Let $F$ be an infinite field and $f$ a homogeneous form on $F$ such that $f$ has no nontrivial zero in $F$. Let $F'$ be a finite extension of $F$ such that $f$ has a nontrivial zero in $F'$. Is it true there exists a simple extension of $F$ of the form $F(\alpha)$ contained in $F'$ which contains a nontrivial zero of $f$?

$\begingroup$ Isn't every finite separable extension $F'$ of $F$ of the form $F'=F(\alpha)$ for some $\alpha\in F'$ ? $\endgroup$ – Chandan Singh Dalawat Mar 31 '13 at 6:15

$\begingroup$ @Dalawat: Ofcourse but in our case $F'$ is not necessarily separable. $\endgroup$ – Jana Mar 31 '13 at 7:01

$\begingroup$ $x^p+sy^p+tz^p \in \mathbb{F}_p(s,t)[x,y,z]$? $\endgroup$ – M P Mar 31 '13 at 11:05

$\begingroup$ @MP: if I've understood correctly then adjoining a $p$th root $\tau$ of $t$ would be enough to give you the point $[\tau:0:1]$ so this doesn't look like a counterexample to me unless I've misunderstood what you're suggesting. $\endgroup$ – user30035 Mar 31 '13 at 15:03

$\begingroup$ @wccanard: you have interpreted my comment correctly, and I was wrong! $\endgroup$ – M P Mar 31 '13 at 16:52
I believe the following is a negative example: Let $s,t,u,v$ be variables over $\mathbb F_p$. Set $F=\mathbb F_p(s,t,u,v)$ and $F'=F(\sigma,\tau)$ with $\sigma^p=s$, $\tau^p=t$.
Set $$f(X,Y,Z)=(X^psZ^p)u+(Y^ptZ^p)v.$$
Then $f(\sigma,\tau,1)=0$.
We show that any solution of $f=0$ over $F'$ has this form up to a scalar factor: Let $x,y,z\in F'$ with $f(x,y,z)=0$. As $F'=\mathbb F_p(u,v,\sigma,\tau)$, we get $$x^psz^p, y^ptz^p\in\mathbb F_p(u^p,v^p,\sigma^p,\tau^p)=\mathbb F_p(u^p,v^p,s,t),$$ hence $$A(u^p,v^p,s,t)u+B(u^p,v^p,s,t)v=0$$ for rational functions $A,B$ over $\mathbb F_p$ with $x^psz^p=A(u^p,v^p,s,t)$ and $y^ptz^p=B(u^p,v^p,s,t)$.
This implies $A(u^p,v^p,s,t)=0$, for otherwise $u$ were a rational function in $u^p$. For the same reason $B(u^p,v^p,s,t)=0$. We get $x^ptz^p=0=y^ptz^p$, and the claim follows.

$\begingroup$ May I ask how one sees that $x^psz^p$ equals $A(u^p, v^p,\sigma,\tau)$ for some $A$ over $\mathbf F_p$? $\endgroup$ – Ariyan Javanpeykar Mar 31 '13 at 23:35

$\begingroup$ I extended the answer, hope it is clearer now. $\endgroup$ – Peter Mueller Apr 1 '13 at 8:12

$\begingroup$ This is very nice, I must say! If I understand correctly, you show that for the polynomial $f$ we must add at least $two$ elements to $F$ to get a root of $f$ inside $F^\prime$. I think, but I might be wrong, that one can slightly alter your construction (by adding more variables, and writing down $F$, $f$ and $F^\prime$ analogous to your construction) to obtain examples where one must add at least $n$ (with $n$ fixed, but arbitrarily large) variables to $F$ to obtain a root of $f$ within $F^\prime$. $\endgroup$ – Ariyan Javanpeykar Apr 1 '13 at 17:21