3
$\begingroup$

I would like to know, if there is a 4 dimensional anisotropic quadratic form over the 2-adic Integers $\mathbb{Z}_2$, that satisfies the following property:

It is in diagonal form and 2 does not divide a,b,c and d.

My guess: I think that the problem is equivalent to the problem over $\mathbb{Q}_2$. If you have a form with diagonal entries in $\mathbb{Q}_2\setminus \mathbb{Z}_2$, you can scale the form with a suitable power of $2$ to get a form over $\mathbb{Z}_2$. Furthermore I know that the Witt Ring of $\mathbb{Q}_2$ has 32 elements, but I do not know the representatives. Does anyone know the answer or good literature? The books I found exclude the case $p=2$.

Best regards Laura

$\endgroup$
  • 1
    $\begingroup$ Won't $Q=x_1^2+x_2^2+x_3^2+x_4^2$ work? By the same argument you can assume that all the $x_i$ are 2-adic integers and at least one of them is a 2-adic unit. Then modulo 8 brute force check shows that $Q$ has no non-trivial zeros. Essentially because the square of any 2-adic unit is $\equiv 1\pmod 8$. $\endgroup$ – Jyrki Lahtonen Dec 14 '15 at 19:08
  • $\begingroup$ Thank you, that one is a perfect example. I made an mistake, so I rejected it in the first place. $\endgroup$ – Laura Dec 14 '15 at 23:52
1
$\begingroup$

The form $$ Q(x_1,x_2,x_3,x_4)=x_1^2+x_2^2+x_3^2+x_4^2 $$ has no non-trivial zeros with $x_i\in\Bbb{Q}_2, i=1,2,3,4$. This is seen as follows. Because $Q$ is homogeneous w.l.o.g. we can assume that all the $x_i$:s are 2-adic integers, and at least one of them is a 2-adic unit. Permuting and scaling the variables with the inverse of that unit we can further assume that $x_1=1$. Reducing $Q$ and the variables modulo 8 leaves us a brute force check that $$ 1+x_2^2+x_3^2+x_4^2\neq0_{\Bbb{Z}/8\Bbb{Z}} $$ for all combinations of $x_2,x_3,x_4\in\Bbb{Z}/8\Bbb{Z}$. The key point is that all odd squares are $\equiv1\pmod 8$, and the even squares are $\equiv0,4\pmod 8$. The claim follows.


I think it is worth pointing out that the same $2$-adic scaling and modulo $8$ calculation gives the easy direction of the three squares theorem.

$\endgroup$
  • 1
    $\begingroup$ I think I got it. Thank you. $\endgroup$ – Laura Dec 15 '15 at 22:24

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.