Bimodule structure on $\mathrm{Ext}_R^n(M,N)$

Question

Throughout the question, $\mathbb{k}$ is a commutative ring, and $R$, $S$, and $T$ are $\mathbb{k}$-algebras.

Let $M$ be a left $R$-module. The functor $\mathrm{Hom}_R(M,-):R\text{-}\mathrm{Mod}\to \mathbb{k}\text{-}\mathrm{Mod}$ is covariant and right-exact and therefore has (covariant) left derived functors $\mathrm{Ext}_R^n(M,-)$. If $M$ happens to be an $(R,S)$-bimodule (for a $\mathbb{k}$-algebra $S$), then we actually have a covariant right-exact functor $\mathrm{Hom}_R(M,-):R\text{-}\mathrm{Mod}\to S\text{-}\mathrm{Mod}$ and the corresponding left derived functors $\mathrm{Ext}_R^n(M,-)$ also land in $S\text{-}\mathrm{Mod}$.

If $N$ is a left $R$-module, then the functor $\mathrm{Hom}_R(-,N):R\text{-}\mathrm{Mod} \to \mathbb{k}\text{-}\mathrm{Mod}$ is contravariant and left-exact and therefore has (contravariant) right derived functors $\mathrm{Ext}_R^n(-,N)$. If $N$ happens to be an $(R,T)$-bimodule (for a $\mathbb{k}$-algebra $T$), then we actually have a contravariant left-exact functor $\mathrm{Hom}_R(-,N):R\text{-}\mathrm{Mod} \to \mathrm{Mod}\text{-}T$ and the corresponding right derived functors $\mathrm{Ext}_R^n(-,N)$ also land in $\mathrm{Mod}\text{-}T$.

For the moment, let $M$ and $N$ be left $R$-modules (not bimodules). There are two operations we can do: (1) Plug in $N$ to the functor $\mathrm{Ext}_R^n(M, -)$ and get the $\mathbb{k}$-module $\mathrm{Ext}_R^n(M, N)$, or (2) Plug in $M$ to the functor $\mathrm{Ext}_R^n(-,N)$ and get the $\mathbb{k}$-module $\mathrm{Ext}_R^n(M, N)$. It is well-known and suggested by the notation that above two operations are naturally isomorphic.

Here's where things get interesting.

Now let $M$ be an $(R,S)$-bimodule and $N$ be an $(R,T)$-bimodule. Again, there are two operations we can do: (1) Apply the forgetful functor $R\text{-}\mathrm{Mod}\text{-}T \to R\text{-}\mathrm{Mod}$ to $N$ and plug the result in to the functor $\mathrm{Ext}_R^n(M, -)$ to get the left $S$-module $\mathrm{Ext}_R^n(M, N)$, or (2) Apply the forgetful functor $R\text{-}\mathrm{Mod}\text{-}S \to R\text{-}\mathrm{Mod}$ to $M$ and plug the result in to the functor $\mathrm{Ext}_R^n(-, N)$ to get the right $T$-module $\mathrm{Ext}_R^n(M, N)$.

We see that $\mathrm{Ext}_R^n(M,N)$ has both the structure of a left $S$-module and a right $T$-module which leads one to ask: Can we consider $\mathrm{Ext}_R^n(M,N)$ to be an $(S, T)$-bimodule? That is, do the left and right module structures commute?

Answer 1

Just start with either the functor $\mathrm{Hom}_R(M, -): R\text{-}\mathrm{Mod}\text{-}T \to S\text{-}\mathrm{Mod}\text{-}T$ or the functor $\mathrm{Hom}_R(-, N): R\text{-}\mathrm{Mod}\text{-}S \to S\text{-}\mathrm{Mod}\text{-}T$. This works because the categories $R\text{-}\mathrm{Mod}\text{-}T$ and $R\text{-}\mathrm{Mod}\text{-}S$ both have enough projectives (so one can form the derived functors); in fact $R\text{-}\mathrm{Mod}\text{-}T$ is the same as $R\otimes T^{\text{op}}\text{-}\mathrm{Mod}$ (see this question).