Generation, Direct Sums, and Free Modules

Three notions describe how a module is built from a handful of elements: generation (spanning by $R$ -linear combinations), direct sums (assembling a module from independent pieces), and free modules (those with a basis). The last separates module theory from linear algebra: over a general ring a module need not have a basis, and the modules that do — the free ones — are the only ones where the familiar coordinate arguments still run.¹

Generation

$R A$ is the smallest submodule containing $A$ : it is a submodule by the submodule criterion, and any submodule containing $A$ is closed under linear combinations, so contains $R A$ . For a family of submodules $N_{1}, \dots, N_{k}$ , the submodule they generate is their sum $N_{1} + \dots + N_{k} = {a_{1} + \dots + a_{k} : a_{i} \in N_{i}}$ , the smallest submodule containing every $N_{i}$ . Generation for modules is tamer than for groups: because $M$ is abelian, like terms in a combination always collect, $r_{1} a_{1} + r_{2} a_{2} + s_{1} a_{1} = (r_{1} + s_{1}) a_{1} + r_{2} a_{2}$ , and no word problem arises.

Over $Z$ , $Z a = ⟨ a ⟩$ is the cyclic subgroup generated by $a$ ; finitely generated $Z$ -module means finitely generated abelian group.
$R$ over itself is cyclic, generated by $1$ , since $R = R \cdot 1$ . Its cyclic submodules are the principal ideals; its finitely generated submodules are the finitely generated ideals.
Generation is not inherited by submodules. Let $R = F [x_{1}, x_{2}, \dots]$ be the polynomial ring in infinitely many variables. Then $R$ is cyclic over itself, yet its submodule (ideal) generated by all the $x_{i}$ is not finitely generated. A finitely generated module can have a submodule that is not.

A finitely generated module has a smallest generating size $d$ ; any generating set of size $d$ is a minimal generating set. Unlike in a vector space, minimal generating sets can have different relations among them, and $d$ is not a complete invariant: the relations matter, and the free-module theory below tracks them.

Direct sums

The external direct sum combines separate modules. The more useful question is when a module already is a direct sum of submodules it contains, an internal direct sum. The following three conditions, each equivalent, pin this down.²

When these hold, $M = N_{1} \oplus \dots \oplus N_{k}$ (internally): the external direct sum of the $N_{i}$ is isomorphic to the submodule they span, and each element has one representation. The uniqueness is of the module elements $a_{i}$ , not of any scalars.

An internal direct sum

M = N_{1} \oplus N_{2}

resolves every element into a unique pair of components, one from each summand, exactly as coordinates resolve a point of the plane along two independent axes.

Free modules

The standard basis $e_{i} = (0, \dots, 0, 1, 0, \dots, 0)$ generates $R^{n}$ , and every tuple $(s_{1}, \dots, s_{n}) = \sum s_{i} e_{i}$ has a unique such expression — unique in the ring coefficients, not merely in the module components. That extra uniqueness is the definition of a basis.

The uniqueness here is stronger than in a direct sum. In $Z /2 Z \oplus Z /2 Z$ each element is a unique sum $n_{1} + n_{2}$ of components, but the component $n_{1} = (1, 0)$ equals $1 \cdot (1, 0) = 3 \cdot (1, 0) = \dots$ , so the scalar is not determined. That module is not free on ${(1, 0), (0, 1)}$ , nor on any set. Freeness requires the scalars to be determined too.

A free module of rank

n

is a direct sum of

n

copies of the ring; a basis assigns to each element a unique coordinate tuple of ring scalars, one per copy.

The universal property

Free modules are characterized by how maps out of them behave: a homomorphism from a free module is determined by arbitrary values on a basis, with no relations to respect.

Concretely $F (A)$ is the set of functions $A \to R$ vanishing at all but finitely many points, identified with formal finite sums $\sum r_{i} a_{i}$ . The extension is forced: $Φ (\sum r_{i} a_{i}) = \sum r_{i} φ (a_{i})$ , well defined precisely because the expression $\sum r_{i} a_{i}$ is unique. This is the same universal property free groups have, and it is why one may define a homomorphism on a basis and extend by linearity: the extension exists and is unique.³

The universal property: any set map from a basis

A

into a module

M

extends along the inclusion to one and only one homomorphism out of the free module

F (A)

Invariance of rank

For finite-dimensional vector spaces, any two bases have the same size. The same holds for free modules over a commutative ring, so rank is an invariant.

Commutativity is a real hypothesis: over suitable noncommutative rings one can have $R ≅ R^{2} ≅ R^{n}$ for all $n$ , so rank is meaningless there, and the invariant-basis-number property is a genuine restriction.

Not every module is free

A module can have no basis at all.

Torsion is the obstruction. A basis element must be free of relations, but every element of a nontrivial finite abelian group is annihilated by its order. More generally, any module with a nonzero torsion element is not free. Over a field there is no torsion, which is why every vector space is free; over $Z$ or $F [x]$ torsion is generic, and measuring it is the content of the structure theorem for modules over a PID.

Torsion is not the only obstruction: a torsion-free module can still fail to be free.

Z / n Z

the multiples of any element

a

close up after

n

steps, so the relation

n a = 0

holds and

{a}

can never be a linearly independent set: torsion blocks a basis.

Constructing modules by generators and relations

Every module is a quotient of a free one. Given a generating set $S$ for $M$ , the universal property produces a surjection $F (S) ↠ M$ (the map that is the identity on $S$ ), and its kernel $K$ records the relations among the generators, giving a short exact sequence:

0 ⟶ K ⟶ F (S) ⟶ M ⟶ 0.

This exhibits $M ≅ F (S) / K$ — a presentation of $M$ by generators (a basis of $F (S)$ ) and relations (generators of $K$ ). The worked example above is exactly such a presentation: $Z /6$ presented by two generators mapping to $2$ and $3$ and the relations $(3, 0)$ and $(0, 2)$ .

Notion	Uniqueness enforced on	Field analogue	Present over $Z$ ?
Generating set	none	spanning set	always
Direct sum $⨁ N_{i}$	components $a_{i}$	complementary subspaces	always
Free / basis	components and scalars	basis	only for torsion-free special cases

Dummit & Foote, Abstract Algebra, 3rd ed., §10.3 — generation of modules, finitely generated and cyclic modules, and the submodule generated by a subset as the set of finite $R$ -linear combinations. ↩
Dummit & Foote, §10.3, Proposition 5 — the equivalence of the three characterizations of an internal direct sum of submodules. ↩
Dummit & Foote, §10.3, Theorem 6 and Corollary 7 — existence of the free module $F (A)$ , its universal property with respect to set maps out of $A$ , and the resulting define on a basis, extend by linearity principle; the invariance-of-rank statement is Exercise 2 of the same section. ↩