Discussion:Théorème de Davenport-Cassels

Forgive me for writing in English -- if I were confident enough in my French, I would modify the article myself.

I noticed that the article references my lecture notes on the Davenport-Cassels Lemma. Up until today, these notes simply recopied the proof from Serre's Course of Arithmetic: in particular, there was the additional hypothesis that the quadratic form be positive definite. I also included, as a remark, that I believed this positive definite hypothesis to be superfluous.

When I read your article today, I noticed the reference to Weil's text, which states and proves the result in the most general form: let $f$ be an integral quadratic form (in the more restrictive, classical sense: the matrix entries are integral) which satisfies the following condition: for every $x \in \mathbb{Q}^n \setminus \mathbb{Z}^n$, there exists $y \in \mathb{Z}^n$ such that $0 < |f(x-y)| < 1$. Then every integer which is rationally represented by $f$ is also integrally represented by $f$.

If $f$ is assumed to be anisotropic, then since $x$ does not have integer entries and $y$ does, $x-y \neq 0$ so $|f(x-y)| > 0$ necessarily. Thus you don't need the first inequality in this case. However, you do need it in general -- and hence the current statement of the theorem is false -- as one can see by considering the form $x^2-y^2$.

I have modified Serre's proof to yield the results as stated above and updated the file. The link now directs to the new version. -- Pete L. Clark

Thanks a lot for pointing out this mistake. I repaired it according to your very clear explanations. Anne Bauval (d) 5 juin 2010 à 01:02 (CEST)