Does the statement "If $A$, $B$ are modules over a commutative ring $R$, then $B$ is a maximal submodule of $A$ if and only if $A/B$ is a simple module" generalize to the setting of abelian categories? That is, is it true to say "If C is an abelian category and $A, B \in$ C, then $B$ is a maximal subobject of $A$ if and only if $A/B$ is a simple object?"

My hunch is that this statement does generalize, since I've seen it stated that the Jordan-Holder theorem for abelian categories is a "straightforward generalization" of the version for modules (wherein this fact is used), but I've had a lot of trouble finding anything on maximal subobjects at all and I haven't figured out the details yet.


An attempt (everything happens inside abelian categories here and sorry for not being able to draw commutative diagrams with named arrows):

$\textbf{Definition:}$ A monomorphism $\iota: B \rightarrow A$ is called a maximal subobject of $A$ if $\iota$ is not an isomorphism and the existence of two monomorphisms $B \rightarrow B'$ and $B' \rightarrow A$ sitting in a commutative diagram with $\iota$ implies that one of them is an isomorphism.

$\textbf{Definition:}$ An object $S$ is called simple if $0 \rightarrow S$ is a maximal subobbject of $S$.

$\textbf{Proposition:}$ Let $\iota: B \rightarrow A$ be a morphism with cokernel $\pi: A \rightarrow S$. If $\iota$ is a maximal subobject of $A$, then $S$ is simple.
Proof. Note that since $\iota$ is not an isomorphism, we have $S \neq 0$ so we have the non-triviality condition. Let $\alpha: T \rightarrow S$ be a monomorphism. We want to show that either $\alpha = 0$ or $\alpha$ is an isomorphism. Form the pull-back along $\alpha$ and $\pi$, with parallel arrows to $\alpha$ and $\pi$ being $\alpha':P \rightarrow A$ and $\pi': P \rightarrow T$, respectively.
First, $\alpha'$ is a monomorphism since pull-backs preserve kernels.
Second, since $\pi$ is an epimorphism, the pull-back square is also a push-out; hence $\pi'$ is an epimorphism.
Third, since $\iota$ is a monomorphism, it is a kernel, hence a kernel of its cokernel $\pi$. Again since pull-backs preserve kernels, there is a morphism $\iota' : B \rightarrow P$ which is a kernel of $\pi'$ and makes the relevant triangle commute ($\alpha' \circ \iota' = \iota$).
Now, since $\iota$ is a maximal subobject, there are two possibilities: either $\iota'$ or $\alpha'$ is an isomorphism. The former yields $$\alpha \circ \pi' = \pi \circ \alpha' = \pi \circ \iota \circ (\iota')^{-1} = 0$$ hence $\alpha = 0$ as $\pi'$ is an epimorphism. And the latter yields that $\alpha$ is an isomorphism since $\alpha$ and $\alpha'$ are parallel in a push-out square.

I haven't tried but I believe the converse statement can be proved similarly.

  • $\begingroup$ Thanks very much for the answer. Looks like the proof is much more complicated than I thought it would be...! $\endgroup$ – Omnivium Jul 14 '13 at 20:03
  • $\begingroup$ You are welcome! I agree that chasing all these diagrams feels more complicated than it should be, but the general principle I used here is simple once you get used to it: When we have a surjection of modules we take inverse images of elements (or submodules), which is not an option for an epimorphism in an arbitrary abelian category. In such situations, forming a pull-back along that epimorphism (which is immediately also a push-out etc.) usually serves well. $\endgroup$ – Cihan Jul 14 '13 at 22:30

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