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It is said that the category of sheaves of abelian groups on a Grothendieck site(topology) is an abelian category. On the other hand, it is known that in usual algebraic geometry, given an variety (say, over $\mathbb C$), the category of coherent sheaves of $\mathcal O_X$-modules is abelian.

Now, a rigid analytic space $X$ inherits a Grothendieck topology and we can similarly define coherent sheaves of $\mathcal O_X$-modules with respect to this. So I would like to ask

Question: Do coherent sheaves on rigid analytic spaces form an abelian category?

In the first place, we know from BGR's book that taking kernels and images will preserve coherence in this setting. So the answer might be positive. But the axioms of abelian categories are too abstract for me, and I have no idea about how to prove or disprove this.

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  • $\begingroup$ For schemes, you need local Noetherianity or something like that (the category of finitely presented modules over a non-Noetherian commutative unital ring is not abelian in an obvious way, see mathoverflow.net/questions/327961/…). I am not familiar with rigid analytic geometry. Is some finiteness hypothesis a part of the definition? $\endgroup$
    – user140765
    Commented May 31, 2019 at 19:12
  • $\begingroup$ @kartop_man Thanks for the point. Let's say we are in good enough situations and assume Noetherianity or something similarly. $\endgroup$
    – Hang
    Commented May 31, 2019 at 19:56
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    $\begingroup$ In non-noetherian situations, coherent sheaves are still an abelian category (but they don't coincide with finitely presented ones). See stacks.math.columbia.edu/tag/01BU $\endgroup$
    – Mattia Talpo
    Commented Jun 1, 2019 at 22:28

1 Answer 1

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The answer is yes.

An abelian category is a category $\mathcal C$ with the following properties:

  1. $\mathcal C$ is additive.

  2. $\mathcal C$ has kernels and cokernels.

  3. Images and coimages coincide. That is, for every morphism $f$ in $\mathcal C$, the canonical map $coker(ker(f)) \to ker(coker(f))$ is an isomorphism.

(1) is obvious in our case. (2) follows because $Coh(X) \subseteq Sh(X)$ is a full subcategory closed under kernels and cokernels, and $Sh(X)$ has all kernels and cokernels. (3) is then automatic: because this is an isomorphism in $Sh(X)$ and $Coh(X)$ is a full subcategory, it is also an isomorphim in $Coh(X)$.

In conclusion: if $\mathcal C \subseteq \mathcal D$ is a full subcategory containing 0, closed under finite direct sums, and closed under kernels and cokernels, and if $\mathcal D$ is abelian, then so is $\mathcal C$.

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