1
$\begingroup$

Let $q$ denote a quadratic form over a field $k$.

The u-invariant of a field $u(k)$ is defined by $u(k):=\{ max (\mathrm{rank}(q)) $ | $ q $ is anisotropic over $k\}$.

Let $k = \mathbb{Q}_p$ for any prime $p$ and set

$L = k(t_1,..,t_n)$.

It is known that $u(k)=4$ and newer results by David B. Leep state that

$u(L) = 4\cdot2^n = 2^{n+2}$.

As a consequence from the Arason-Pfister Hauptsatz we have that

$2^{cd(L)} \leq u(L) = 2^{n+2}$, while $cd(L)$ denotes the cohomological dimension of $L$.

Is $cd(L) = n+2$ i.e. does equality hold in the upper equation?

This question is not trivial in general as Serre points out by mentioning results of Merkurjev in Galois Cohomology. Merkurjev constructs fields $k$ with $cd(k)=2$ having any desired even $u(k) \geq 2$.

$\endgroup$
2
  • $\begingroup$ Can you say what $u$ is? $\endgroup$ Oct 23, 2014 at 0:28
  • 1
    $\begingroup$ Its the u-invariant of k. I will include its definition in my first post. $\endgroup$
    – nxir
    Oct 23, 2014 at 0:37

1 Answer 1

2
$\begingroup$

I think the answer to your question can be found in Serre's "Galois cohomology" book, Section II.4.2, where a more general result is proved for transcendental field extensions.

$\endgroup$
1
  • $\begingroup$ Thanks! I cant believe i overlooked it. Edit: And the answer is "Yes"! $\endgroup$
    – nxir
    Apr 20, 2015 at 23:08

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.