Let $A,B$ be commutative rings with unit.
I'm trying to prove that if $f:A\to B$ is flat and $q\subset B$ is a prime ideal with $p=f^{-1}(q)$, then $\hat{f}:A_p\to B_q$ is faithfully flat, where $\hat{f}\left(\frac{a}{s}\right)=\frac{f(a)}{f(s)}$.
Here is my attempt:
Let $M,N$ $A_p$-modules and $g:M\to N$ injective. If $i:A\to A_q$ is the natural inclusion, let $i_*$ be the localization functor and $i^*$ the forgetful functor. We have $i^*(g):i^*(M)\to i^*(N)$ injective and, since $f:A\to B$ is flat, we get that $i^*(g)\otimes id_B:i^*(M)\otimes_{A}B\to i^*(N)\otimes_{A}B$ injective. Now applying $i_*$ (and since localization is exact), we have $g\otimes id_B:M\otimes_{A_q}B_q\to N\otimes_{A_q}B_q$ injective. Now $M\otimes_{A_q}B_q\simeq M\otimes_{A_p}B_q$ as $B_q$-modules, via $\phi(m\otimes b/s)=m\otimes b/s$, so $g\otimes id_B:M\otimes_{A_p}B_q\to N\otimes_{A_p}B_q$ injective. Therefore $\hat{f}:A_p\to B_q$ is flat.
To prove faithfully flatness, I use the fact that if $\hat{f}:A_p\to B_q$ is flat, then it is faithfully flat $\iff \hat{f}_*(\mathfrak{m})\neq B_q$ for all maximal ideals $\mathfrak{m}\subset A_p$ (exercise $16$, chapter $3$ from Atiyah Macdonald). Since $A_p$ is local, its only maximal ideal is $pA_p$, and $\hat{f}_*(pA_p)=pA_p\otimes_{A_p}B_q=A_p\otimes_{A_p}f(p)B_q\simeq f(p)B_q\subset qB_q\neq B_q$.
I belive the second part is ok, but the first one felt very uncomfortable to me. For example, I don't know how to justify that $i^*$ preserves injectiveness. Also, when I apply $i_*$, I assume that $i_*(i^*(g)\otimes id_B)=g\otimes id_B$ and that $i_*(i^*(M))=M$, which I also don't know how to justify. At last, the argument $M\otimes_{A_q}B_q\simeq M\otimes_{A_p}B_q$ seemed a little clumsy, and gave me the impression that this was not the way to go.
Are any of these steps wrong? If they are right, how could I formalize them?
