Let $k$ be a field and $R$ a commutative associative unital $k$-algebra with augmentation $\epsilon:R\rightarrow k$. Assume that $R$ is either a local Artinian ring or a complete local ring (i.e. it is $\mathfrak{m}$-adically complete, for $\mathfrak{m}$ the unique maximal ideal of $R$).
Consider the induction aka extension of scalars functor with respect to the augmentation $\epsilon$
$$\operatorname{Ind}_R^k: R\mathsf{-mod}\rightarrow k\mathsf{-mod}$$
given by $\operatorname{\operatorname{Ind}}_R^k(B)=k\otimes_RB$. Here we consider $k$ as a right $R$-module by restriction of scalars along $\epsilon$. The abelian group $k\otimes_RB$ becomes a left $k$-module via the left multiplication of $k$. Restrict the functor $\operatorname{Ind}_R^k$ to the full subcategory of $R\mathsf{-mod}$ given by the free $R$-modules. Call this restricted functor $F$.
I am trying to show that $F$ reflects isomorphisms. Is this even true?
Proof when $R$ is commutative and local: The restriction to free modules is important. It lets us work with matrices. The problem then becomes: if a matrix over $R$ becomes invertible over $k$, then it is already invertible over $R$. But this is clear using the determinant and since the invertible elements of $R$ are exactly the ones that become invertible in $k$.