Role of stably free modules in algebraic geometry

Question

For any ring $R$, a non-zero module $S$ is stably free if $S\oplus R^a$ is free ($a\geq 1$). This may be an overly vague question, but I am wondering in what contexts do stably free modules arise in algebraic geometry?

I know Serre asked whether projective modules (that turn out to be stably free) over $k[x_1,\ldots, x_n]$ ($k$ field) are free. This, of course, corresponds to asking whether vector bundles over the affine $n$-space $\mathbb{A}^n_k$ are trivial. However, this question was asked back in the 50s (eventually being answered that all projective modules are free!) and I am wondering what role stably free modules now play (or can play) in algebraic geometry.

Answer 1

This is just a long comment.

Stably free modules appear often in many questions regarding number of generators of modules over rings. For example, if $I\subset R$, a polynomial ring over a field in $n$ variables and $I$ an ideal, a question of Murthy is, can one say $I$ is generated by the same number of elements which generates $I/I^2$. Most of the known results use certain stably free module being free.

The universal example of a stably free module is that given by the presentation, $0\to R\to R^n\to P\to 0$, where $R=k[x_i,y_i, 1\leq i\leq n]/\sum x_iy_i=1$ and the map $R\to R^n$ given by the unimodular row $(x_1,\ldots, x_n)$. If $n\geq 3$, $P$ is stably free and not free. Suslin proved that $(x_1^{a_1},\ldots, x_n^{a_n})$ gives a free module if $(n-1)!$ divides $\prod a_i$. There are topological proofs for the first statement, an algebraic proof of both these (the necessity of the condition of Suslin) uses Grothendieck-Riemann-Roch without denominators (the $(n-1)!$ should be a clue.) Similar principles can be used to show that there are affine varieties (say over complex numbers) of dimension $p+2$, $p$ a prime and stably free modules of rank $p$ which are not free, very far from what you can do with topology, these being all topologically trivial.