Let $(R, \mathfrak m,k)$ be an Artinian local Gorenstein ring, hence $\text{Hom}\_R(k, R)\cong k$, and so


$\text{Hom}\_R(k, R^{\oplus n})\cong k^{\oplus n} \cong k \otimes\_R R^{\oplus n} , \forall n \ge 0.$ Let $\mod (R) $ be the category of finitely generated $R$-modules, and $\mathcal F(R)$ be the subcategory of all finitely generated free $R$-modules.


Consider the two covariant functors $F, G: \mathcal F(R)\to \mod(R)$ defined as $F(-):=\text{Hom}\_R(k,-)$ and $G(-):=k \otimes\_R -$. From what we have mentioned above, we see that $F(M)\cong G(M), \forall M \in \mathcal F(R) $.


My question is: Are the two functors $F$ and $G$ naturally Isomorphic?



On the functors $\text{Hom}_R(k,-)$ and $k \otimes_R ( -)$ for Artinian local Gorenstein ring $R$
Yes.


We may as well take $\mathcal F(R)$ to be the category with objects $R^n$ for $n\in \mathbb N$ and take morphism $\operatorname{Hom}(R^m,R^n) =M\_{n,m}(R) $.


Both functors take $R^n$ to $k^n$ and take a homomorphism in $M\_{n,m}(R)$ to that matrix modulo the maximal ideal, so are naturally isomorphic.


Alternatively, we can express this functor as follows: Given $ a \in k \otimes\_R M$, we map $b \in k$ to $M$ by taking $ab \in k \otimes\_R M$ and applying the map $k \otimes\_R M \to R \otimes\_R M =M $ given as the tensor product of an injection $k\to R$ and the identity. This is an isomorphism for $R$, hence for all free modules.


