Calculating dimension of a tensor product of algebras 
Problem:

Let $\phi:(R,m)\to (S,n)$ be a local homomorphism of Noetherian local rings with $S$ formally equidimensional and so that $\dim S=\dim R+\dim S/mS$. Let $q\in \text{Spec} (R)$ and suppose that $Q\in \text{Spec}(S)$ contracts to $q$. We claim that: ($\ast$) $\dim S_Q/qS_Q\geq \text{ht} Q-\text{ht}(q)$.

I have:

($\ast \ast$) $\text{ht}(Q)\leq \text{ht}(q)+\dim S\bigotimes_R \kappa(q)$, and hence ($\ast$) holds if $\dim S_Q/qS_Q\geq\dim S\bigotimes_R \kappa(q)$ (where $\kappa(q)$ is the field $R_q/q_q$). Moreover, $S\bigotimes_R \kappa(q)\cong S\bigotimes_R (R/q)_q$ as $R-$algebras (since $\kappa(q)$ and $(R/q)_q$ are isomorphic $R-$algebras); and certainly $S_Q/qS_Q\cong S_Q\bigotimes_R (R/q)$ as $R-$algebras, which is almost what I want, but not quite (note that this tensor product is over $R$ instead of $S$).
Edit: I might have a proof but it is still fuzzy. The idea is that I think I can get a ring iso $S_Q/qS_q\to (S/\phi(q)S)/(Q/\phi(q)S)$, and I think it's not difficult to get that the RHS has dimension $\text{ht} Q-\text{ht} (\phi(q)S)$ (note that $S$ is catenary), and maybe it it somehow true that $\text{ht}(\phi(q)S) = \text{ht}(q)$. If all of that works, then I win, I will put an update or answer if I get that to work.

By theorem 15.1 in Matsumara,
$\dim (S_Q/qS_Q) \geq \text{ht}(Q)-\text{ht}(q)$.
