelian subcategory (by which I mean sub-abelian category) generated by an object $X$ is all subquotients of finite sums of $X$. It is contained in these subquotients, but it might not be all of them. This is because, for instance, not every subobject of $X$ is the kernel of an endomorphism of $X$ (or more generally, a map from $X$ to a sum of copies of $X$).

As an example, consider the abelian subcategory of $\mathbb{Z}$-modules generated by $\mathbb{Q}$. Because any $\mathbb{Z}$-homomorphism of $\mathbb{Q}$-vector spaces is automatically $\mathbb{Q}$-linear, you get only the finite dimensional $\mathbb{Q}$-vector spaces, and don't, for instance, get the subgroup $\mathbb{Z}$ of $\mathbb{Q}$.

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Good point. The closure with respect to subquotients yields the smallest localizing/thick (...) subcategory containing $X$. –  Martin Brandenburg Mar 13 '12 at 20:07
    
Thank you! That makes sense. –  JvW Mar 14 '12 at 8:30

If $\mathcal{A}$ is some abelian category and $S$ is some set of objects in $\mathcal{A}$, then its "abelian closure" $\overline{S}$ should be the smallest abelian full subcategory of $\mathcal{A}$ containing $S$ such that the inclusion functor is exact (in other words, it's a "sub-abelian category" as Qiaochu calls it). Note that it is automatically closed under isomorphisms (consider $0 \to A \cong A' \to 0$).

It always exists: Consider the set of all full subcategories $T$ of $\mathcal{A}$ containing $S$, which also contain $0$ and are closed with respect to kernels, cokernels and direct sums. By this I mean that if $f$ is some morphism between objects in $T$, then every kernel/cokernel of $f$ also lies in $T$ (similarly with direct sums). Now take their intersection.

If you work with classes instead of universes, this intersection will cause some set-theoretic problems. One can avoid them as follows, thereby giving another construction of $\underline{S}$:

Define full subcategories $S_{\alpha}$ for ordinals $\alpha$ as follows: Let $S_{0} = S \cup \{0\}$. In the limit step, take the union. If $S_{\alpha \omega + 3n}$ is already defined, then adjoin all kernels of all morphisms in this set and optain $S_{\alpha \omega + 3n+1}$. Then, let $S_{\alpha \omega + 3n+2}$ be the closure under cokernels, and let $S_{\alpha \omega + 3n+3}$ be the closure under binary direct sums. Finally, define $\overline{S} := \cup_{\alpha} S_{\alpha}$.

As for a specific example, $R$ is "abelian dense" in $\mathrm{Mod}_{fg}(R)$, when $R$ is noetherian.

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$R$ is not "abelian dense" in $\mathrm{Mod}(R)$, as this construction does not allow you to take infinite sums. If $R$ is Noetherian, the abelian subcategory generated by $R$ will be the finitely generated $R$-modules. –  Eric Wofsey Mar 13 '12 at 19:17
    
Sure, I've edited it. –  Martin Brandenburg Mar 13 '12 at 19:59
    
Hi Martin. Thank you. But how does your construction imply that your intersection is closed under direct sums? –  JvW Mar 14 '12 at 8:57
    
@JvW: Of course we also have to adjoin them, thanks. I've corrected it. –  Martin Brandenburg Mar 14 '12 at 9:03
    
Why should one want the abelian closure to be closed under taking extensions? If this condition is not required, it seems to me that one gets a sub-abelian category with exact inclusion (which is fine enough) but maybe not closed under isomorphisms. –  Matthieu Romagny Apr 4 at 7:50

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