$\operatorname{Hom}(K,R)\otimes\operatorname{Hom}(C,R)\simeq\operatorname{Hom}(K \otimes C,R)$ Is the following statement true when $K$ is a chain complex of finitely generated free module over $R$ PID and $C$ is a chain complex over $R$ ?
$$\operatorname{Hom}(K,R)\otimes\operatorname{Hom}(C,R)\simeq\operatorname{Hom}(K \otimes C,R)$$
I thought to prove the statement index by index since the tensor product of two chain complex is defined $C \otimes D = \bigoplus\limits_{p+q = n}C_p \otimes C_q$.
I don't know how to use the fact that $K_q$ is finitely generated, maybe should I use structure theorem on PID to assert that $\exists I_1,\cdots,I_m : K_q = \bigoplus\limits_{i = 1}^m R/I_i $?
 A: Let's first consider the case of modules over a PID. In this case, we can use the structure theorem to see that if $A,B$ are finitely generated modules, we can write them as a direct sum of free modules and torsion modules. Taking duals (which is what taking morphisms into the base ring is), kills torsion modules, so we can identify $A^* \otimes B^*$ with $(R^n)^* \otimes (R^m)^*$, where $n,m$ come from the decompositions of $A,B$. Similarly, we see that $(A \otimes B)^*$ can be identified with $(R^{nm})^*$, and the map $(R^n)^* \otimes (R^m)^* \rightarrow (R^{nm})^*$ is given by the map which sends $f \otimes g$ to the element of $(R^{nm})^*$ corresponding to the bilinear map $(x,y) \rightarrow f(x)g(y)$ from $R^n \times R^m$ to $R$.
This map is clearly injective since if both $f$ and $g$ are nontrivial, this will yield a nontrivial bilinear map. Now, since we are finitely generated, dimension counting yields this is an isomorphism.
From here it is not difficult to see the chain complex case. Presumably, you mean for your chain complexes to be bounded below since this is another common finiteness condition (otherwise this is not true). In this case, after using the fact that duals commute with finite direct sums, your map $K^* \otimes C^* \rightarrow (K \otimes C) ^*$ will levelwise consist of a direct sum of finitely many copies of the type of map we just proved the result for. Since a direct sum of isomorphisms is an isomorphism, you will see that the map involved is an isomorphism.
Where does the differential come in you might ask, well all you need to do is ensure the map described is a map of chain complexes. This is an instructive exercise, especially if you have not done much work with chain complexes before.
