**This is a cross-post from [math.stackexchange](https://math.stackexchange.com/questions/4220444/proof-of-derived-tensor-hom-adjunction), since I didn't get any answers there.**


As far as I know, for $R,S,V,W$ rings and $M$ an $(R,W)$-bimodule, $N$ an $(R,S)$-bimodule and $L$ an $(S,V)$-bimodule, we have an isomorphism in $D(V$-lmod$)$
$$
\text{RHom}\_R(N \otimes\_S^L L, M) \cong \text{RHom}\_S(L,\text{RHom}\_R(N,M)).
$$


The only proof I can imagine would take flat / injective resolutions $P^\bullet$ and $Q^\bullet$ of $N$ and $M$ to yield
$$
\text{Hom}^\bullet\_R(P^\bullet \otimes\_S L, Q^\bullet) \cong \text{Hom}^\bullet\_S(L,\text{Hom}^\bullet\_R(P^\bullet,Q^\bullet)).
$$
However, it is not clear to me why there exists a resolution of $N$ consisting of $(R,S)$-bimodules that are flat right-$S$-modules.
For instance let $S=\mathbb Z$, $R = \mathbb Z/p$, in which case all flat $S$-modules are torsion free, so none of them can have an $R$-module structure.


Does the 'derived adjunction' still hold? If yes, how do you prove it? I would prefer a constructive proof which allows me to understand the map.



Proof of derived tensor-hom adjunction

> 
> Does the 'derived adjunction' still hold? If yes, how do you prove it? I would prefer a constructive proof which allows me to understand the map.
> 
> 
> 


Yes. The easiest way to get the derived adjunction is to specialize the ordinary adjunction to appropriate resolutions of the given bimodules.


For example, using projective resolutions, we can


* projectively resolve $N$ as an $R$-$S$-bimodule, i.e., as a left $R⊗S^{\rm op}$-module;
* projectively resolve $L$ as a left $S$-module.


This follows immediately from the fact
that the underlying tensor-hom adjunction in two variables
is a Quillen adjunction in two variables,
using the indicated projective model structures.


Alternatively, using injective resolutions, we can


* injectively resolve $M$ as a left $R$-module;
* projectively resolve $N$ as a right $S$-module or $L$ as a left $S$-module.


