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.



Ring $R$ such that $R^n$ contains unimodular elements that are not part of a free basis for all $n \geq 2$
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.


