Let $R$ be a commutative ring and let $R^n$ denote the free $R$-module of rank $n$. A unimodular column is an element $\alpha = (a_1, \dots, a_n) \in R^n$ for which there exist $b_i \in R$ with $a_1b_1 + \cdots +a_nb_n=1$. A commutative ring has the unimodular column property if, for every $n$, every unimodular column is the first column of some $n\times n$ invertible matrix over $R$.
In Rotman's "An Introduction to Homological Algebra", on page 204, he states that $\mathbb{Z}_6$ does have the unimodular column property. I am having difficulty trying to show that. (He uses $\mathbb{I}_6$ to denote $\mathbb{Z}_6$).
$\mathbb{Z}_6$ has nonfree projectives and $\mathbb{Z}_6$ is not a local ring. So the only other thing I can think of is to try to extend the unimodular column $(a_1, \dots, a_n)$ to a basis of $GL_n(\mathbb{Z}_6)$. Permuting the $a_i$ will also lead to another unimodular column. But $(1, 1, \dots, 1)$ is a unimodular column and it won't extend to basis by permutation alone. (I got this idea from Lang's "Algebra", Chapter XXI, Section 3, page 847.)\
Any suggestions on how to proceed? I must be missing something very obvious.