# A two-variable quadratic form over a field of characteristic 2 with no nontrivial roots

I'm looking for a quadratic form of the form $q(x,y)=ax^2 + bxy + cy^2 \in F[x,y]$, where $F$ has characteristic 2, and $q(x,y)$ has no roots besides the obvious one, $x=y=0$. I've proved the case of finite fields with characteristic $>2$ by group-theoretic arguments, but this case seems like it will require much more subtle techniques. I'm not so much interested in a constructive proof, I mainly want to show the existence of such a quadratic form (I'm reasonably sure that, at least in the case of finite fields, they do exist).

• For a finite field of char. 2 let $x^2 +bx+c$ in $F[x]$ be irreducible. Then $x^2 + bxy + cy^2$ on $F^2$ vanishes only at $(0,0)$. For example, if $F = F_2$ then you can use $b=c=1$. – KCd Feb 4 '16 at 6:43
• Thanks! Do you have any idea of whether or not the statement would be true over infinite fields? – Monstrous Moonshine Feb 4 '16 at 6:56
• The only thing I used about $F$ being finite is that we are guaranteed an irreducible quadratic in $F[x]$ exists. (I'm not even using anything about the characteristic being 2 or not.) Even if $F$ is infinite, as long as $F$ has a quadratic irreducible the same construction works. There are certainly $F$ for which there are no irreducible quadratics, such as algebraically closed fields, which exist in every characteristic. – KCd Feb 4 '16 at 9:35

To the case of infinite fields: If you search for a field, in which you cannot solve some equation, you should always start with a function field, since they are very far from being algebraically closed. So lets take $F=\mathbb F_2(t)$ and the equation

$$x^2+fy^2=0.$$

Cleary if one of $x,y$ is non-zero, both are non-zero, hence after multiplication with $(xy)^{-1}$ we have

$$\frac{x}{y}+f\frac{y}{x}=0,$$

hence we have to solve

$$z+fz^{-1}=0 \Longleftrightarrow z^2+f=0.$$

So you should find some $f$, such that this equation does not admit a solution. Hence, in fact, you need a non-square in $F$. This should be easy.