Let $A \subseteq B$ and $A \subseteq C$ be commutative noetherian domains.
Assume that $A$ and $C$ are regular rings (=every localization at a maximal ideal is a regular local ring).
Assume that $B$ is a free $A$-module of finite rank $r$, so $B \cong \oplus\_{i=1}^{r} A$ as $A$-modules.
Is it possible to decide whether the ring $C \otimes\_A B$ is regular or not? 



Regularity of a tensor product
Alright, I'm posting this as an answer.


As in the comments above, there are counterexamples to your general question.


As for the specific situation, unless you want to assume something more on $p$, $q$ and $w$ I guess it's still false: assume $p=x$ and $q=y$, so that $A=k[x,y]$, and take $B=k[x,y,w]/(w^2-x)$, $M=(x,y,w)$, so $m=(x,y)$. So $C$ is the residue field of $(x,y)$, and the (spectrum of the) tensor product $B\otimes\_A C$ is the fiber of the map $\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ over $(x,y)$. You can easily see that the fiber is $\mathrm{Spec}(k[w]/(w^2))$, which is not regular.


Since you're basically looking at regularity of fibers, it seems that what you really want is smoothness of $\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ to get your conclusion.


