“Complexity that works is built up out of modules that work perfectly, layered one over the other.”
―Kevin Kelly
Modules are a generalization of the concept of a vector space to any ring of scalars. But while vector spaces make for a great first example of modules, many of the basic facts we are used to from linear algebra are often a little more subtle over a general ring. These differences are features, not bugs. We will introduce modules, study some general linear algebra, and discuss the differences that make the general theory of modules richer and even more fun.
Definition12.1.Module.
Let \(R\) be a ring (with \(1\)).
A left\(R\)-module is an abelian group \((M,+)\) together with a pairing \(R \times M \to M\text{,}\) written \((r,m) \mapsto rm\text{,}\) such that for all \(r,s \in R\) and \(m,n \in M\)
\((r + s)m = rm + sm\text{,}\)
\((rs)m = r(sm)\text{,}\)
\(r(m + n) = rm + rn\text{,}\) and
\(1m = m\text{.}\)
A right\(R\)-module is an abelian group \((M,+)\) together with a pairing \(M \times R \rightarrow M\text{,}\) written \((m,r) \mapsto mr\text{,}\) such that for all \(r,s \in R\) and \(m, n \in M\)
\((\mathrm{m}+\mathrm{n}) \mathrm{r}=\mathrm{mr}+\mathrm{nr}\text{,}\) and
\(\mathrm{m} 1=\mathrm{m}\text{.}\)
Often, the elements of the ring \(R\) are referred to as scalars, and the four rules given above are referred to as an \(R\)-action.
Typically, one first encounters modules in an undergraduate linear algebra course: the vector spaces from linear algebra are modules over fields. Later we will see that vector spaces are much simpler modules than modules over other rings. So while one might take linear algebra and vector spaces as an inspiration for what to expect from a module, be warned that this perspective can often be deceiving.
Definition12.2.Vector Space.
Let \(F\) be a field. An \(F\)-vector space is a (left) \(F\)-module.
Exercise12.3.Modules in Commutative Rings.
If \(R\) is a commutative ring, then any left \(R\)-module \(M\) may be regarded as a right \(R\)-module by setting \(m r = r m\text{.}\) Likewise, any right \(R\)-module may be regarded as a left \(R\)-module.
Remark12.4.
For non-commutative rings, left and right modules are not the same: trying to make a left \(R\)-module \(M\) into a right one by setting \(mr = rm\) fails to satisfy the second axiom, since \(m(rs) = (rs)m\) and \((mr)s = (rm)s = s(rm) = (sr)m\text{,}\) and, unless \(rs =sr\text{,}\) we cannot conclude that \(m(rs) = (mr)s\text{.}\)
Exercise12.5.Module and Opposite Ring.
For a ring \(R\text{,}\) recall the opposite ring, \(R^{op}\text{,}\) as defined in Item 8.6.1. Given a left \(R\)-module \(M\text{,}\) prove that \(M\) is a right \(R^{op}\)-module via the same rule for addition but with the rule for scaling on the right defined to be \(m r := rm\) for any \(r \in R\) and \(m \in M\text{.}\)
Lemma12.6.Arithmetic in Modules.
Let \(R\) be a ring and let \(M\) be a (left) \(R\)-module. Then for all \(r \in R\) and \(m \in M\) we have
\(0_Rm = 0_M\text{,}\)
\(r 0_M = 0_M\text{,}\)
\((-1_R)m = -m\text{,}\) and
\((-r)m = -(rm)\text{.}\)
Proof.
First, note \(0_Rm\in M\text{,}\) which is an ableian group. Observe
For any ring \(R\text{,}\) the zero module is \(0=\{0\}\) with \(r0=0\) for any \(r\in R\text{.}\)
Every ring \(R\) is a left module over itself with the rule for scaling given by the ring multiplication rule. It is also a right module over itself.
More generally, if \(R\) is any ring and \(I\) is a left ideal, then \(I\) is a left-\(R\)-module.
Let \(F\) be a field and \(R = \Mat_{n \times n}(F)\) (the ring of \(n \times n\) matrices with entries in \(F\)). Let \(M\) be the collection of column vectors with entries in \(F\) having \(n\) entries. The usual rules for adding column vectors and multiplying column vectors on the left by matrices make \(M\) into a left \(R\)-module. Likewise if \(N\) is the collection of all row vector with \(n\) entries in \(F\text{,}\) the \(N\) is a right \(R\)-module via addition and matrix multiplication.
