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Is there an example of a full subcategory of $\mathsf{Mod}_R$ (namely left $R$-modules) which are not abelian categories, but are "almost" abelian -- that is, they satisfy all of the prerequisites of an abelian category (additive + kernels/cokernels) except that the canonical map $coim(f) \to im(f)$ is not an isomorphism.

There are of course many examples of those that are, e.g. torsion $R$-modules form a Serre subcategory of $\mathsf{Mod}_R$ and coherent $R$-modules form a weak Serre subcategory of $\mathsf{Mod}_R$.

MT_
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1 Answers1

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For the case where $R=\Bbb Z$ : the full subcategory $\mathbf{Ab}_{t-f}$ of torsion-free abelian groups is additive, has kernels (because a subgroup of a torsion-free abelian group is torsion-free), and it also has cokernels : given a morphism $f:A\to B$, its cokernel in $\mathbf{Ab}_{t-f}$ is the quotient of $\frac{B}{f(A)}$ by its torsion subgroup $T\left(\frac{B}{A}\right)$, or equivalently the quotient $\frac{B}{\widehat{f(A)}}$, where $$\widehat{f(A)}=\{b\in B\mid \exists m\in \Bbb Z\setminus0 \text{ such that } m\cdot b\in f(A) \}.$$ Indeed, any morphism $g:B\to C$ such that $gf=0$ factorizes in $\mathbf{Ab}$ through $\frac{B}{f(A)}$, and then if $C$ is torsion-free the corresponding morphism $\frac{B}{f(A)}\to C$ should be $0$ on the torsion of $\frac{B}{f(A)}$.

Now this is not an abelian subcategory because for example the inclusion $2\Bbb Z\to \Bbb Z$ has both its kernel and cokernel equal to $0$, but is not an isomorphism.

darij grinberg
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Arnaud D.
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  • Great example. You could even take the subcategory to be finitely generated, since the ring is Noetherian the kernel will be the same. – MT_ Oct 11 '17 at 13:47