3

Let $F$ be a simple transcendental extension of the field with three elements. How can I find $a, b \in F$ such that the quadratic form $x^2-ay^2-bz^2$ is anisotropic?

Rupert
  • 928
  • Not very much at this stage. Should I just try a few examples and try to work out whether the quadratic form is anisotropic or not and see if I can get some intuition for the problem that way? – Rupert Sep 07 '16 at 04:31
  • Probably not a bad idea. Also this might not help, but maybe if you phrase the question as finding out for which $a,b$ does $x^2−ay^2−bz^2$ have roots over $\mathbb{F}_3(\alpha)$, you might be able to entice an arithmetic geometry type to answer? – A. S. Sep 07 '16 at 05:09

1 Answers1

1

Let $k$ be any finite field, or even any field with characteristic different from 2 where not every element is a square. Let $F = k(t)$, so that $F$ is a "simple transcendental extension of $k$".

Let $\langle 1, -a, -b \rangle$ stand for the quadratic form $x^2 - ay^2 - bz^2$. Let $a \in k$ where $a$ is not a square in $k$ and let $b = t$. Then $\langle 1, - a, -t\rangle$ is anisotropic over $F$.

To show this, suppose that $f(t)^2 - ag(t)^2 -th(t)^2 = 0$, where $f,g,h \in F$, not all zero. We can multiply $f,g,h$ by a common denominator so that we can assume from the start that $f,g,h \in k[t]$. Now look at the highest power of $t$ that appears. Because $a$ is not a square in $k$, it follows that $f(t)^2 - ag(t)^2$ has even degree and it is clear that $th(t)^2$ has odd degree. This is impossible, and so there can be no solution unless $f = g = h=0$.

There are many other ways to construct a $3$-dimensional anisotropic form.

David Leep
  • 196