Extension, restriction, and coextension of scalars adjunctions in the case of noncommutative rings?

Solution 1:

The adjunction is the same in the non-commutative case. To construct it quickly, use the tensor-hom adjunction (which you'll certainly find stated and proven for non-commutative rings too).

On one side, you can see that the restriction $f^*$ is obviously isomorphic to $f^*S ⊗_S -$, where $f^*S$ is $S$ equipped with the $(R, S)$ structure you mentioned, so it follows that $f^*$ is left adjoint to $f_* ≅ \mathrm{Hom}_R(f^*S, -)$.

On the other, denote with $f_!R$ the $(S, R)$ structure on $S$, and notice that $f^*$ is isomorphic to $\mathrm{Hom}_S(f_!R, -)$, and thus $f_!R ⊗_R - ⊣ f^*$.

Another argument is an overkill, but it's good as a reality check. Every cocontinuous functor between categories of modules is isomorphic to tensoring by a bimodule and in particular left adjoint (Eilenberg-Watts theorem), and every continuous functor is right adjoint and moreover isomorphic to a hom-functor (by the special adjoint functor theorem and the previous result).

Solution 2:

As mentioned by Qiaochu Yuan in the comments, one nice thing about the case where $f: C \to K$ is a morphism of commutative rings is that scalar extension $f_!$ is strong monoidal.

In particular, for $f: C \to K$ a morphism of commutative rings, \begin{align} f_!(V) \otimes_K f_!(W) =& (K \otimes_C V) \otimes_K (K \otimes_C W)\cong (V \otimes_C K) \otimes_K (K \otimes_C W) \\ \cong& V \otimes_C K \otimes_C W \cong K \otimes_C V \otimes_C W = f_!(V \otimes_C W). \end{align}

This is given as Proposition 3 in section II.5.1 of Bourbaki's Algebra I.

As a corollary, it follows that restriction of scalars is lax monoidal.