Free module ($M\cong {⨁}_{i\in I}R$)

A free module is a module with a basis of a module $(e_i{)}_{i\in I}$: every element has exactly one expression $m={\sum }_{i\in I}{r}_i{e}_i$ with $r_i\in R$ and only finitely many $r_i$ nonzero. Equivalently, the coefficient map ${⨁}_{i\in I}R\to M$ is a module isomorphism. Existence means generation; uniqueness means linear independence in a module. For finite $I$ of size $n$, this says $M\cong R^n$, a finite free module. See Neil Strickland's Rings, Modules and Linear Algebra, section 11.

The integer lattice ${\mathbb{Z}}^2$ in the opening diagram is a free module over $\mathbb{Z}$ with basis of a module $e_1=(1,0)$, $e_2=(0,1)$. For example, $(3,2)=3e_1+2e_2$, and comparing the two coordinates makes the coefficients unique. A different basis of a module is $u=(1,0)$, $v=(1,1)$, because $(a,b)=(a-b)u+bv$; in particular $(3,2)=u+2v$.

By contrast, $w_1=(2,0)$ and $w_2=(0,1)$ are linearly independent in a module, but generate only the submodule $2\mathbb{Z}\times \mathbb{Z}$. They miss $(1,0)$ because $2a=1$ has no integer solution. They would form a basis over the rational numbers. Even generation need not let one extract a basis of a module: the elements $2,3$ generate $\mathbb{Z}$, since $1=3-2$, but neither singleton generates it, and $3\cdot 2-2\cdot 3=0$ is a nontrivial linear relation between them. These are concrete failures of the usual vector space argument that divides by any nonzero coefficient.

The cyclic module $\mathbb{Z}/6\mathbb{Z}$ has no basis of a module over $\mathbb{Z}$: every element has the nontrivial relation $6m=0$, so no nonempty subset is linearly independent in a module, while the empty subset cannot generate this nonzero module. Specifying the scalar ring is essential: over $R=\mathbb{Z}/6\mathbb{Z}$, the same additive object is a free module with the one-element basis of a module $\{\overline{1}\}$, since each coefficient is itself a unique class in $R$.

For a free module $F$, a surjective module homomorphism $\varphi :P\to M$, and a module homomorphism $\psi :F\to M$, there is a module homomorphism $q:F\to P$ satisfying $\varphi q=\psi$. Choose a preimage under $\varphi$ of the image of each element of a basis of a module of $F$, and extend by linearity. Thus every free module is a projective module.

A submodule $N$ of a free module with basis of a module $E$, generated by a countable set, is contained in the submodule generated by some countable subset $E^{\prime }\subseteq E$. Take $E^{\prime }$ to be the union of the finite coordinate supports in a free module of a countable generating family for $N$. This union is countable, and every generator belongs to $⟨E^{\prime }⟩$, which proves the inclusion. Combining this with basis partition gives a module direct sum places $N$ inside a countably generated direct summand.

Partition a basis of a module $E$ into disjoint subsets $E^{\prime },E^{\prime \prime }$. Splitting the finite expression of each element gives $M=⟨E^{\prime }⟩+⟨E^{\prime \prime }⟩$. An element in both generated submodules would give, by subtraction, a relation among the disjoint parts of $E$; linear independence in a module forces every coefficient to vanish. The intersection is zero, proving the direct sum of modules decomposition. Each part is itself a basis of a module for the submodule it generates.

Order the linearly independent subsets of a vector space by inclusion. A chain's union is independent, since every finite relation is contained in one chain member. Zorn's lemma gives a maximal independent subset. A vector outside its span could be adjoined while preserving independence, so it spans. Sending finite-support coefficient families to their linear combinations gives an isomorphism from the free module on that subset to $V$. This supplies an arbitrary, possibly infinite-dimensional basis using the axiom of choice; no pre-existing basis theorem is invoked.

Elements of a module are linearly independent if every finite linear relation between them has all coefficients zero. An independent finite set freely generates its span, a finite free module. Over an integral domain, maximal independence within a finite generating set enables clearing denominators relative to an independent module subset.

A finite free module is a free module with a finite basis, equivalently a module isomorphic to $R^n$ for some nonnegative integer $n$.

A basis $S$ of an $R$-module $M$ is a subset such that every element of $M$ has a unique expression as a finite $R$-linear combination of elements of $S$.

For an element $m={\sum }_{e\in E}{r}_{e}e$ of a free module with specified basis of a module $E$, its coordinate support is the finite set of basis of a module elements having nonzero coefficients. This is well-defined by uniqueness of the expression. The coordinate support of a sum is contained in the union of the two coordinate supports, and scalar multiplication cannot enlarge it. It is distinct from the prime-ideal-valued support of a module.

A module $F$ is free on a set $S$ exactly when every function from $S$ to any module $N$ extends uniquely to an $R$-module homomorphism $F\to N$.

For a nonzero commutative scalar ring, every two bases of a finite-rank free module have the same cardinality. This cardinality is the rank.

Over a nonzero commutative ring, an injection $R^m\hookrightarrow R^n$ implies $m\le n$. If $m>n$, appending zero coordinates gives an injective endomorphism of $R^m$ with zero determinant; the Cayley--Hamilton theorem and cancellation of that injection eventually force the identity to vanish.

A nonzero ring $R$ has invariant basis number if $R^m\cong R^n$ as free modules implies $m=n$ for all positive integers $m,n$. The property is not automatic for arbitrary rings; the commutative-ring case is proved below.

If $R$ is a nonzero commutative ring and $R^m\cong R^n$, reduction modulo a maximal ideal produces isomorphic vector spaces of dimensions $m$ and $n$. Hence $m=n$.