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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 $u(k) \geq 2$.

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

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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.

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

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