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Let $\mathcal C$ be an abelian category and consider its ind-category $Ind(\mathcal C)$:

(1) Is $Ind(\mathcal C)$ always abelian? (If not, what conditions are needed?)

(2) Is $\mathcal C\subseteq Ind(\mathcal C)$ a Serre subcategory? In other words, if we have a short exact sequence:

$$0\longrightarrow M'\longrightarrow M\longrightarrow M'' \longrightarrow 0$$ in $Ind(\mathcal C)$, then $M\in \mathcal C$ iff $M'$ and $M''$ are in $\mathcal C$?

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    $\begingroup$ The ind-completion of an abelian category is an abelian category, but the converse is not true. $\endgroup$
    – Zhen Lin
    Jan 1, 2016 at 14:17
  • $\begingroup$ arxiv.org/abs/1211.3678 might be relevant. $\endgroup$
    – HeinrichD
    Nov 11, 2016 at 21:35

1 Answer 1

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(1) is proved in Kashiwara-Schapira [KS], Theorem 8.6.5.

For (2), [KS], Proposition 8.6.11 prove that $\mathcal C$ is closed under kernels, cokernels, and extensions in $Ind(\mathcal C)$.

I believe the following is true, but I was not able to find any reference, unfortunately. The condition is certainly often satisfied in practice.

Claim: We have that $\mathcal C$ is a Serre subcategory of $Ind(\mathcal C)$ if and only if every filtered colimit of subobjects of a fixed object $X \in \mathcal C$ exists in $\mathcal C$.

First, suppose the condition holds. By above, it suffices to show that $\mathcal C$ is closed under subobjects in $Ind(\mathcal C)$. Suppose we have a subobject $0 \to "\varinjlim_I\!\!" X_i \to X$. The map is $"\varinjlim_I\!\!" (X_i \to X)$, and it's a general fact that we can compute finite (co)limits ``pointwise'', in particular the image of this morphism is given by $"\varinjlim_I\!\!" X_i'$, where $X_i'$ is the image of $X_i$ in $X$ (see e.g. [KS], Lemma 8.6.4). By assumption, $\varinjlim_I X_i'$ exists in $\mathcal C$, and finally

$$"\varinjlim_I\!\!" X_i \cong "\varinjlim_I\!\!" X_i' \cong \varinjlim_I X_i' \in \mathcal C,$$

as desired. In more detail, the second isomorphism holds because for any $Y \in Ind(\mathcal C)$ we have

$$Hom_{Ind(\mathcal C)}("\varinjlim_I\!\!" X_i',Y) = \varprojlim_I Hom_{Ind(\mathcal C)}(X_i',Y) = \varprojlim_I Hom_{\mathcal C}(X_i',Y) = Hom_{\mathcal C}(\varinjlim_I X_i',Y) = Hom_{Ind(\mathcal C)}(\varinjlim_I X_i',Y).$$

The converse follows easily, using the last displayed equation (the first two equalities, and restrict to $Y \in \mathcal C$).

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    $\begingroup$ What's a counterexample? Finitely presented modules of a ring that is coherent but not noetherian? Say, $k[x_1,x_2,...]$, the maximal ideal exists in the ind-category as a submodule? $\endgroup$
    – Ben Wieland
    Jun 16, 2023 at 18:00
  • $\begingroup$ Good question! I think you are right, we always have $Ind(f.p. R-mod) = R-mod$, so we just need any ring which has a f.p. module that admits a non-f.p. submodule. $\endgroup$
    – alpha101
    Jun 16, 2023 at 18:31
  • $\begingroup$ In particular, if $M$ is finitely presented, with non-f.p. submodule $M'$, then $M'$ is a filtered colimit of all f.p. submodules of $M'$ in the Ind-category $R-Mod$; this particular limit can then not exist in (f.p. R-modules). $\endgroup$
    – alpha101
    Jun 16, 2023 at 19:05

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