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I'll be quite fuzzy with the prerequisites, as I don't know myself in what generality the statement I want to understand holds.

Let $\mathcal{C}$ be a category of modules over a non-commutative ring $R$. We have a contravariant exact functor $\mathcal{C}\to\mathcal{C}$ that maps a module $M$ to $M^{\vee}$.

Now, the text I'm reading states that $M$ is simple if $M^{\vee}$ is simple.

How can I prove this? I basically don't know how to express simplicity of a module in terms of short exact sequences...

Thank you!

Edit: I forgot to tell you that the functor induces a self-equivalence of $\mathcal{C}$: its square is naturally isomorphic to the identity functor.

Edit 2: Also, $\mathcal{C}$ is stable under submodules and quotients.

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Your functor is not merely exact but an equivalence of categories. Being a simple module is something that can be detected purely in terms of the category of modules, so assuming $\mathcal{C}$ has enough submodules, your functor will preserve simple modules. –  Zhen Lin Jan 27 '13 at 18:32
    
@ZhenLin: That is exactly what I'm looking for: how to detect that a module is simple in categorial terms? Also, what does 'enough submodules' mean? –  Sh4pe Jan 27 '13 at 18:35
    
A simple module is a module $M$ such that every submodule is either $0$ or $M$. There's not much more to it than that! The trouble is that $\mathcal{C}$ may not contain all the submodules of $M$, so from the perspective of $\mathcal{C}$, $M$ may appear to be simple but not actually be simple. –  Zhen Lin Jan 27 '13 at 18:40
    
@ZhenLin: $\mathcal{C}$ is stable under submodules and quotients in my case. I'm still not able to see a proof for the statement in my question. Could you write one down? –  Sh4pe Jan 27 '13 at 18:46
    
The answer to your question is extremely simple and is contained in my previous comments. If you do not see it, then most likely this is because your understanding of category theory is incomplete. For instance, is it obvious to you that submodules are the same thing as monomorphisms? –  Zhen Lin Jan 27 '13 at 18:50
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1 Answer 1

up vote 2 down vote accepted
  1. Being the zero module is a categorical property: $M$ is the zero module if and only if $\textrm{Hom}(M, M) = 0$.

  2. We identify submodules $M' \subseteq M$ with the inclusion homomorphisms $M' \hookrightarrow M$. Being an inclusion is not a categorical property, but being a monomorphism is: $M' \to M$ is a monomorphism if and only if $\textrm{Hom}(N, M') \to \textrm{Hom}(N, M)$ is injective for "enough" modules $N$. (Exactly what "enough" means will depend on how big your category is.)

  3. Therefore being a simple module is a categorical property: $M$ is a simple module if and only if, for all monomorphisms $M' \to M$, either $M' = 0$ or $M' \to M$ is an isomorphism.

  4. Moreover, being a simple module is a self-dual property: $M$ is a simple module if and only if, for all epimorphisms $M \to M''$, either $M'' = 0$ or $M \to M''$ is an isomorphism. (Just take kernels!)

  5. Putting all these together, we see that a contravariant auto-equivalence will take simple modules to simple modules.

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Awesome! Thank you very very much! :) –  Sh4pe Jan 27 '13 at 19:15
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