Q:

Given a generating set $S$ of a free $A$-module $M$, must $S$ contain an $A$-basis for $M$?

Given a generating set $S$ of a free $A$-module $M$, must $S$ contain an $A$-basis for $M$?
I'm not sure if this is true or not. I've tried using a Zorn's Lemma argument which failed. 

A:

No, a generating set for a free $A$-module $M$ need not contain a basis.
A simple counterexample is that $\{2,3\}$ is a generating set for the free $\mathbb{Z}$-module $\mathbb{Z}$. That can be generalised of course, whenever you have non-units $x_i \in A, \; 1 \leqslant i \leqslant k$ such that the ideal generated by the $x_i$ is the entire ring, you have a generating set that does not contain a basis. Of course, if $A$ is a field, then every generating set contains a basis.

