Why not _co_free modules?

2010-09-08T18:36:47Z

Let $R$ be a ring, and $R\text{-Mod}$ its category of all left modules. There is a "forgetful" functor $\operatorname{Forget}: R\text{-Mod} \to \text{AbGp}$, which is additive, continuous, and cocontinuous (in particular, exact). Since $R\text{-Mod}$ is both complete and cocomplete, $\operatorname{Forget}$ has both a left adjoint $\operatorname{Free}: \text{AbGp} \to R\text{-Mod}$ and a right adjoint $\operatorname{Cofree}: \text{AbGp} \to R\text{-Mod}$.

You can see what these functors are explicitly. Let me write $_R R_{\mathbb Z}$ for "$R$ as a left module" and $_{\mathbb Z} R _R$ for "$R$ as a right module". The $\operatorname{Forget}$ functor is (isomorphic to) the functor $\operatorname{Hom}_R({_R R_{\mathbb Z}},-)$ — this description makes it clearly continuous, and its left adjoint is $\operatorname{Free} \cong {_R R_{\mathbb Z}} \otimes_\mathbb Z (-)$ . But we also have $\operatorname{Forget} \cong {_{\mathbb Z} R _R}\otimes_R (-)$ , whence its right adjoint is $\operatorname{Cofree} \cong \operatorname{Hom}_{\mathbb Z}({_{\mathbb Z} R _R},-)$ .

I feel like I have some positive amount of experience with free modules. (I would say, given the above, that the correct definition of "free module" is "object in the essential image of $\operatorname{Free}$", although what's actually used is "object of the form $\operatorname{Free}(\mathbb Z^{\oplus \kappa})$ for some cardinal $\kappa$.) But I hardly ever come across the essential image of $\operatorname{Cofree}$, or indeed the cofree functor at all. (Again, maybe the "standard" definition of "cofree module" is "module isomorphic to $\operatorname{Cofree}((\mathbb Q/\mathbb Z)^{\times \kappa})$," or something.) The functors are not the same: when $ R = \mathbb Z/2$, then $\operatorname{Free}(\mathbb Z) = \mathbb Z/2$, whereas $\operatorname{Cofree}(\mathbb Z) = 0$. If you would rather replace $\mathbb Z$ by a field throughout, then they are still not the same when $R$ is infinite-dimensional (for example).

So: Do people use cofree modules? If so, how? If not, why not? Are free modules just a lot nicer than cofree ones, and if so, how?

Answer by Emerton for Why not _co_free modules?

2010-09-08T19:10:06Z

This construction is used frequently (at least, I use it frequently in my work). For example, it appears in the usual proof that module categories have enough injectives. (In this case one studies $Cofree(\mathbb Q/\mathbb Z)$, as you anticipated.)

If we generalize slightly, and replace $\mathbb Z$ by the group ring $k[H]$ and $R$ by the group ring $k[G]$ (with $H$ being a subgroup of $G$), then $Hom_{k[H]}(k[G],\text{--})$ is precisely the functor of induction from $H$-representations to $G$-representations, and the adjointness you note is a form of Frobenius reciprocity.

If $R$ is a Hecke algebra (over $\mathbb Z$) on a space of weight $k$-cuspforms of some level, then $Cofree(\mathbb Z)$ is the space of modular forms of weight $k$ with coefficients in $\mathbb Z$. (This technical relationship between Hecke operators and the space of modular forms on which they operate is used frequently by number theorists working on the arithmetic of modular forms.)

There are lots of other contexts in which this functor (and its variants, replacing $\mathbb Z$ by other rings) appear, but maybe I've said enough for now.

Why not _co_free modules? - MathOverflow

Why not _co_free modules?

Answer by Emerton for Why not _co_free modules?