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

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    $\begingroup$ 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

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.

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

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