Is a vector of coprime ring elements column of an invertible matrix? 
Given a commutative ring $R$ with unit and $a_1=(r_1,\ldots,r_n)^T \in R^n$ with coprime entries (i.e. $\sum_i Rr_i=R)$. Are there $a_2,\ldots,a_n \in R^n$ such that the matrix $A = (a_1,\ldots,a_n) \in R^{n \times n}$ is invertible? 

The answer is "yes" if $R$ is a PID by "Is a vector of coprime integers column of a regular matrix?". 
However none of the solutions there generalizes because the first answer uses division by a gcd and the second one (to my understanding) the elementary divisors theorem. 
But because the answer is always "yes" in case $n=2$ (write $r_1s_1+r_2s_2=1$ and take $a_2=(-s_2,s_1)^T)$ I wonder if this is also true for $n > 2$ ? 
 A: No, this is not true, but it is quite non-trivial to find examples. At least at the top of my head I only know the one below - does anyone know a simpler one?
The entries of the vector $a_1 := (r_1,...,r_n)$ being coprime means that the map $$\alpha: R^n\xrightarrow{(r_1,...,r_n)} R$$ 
is surjective, so $P := \text{ker}(\alpha)$ is a rank projective $R$-module with $P\oplus R\cong R^{n}$. Conversely, any such module arises in this way. 
The existence of the vectors $a_2,...,a_n$, however, is equivalent to the existence of a commutative diagram
$$\begin{array}{ccc} R^n\ \  & \xrightarrow{\alpha} & R \\ \downarrow{\small\cong} && \ \ \downarrow{\small 1} \\ R^n \ \ & \xrightarrow{p_1} & R\end{array}$$
which is equivalent to $\text{ker}(\alpha)$ being free on $(n-1)$-generators.
Over a PID, any projective module is free, confirming your observation that any vector of coprime elements can be extended to a regular matrix. In general, however, there are non-free stably free modules: Taking $R:={\mathbb R}[x,y,z]/(x^2+y^2+z^2-1)$ and $a_1 := (x,y,z)$ gives such an example. See Is a stably free module always free? for details.
