Definition of pushforward and pullback for modules

Question

Let $R$ be a commutative ring with unit and $\sigma:R\longrightarrow R$ an endomorphism of the ring $R$. Let $M$ be an $R$-module.

1. What are the definitions of $\sigma^*M$ and $\sigma_* M$?
2. How do they behave under tensor product?

Many thanks!

Answer 1

There are actually two somewhat reasonable notions of a module pushforward. This is outlined in J.P. May's "Notes on Ext and Tor" around page 13 which you can read here (as of Aug 30, 2021).

The set-up is a ring homomorphism $f\colon R\to S$, a left $R$-module $M$, and a left $S$-module $P$.

The pushforward $f^*$ is a functor $S$-$\textbf{Mod}\Rightarrow R$-$\textbf{Mod}$ defined in the obvious way: $f^*P$ is the set $P$ with action $r\cdot p=f(r)\star p$ (using $\cdot$ and $\star$ for the respective $R$ and $S$ actions).

This functor has a left adjoint $f_!\colon R$-$\textbf{Mod}\Rightarrow S$-$\textbf{Mod}$ given by $f_!M=S\otimes_R M$ with the usual left $S$-action.

But $f^*$ also has a right adjoint $f_*\colon R$-$\textbf{Mod}\Rightarrow S$-$\textbf{Mod}$ which is less obvious. On objects,

$$f_*M=\operatorname{Hom}_R(f^*S,M).$$

An element of $f_*M$ is thus an $R$-module homomorphism $g\colon f^*S\to M$. In other words, $g(f(r)s)=r\cdot g(s)$. Now $f_*M$ is a left $S$-module as follows: $s\star g$ is the map $(s\star g)(t)=g(ts)$ (here $ts$ is just ring multiplication). To verify $s\star g$ is another $R$-module homomorphism, $$(s\star g)(r\cdot t)=(s\star g)(f(r)t)=g(f(r)ts)=r\cdot g(ts)=r\cdot (s\star g)(t).$$

Anyway, this definition is a bit clunky, but it's justified by the adjunctions $f_!\dashv f^*\vdash f_*$.

Answer 2

More generally, if $\sigma:R\to S$ is a homomorphism, and $N$ is an $S$-module, then the pullback $\sigma^*N$ is an $R$-module where the multiplication is given by $r\cdot_Rn = \sigma(r)\cdot_S n$.

I don't know that there is a nice push-forward, even for an endomorphism $R\to R$. Say $R = \Bbb Z^2$ and $\sigma(a, b) = (a, 0)$. What would be a natural way to define product by $(0,1)\in R$ on $\sigma_*M$?

Answer 3

Given a morphism of commutative unital rings $\sigma : R\to S$, if you take the pullback $\sigma^* N$ of an $S$-module $N$ to be $N$ considered as an $R$-module via $\sigma, $ then the pushforward $\sigma_* M$ of an $R$-module $M$ should be $M\otimes_R S.$ Then you get the adjunction $$\operatorname{Hom}_R(M,\sigma^*N) \cong \operatorname{Hom}_S(\sigma_* M, N).$$

However, this notation is a little at odds with the geometric version: let $f : X\to Y$ be a morphism of schemes, and let $\mathcal{F}$ be a sheaf oof $\mathcal{O}_X$-modules and $\mathcal{G}$ be a sheaf of $\mathcal{O}_Y$-modules. Then we have $$\operatorname{Hom}_Y(\mathcal{G},f_*\mathcal{F})\cong\operatorname{Hom}_X(f^*\mathcal{G},\mathcal{F}).$$

As such, given a morphism of commutative unital rings $\sigma : R\to S,$ I would prefer to define $\sigma_* N$ for an $S$-module $N$ to be $N$ considered as an $R$-module via $\sigma, $ and $\sigma^* M$ for an $R$-module $M$ to be $M\otimes_R S.$ Then we have $$\operatorname{Hom}_R(M,\sigma_*N) \cong \operatorname{Hom}_S(\sigma^* M, N).$$ If you let $f$ denote the morphism of affine schemes $f : X = \operatorname{Spec}S\to Y = \operatorname{Spec}R$ corresponding to $\sigma,$ and let $\widetilde{M}$ denote the sheaf of $\mathcal{O}_Y$-modules corresponding to an $R$-module $M$ (and similarly for an $S$-module $N$), then we have $\widetilde{\sigma^* M}\cong f^*\widetilde{M}$ and $\widetilde{\sigma_* N}\cong f_*\widetilde{N}.$