# nLab extension of scalars

### Context

#### Algebra

higher algebra

universal algebra

# Contents

## Idea

The extension of scalars of a module along a homomorphism of rings is the algebraic dual of what geometrically is the pullback of bundles along a map of their base spaces (with respect to the discussion at modules - as generalized vector bundles).

Explicitly, extension of scalars along a ring homomorphism $f : R \to S$ is the operation on $R$-modules given by forming the tensor product of modules with $S$ regarded as an $R$-module via $f$.

There are similar functors for bimodules and in some other categories.

## Definition

Let $R$ and $S$ be commutative rings and let $f \colon R\to S$ be a homomorphism of rings.

We discuss extension of scalars along $f$ first general abstractly and then explicitly in components.

### General abstract

Write $R$Mod and $S$Mod for the categories of modules over $R$ and $S$, respectively.

###### Definition

Given a ring homomorphism $f : R \to S$ the restriction of scalars functor

$f^* : S Mod \to R Mod$

is the functor that takes an $S$-module $N$ to the $R$-module $f^*N$ whose underlying abelian group is that of $N$ and whose $R$-action is given by

$r \cdot n \coloneqq f(r)\cdot n \;\;\;\; for r \in R, n \in N \,.$
###### Proposition

The restriction of scalars functor, def. 1, is the right adjoint in a pair of adjoint functors

$( f_! \dashv f^* ) : S Mod \stackrel{\overset{f_!}{\leftarrow}}{\underset{f^*}{\to}} R Mod \,.$
###### Definition

The left adjoint $f_! \colon R Mod \to S Mod$ in prop. 1 is called extension of scalars along $f$.

###### Remark

A further right adjoint $f_*$ would be called coextension of scalars along $f$.

### In components

###### Proposition

Given a ring homomorphism $f : R \to S$, the extension of scalars functor $f_!$ of def. 2 is the functor

$f_! \coloneqq S \otimes_R (-) \,:\, R Mod \to S Mod$

given by tensor product of modules with $S$ regarded as an $S$-$R$-bimodule: the left action being the canonical action of $S$ on itself, the right being the restriction of scalars-action along $f$.

Explicitly, for $N \in R Mod$

• the elements of $f_! N$ are equivalence classes of pairs $(s,n) \in S \times N$ under the equivalence relation $(s \cdot f(r), n) = (s, r\cdot n)$ for all $s \in S$;

• the left $S$-action is given by $s' \cdot(s,n) = (s' \cdot s,n)$.

## Properties

### Geometric interpretation

Under Isbell duality extension of scalars turns into a statement about geometry.

By definition the category

$Ring^{op} \underoverset{Spec}{\colon \simeq}{\to} Aff$

of (absolute) affine schemes is the opposite category of Ring.

Hence for $f : R \to S$ a ring homomorphism, we have equivalently a morphism

$Spec(f) : Spec(S) \to Spec(R)$

of affine schemes.

An $R$-module $N$ corresponds to the collection of sections of a “generalized vector bundle” over $Spec(R)$: something that has a quasicoherent sheaf of sections.

The pullback of this “bundle” along $Spec(f)$ has sections forming the module $f_! N$.

Generally, for any fibered category like Mod$\to Aff$ we may regard the inverse image functor as the extension of scalars.

For that reason if there is some other fibered category $\mathcal{F}$ over the opposite of some algebraic category $\mathcal{A}$ whose objects are considered “objects of scalars” one is inclined to call the inverse image functor, the extension of scalars.

## Examples

Revised on July 2, 2016 04:54:00 by Bartek (219.88.237.29)