Change of rings

In algebra, given a ring homomorphism $f: R \to S$, there are three ways to change the coefficient ring of a module; namely, for a left ''R''-module ''M'' and a left ''S''-module ''N'',
*$f_! M = S\otimes_R M$, the induced module.
*$f_* M = \operatorname_R(S, M)$, the coinduced module.
*$f^* N = N_R$, the restriction of scalars.
They are related as adjoint functors:
:$f_! : \text_R \leftrightarrows \text_S : f^*$
and
:$f^* : \text_S \leftrightarrows \text_R : f_*.$
This is related to Shapiro's lemma.

Operations

Restriction of scalars

Throughout this section, let $R$ and $S$ be two rings (they may or may not be commutative, or contain an identity), and let $f:R \to S$ be a homomorphism. Restriction of scalars changes ''S''-modules into ''R''-modules. In algebraic geometry, the term "restriction of scalars" is often used as a synonym for Weil restriction.

Definition

Suppose that $M$ is a module over $S$. Then it can be regarded as a module over $R$ where the action of $R$ is given via
:$\begin M\times R &\longrightarrow M \\ (m,r) &\longmapsto m\cdot f(r) \end$
where $m\cdot f(r)$ denotes the action defined by the $S$-module structure on $M$.

Interpretation as a functor

Restriction of scalars can be viewed as a functor from $S$-modules to $R$-modules. An $S$-homomorphism $u : M \to N$ automatically becomes an $R$-homomorphism between the restrictions of $M$ and $N$. Indeed, if $m \in M$ and $r \in R$, then

: $u(m\cdot r) = u(m\cdot f(r)) = u(m)\cdot f(r) = u(m)\cdot r\,$.

As a functor, restriction of scalars is the right adjoint of the extension of scalars functor.

If $R$ is the ring of integers, then this is just the forgetful functor from modules to abelian groups.

Extension of scalars

Extension of scalars changes ''R''-modules into ''S''-modules.

Definition

Let $f : R \to S$ be a homomorphism between two rings, and let $M$ be a module over $R$. Consider the tensor product $M^S = M\otimes_R S$, where $S$ is regarded as a left $R$-module via $f$. Since $S$ is also a right module over itself, and the two actions commute, that is $r\cdot (s\cdot s') = (r\cdot s)\cdot s'$ for $r \in R$, $s,s' \in S$ (in a more formal language, $S$ is a $(R,S)$- bimodule), $M^S$ inherits a right action of $S$. It is given by $(m\otimes s)\cdot s' = m\otimes ss'$ for $m \in M$, $s,s' \in S$. This module is said to be obtained from $M$ 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 ''R''-module with an $(R,S)$-bimodule 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, $x^2+1,$ 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 $R$-modules to $S$-modules. It sends $M$ to $M^S$, as above, and an $R$-homomorphism $u : M \to N$ to the $S$-homomorphism $u^S : M^S \to N^S$ defined by $u^S = u\otimes_R\text_S$.

Co-extension of scalars (coinduced module)

Relation between the extension of scalars and the restriction of scalars

Consider an $R$-module $M$ and an $S$-module $N$. Given a homomorphism $u \in \text_R(M,N_R)$, define $Fu : M^S \to N$ to be the composition
:$M^S = M \otimes_R S \xrightarrow N_R \otimes_R S \to N$,
where the last map is $n\otimes s\mapsto n\cdot s$. This $Fu$ is an $S$-homomorphism, and hence $F : \text_R(M,N_R) \to \text_S(M^S,N)$ is well-defined, and is a homomorphism (of abelian groups).

In case both $R$ and $S$ have an identity, there is an inverse homomorphism $G : \text_S(M^S,N) \to \text_R(M,N_R)$, which is defined as follows. Let $v \in \text_S(M^S,N)$. Then $Gv$ is the composition
:$M \to M \otimes_R R \xrightarrow M \otimes_R S \xrightarrow N$,
where the first map is the canonical isomorphism $m\mapsto m\otimes 1$.

This construction shows that the groups $\text_S(M^S,N)$ and $\text_R(M,N_R)$ 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.

See also

*Six operations
*Tensor product of fields
*Tensor-hom adjunction

References

* {{Cite book, title=Abstract algebra, url=https://archive.org/details/abstractalgebra00dumm_304, url-access=limited, last=Dummit, first=David, date=2004, publisher=Wiley, others=Foote, Richard M., isbn=0471452343, edition=3, location=Hoboken, NJ, oclc=248917264, p
359
€“377
*J.P. May
Notes on Tor and Ext
*NICOLAS BOURBAKI. Algebra I, Chapter II. LINEAR ALGEBRA.§5. Extension of the ring of scalars;§7. Vector spaces. 1974 by Hermann.

Further reading

Induction and Coinduction of Representations

Commutative algebra
Ring theory
Adjoint functors