5
$\begingroup$

Im sure this is a beginners question.

Let $k$ be a field and $I(k)$ the fundamental ideal in the Witt-ring W(k). The Arason-Pfister-Hauptsatz states:

"If $\varphi$ is any anisotropic class in $I^n(k)$, then $rank(\varphi) \geq 2^n$."

It is well known that for the kernel of the discriminant $ker(e_1) = I^2$ holds.

Since there is a filtration $W(k) \supseteq I(k) \supseteq I^2(k) \supseteq..$ , the set of classes of elements in $I(k)$ with non trivial discriminant is the complement $I(k)\backslash I^2(k)$. Since the Arason-Pfister-Hauptsatz only gives a lower bound on the rank of elements in $I(k)$ it might be possible that there is a class $\varphi \in I(k)\backslash I^2(k)$ with $rank(\varphi) = 4$ or even higher.

Do you have an example such that this happens?

In "Galois Cohomology" Serre referres to some paper of Alexander Merkurjev, in which he shows that it is possible for "every" even $N\geq1$ to find a $k$ having $u$-invariant $u(k)=N$ ,such that $I^3(k)$ vanishes i.e. $k$ has cohomological depth $cd(k)=2$. This is basically the answer to an analog of my question in case of finding some anisotropic rank $8$ class in $I^2(k)\backslash I^3(k)$.

$\endgroup$

1 Answer 1

3
$\begingroup$

The answer depends on your field $k$. For example if $k$ is the $p$-adic field $\mathbb{Q}_p$, $p\neq 2$, it is known that the only anisotropic form of dimension $4$ over $k$ is isomorphic to the norm form of the unique quaternion algebra over $k$ which is of course in $I^2$ since it is a Pfister form. By contrast you can construct many fields $k$ with your desired property, for instance consider the field $k=\mathbb{Q}(x,y,z,t)$ and the form $\varphi=\langle x,y,z,t\rangle\in I\backslash I^2$.

Added: regarding your question in the comments about $I^3$:

By a theorem of Merkurjev, the Hasse-Witt invariant induces an isomorphism between $I^2/I^3$ and the two torsion part of the Brauer group $Br_2(k)$. Hence it suffices to give an example of a field $k$ and an anisotropic form $\varphi\in I^2$ of dimension $8$ such that the Hasse-Witt invariant of $\varphi$ is nontrivial. For this consider an anisotropic Albert form, e.g., the form $\alpha=\langle x,y,-xy,-z,-t,zt\rangle$ over the transcendental field $\mathbb{Q}(x,y,z,t)$. Now consider the transcendental field $k=\mathbb{Q}(x,y,z,t,u,v)$ and the eight dimensional anisotropic form $\varphi=-uv\cdot\alpha\perp\langle u,v\rangle\in I^2$. The Hasse-Witt invariant of $\varphi$ is isomorphic to the Clifford algebra $Cl(\varphi)\simeq C(\langle u, v\rangle)\otimes Cl(\alpha)$ which is not split.

$\endgroup$
2
  • $\begingroup$ Thanks for this easy understandable example. Lets raise the bar a bit. If we consider the Arason-Invariant with Kernel $I^3$, is the same thing possible the?. Meaning is there a field $k$ such that theres an anisotropic form over k with rank at least $8$,which is not in $I^3$? What is known about the generalization of this problem to higher $n$ for $I^n$ ? $\endgroup$
    – nxir
    Jun 18, 2014 at 2:39
  • $\begingroup$ This is basically a concrete example for the result of Merkurjev on the u-invariant and galois cohomology i mentioned in my first post. Thanks. I i guess that finding such examples for every power of $I$ is harder,since the invariants $e_n$ which give the isomorphism between $I^n/I^{n+1}$ and $H^n(k,\mu_2)$ are not as intuitive as the classical three invariants. $\endgroup$
    – nxir
    Jun 18, 2014 at 17:28

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.