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If we have a ring homomorphism $f: R \to S$, then we can put an $(R,S)$ or $(S, R)$ bimodule structure on $S$ and define the extension $f_!:= S \otimes_R (-) : R \text{Mod} \to S \text{Mod}$, restriction $f_*: S \text{Mod} \to R \text{Mod}$ and coextension $f_*:= \text{Hom}_{R}(S, -) : R \text{Mod} \to S \text{Mod}$ of scalars functors. These constructions do not depend at all on the commutativity of the rings $R$ and $S$, but in every source I've seen that mentions extension and coextension of scalars, it is assumed that the rings are commutative, such as in the nLab article.

Is there any reason that the rings are assumed to be commutative? Does the adjunction still hold in the noncommutative case? Are there any properties of these constructions that are "less nice" in the noncommutative case? Or is it simply for convenience?

Most of my algebra knowledge is self-taught, so I apologize if I'm missing something obvious.

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    They are assumed to be commutative simply because people writing those references tend to be thinking about (commutative!) algebraic geometry – Mariano Suárez-Álvarez Jul 02 '16 at 20:46
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    I would look in Section II.5 of Bourbaki's Algebra for stuff like this. I don't think the proofs are any harder for these particular facts, but if you want an authority here it is! – Hoot Jul 02 '16 at 20:55
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    In the commutative case we have more structure: namely, extension of scalars is symmetric monoidal. – Qiaochu Yuan Jul 02 '16 at 21:05
  • @Qiaochu Yuan, thanks! That was the sort of thing I was wondering about. – ಠ_ಠ Jul 02 '16 at 23:12

2 Answers2

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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.

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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).

user54748
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  • Could you please explain how to see restriction $S\mathsf{Mod}(f!R,-)$ is isomorphic to $f^\ast S \otimes_S -$? Is the natural isomorphism given by $g\mapsto 1\otimes_Sg(1)$? Also, I don't understand your reasoning in the second and third paragraphs: why $f^*S ⊗_S -;\dashv;_R\mathsf{Mod}(f^\ast S,-)$? – Arrow Jun 17 '18 at 14:32
  • @Arrow Yes, it is g ↦ 1 ⊗ g(1), but you don't really need that. You want the chain $S\mathrm{Mod}(f!R, -) ≅ f^* ≅ f^S ⊗_S -$, where the first isomorphism is g ↦ g(1), and the second n ↦ 1 ⊗ n. These are both isos on the level of Abelian groups, so you only need to check that they preserve the R-action too. When you know that $f^$ is isomorphic both to a hom functor and to a tensor product you can use the tensor-hom adjunction to get both its left and right adjoint.

    Tensor-hom adjunction for left modules says that $P ⊗_S - ⊣ {}_R\mathrm{Mod}(P, -)$ for every (R, S)-bimodule P.

     – user54748 Jun 17 '18 at 16:16
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    Is it written somewhere for R and S graded algebras?For R and S differential graded algebras? – Jim Stasheff Aug 07 '21 at 13:40