For a non-negative integer \(n\text{,}\) the “standard" free (left) \(R\)-module of rank \(n\) is the set
Prove that an \(R\)-module \(M\) is the zero module if and only if the zero and identity maps on \(M\) are equal.
Exercise12.9.Closure Properties of Modules.
Given \(R\)-modules \(M\) and \(N\) and an ideal \(I\text{,}\) the following are \(R\)-modules:
\(\displaystyle M+N\)
\(\displaystyle M\cap N\)
Solution.
From 817 we already know they are subgroups under \(+\text{,}\) and it is evident from the definitions that each is closed under scaling by elements of \(R\text{.}\)
Theorem12.10.Abelian Groups are \(\Z\)-modules.
Let \(M\) be an abelian group under \(+\text{.}\) Then \(M\) becomes a \(\Z\)-module upon defining the rule for scaling to be \(nm :=\)
\begin{equation*}
0 \textrm{ if } n=0,
\end{equation*}
\begin{equation*}
\overbrace{m + \cdots + m}^{\text{n times}}, \text{ if } n > 0, \text{ and }
\end{equation*}
Let \(n\) be a positive integer and recall that \(\Z/n\) denotes the ring of integers modulo \(n\) (whose elements I will write as \(\ov{0}, \ov1, \cdots, \ov{n-1}\)).
Show that if \(M\) is any abelian group (under the operation \(+\)) such that \(nx = 0\) for all \(x \in M\) (where \(nx := \overbrace{x + x + \cdots + x}^{\text{n times}}\)), then the pairing \(\Z/n \times M \to M\) given \((\ov{j}, m) \mapsto jm\) makes \(M\) into a \(\Z/n\)-module. (Be sure to check this pairing is well-defined.)
Conversely, show that if \(M\) is a \(\Z/n\)-module, then the underlying abelian group \((M, +)\) has the property that \(nx = 0\) for all \(x \in M\text{.}\)
Definition12.12.Submodule.
Let \(R\) be a ring and let \(M\) be a left \(R\)-module. An \(R\)-submodule of \(M\) is a subset \(N \subseteq M\) such that
\(N\) is a subgroup of \(M\) under \(+\) (so, we have \(0 \in N\text{,}\) if \(n \in N\) then \(-n \in N\text{,}\) and if \(n_1, n_2 \in N\) then \(n_1 + n_2 \in N\)), and
\(rn \in N\) for all \(r \in R\) and \(n \in N\text{.}\)
We could equivalently define “submodule” to be a subset of \(M\) that is an \(R\)-module using the same operations of addition and scaling as in \(M\text{.}\)
Example12.13.Submodules.
A subset \(I\) of a ring \(R\) is a (left) submodule of \(R\) if and only if it is a (left) ideal.
Let \(R\) be a commutative ring with \(1 \neq 0\text{,}\) let \(I\) be an ideal of \(R\) and let \(M\) be an \(R\)-module. Then \(I M:=\left\{\sum_{k=1}^{n} j_{k} m_{k} \mid n \geq 0, j_{k} \in I, m_{k} \in M\right.\) for \(\left.1 \leq k \leq n\right\}\) is a submodule of \(M\text{.}\)
Let \(M\) be an \(R\)-module and \(I\) be an ideal in \(R\text{.}\)
\begin{equation*}
N = \{ m \in M \mid am = 0 \textrm{ for all } a \in I \}
\end{equation*}
is an \(R\)-submodule of \(M\text{.}\)
Exploration12.1.Submodules and Zorn’s Lemma.
Let \(N\) be a submodule of an \(R\)-module \(M\text{.}\) Using Zorn’s Lemma, prove that there is a submodule \(N'\) such that
\(N\cap N' = (0)\text{,}\) and
\(N''\cap(N + N')\neq(0)\) for every non-zero submodule \(N''\) of \(M\text{.}\)
Qual Watch.
Proving Exploration 12.1 was [cross-reference to target(s) "jan-2012-8" missing or not unique] on the [cross-reference to target(s) "jan-2012" missing or not unique] qualifying exam.
SubsectionRestriction of Scalars and Algebras
“Restrictions will set you free.”
