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 $.