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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 $f$?

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Isn't every finite separable extension $F'$ of $F$ of the form $F'=F(\alpha)$ for some $\alpha\in F'$ ? –  Chandan Singh Dalawat Mar 31 '13 at 6:15
    
@Dalawat: Ofcourse but in our case $F'$ is not necessarily separable. –  Jana Mar 31 '13 at 7:01
    
$x^p+sy^p+tz^p \in \mathbb{F}_p(s,t)[x,y,z]$? –  M P Mar 31 '13 at 11:05
    
@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. –  user30035 Mar 31 '13 at 15:03
    
@wccanard: you have interpreted my comment correctly, and I was wrong! –  M P Mar 31 '13 at 16:52
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1 Answer

up vote 5 down vote accepted

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.

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May I ask how one sees that $x^p-sz^p$ equals $A(u^p, v^p,\sigma,\tau)$ for some $A$ over $\mathbf F_p$? –  Ari Mar 31 '13 at 23:35
    
I extended the answer, hope it is clearer now. –  Peter Mueller Apr 1 '13 at 8:12
    
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 $F^\prime$. –  Ari Apr 1 '13 at 17:21
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