―W.A. Mathieu
For an \(R\)-module \(M\) the ring \(R\) is often referred to as the “ring of scalars” for the module (by analogy to the vector space case). Given an action of a ring of scalars on a module, we can sometimes produce an action of a different ring of scalars on the same set, producing in effect a new module structure.
Theorem12.14.Restriction of Scalars.
Let \(\phi: R \to S\) be a ring homomorphism. Any left \(S\)-module \(M\) may be regarded via restriction of scalars as a left \(R\)-module with the following structure:
the rule for addition \(+\) on \(M\) is the same as in the original structure and
the rule for scaling by elements of \(R\) is
\begin{equation*}
r \cdot m := \phi(r)m \text{ for any }r \in R \text{ and }m\in M.
\end{equation*}
Proof.
Let \(x,y \in R\) and \(m,n \in M\text{.}\) One checks that the properties in the definition of module hold for the given action using properties of ring homomorphisms. In detail,
As a special case of Theorem 12.14, if \(R\) is a subring of \(S\text{,}\) then every left \(S\)-module becomes a left \(R\)-module via restriction of scalars along the inclusion map of \(R\) into \(S\text{.}\) This explains the use of the phrase “restriction of scalars”. That being said, in generality the phrase “translation of scalars” would be more appropriate.
Definition12.15.Algebra.
Given a ring \(R\text{,}\) an \(R\)-algebra is a ring \(S\) equipped with a ring homomorphism \(\phi: R \to S\text{.}\) This defines an \(R\)-module structure on \(S\) given by restriction of scalars: for each \(r\in R\) and \(s\in S\text{,}\)\(rs := \phi(r)s\text{.}\) This \(R\)-module structure on \(S\) is compatible with the internal multiplication of \(S\) i.e.,
\begin{equation*}
r(st)=(rs)t=s(rt) \text{ for all } r\in R, s, t\in S.
\end{equation*}
We will call \(\phi\) the structure homomorphism of the \(R\)-algebra \(S\text{.}\)
Example12.16.Polynomial Rings are Algebras.
If \(A\) is a ring and \(x_1,\dots,x_n\) are indeterminates, the inclusion map \(A \hookrightarrow A[x_1,\dots,x_n]\) makes the polynomial ring into an \(A\)-algebra.
Example12.17.Inclusion Maps Give Algebra Structure.
More generally, any inclusion map \(R \subseteq S\) gives \(S\) an \(R\)-algebra structure. In this case the \(R\)-module multiplication coincides with the internal (ring) multiplication on \(S\text{.}\)
Example12.18.Rings Have Unique Structure as \(\Z\)-algebra.
Any ring comes with a unique structure as a \(\Z\)-algebra, since there is a unique ring homomorphism \(\Z \to R\text{:}\) the one given by \(n \mapsto n \cdot 1_R\text{.}\)
Example12.19.Restrictions of Scalars.
Complex Vector Spaces are Real.
Since \(\R\) is a subring of \(\C\text{,}\) every complex vector space may be regarded as a real vector space, by restriction of scalars from \(\C\) to \(\R\text{.}\) Likewise, any real vector space may be regarded as a rational vector space, etc.
Polynomial Rings are Modules.
The polynomial ring \(R[x_1,\ldots,x_n]\) is a left \(R\)-module for any \(n\geq 0\) via the evident injective ring homomorphism \(R \into R[x_1, \ldots, x_n]\text{.}\)
\(\Mat_{n \times n}(R)\) is a left \(R\)-module for \(n\geq 1\) given by the ring map \(R \to \Mat_{n \times n}(R)\) sending \(r\) to \(r \cdot I_n\text{.}\)
\(R/I\) is an \(R\)-module.
If \(I\) is a (two-sided) ideal of a ring \(R\) then applying restriction of scalars along the quotient homomorphism \(q:R\to R/I\) gives that any left \(R/I\)-module is also a left \(R\)-module. The rule for scaling is \(r \cdot (r' + I) = rr' + I\text{.}\)
Summary
A Module is a generalization of the concept of a vector space to any ring of scalars.
Given a ring \(R\) and an ideal \(I\text{,}\) each of \(R\text{,}\)\(I\text{,}\) and \(R/I\) can be viewed as \(R\)-modules.