I have a question about Exercise 2.18 on Introduction to Commutative Algebra by Atiyah and MacDonald.
Let $f:A\to B$ be a flat homomorphism of rings, let $Q$ be a prime ideal of $B$ and let $P=Q^c,$ the contraction of $Q.$ Then I am asked to prove that $f^*:\text{Spec}(B_Q)\to\text{Spec}(A_P)$ is surjective. The hint given in the book states that $B_Q$ is a local ring of $B_P$ but I don't really understand this. (I think I'm especially uncomfortable with working with modules, as I need to be careful about which ring they are modules in.)
Let $S=A\setminus P.$ Then I know that $S^{-1}B\cong [f(S)]^{-1}B$ as $S^{-1}$-modules, i.e. $B_P\cong [f(S)]^{-1}B$ as $A_P$ modules. Now take $T=B\setminus Q.$ Then we know that $f(S)\subset T.$ Does this mean we can create an isomorphism $$B_Q=T^{-1}B\cong (f(S)^{-1}T)^{-1}(f(S)^{-1}B)=(f(S)^{-1}T)^{-1}B_P?$$ But I cannot deduce where they are isomorphic in, i.e. are they isomorphic as $B_Q$-modules? Also, is it possible to conclude from this that $B_Q$ is a localization of $B_P,$ i.e. that $f(S)^{-1}T$ is the complement of a prime ideal over whichever ring that we are working in in the above equation?
I can sort of "see" in my head that after localizing $B$ at $P,$ we can localize $B_P$ again with respect to $Q:$ the prime ideal $Q$ which is in a sense "bigger" than the prime ideal $P$ because $f(P)\subset Q$ and so it won't be "killed" when we are localizing by $P.$ (Does this mean we are considering $B_P$ as a $B$-module?) It would be much appreciated if someone could verify this "picture" and if correct, make it a bit more concrete as I have tried above.
*Note: I did find the same question on SE (the answers to which I have looked at) but they haven't helped me answer the above questions, hence the new question.