TITLE: Is there always exist a polynomial with a unique zero point?
QUESTION [6 upvotes]: Assume K is a field that is not algebraically closed, then how to prove that for any m>0, there is a polynomial with m variables over K that possess a unique zero point?

REPLY [5 votes]: As $K$ is not algebraically closed, there exists $p\in K [X]$ with no root in $K$ and $\deg p\ge1$. Let $f(X,Y)=Y^{\deg p}p\left(\frac XY\right)$, that is, if $\deg p=d$ and
$$p(X)=a_0+a_1X+\ldots +a_dX^d $$
we let
$$ f(X,Y)=a_0Y^d+a_1XY^{d-1}+\ldots +a_dX^d. $$
Then clearly $f(0,0)=0$. If $v\ne 0$ then $f(u,v)=v^dp(\frac uv)\ne 0$ and if $v=0, u\ne 0$ then $f(u,v)=a_du^d\ne 0$. Thus $f(u,v)=0\iff u=v=0$.
For $m\ge 2$ let $f_2=f$ and recursively $f_{m+1}(X_1,\ldots, X_{m+1}) = f(f_m(X_1,\ldots,X_m),X_{m+1})$. Then it follows quickly by induction that for $u_1,\ldots ,u_m\in K$ we have 
$f_m(u_1,\ldots ,u_m)=0\iff u_1=\ldots u_m=0$.