Exact functors preserve free modules? Let $R$ be a principal ideal domain. Suppose that $F$ be an exact functor from the category of $R$-modules to itself. If $M$ is a free $R$-module, is $F(M)$ still a free $R$-module?
I do not know how to prove it, but a note says that this is true without proof.
I shall be grateful if someone helps to give any idea.
 A: This is false.
Taking these two facts:


*

*tensoring $- ⊗_R M$ is exact iff M is a flat module

*a module over a PID is flat iff it is torsion free
we see that $- ⊗ \mathbb Q : \mathrm{Ab} → \mathrm{Ab}$ is an exact functor, but it doesn't preserve free objects: $\mathbb Z ⊗ \mathbb Q = \mathbb Q$, and $\mathbb Q$ is not free over $\mathbb Z$ (any two elements in $\mathbb Q$ are $\mathbb Z$-linearly dependent).
edit: Maybe some extra comments will be useful. A module over a PID is free iff it's projective, so what you would need is a functor that preserves projectives. Only condition for preserving projectiveness I know of is this:
A functor $F : \mathcal C → \mathcal D$ preserves projectives if it's a left adjoint, and if the right adjoint $G$ preserves epis.
If $\mathcal C$ and $\mathcal D$ are Abelian, this translates to $F$ being right exact and preserving arbitrary sums, and having an exact right adjoint.
[edit2: (appologies for the constant edits) if these conditions hold and if $\mathcal C$ and $\mathcal D$ are categories of modules, by Eilenberg-Watts theorem it follows that $F ⊣ G$ is the tensor-hom adjunction, and since $G$ is exact, that $F ≅ - ⊗_R P$ for a projective $P$. If $R$ is a PID, then the only functors (up to isomorphism) that satisfy the above condition are the obvious ones: $- ⊗_R P$ for a free $P$. I don't know if there are any others.]
In general, if $R$ is not a PID, there probably can't be a categorical condition for preserving freeness even for endofunctors, because freeness is not a purely categorical concept: it is defined only wrt. a forgetful functor, and can't be characterized in terms of the abstract category $\mathrm{Mod}_R$ alone. To see this, take the canonical example of Morita equivalence: $\mathrm{Mod}_k = \mathrm{Vect}_k ≅ \mathrm{Mod}_{M_n(k)}$, for a field $k$ (or in fact any ring). Assuming choice, every module over $k$ is free, but this is not true for modules over $M_n(k)$ in general.
(Of course, assuming on the other hand that $F : R ↦ R$ and preserves arbitrary sums is enough for any $R$.)
