7
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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