Is $\operatorname{Hom}(A,\mathbb{Z})\otimes_{\mathbb{Z}}\mathbb{R} \simeq\operatorname{Hom}(A\otimes_{\mathbb{Z}}\mathbb{R},\mathbb{R})$? Question is above: Given a ring $A$ (a $\mathbb{Z}$-module) is $\operatorname{Hom}(A,\mathbb{Z})\otimes_{\mathbb{Z}}\mathbb{R} \simeq \operatorname{Hom}(A\otimes_{\mathbb{Z}}\mathbb{R},\mathbb{R})$ using the fact that $\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{R} = \mathbb{R}$. If so, does this hold for general rings ($\mathbb{Z}$-modules) $B,C$ instead of $\mathbb{Z},\mathbb{R}$ or even for general $R$-modules $M,N,P$:
$$
\operatorname{Hom}_R(M,N)\otimes_{R}P \simeq \operatorname{Hom}_R(M\otimes_R P,N \otimes_R P)
$$
? (Some sort of distributivity between $\operatorname{Hom}$-functor and Tensor product). If not, are there any constraints on $R,M,N,P$ so that this isomorphism holds?
I tried using the universal property of the tensor product, but I cannot specify those bilinear maps concretely. Maybe that's the case because this equality is wrong. If so, is $\operatorname{Hom}(A,\mathbb{Z})\otimes_{\mathbb{Z}}\mathbb{R} \simeq \operatorname{Hom}(A\otimes_{\mathbb{Z}}\mathbb{R},\mathbb{Z})$ or $\operatorname{Hom}(A,\mathbb{Z})\otimes_{\mathbb{Z}}\mathbb{R} \simeq \operatorname{Hom}(A,\mathbb{R})$ correct so that the tensor product is only used on one of the arguments of $\operatorname{Hom}$?
Any help would be appreciated as I'm not too familiar with using the universal property to show isomorphism.
 A: Let $R$ be a commutative ring, with $M,N,P$ being $R$ modules. In what follows, "linear,bilinear", Hom and tensor products all refer to the ring $R$.
The bilinear mapping
\begin{align}
(T,p)\mapsto [m\mapsto T(m)\otimes p] 
\end{align}
from $\text{Hom}(M,N)\times P$ into $\text{Hom}(M, N\otimes P)$ gives rise a unique linear mapping (by the universal property)
\begin{align}
\text{Hom}(M,N)\otimes P \to \text{Hom}(M,N\otimes P) 
\end{align}
such that the appropriate diagram commutes (meaning $T\otimes p$ is mapped to the expression above in square brackets).
If we further assume $M$ is finitely generated and free, then this mapping is even an isomorphism (try proving this). I'm not an algebra wizard, so I'm not sure if/how much we can weaken the assumptions on $M$. In short, tensoring a Hom puts the tensor product in the target space.
If you want to know what happens when the tensor product is in the domain of Hom, then we have (no assumptions needed on $M,N,P$)
\begin{align}
\text{Hom}(M\otimes P,N)\cong \text{Hom}^2(M\times P,N)\cong \text{Hom}\left(M,\text{Hom}(P,N)\right).
\end{align}
The first is by the universal property (a linear mapping out of the tensor product is the same as a bilinear mapping). The next is clear because if $f$ is bilinear we can map it to $m\mapsto f(m,\cdot)$.
By repeatedly the two observations above, we see that if $M_1,M_2,N_1,N_2$ are $R$ modules with $M_1,N_1$ finitely generated and free, then we have
\begin{align}
\text{Hom}(M_1,M_2)\otimes \text{Hom}(N_1,N_2)&\cong
\text{Hom}(M_1\otimes N_1,M_2\otimes N_2)
\end{align}
