Proof of derived tensor-hom adjunction

Question

This is a cross-post from math.stackexchange, 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.

Answer 1

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.