Confusion about direction of arrow in general extension of scalars (base change)

Question

I am confused about the direction of an arrow in the following situation: Given a ring homomorphism $f: R \to S$, the extension of scalars or base change along $f$ is the functor that maps an $R$-module $X$ to an $S$-module $X_S := X \otimes_R S$. But as described here: https://www.encyclopediaofmath.org/index.php/Base_change, a base change along $f: R \to S$ goes from $S$-mods to $R$-mods. How does this fit with lets say the canonical example where $R \hookrightarrow S$?

Answer 1

The correct framework to understand that is Grothendieck (op)fibrations. I'll give a quick recap.

Definition 1. Let $F : \mathbf E \to \mathbf B$ be a functor. A morphisms $f : x \to y$ in $\mathbf E$ is said to be cartesian (relatively to $F$) if for all $g: z \to y$ such that $F(g) = F(f) \circ u$ for some morphism $u$ in $\mathbf B$, there exist a unique $h : z \to x$ such that $$ F(h) = u \quad \text{and} \quad g = fh. $$

Dually, $f$ is said to be cocartesian (relatively to $F$) if it is cartesian in $\mathbf E^{\rm op}$ relatively to $F^{\rm op}$.

[A good exercise to get a grasp at the notions is first to rewrite the definition of cartesian morphisms with diagrams and next to rewrite the definition of cocartesian morphisms without summoning the gods of duality.]

Definition 2. A Grothendieck fibration is a functor $F:\mathbf E \to \mathbf B$ such that for any given $u : a\to b$ in $\mathbf B$ and $y \in \mathbf E$ with $Fy=b$, there exists a cartesian morphism $f$ with $F(f) = u$.

A Grothendieck opfibration is a functor $F:\mathbf E \to \mathbf B$ such that $F^{\rm op}$ is a Grothendieck fibration.

[Again, you should try to write a plain definition for opfibrations.]

Now, for any category $\mathbf B$, you get a functor $\mathrm{cod} : \mathbf B^{\mathbf 2} \to \mathbf B$ from the category of arrows of $\mathbf B$ to $\mathbf B$ itself that maps an arrow $a \to b$ to its codomain $b$. Remember that morphisms from $f$ to $g$ in $\mathbf B^{\mathbf 2}$ are the commutative square of $\mathbf B$ with $f$ on top and $g$ on the bottom. So you can check that cartesian morphisms relatively to $\mathrm{cod}$ are precisely the pullback square of $\mathbf B$. Hence ${\rm cod}$ is fibration whenever $\mathbf B$ has all pullbacks. So "base change" (in the sense of your link) of $y$ along $u : a \to b$ denotes the domain of a cartesian morphism with codomain $y$ above $u$.

We now have a general definition of base change.

Definition 3. Let $F : \mathbf E \to \mathbf B$ be a Grothendieck fibration. A base change of $y \in \mathbf E$ along $u : a \to b \in \mathbf B$ is a object $x \in \mathbf E$ (together) with a cartesian morphism $f: x \to y$ such that $F(f) = u$.

The existence of base changes is more or less the definition of a fibration. A base change of $y$ along $u: a \to b$ is unique up to unique isomorphism. [You should try to make that precise and to prove it.]

Of course, it can be extended to the Grothendieck opfibration: if $F : \mathbf E \to \mathbf B$ is an opfibration, "base change" makes sense for the fibration $F^{\rm op}$. Translating this back to $\mathbf E$, a base (co)change of $x \in \mathbf E$ along $u : a \to b \in \mathbf B$ is an object $y$ together with a cocartesian morphism $f : x \to y$ with $F(f) = u$.

Now take the category $\mathbf{Mod}$ of all modules (I ignore size issues here, insert "small" or "$\mathbb U$-small" whenever needed): objects are modules over any commutative rings, and morphisms from a $R$-module $M$ to a $S$-module $N$ are couples $(\varphi,f)$ with $\varphi : R \to S$ ring morphism and $f: M \to N$ morphism of $R$-modules (where $N$ is view as an $R$-module). More or less by definition of $-\otimes_R S$, you can see that $\mathbf{Mod} \to \mathbf{Ring}$ is a Grothendieck opfibration with cocartesian morphisms those $(\varphi : R \to S, M \to M \otimes_R S)$. Hence the terminology.