$\operatorname{Hom}_B$ of flat modules is flat over $A$? Let $(A, \mathfrak{m}) \rightarrow (B,\mathfrak{n})$ be a local homomorphism of noetherian local rings and let $M$ be a finitely generated $B$-module flat over $A$. Suppose moreover that $B$ is also of finite type and flat over $A$.
Q: Is it true that $\operatorname{Hom}_B(M,B)$ is flat over $A$?
I've tried to use the flatness of $M$ and $B$ to show that the natural map $$\mathfrak{m}\otimes_A \operatorname{Hom}_B(M,B) \longrightarrow \operatorname{Hom}_B(M,B)$$ is injective (to apply the Local Criterion for flatness) but I could not prove that neither find a counterexample.
ADDED: Here follows an naive idea I had.
Consider the free resolution of $A/\mathfrak{m}$:
$$
\cdots\longrightarrow A^k  \overset{R}{\longrightarrow} A^n  \overset{G}{\longrightarrow} A \longrightarrow A/\mathfrak{m} \longrightarrow 0
$$
where $G= (a_1, \dots, a_n)$ is given by the generators of $\mathfrak{m}$ and $R= (r_{ij})$ is given by the first syzygies. Then we have
$$
\operatorname{Hom}_B(M,B)^k  \overset{R}{\longrightarrow} \operatorname{Hom}_B(M,B)^n  \overset{G}{\longrightarrow} \operatorname{Hom}_B(M,B)
$$
and $\operatorname{Tor}^A_1 \left( A/\mathfrak{m}, \operatorname{Hom}_B(M,B)\right)= \ker G / \operatorname{im} R$. We just need to see if $\ker G \subset \operatorname{im} R$. Let $(f_1, \dots, f_n)\in \ker G$ i.e. $\sum_j a_jf_j =0$. Then for every $x\in M$ we have $\sum_j a_jf_j(x) =0$ and since $B$ is flat there exist $(g_1(x), \cdots, g_k(x)) \in B^k$ such that 
$$
f_i(x) = \sum_j r_{ij}g_j(x)
$$
and we have $g_j \colon M \longrightarrow B$ maps of sets. 
The problem is reduced to show whether we can produce $B$-homomorphisms this way. 
Note that when $M$ is free (as a $B$-module) then we only need to define $g_j$ on generators but in general it is not true. Also note that the flatness of $M$ was not used so far. 
 A: This is my idea:
Since you have the morphism $\alpha:A\rightarrow B$ you have the next isomorphism
$$Hom_{B}(M,Hom_{A}(B,A))\cong Hom_{A}(M,A)$$
With $M$ a $B-$module that can be seen as an $A-$module. 
Now we have that $B$ is a finitely generated flat $A$-module where $A$ is a noetherian local ring. It follows that $B$ is a free $A$-module therefore $B\cong\displaystyle\bigoplus_{i\in I}A$ for some finite set $I$. 
From this we have that
\begin{eqnarray}
Hom_{B}(M,B)&\cong&Hom_{B}\left(M,\displaystyle\bigoplus_{i\in I}A\right)\\
&\cong&Hom_{B}\left(M,\displaystyle\bigoplus_{i\in I}Hom_{A}(A,A)\right)\\
&\cong& Hom_{B}\left(M,Hom_{A}\left(\displaystyle\bigoplus_{i\in I}A,A\right)\right)\\
&\cong&Hom_{B}(M,Hom_{A}(B,A))\\
&\cong&  Hom_{A}(M,A).
\end{eqnarray}
Since $M$ is a finitely generated $B-$module and $B$ is a free module over $A$ you have that $M$ is a finitely generated $A-$module which is flat as an $A-$module therefore $M$ is a free $A$-module. 
It follows that $M\cong\displaystyle\bigoplus_{j\in J}A$  for some finite set $J$. 
Thus
$$Hom_{A}(M,A)\cong \displaystyle\bigoplus_{j\in J}A $$
whence it follows that $Hom_{B}(M,B)$ is a free $A-$module and hence it is a flat $A-$module 
