# Extension of scalars

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

Informally, extension of scalars is "the tensor product of a ring and a module"; more formally, it is a special case of a tensor product of a bimodule and a module – the tensor product of an ${\displaystyle (S,R)}$ bimodule with an R-module is an S-module.

## Examples

One of the simplest examples is complexification, which is extension of scalars from the real numbers to the complex numbers. More generally, given any field extension K < L, one can extend scalars from K to L. In the language of fields, a module over a field is called a vector space, and thus extension of scalars converts a vector space over K to a vector space over L. This can also be done for division algebras, as is done in quaternionification (extension from the reals to the quaternions).

More generally, given a homomorphism from a field or commutative ring R to a ring S, the ring S can be thought of as an associative algebra over R, and thus when one extends scalars on an R-module, the resulting module can be thought of alternatively as an S-module, or as an R-module with an algebra representation of S (as an R-algebra). For example, the result of complexifying a real vector space (R = R, S = C) can be interpreted either as a complex vector space (S-module) or as a real vector space with a linear complex structure (algebra representation of S as an R-module).

### Applications

This generalization is useful even for the study of fields – notably, many algebraic objects associated to a field are not themselves fields, but are instead rings, such as algebras over a field, as in representation theory. Just as one can extend scalars on vector spaces, one can also extend scalars on group algebras and also on modules over group algebras, i.e., group representations. Particularly useful is relating how irreducible representations change under extension of scalars – for example, the representation of the cyclic group of order 4, given by rotation of the plane by 90°, is an irreducible 2-dimensional real representation, but on extension of scalars to the complex numbers, it split into 2 complex representations of dimension 1. This corresponds to the fact that the characteristic polynomial of this operator, ${\displaystyle x^{2}+1,}$ is irreducible of degree 2 over the reals, but factors into 2 factors of degree 1 over the complex numbers – it has no real eigenvalues, but 2 complex eigenvalues.

## Interpretation as a functor

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

## Connection with restriction of scalars

${\displaystyle _{S}M=S\otimes _{R}M\xrightarrow {{\text{id}}_{S}\otimes u} S\otimes _{R}N\to N}$,

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

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

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

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

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

## References

NICOLAS BOURBAKI. Algebra I, Chapter II. LINEAR ALGEBRA.§5. Extension of the ring of scalaxs;§7. Vector spaces. 1974 by Hermann.