Countability of the zero set of a real polynomial

This is the question from my calculus homework: Is it possible for a polynomial $f\colon\, \mathbb{R}^{n}\to \mathbb{R}$ to have a countable zero-set $f^{-1}(\{0\})$? (By countable I mean countably infinite).

Of course, I claim that it's impossible.

Surely, if the zero $z$ isn't shared with at least one of the partial derivatives, say $\frac{\partial f}{\partial x_{1} }(z)\neq 0$, by Implicit Function Theorem we get (locally) a smooth curve $\{f(x_{1},t)=0\}=\{(x_{1},\gamma(x_{1}) )\}$, so it's surely uncountable. However, it may be the case that all of the real zeros are shared with all the derivatives. My thought is to proceed somehow inductively, but I have no idea how to do it.

Our proffesor gave me a "hint" - to show that locally, the roots of a polynomial (in one variable) vary analytically as a function of coefficents. Is it true? Is there any elementary proof of that fact? I know that they vary continously (as discussed here), but those multiple roots are driving me crazy...

If you were so kind and help me, I would be very grateful. Also, it's my first question, so please forgive any mistakes made.

Thanks in advance.

Solution 1:

The only proof I could find so far does not use analycity. Say $f$ is a polynomial in the variables $x_1,x_2, \ldots ,x_n$. Denote by $Z(f)$ the set of zeroes of $f$, let $p_i$ be the projection on the $i$-th coordinate axis :

$$ p_i(x_1,x_2, \ldots ,x_n)=x_i $$

and let $Z_i=p(Z(f))$. By the Tarski-Seideberg theorem, each $Z_i$ can be defined by a set of univariate polynomial equalities or inequalities. So each $Z_i$ is a finite union of intervals or points of $\mathbb R$. If $Z_i$ contains an interval of positive length, $Z_i$ is uncountable ; otherwise $Z_i$ is finite.

If some $Z_i$ is uncountable, so is $Z(f)$. If all the $Z_i$ are finite, so is $Z(f)$. QED

Here are some examples of how this works : for a univariate polynomial $f$,

-When $f$ has degree $1$, $f$ always has a unique simple root.

-When $f$ has degree $2$, $f$ has two simple roots if ${\sf disc}(f)>0$, one double root if ${\sf disc}(f)=0$, and no root at all if ${\sf disc}(f)<0$.

-When $f$ has degree $3$, $f=a_3x^3+a_2x^2+a_1x+a_0$,

$$ \begin{array}{|l|l|l|} \hline \text{Case} & & \text{Roots of } \ f \\ \hline {\sf disc}(f') < 0 & & \text{one simple root}\\ \hline {\sf disc}(f')= 0 & {\sf disc}(f) =0 & \text{one triple root} \\ \hline {\sf disc}(f')= 0 & {\sf disc}(f) \lt 0 & \text{one simple root} \\ \hline {\sf disc}(f')\gt 0 & G(f)\lt 0 & \text{three simple roots}\\ \hline {\sf disc}(f')\gt 0 & G(f)= 0 & \text{one simple root, one double root} \\ \hline {\sf disc}(f')\gt 0 & G(f)\gt 0 & \text{one simple root} \\ \hline \end{array} $$

where $G(f)=4a_3a_1^3 - a_2^2a_1^2 - 18a_0a_3a_2a_1 + 4a_0a_2^3 + 27a_0^2a_3^2 $