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



Simple field extension and rational points
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 $$f(X,Y,Z)=(X^p-sZ^p)u+(Y^p-tZ^p)v.$$


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
$$x^p-sz^p, y^p-tz^p\in\mathbb F\_p(u^p,v^p,\sigma^p,\tau^p)=\mathbb F\_p(u^p,v^p,s,t),$$
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 $x^p-sz^p=A(u^p,v^p,s,t)$ and $y^p-tz^p=B(u^p,v^p,s,t)$.


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 $x^p-tz^p=0=y^p-tz^p$, and the claim follows.


