Free module

A free module has a basis: every element has a unique finite linear combination in that basis with coefficients in the scalar ring.

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 $S$.

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.

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 $m=n$.