5
$\begingroup$

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.

$\endgroup$

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.