Finding Bases for Free Modules Let $A$ be a commutative ring with unity. If we have an element $v = (a_1,a_2,\dots,a_n)$ of $A^n$ such that the ideal generated by the entries of $v$ is the whole ring, can we extend the singleton set containing $v$ to a basis for $A^n$? This condition is necessary, because if we write the basis as a matrix, then it must have an inverse, and then by matrix multiplication we will get the above condition.
I know this to be true if the ring is a PID. If this is not true, can we say anything about vectors which can occur as part of a basis other than the above observation?
 A: Such vectors $v$ are called "unimodular rows". This is an active area of research, which is related to stably free modules.
Recall that a finitely generated module $M$ is stably free if there exists a free module $L$ of finite rank such that $M\times L$ is free.
In fact, one can show the equivalence of the following properties for any commutative ring $A$ with unit:
$(1)$ Every unimodular row of$A^n$ is part of a basis
$(2)$ $GL_n(A)$ acts transitively on the right on the set of unimodular rows
$(3)$ any $A$-module $M$ such that $M\times A\simeq A^n$ is free.
There are also "global" equivalences: the following properties  are equivalent: 
$(1')$ For all $n\geq 1$, every unimodular row of$A^n$ is part of a basis
$(2')$ For all $n\geq 1$ $GL_n(A)$ acts transitively on the right on the set of unimodular rows
$(3)$ any stably free $A$-module is free.
For your information, a big part of the proof of Quillen-Suslin theorem, which asserts that any finitely projective $K[X_1,\ldots,X_n]$-module is free, consists of proving that any stably free $K[X_1,\ldots,X_n]$-module is free. They did it by proving that any unimodular row is part of a basis.
To go back to your question, not any unimodular row is part of a basis if $n\geq 3$ in general (however, you can easily show that it is true for $n=1,2$). 
The classic counterexample is $A=\mathbb{R}[X,Y,Z]/(X^2+Y^2+Z^2-1)$ and $v=(\bar{X},\bar{Y},\bar{Z})$. The key argument is the hairy ball theorem.
Sketch of proof. Assume that $v$ is a part of a basis $(v,w,t)$. It is easy to check that one may assume that $vw^t=0$.
Now  the vector $w\in A^3$ induces a continuous map from the unit sphere $u\in \mathbb{S}^2\mapsto w(u)\in \mathbb{R}^3$ , which happens to be nonzero everywhere (this follows from the fact that  $w$ is part of a basis)
Now $v w^t=0$. But $v(u)=u$ for all $u$, so it means that $w(u) and (u)$ are orthogonal for all $u\in\mathbb{S}^2$, contradicting the hairy ball theorem.
