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Let $A$ be a commutative ring. I thought that an exact functor (from the category of $A$-modules to itself) is defined to be a functor which sends every exact sequence to an exact sequence. But many books seem to define it to be a functor which sends every short exact sequence to a short exact sequence. But I don't think they are equivalent unless it sends a zero module to a zero module.

Is it true that every functor on the category of $A$-modules to itself sends a zero module to a zero module?

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The "correct" definitions (in that they generalize the most smoothly to non-additive contexts) are that a functor is

  • left exact if it preserves finite limits,
  • right exact if it preserves finite colimits, and
  • exact if it preserves both.

For a functor between abelian categories, the last condition is equivalent to being additive and preserving short exact sequences (exercise). In any case, by default people usually only consider additive functors between abelian categories anyway.

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The answer is already given by Sdevalapurkar and Qiaochu Yuan, but here is a deceptively easy proof which doesn't use category theoretic terminology:

Proposition An exact functor $F:A\operatorname{Mod}\to A\operatorname{Mod}$ sends a zero module to a zero module.

Proof The exact sequence $$0\to0\to0$$ is sent to an exact sequence $$F(0)\xrightarrow{\operatorname{id}_{F(0)}} F(0)\xrightarrow{\operatorname{id}_{F(0)}}F(0),$$ which implies that $F(0)=0$.

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    $\begingroup$ It's not "deceptively" easy. It is simply easy. $\endgroup$
    – Zhen Lin
    Sep 9 '14 at 8:18
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An exact functor is additive. It can be shown that the condition that a functor is additive is equivalent to the condition that it preserves the zero object and binary direct sums; see here. Hence, any exact functor preserves the zero object. It is not necessary that every such functor the zero object. If you consider the $\mathrm{Ab}$-enriched category $R\mathrm{Mod}$ with finite firect sums, then you get the fact that exact functors from $R\mathrm{Mod}$ to $R\mathrm{Mod}$ take the zero module to itself.

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  • $\begingroup$ Why is an exact functor additive? $\endgroup$
    – ashpool
    Sep 9 '14 at 3:41
  • $\begingroup$ @ashpool Between additive categories, which is the context in which you're working, a functor is exact if it is additive and preserve all kernels and cokernels. $\endgroup$
    – user122283
    Sep 9 '14 at 4:05
  • $\begingroup$ I wonder if there is a concrete proof that shows "every exact functor $A\operatorname{Mod}\to A\operatorname{Mod}$ sends a zero module to a zero module," without any reference to the category theory? Here by an "exact functor" I mean it sends every exact sequence to an exact sequence. $\endgroup$
    – ashpool
    Sep 9 '14 at 6:06
  • $\begingroup$ @ashpool This can be interpreted as a joke - see your own answer. :-) $\endgroup$
    – user122283
    Sep 9 '14 at 18:34

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