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# Extension of scalars

In abstract algebra, extension of scalars is a means of producing a module over a ring $S$ from a module over another ring $R$, given a homomorphism $f : R \to S$ between them. Intuitively, the new module admits multiplication by more scalars than the original one, hence the name extension.

## Definition

In this definition the rings are assumed to be associative, but not necessarily commutative, or to have an identity. Also, modules are assumed to be left modules. The modifications needed in the case of right modules are straightforward.

Let $f : R \to S$ be a homomorphism between two rings, and let $M$ be a module over $R$. Consider the tensor product $_SM = S \otimes_R M$, where $S$ is regarded as a right $R$-module via $f$. Since $S$ is also a left module over itself, and the two actions commute, that is $s \cdot (s' \cdot r) = (s \cdot s') \cdot r$ for $s,s' \in S$, $r \in R$ (in a more formal language, $S$ is a $(S,R)$-bimodule), $_SM$ inherits a left action of $S$. It is given by $s \cdot (s' \otimes m) = ss' \otimes m$ for $s,s' \in S$ and $m \in M$. This module is said to be obtained from $M$ through extension of scalars.

## Interpretation as a functor

Extension of scalars can be interpreted as a functor from $R$-modules to $S$-modules. It sends $M$ to $_SM$, as above, and an $R$-homomorphism $u : M \to N$ to the $S$-homomorphism $u_S : _SM \to _SN$ defined by $u_S = \text{id}_S \otimes u$.

## Connection with restriction of scalars

Consider an $R$-module $M$ and an $S$-module $N$. Given a homomorphism $u \in \text{Hom}_R(M,N)$, where $N$ is viewed as an $R$-module via restriction of scalars, define $Fu : _SM \to N$ to be the composition

$_SM = S \otimes_R M \xrightarrow{\text{id}_S \otimes u} S \otimes_R N \to N$,

where the last map is $s \otimes n \mapsto sn$. This $Fu$ is an $S$-homomorphism, and hence $F : \text{Hom}_R(M,N) \to \text{Hom}_S(_SM,N)$ is well-defined, and is a homomorphism (of abelian groups).

In case both $R$ and $S$ have an identity, there is an inverse homomorphism $G : \text{Hom}_S(_SM,N) \to \text{Hom}_R(M,N)$, which is defined as follows. Let $v \in \text{Hom}_S(_SM,N)$. Then $Gv$ is the composition

$M \to R \otimes_R M \xrightarrow{f \otimes \text{id}_M} S \otimes_R M \xrightarrow{v} N$,

where the first map is the canonical isomorphism $m \mapsto 1 \otimes m$.

This construction shows that the groups $\text{Hom}_S(_SM,N)$ and $\text{Hom}_R(M,N)$ are isomorphic. Actually, this isomorphism depends only on the homomorphism $f$, and so is functorial. In the language of category theory, the extension of scalars functor is left adjoint to the restriction of scalars functor.

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