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.
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 ${\sf disc}(f)&gt;0$, one double root if ${\sf disc}(f)=0$, and no root at all if ${\sf disc}(f)&lt;0$. 

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

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

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
$
Assume $p(x)$ is a polynomial of degree $n\geq 1$ (with leading coefficient not zero, that is, $p$ is not also a polynomial of degree $n-1$) and assume that the zero set of $p$ is not finite (possibly even uncountable).

First: it is obvious that $p'(x)$ has degree $n-1$, $p''(x)$ has degree $n-2$ and so on.

Second: take a countable subset of the zero set of $p$ and label its elements as $z_0,z_1,z_2,\ldots$ such that $z_i\leq z_{i+1}$. Then, by Rolle's thm, it follows that in each of the intervals $(z_i,z_{i+1})$ there is at least one zero of $p'(x)$. Hence also $p'(x)$ has a not finite zero set.

Third (and last): iterating the previous step $n-1$ times, we obtain that $p^{(n-1)}(x)$ also has a not-finite zero set. However, $p^{(n-1)}(x)$ is a linear function, with exactly one zero (notice that $p^{(n-1)}(x)$ can't be the null function, cause we assumed that the leading coefficient is not zero). Contradiction.