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I have the following question:

If a functor between two categories sends exact sequences to exact sequences, how does it follow that it preserves simple modules as well?

Thanks for the help.

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categories of modules, presumably? –  rschwieb Apr 18 '13 at 16:19
    
The zero functor is a counterexample (simple modules are nontrivial by definition). –  Martin Brandenburg Apr 21 '13 at 18:39

1 Answer 1

up vote 2 down vote accepted

I don't think this is true. For example take the category of abelian groups (i.e. $\mathbb{Z}$ -modules). Then $\mathbb{Z}^{2}$ is free and hence projective, hence flat, so tensoring with $\mathbb{Z}^{2}$ preserves exact sequences. Now $\mathbb{Z}/2\mathbb{Z}$ is a simple $\mathbb{Z}$-module but $\mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}^{2} = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ is not simple.

I think the property you are looking for is that it reflects exactness. That is if the image of a sequence is exact, then the original sequence was exact.

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Although that only works if the functor is essentially surjective. What's the context of the question? –  tharris Apr 18 '13 at 16:04

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