# 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.

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What does analytically mean here? There is no neighborhood of the origin on which the square root function is analytic, and yet $x^2 - \epsilon$ is coefficientwise close to $x^2$. I think the claim is true locally away from multiple roots. –  user29743 Apr 23 '12 at 21:02
Yes - you're right. I'm sorry for the latter - I think I meant the complexification. However, I don't know very much about complex analysis, so I might have mistaken something. –  L_b Apr 23 '12 at 21:09
I think you can complete your proof using an inductive argument on the degree. Roots of $f$ fall into two categories : those who are also roots of $\frac{\partial f}{\partial x_1}$, and those who aren't. Because of our inductive hypothesis, the set of roots of the first kind is either finite or uncountable. And the Implicit Function Theorem shows that the set of roots of the second kind is either empty or uncountable (your argument above). Which concludes. –  Joel Cohen Apr 23 '12 at 21:35
@L_b: They vary analytically in a very obvious sense: take the zero locus $V$ of $x_0 + x_1 y + \cdots + x_n y^n = 0$ in $\mathbb{C}^{n+2}$. This is clearly a complex algebraic variety, and there is a dense open subset which is even a complex analytic manifold. There is an evident projection $V \to \mathbb{C}^{n+1}$ (forget $y$), and away from the branching locus, this will be a local analytic isomorphism. –  Zhen Lin Apr 23 '12 at 21:39
@Joel Cohen: Well.. really? I know that there is only finitely many roots of the first category, which are of the first category for the derivative. Hence, I've got smooth curves for the zeros of the derivative. Although this vanishes, I don't see any guarantee that my polynomial would vanish on theese curves as well. Therefore, I know something about theese roots, but why, still, I cannot have countably many roots of the first category, which are of the second for the derivative? (I know it's quite complicated, but I hope you see my point) –  L_b Apr 23 '12 at 21:41

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

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Nice! I didn't know this theorem, and that's why I was stuck. I think in the definition of $Z_i$, it's p_{\color{red}i}(Z(f))$. – Davide Giraudo Oct 1 '12 at 11:19 @Davide : By the way, I just thought of a different method : it suffices to show the result when$f$is irreducible, and then Bezout’s theorem (the one from algebraic geometry) ensures that$f$has at most$n(n-1)^n$critical points, where$n$is the number of variables. But Bezout’s theorem is fairly advanced and nontrivial, just like the Tarski-Seidenberg theorem. I wonder if there is a more “elementary” proof. – Ewan Delanoy Oct 1 '12 at 12:16 Thank you! It happens I've started studying algebraic geometry; and it seems a splendid application of Bezout theorem. I hope I'll learn more (BTW: would you recommend some books for starters?) to invent such solutions myself... (though Tarski-Seidenberg seems still mysterious to me). Anyway, congatulations, and thanks once more. (Of course, it's still an interesting problem whether one can find an elementary solution) – L_b Jan 9 '13 at 0:03 Assume$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)$can't be the null function, cause we assumed that the leading coefficient is not zero). Contradiction. - Correct, but this is only for polynomials of one variable. In the one-variable case one can even avoid any calculus and just use the fact that the quotient$p(z)/(z-z_0)$is a polynomial of degree$\mathrm{deg }\,p-1$. – user31373 Sep 26 '12 at 23:02 Oh c**p, I didn't see that it was in dimension$n\$. My bad. =) –  bartgol Sep 26 '12 at 23:41