Basis of Free Module is No Greater than Generator

Theorem
Let $R$ be a commutative ring with unity.

Let $M$ be a free $R$-module with basis $B$.

Let $S$ be a generating set for $M$.

Then:
 * $\size B \le \size S$.

That is, there exists an injection from $B$ to $S$.

Outline of Proof
Because $S$ is a spanning set, we can construct a surjective homomorphism $R^{\paren S} \to R^{\paren B}$.

Using Krull's Theorem we divide through by a maximal ideal of $R$ to reduce it to the case where $R$ is a division ring, that is, the case of vector spaces.

We then conclude by comparing cardinalities and using Basis of Vector Space Injects into Generator.

Proof
Because $S$ is a generating set, there is a surjective homomorphism
 * $\phi : R^{\paren S} \to M$

where $R^{\paren S}$ is the free $R$-module on $S$.

Because $B$ is a basis, there is an isomorphism
 * $\psi : R^{\paren B} \to M$

Thus $f = \psi^{-1} \circ \phi: R^{\paren S} \to R^{\paren B}$ is a surjective module homomorphism.

By Krull's Theorem, there exists a maximal ideal $M \subset R$.

By Maximal Ideal iff Quotient Ring is Field, $R / M$ is a field.

Let $k = R / M$.

Let $\pi: R \to k$ denote the quotient mapping.

By some universal properties, there exists a $k$-module homomorphism $\bar f: k^{\paren S} \to k^{\paren B}$ such that:
 * $\pi^B \circ f = \bar f \circ \pi^S$

Where $\pi^B$ denotes direct sum of module homomorphisms.

Because $\pi^B \circ f$ is surjective, so is $\bar f$.

Let:
 * $C$ be the canonical basis of $k^{\paren S}$
 * $D$ be the canonical basis of $k^{\paren S}$

Because $C$ is a generator of $k^{\paren S}$ and $\bar f$ is surjective, $\map {\bar f} C$ is a generator of $k^{\paren B}$.

We have:

Also see

 * Bases of Free Module have Equal Cardinality