Let $R$ be a commutative ring. A vector $(c_1,\ldots,c_n) \in R^n$ is **unimodular** if $Rc_1 + \cdots + Rc_n = R$. Say that a vector $\vec{v} \in R^n$ is a **basis element** if there exists a free basis for $R^n$ containing $\vec{v}$. It is clear that all basis elements of $R^n$ are unimodular. Moreover, if $\vec{v} \in R^n$ is unimodular and $V = R \cdot \vec{v}$, then there exists an $R$-submodule $W \subset R^n$ such that $R^n = V \oplus W$ (proof : if $\vec{v} = (c_1,\ldots,c_n)$ and $1 = a_1 c_1 + \cdots + a_n c_n$ with $a_i \in R$, then we can define a surjection $\phi : R^n \rightarrow R$ via the formula $\phi(x_1,\ldots,x_n) = a_1 x_1 + \cdots + a_n x_n$; the map $\phi$ is split via the inclusion $R \hookrightarrow R^n$ that takes $1$ to $\vec{v}$). Clearly $V \cong R^1$, but it is not necessarily true that $W$ is a free $R$-module, so it does not follow that $\vec{v}$ is a basis element.

It is clear that unimodular vectors in $R^1$ are basis elements.

Question : Can someone give me an example of a ring $R$ such that for all $n \geq 2$, there exist unimodular vectors in $R^n$ that are not basis elements?

Such rings have to be pretty weird; for instance, it is standard that if a ring satisfies Bass's stable range condition $SR_{d+2}$, then for $n \geq d+2$ all unimodular vectors in $R^n$ are basis elements. This means that rings $R$ as in our question must either be non-Noetherian or have infinite Krull dimension.