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Setting: let $R$ be a ring, $f: R \to S$ a ring homomorphism, $A$ a $R$-module and $B$ a $S$-module. Sometimes, when I compute by hand some $Tor$ groups, I use the property of tensor product:

$ A \otimes_{R} B \simeq A \otimes_{R} ( S \otimes_{S} B) \simeq (A \otimes_{R} S) \otimes_{S} B$

My question is: what is the "analogue" for the $Hom$ functor, i.e. the the analogue law in the $Ext$ context?

I have tried something intuitively without succeed, and there is no reference on my book (I think it should be too basic for deserve attention...but I have just discovered to have this lack in my knowledge and I don't want to proceed without having clarified this point).

Thank you in advance! Cheers

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    By tensor-Hom adjunction, you have two isomorphisms $Hom_R(A,B)=Hom_S(S\otimes_R A, B)$ and $Hom_R(B,A)=Hom_S(B,Hom_R(S,A))$. Is this what you're looking for ? – Roland Jul 03 '15 at 08:15
  • @user113969 how do you define $Hom_R(B,A)$ when $B$ is an $S$-mod? Do we need to use the map $f$? And even with this map, does the adjunction still hold? (I'm confused right now, that's why I'm asking, because I wanted to answer with an analogue answer as yours, but then got stuck in these details) – Riccardo Jul 03 '15 at 15:45
  • @Riccardo Yes we need the map $f$ (the same way we need the map $f$ for the isomorphism in your question). In fact, if $B$ is a $S$-module, it becomes a $R$-module by : $r.b=f(r)b$. This is the so called restriction of scalar functor. Now that $B$ is a $R$-module we can take tensor product over $R$ (hence the $A\otimes_R B$ in your question) or module homomorphism $Hom_R(B,A)$. – Roland Jul 03 '15 at 15:50
  • @user113969 thanks for the clarification! I'll work on it right now. Let me just ask you another thing, when you write $Hom_R(S,A)$ you are considering $S$ as an $S$ and $R$ module at the same time right? Is it in general (so any module) possible or what kind of compatibility do we need to impose on it? (I'm not speaking of a bi module, but a "bi-left module" let's say – Riccardo Jul 03 '15 at 16:13
  • @Riccardo Yes, this is general for bimodule (left and right though). More precisely : if $A$ is a $(S,R)$-bimodule, $B$ is a $(S,T)$-bimodule, then $Hom_S(A,B)$ is a $(R,T)$-bimodule. If $C$ is a $(R,U)$-bimodule, then $A\otimes_R C$ is a $(S,U)$-bimodule. Finally, $Hom_R(C,Hom_S(A,B))=Hom_S (A\otimes_R C, B)$ as $(U,T)$-bimodule. – Roland Jul 03 '15 at 16:29
  • The isomorphisms of my first comment are special cases of this last isomorphism when $S$ is considered as a right $R$-module for the first one, as a left $R$-module for the second, and using the obvious identifications $Hom_S(S,B)=B$ and $S\otimes_S B=B$. – Roland Jul 03 '15 at 16:34
  • @Roland yes, thank you very much! –  Jul 18 '15 at 18:54
  • You may want to google "change-of-rings spectral sequence". – Pedro Sep 25 '17 at 21:29

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