(Inspired by an answer given here: does monomorphism of sheaves of abelian groups imply it's injective on the level of open sets? )
I wonder if, in a general (elementary) topos $\mathcal{T}$, if you have a ring object $R$, then the underlying set functor $R{-}\mathbf{Mod} \to \mathcal{T}$ has a left adjoint.
Out of the general constructions I can think of:
- The construction as a submodule of $R^S$ requires decidability of equality on $S$, in order to get the coprojections $S \to R^S$.
- The construction in terms of formal linear combinations $\sum_{i=1}^n r_i e_{s_i}$ requires a natural numbers object.
- The construction a la Freyd's adjoint theorem (take the collection of all possible module structures on a subset of a "large enough" base set, with maps $S$ to the subset; take the product of all these modules; then take the submodule generated by $S$) typically requires some cardinality arguments to ensure the base set is indeed "large enough". These arguments might not be valid in a general topos, as they might require excluded middle and/or axiom of choice. On top of that, the cardinality argument that immediately comes to my mind also involves a natural numbers object.
So, to try to come up with a counterexample, I looked at the case of the topos of finite sets (to avoid the topos having a natural numbers object). However, in this topos, free modules do exist, since this topos has decidable equality on every object. To avoid this, I considered passing to the topos of "pointwise finite" presheaves on a finite category. However, in this topos, you can still form free modules "pointwise".
(Of course, if a general topos does have free modules over ring objects, then it would have to be something special to this case. For example, certainly free abelian groups don't exist in the topos of finite sets.)