Let $R$ be a commutative ring with unity and $M$ be a finitely generated free $R$-module.
Let $S$ be a finite subset of $M$ generating $M$ as $R$-module.
From this, can we say that $M$ has finite basis?
If $R$ is a field, it is the case because the maximal linearly independent set in $S$ gives rise to the basis of $M$.
But when $R$ is just commutative ring not being a field, we cannot apply this argument.
If $M$ also has finite basis, can we say further that the number of basis is equal or less than $|S|$?
Best Answer
No, in general you can't extract a basis from a set of generators. The simplest counterexample is $R=\mathbb{Z}$ (the integers) and $M=\mathbb{Z}$. The subset $S=\{2,3\}$ generates $M$, because $1=-1\cdot2+1\cdot 3$, but no subset of $S$ is a set of generators.
It is however true that, if $M$ is a finitely generated free module over a commutative ring, then the rank (number of elements in a basis) is less than or equal to the number of elements in any finite set of generators.
How can we prove it? It's quite simple. Consider $M$ a finitely generated free module and $\mathfrak{m}$ a maximal ideal of $R$. Then $M/\mathfrak{m}M$ is a vector space over $R/\mathfrak{m}$. Any set of generators in $M$ projects to a set of generators of $M/\mathfrak{m}M$ and a basis projects to a basis.
Over non commutative ring, the notion of rank is not well defined for finitely generated free modules: there are non commutative rings $R$ such that, as $R$-modules, $R\cong R\oplus R$, for instance.