# Isotropic forms in field of p-adic numbers

• Nov 22nd 2012, 05:41 AM
AlexanderW
Isotropic forms in field of p-adic numbers
Hi,

for which primes $p$ are the quadratic forms

$7x^2-y^2-5z^2$

and

$w^2+x^2+3y^2+11z^2$

isotropic over the field over $p$-adic numbers $\mathbb Q_p$ ?

A abbreviated form for the two quadratic forms above is $\left \langle 7,-1,5 \right \rangle$ and $\left \langle 1,1,3,11 \right \rangle$.

Define $\varepsilon (\left \langle a_1,...,a_n \right \rangle):=\prod_{1\leq i< j\leq n} (a_i,a_j) \in \left \{ \pm 1 \right \}$.

I can use the fact that a regular quadratic form $\phi$ over a field $F$ is isotropic if $dim(\phi)=3$ and
$\varepsilon(\phi)=(-1,-d)$.

[ $(. , .)$ denotes the Hilbert symbol and $d:=disc(\phi)$ the discriminant of $\phi$.]

Moreover, a regular quadratic form $\phi$ over a field $F$ is isotropic if $dim(\phi)=4$ and
$d \neq 1$ or { $d=1$ and $\varepsilon(\phi)=(-1,-1)$}.

$d \neq 1$ means actually that the residue classes of $d$ and $-1$ are not equal modulo $F^*^2$.

Let's come to the first form:
$\varepsilon (\left \langle 7,-1,-5 \right \rangle)=(7,-1)(7,-5)(-1,-5)$

$d=disc(\left \langle 7,-1,-5 \right \rangle)=35$, therefore $-d=-35$.

Now I have to check for which primes $p$
$(7,-1)(7,-5)(-1,-5) = (-1,-35)$ in $\mathbb Q_p$.

But I don't know how to solve it.

Thanks!

Regards

Alexander
• Nov 28th 2012, 09:48 AM
AlexanderW
Re: Isotropic forms in field of p-adic numbers
Hello,

this problem is still unsolved.
If you have any hints or ideas, it would be great if you would answer.

Bye,
Alexander