Hom of tensor products and change of algebra Motivated by the question in this post here, I'm wondering if a similar statement is true in the following sense:

Let $A$ be a commutative algebra over a field $\mathbb{K}$, and let $0\not=I\subset A$ be an ideal. Suppose $P,Q$ are $A$-modules. Is it true that
$$ \operatorname{Hom}_{A/I}(A/I\otimes_{A}P,A/I \otimes_{A}Q)\cong A/I\otimes_{A}\operatorname{Hom}_{A}(P,Q)?$$

 A: The have the usual sequence of natural isomorphisms
$$ \mathrm{Hom}_{A/I}(A/I\otimes P,A/I\otimes Q)\cong
\mathrm{Hom}_A(P,\mathrm{Hom}_{A/I}(A/I,A/I\otimes Q))\cong\mathrm{Hom}_A(P,A/I\otimes Q) $$
as well as a natural transformation
$$ A/I\otimes\mathrm{Hom}_A(P,Q) \to \mathrm{Hom}_A(P,A/I\otimes Q), \quad
(\bar a\otimes f)(p) = \bar a\otimes f(p). $$
We have seen several examples of how this can fail to be an isomorphism. Here's another, with everything being finite dimensional: take $A=k[x,y]/(x,y)^2$, a local ring with maximal ideal $\mathfrak m=(x,y)$, $P=k$, and $Q=A$. Then $A/\mathfrak m\otimes Q\cong Q/\mathfrak mQ\cong k$, and we have $\mathrm{Hom}_A(P,Q/\mathfrak mQ)\cong k$. On the other hand, we have $\mathrm{Hom}_A(P,Q)\cong k^2$ (the socle of $Q$). So there is some surjection $A/\mathfrak m\otimes\mathrm{Hom}_A(P,Q)\to\mathrm{Hom}_A(P,Q/\mathfrak mQ)$. However, the natural map is actually zero, since every map $P\to Q$ lands in the socle of $Q$, which is then killed when tensoring with $A/\mathfrak m$.
When do we have a natural isomorphism? This is clear if $P\cong A$, since then $\mathrm{Hom}_A(P,Q)\cong Q$. Since we are dealing with a natural transformation between additive functors, it follows immediately that we have a natural isomorphism whenever $P$ is finitely generated projective.
Similarly, if $I=0$, then we have a natural isomorphism, and so we have a natural isomorphism whenever $A/I$ is finitely generated projective.
A: The statement in question is false in general, at least if you want the natural map to be an isomorphism. Take for example $A=\mathbb{K}[x]$, $I=(x^2)$, $P=A/(x^2)$ and $Q=A$. Note that we have a natural homomorphism $A/I\otimes_A M\cong M/IM$ for any $A$-module $M$. Now if we want to have the isomorphism in question, we have to have that the map
$$
\operatorname{Hom}_A(P,Q)\to\operatorname{Hom}_{A/I}(P/IP,Q/IQ)\\
\phi\mapsto\left(x+IP\in P/IP\mapsto \phi(x)+IQ\in Q/IQ\right)
$$
is surjective. Now as $P/IP\cong Q/IQ$, we have $\operatorname{id}_{P/IP}\in\operatorname{Hom}_{A/I}(P/IP,Q/IQ)$. Suppose there is $\phi\in\operatorname{Hom}_A(P,Q)$ such that $\phi\mapsto\operatorname{id}_{P/IP}$ under the above map. Then we must have
$$
\phi(x+(x^2))+IQ=x+IQ\implies\phi(x)=x+a_2x^2+\dots+a_nx^n\text{ for some }a_2,\dots,a_n\in\mathbb{K}
$$
But then we must also have
$$
0=\phi(0)=\phi(x\cdot(x+(x^2))=x\cdot\phi(x+(x^2))=x^2+a_2x^3+\dots+a_nx^{n+1}
$$
which is impossible. Hence the natural map $\operatorname{Hom}_A(P,Q)\to\operatorname{Hom}_{A/I}(P/IP,Q/IQ)$ isn't surjective.
