The polynomial $x^2+y^2$ has an isolated zero at the origin. And so do powers $(x^2+y^2)^n$ of this polynomial. I'm wondering if this is a special property of these real polynomials.

Here's the precise question. Suppose a real polynomial $f(x,y)$ is not divisible by $x^2+y^2$. Is it possible for $f$ to have an isolated zero at the origin?

It seems to me that such a real polynomial would necessarily be either nonnegative or nonpositive in a small neighbourhood of the origin.

More generally, I'm also interested in the case where $f(x,y)$ is real analytic.

EDIT: I'm really delighted by the really comprehensive and diverse reponses I am getting from everybody. Feels like we are in a coffee shop dicsussing mathematics!

Here is what we have gathered. Our basic example of polynomials with isolated zero at the origin is of the form $x^{2n}+y^{2m}$ for positive integers $n$ and $m$. By a linear change of coordinates, this essentially includes examples like $ax^{2n}+by^{2m}$ for positive real coefficients $a$ and $b$.

From this basic example, we can apply perturbion (thanks to Bruno!). Suppose $n\le m$, then we can perturb our basic example add any real analytic function $g(x,y)$ whose lowest degree terms are of degree strictly larger than $m$. The resulting $f+g$ also has an isolated zero.

Does our discussion exhaust the possibilities? Can we say that an polynomial that has an isolated zero is a perturbation of $x^{2n}+y^{2m}$, up to a change of coordinates? Thank you everyone for the lively discussion. Let's charge on!

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    $\begingroup$ What about x^2 + y^4? $\endgroup$ – Todd Trimble Nov 6 '10 at 3:42
  • $\begingroup$ or, say, x^2+2y^2? $\endgroup$ – Fedor Petrov Nov 6 '10 at 14:55

$x^{2i}+y^{2j}$ isn't usually divisible by $x^2+y^2$.

$x^2(y^2+y+1)+y^2(x^2+x+2)$ is an example showing that odd powers can be used as well.

  • $\begingroup$ The lowest degree terms of the second polynomial are $x^2+2y^2$. From this observation, we can understand your second polynomial as a perturbation of $x^2+2y^2$ following Bruno and Vivek. $\endgroup$ – user2529 Nov 7 '10 at 8:44

The real cubic


has an isolated point at the origin. A picture of this curve can be found here:



There are plenty of such polynomials and/or analytic functions, it suffices to perturb $x^2+y^2$ by adding a polynomial/function $g(x,y)$ whose Taylor expansion in zero has order at least 3. For instance, take $g(x,y)$ a polynomial whose monomials have degree at least 3. As an example, $g(x,y) = x^3$. You only need to care that $g$ is not divisible by $x^2+y^2$.

Concerning you remark, any continuous real function $f$ having an isolated zero on $(0,0)$ is nonnegative or nonpositive in a small neighbourhood of $(0,0)$. Since the zero is isolated, there is a neighborhood $U$ of $(0,0)$ such that $f(U\setminus\{(0,0)\})\subset \mathbb R$ does not contain 0. Since $U\setminus \{(0,0)\}$ is connected, its image $f(U\setminus \{(0,0)\})$ also is, and is therefore an interval in $\mathbb R$. An interval that does not contain zero consists of either only positive or only negative real numbers.

  • $\begingroup$ I think you mean "only positive or only negative" in the last sentence, rather than "all positive or all negative". $\endgroup$ – Charles Staats Nov 6 '10 at 16:19
  • $\begingroup$ Yes, thanks, I have edited as you suggested. $\endgroup$ – Bruno Martelli Nov 6 '10 at 17:47

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