Complements of real polynomial curves

Question

Working on a problem on matroids I found this small problem to solve. I think it should be easy but I can't solve it.

Let $F$ be a non zero polynomial in $\mathbb{R}[x_{1},\ldots,x_{n}].$ Is it true (and why? maybe with a bibliographic reference?) that $\mathbb{R}^{n}\setminus V(F)$ is non empty? Here $V(F)=\{F=0\}.$

Thanks a lot to everyone will answer me.

Answer 1

We will show by induction on $n\in\mathbf{N}^{*}$ that if $V(F) = \mathbf{R}^n$ then $F = 0$.

• If $n = 1$ this is trivial, as a non zero polynomial as finitely many roots.
• Suppose this is true for some $n$ and let $F\in \mathbf{R}[x_{1},\ldots,x_{n},x_{n+1}]$ such that $V(F) = \mathbf{R}^{n+1}$. Then write $F = \sum_{k=0}^d F_i (x_1,\ldots,x_n) x_{n+1}^k$ where the $F_i \in \mathbf{R}[x_{1},\ldots,x_{n}]$. Fix $z = (z_1,\ldots,z_n)\in \mathbf{R}^n$, as $V(F) = \mathbf{R}^{n+1}$, for all $y\in\mathbf{R}$ you have $F(z,y) = \sum_{k=0}^d F_i (z_1,\ldots,z_n) y^k = 0$. The case $n=1$ implies that all the $F_i (z_1,\ldots,z_n)$ are zero, and this for all $z = (z_1,\ldots,z_n)\in \mathbf{R}^n$. By induction hypothesis you have that all the $F_i$'s are the zero polynomial, which implies that $F = 0$. $\square$

We have shown that $V(F) = \mathbf{R}^n$ then $F = 0$, which is the contraposed of the assertion you want to prove.

Question. How would/could you extend this to other fields than $\mathbf{R}$ ?