Modules over Rings

Modules generalize vector spaces by replacing the scalar field with an arbitrary ring, providing a unified framework for linear algebra over rings, abelian group theory, and representation theory.

Basic Definitions

Definition9.1Module

A left module over a ring $R$ (or left $R$ -module) is an abelian group $(M, +)$ together with a scalar multiplication $R \times M \to M$ , $(r, m) \mapsto rm$ , satisfying:

$r(m + n) = rm + rn$ for all $r \in R$ , $m, n \in M$ .
$(r + s)m = rm + sm$ for all $r, s \in R$ , $m \in M$ .
$(rs)m = r(sm)$ for all $r, s \in R$ , $m \in M$ .
$1_R m = m$ for all $m \in M$ (for unital modules).

ExampleFundamental examples of modules

Vector spaces: Modules over a field $k$ .
Abelian groups: Modules over $\mathbb{Z}$ (with $n \cdot m = m + \cdots + m$ ).
Ideals: $I \subseteq R$ is an $R$ -module via ring multiplication.
$R[x]$ -modules: An $R[x]$ -module is an $R$ -module $M$ with a chosen $R$ -linear endomorphism $T: M \to M$ ( $x$ acts as $T$ ). This connects module theory to linear algebra.
$R^n$ : The free module of rank $n$ , analogous to $k^n$ .

Submodules and Quotients

Definition9.2Submodule, quotient, homomorphism

A submodule $N \leq M$ is a subgroup closed under scalar multiplication. The quotient module $M/N$ consists of cosets $m + N$ with the natural operations. A module homomorphism $f: M \to N$ is a function satisfying $f(rm + sn) = rf(m) + sf(n)$ .

The kernel $\ker f$ is a submodule of $M$ , and $M/\ker f \cong \mathrm{im}(f)$ (first isomorphism theorem for modules).

Free Modules and Bases

Definition9.3Free module

An $R$ -module $M$ is free if it has a basis: a set $\{e_i\}_{i \in I}$ such that every $m \in M$ has a unique expression $m = \sum r_i e_i$ (finite sum) with $r_i \in R$ . Then $M \cong R^{(I)} = \bigoplus_{i \in I} R$ .

For commutative rings $R$ , a free module of rank $n$ is isomorphic to $R^n$ , and the rank is well-defined (the IBN property: invariant basis number).

RemarkNot every module is free

Unlike vector spaces, modules over general rings need not be free. $\mathbb{Z}/2\mathbb{Z}$ is a $\mathbb{Z}$ -module with no basis: $2 \cdot \bar{1} = \bar{0}$ shows no element is "linearly independent." Modules over a field are always free (they are vector spaces), but this fails for modules over $\mathbb{Z}$ , polynomial rings, etc.