Ring $R$ such that $R^n$ contains unimodular elements that are not part of a free basis for all $n \geq 2$ 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.
 A: Start with the integers.  Adjoin variables $X_{in}$ and $Y_{in}$ for all $1\le i \le n$.  Mod out by all relations of the form
$$\sum_{i=1}^nX_{in}Y_{in}=1$$
Call the resulting ring $R$.
Then, by construction, any $(X_{1n},X_{2n},\ldots,X_{nn})$ is a unimodular row over $R$.  I claim it's not a basis element. Equivalently, I claim that the complement of this row (i.e. the cokernel of the linear map $R\rightarrow R^n$ that it defines) is not free.
To see this, let $S$ be any ring over which there exists a unimodular row $(t_i,\ldots,t_n)$ which is not a basis element.  (Such rings are known to exist by results of Raynaud, or, alternatively, of Mohan Kumar and Nori).  Write $\sum_{i=1}^nt_iu_i=1$.  Map $R$ to $S$ by
$$X_{in}\mapsto t_i$$
$$Y_{in}\mapsto u_i$$
and, for each $m\neq n$,
$$X_{1m}\mapsto 1$$
$$Y_{1m}\mapsto 1$$
$$X_{jm}\mapsto 0 \quad (j\neq  1)$$
$$Y_{jm}\mapsto 0 \quad (j\neq 1)$$
Now let $P_n$ be the $R$-module that is the complement of $(X_{i1},\ldots, X_{in})$.  Then $P\otimes_RS$ is the complement of $(t_1,\ldots,t_n)$, and hence not free.  Therefore $P$ cannot be free, so $(X_{i1},\ldots,X_{in})$ cannot be a basis element.
