## ac.commutativa algebra – Location of the Hom module in an advanced configuration (module 0-dualization)

Leave $$A, , B$$ I know Noetherian rings such that and let's leave $$M$$ bean $$A$$-module. Leave $$g: A a B$$ Be a ring of homomorphism that does. $$B$$ in a free finite $$A$$-algebra. Now we can consider the $$A$$-module $$operatorname {Hom} _A (B, M)$$ as a $$B$$-module via
$$(b cdot phi) (b & # 39;) = phi (b cdot b & # 39;) forall , b, , b & # 39; in B text {and } phi in operatorname {Hom} _A (B, M).$$
Now I want to locate the $$B$$-module $$operatorname {Hom} _A (B, M)$$ in a heyday $$P in operatorname {Spec} (B)$$. How do you see the result?

For some cases we have to locate and travel home in the sense of
$$S ^ {- 1} operatorname {Hom} _A (B, M) cong operatorname {Hom} _ {S ^ {- 1} A} (S ^ {- 1} B, S ^ {- 1} } M)$$
where $$S subset A$$ is a multiplicative set in $$A$$ and the above is a location of $$A$$-modules.

Do we obtain in the previous thing establishing a similar result? I guess that (maybe under some additional requirements in $$A, B, M$$) The following is the result:

Leave $$mathfrak p = g ^ {- 1} (P) in operatorname {Spec} (A)$$.
$$left ( operatorname {Hom} _A (B, M) right) _P cong operatorname {Hom} _ {A _ { mathfrak p}} (B_P, M _ { mathfrak p}).$$

Do you have ideas to prove this (do not forget to make additional assumptions) or some direct counterexample?

Thank you very much in advance!

Edit: Changed assumptions: now $$B$$ is finite free about $$A$$.