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.


1 Answer 1


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.

  • $\begingroup$ That's a beautiful argument. Can you give me a reference for those results of Raynaud and Kumar-Nori? $\endgroup$
    – Evan
    Sep 10, 2013 at 2:06
  • $\begingroup$ The Raynaud paper, using etale cohomology, was in Inventiones in, I think, 1968. It's in French, and title translates to "Universal Projective Modules". I'm sorry that I don't know a reference for the work of Mohan Kumar and Nori (using Chern classes), but Mohan occasionally visits MathOverflow, so maybe we'll get lucky and hear from him. Alternatively, there's a paper by Swan that gives a new proof of the Mohan Kumar/Nori results, and I believe it's easily Googlable. $\endgroup$ Sep 10, 2013 at 2:12
  • $\begingroup$ Great, the paper of Swan is here : math.uchicago.edu/~swan/MKN.pdf. It refers to another paper of Swan for the argument of Kumar and Nori; I suppose that it (like so many theorems of Nori!) was never published. $\endgroup$
    – Evan
    Sep 10, 2013 at 2:18
  • $\begingroup$ Evan: I do have an old typewritten manuscript with the Mohan Kumar/Nori argument, but it's at my office and I'm home now. I can have it scanned later this week. $\endgroup$ Sep 10, 2013 at 2:20
  • $\begingroup$ That would be great! I'll email you with my contact info. $\endgroup$
    – Evan
    Sep 10, 2013 at 2:21

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