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I was trying to prove that free modules were projective in the language of abelian categories, but did not succeed. I was missing a good description of what a free module is. So my question is essentially this; is it possible to give a(n easy) definition of what a free module is without referring to a faithful functor from R-mod to Set?

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If you are working with rings with identity, I would think that "is isomorphic to a direct sum of copies of $R$" would be the only definition you'd ever need... Are there complicating factors that motivate this question? –  rschwieb Oct 16 '12 at 20:04
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There is no notion of free module that doesn't invoke a forgetful functor to $\textbf{Set}$. The next best thing is, well, a projective object. –  Zhen Lin Oct 16 '12 at 21:39
    
@rschwieb but for that description to 'work' it would be nice to identify R (in R-mod). –  krey Oct 17 '12 at 11:22

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up vote 4 down vote accepted

There is no notion of "free object" in an abelian category which restricts to "free module" in the case of $\mathsf{Mod}(R)$. Namely, any sensible notion should be invariant under equivalences of categories. But if $P$ is a finitely generated projective $R$-module of rank $1$ (i.e. a line bundle on $\mathrm{Spec}(R)$), then $P \otimes - : \mathsf{Mod}(R) \to \mathsf{Mod}(R)$ is an equivalence of categories which maps $R \mapsto P$. Here, $R$ is free, but $P$ may be not free.

It was asked in the comments if it possible to reconstruct $R$ from $\mathsf{Mod}(R)$. The above shows that this is not possible. See also the notion of Morita equivalence. At least one can construct the center $Z(R)$ from $\mathsf{Mod}(R)$. Namely, for a category $C$ its center $Z(C)$ is defined to be the monoid of natural transformations $\mathrm{id}_C \to \mathrm{id}_C$. It is commutative. If $C$ is abelian, this monoid is a commutative ring in a canonical way. And one can prove (directly, or via the enriched Yoneda lemma) that $Z(\mathsf{Mod}(R)) \cong Z(R)$.

Also, it is possible to characterize module categories amongst abelian categories:

Let $\mathcal{A}$ be a cocomplete abelian category and $P \in \mathcal{A}$ be a projective compact generator, which means that $\hom(P,-) : \mathcal{A} \to \mathsf{Ab}$ is right exact, preserves (infinite) coproducts, and is faithful. Consider the ring $R := \mathrm{End}(P)$. Then $\mathcal{A} \to \mathsf{Mod}(R), X \mapsto \hom(P,X)$ is an equivalence of categories which maps $P \mapsto R$ (good exercise). Conversely, $R$ is a projective compact generator in $\mathsf{Mod}(R)$. But the same is true for $R^n$, which illustrates that $R$ and $M_n(R)$ are Morita equivalent.

Finally let me mention that $\mathsf{Mod}(R)$ is more than just an abelian category. It also carries a tensor structure, making it an abelian $\otimes$-category, the unit being $R$. With this additional data we have a trivial reconstruction from $R$ out of $\mathsf{Mod}(R)$. But even in the setting of abelian $\otimes$-categories it is not true in general that the unit is projective (consider $\mathrm{Qcoh}(X)$ for a projective scheme $X$). Therefore the statement "free objects are projective" fails again.

If you want to prove that free modules are projective in $\mathsf{Mod}(R)$, see Rasmus' answer.

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The statement about $\mathsf{Mod}(R)$ carrying a tensor product only holds for commutative $R$. Since the question is tagged commutative-algebra, this find be the only case the OP cares about. –  Rasmus Oct 21 '12 at 15:50

A free $R$-module is a module of the form $\bigoplus_{i\in I} R$ for some index set $I$.

Using that $$\mathrm{Hom}_R(\bigoplus_{i\in I} R,M)=\prod_{i\in I}\mathrm{Hom}_R( R,M)=\prod_{i\in I}M$$ it is easy to see that the functor $\mathrm{Hom}_R(\bigoplus_{i\in I} R,\_)$ is exact, showing that $\bigoplus_{i\in I} R$ is projective.

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I think you misunderstood his question. His question was about an analogue of the notion of "free modules" in a general abelian category. –  Your Ad Here Oct 16 '12 at 18:44
    
@IHaveAStupidQuestion: I read the question differently. Let's wait for OP to clarify. –  Rasmus Oct 16 '12 at 18:51
    
@Rasmus: yes, my question is essentially, about 'free modules' in a general abelian category. I do however like the argument you give above, could you please clarify the "easy to see" part? (I'm very new to this) –  krey Oct 16 '12 at 19:16
    
@krey : edited. –  Rasmus Oct 16 '12 at 22:51

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