Projective root of homogeneous polynomials

Question

Let $f(x,y,z)$ be a homogeneous polynomial over a finite field $F_p$ such that $f$ has only $1$ projective root (i.e. $f(x,y,z) = 0$ if and only if $x = y = z$). Then I guess that the degree of $f$ should be at least $p$. Is there any one with any idea about this problem? Thank you!

Answer 1

I'm afraid your hunch fails this time. A quadratic homogeneous polynomial with this property always exists. I interpreted the question to be about $[x:y:z]\in\Bbb{P}^2(\Bbb{F}_p)$, i.e. we are only interested in those zeros where all the coordinates are in the prime field. After all, otherwise there will be infinitely many solutions with parameters ranging over an algebraic closure.

Consider the extension field $\Bbb{F}_{p^2}$. Let $N:\Bbb{F}_{p^2}\to\Bbb{F}_p$ be the norm map. We know that $N(w)=0$ if and only if $w=0$. If we fix a basis $\{1,\alpha\}$ for this extension then it follows that the bivariate polynomial $$ Q(x,y)=N(x+y\alpha) $$ is quadratic homogeneous and vanishes only when $x=y=0$.

But, this means that the quadratic polynomial $$ R(x,y,z):=Q(x-z,y-z) $$ vanishes if and only if $x=z=y$.

• Whenever $p\equiv-1\pmod4$ we can use $\alpha=\sqrt{-1}$ and $N(x+\alpha y)=x^2+y^2$, so $$ R(x,y,z)=(x-z)^2+(y-z)^2 $$ will work.
• With $p=5$ we can use a quadratic non-residue $2$ as we did $-1$ above. Hence the polynomial $$ R(x,y,z)=(x-z)^2-2(y-z)^2 $$ vanishes only when $x=y=z$.

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Using norm functions from cubic extension fields the same recipe allows us to construct homogeneous cubics in four variables that vanish only when $x_1=x_2=x_3=x_4$. It is hopefully clear how to generalize to larger numbers of variables.