# A polynomial that is zero on an open set

Suppose that a polynomial $p(x,y)$ defined on $\mathbb{R}^2$ is identically zero on some open ball (in the Euclidean topology). How does one go about proving that this must be the zero polynomial?

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Do you know how to show that a polynomial $p(x)$ which is constant on an open interval must be a constant polynomial? – Matthew Conroy May 4 '11 at 16:24
Implicit function theorem would be helpful. If we denote the zero set of a nonconstant polynomial $p(x, y)$ as $\Gamma$, then $\Gamma$ is locally given by the graph of a smooth function. This in particular implies that $\Gamma$ has measure zero, so $\Gamma$ cannot contain any set of positive measure. – Sangchul Lee May 4 '11 at 16:35
$n$th degree poly has at most $n$ zeros in 1 variable. in 2 variables $x,y$ fix $y$ to see that the poly has at most $n$ zeros on any horizontal line (so the zero set cannot be open). come up with some induction argument – yoyo May 4 '11 at 18:44
In two words: Taylor's formula. – PseudoNeo May 4 '11 at 21:25

WLOG suppose that the center of ball is the origin and write

$$p(x,y)=\sum _{i,j=0}^ma_{i,j}x^iy^j$$

Plug in $x=y=0$. You find that $a_{0,0}=0$. Take the partial derivative with respect to $x$ and set $x=y=0$. You find that $a_{1,0}=0$. You should be able to finish it from here by continuining similarly. . .

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To make notation simpler, let $(a,b)$ be the center of the open ball. Let $g(x,y)=f(a+x,b+y)$. Then the polynomial $g$ is identically $0$ in an open ball containing the origin. We show that $g(x,y)$ is identically $0$.

Consider any line through the origin. We will show that $g(x,y)=0$ at all points on that line. The lines are given by $y=kx$ where $k$ is a constant, and, easily forgotten, $x=0$.

Let $P(t)=g(t,kt)$ (for the line $x=0$, let $P(t)=g(0,t)$).

Then $P(t)$ is a polynomial, and is identically $0$ in an interval. In particular, $P(t)=0$ for infinitely many $t$. Thus $P(t)$ must be identically $0$ (a non-zero polynomial has only finitely many roots).

We conclude that $g$ is identically $0$ on every line through the origin, and hence everywhere.

Note that essentially the same argument works for polynomials in $n$ variables.

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This follows purely algebraically by induction on degree using the fact that a polynomial has no more roots than its degree over a domain - see my prior post.

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This is not clear to me: Consider for example the polynomial $p(x,y) = xy$. Then this has degree one but $|\mathbb R|$ many roots. What am I missing? – Rudy the Reindeer Mar 7 at 7:05

Write $p(x,y) = \sum_{i=0}^n q_i(x)y^i$, where $q_i(x)$ are polynomials in $x$.

Pick a point $(x_0,y_0)$ interior to $U$, where $U$ is your open ball. Then there exists a $r>0$ so that $B_r(x_0,y_0) \subset U$.

Pick any $x_1 \in (x_0-r,x_0+r)$, and chose some $\delta>0$ so that $\{x_1\} \times (y_0-\delta ,y_0+\delta) \subset U$.

Then $p(x_1,y)= \sum_{i=0}^n q_i(x_1)y^i$ is a polynomial in $y$ with constant coefficients $q_i(x_1)$ which is identically zero on on the interval $(y_0-\delta, y_0+\delta)$. Thus $p(x_1,y)$ is the zero polynomial.

Hence $q_i(x_1)=0$. But since $x_1$ was arbitrary in $(x_0-r,x_0+r)$, each $q_i$ has infinitelly many roots, thus each $q_i=0$.

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A stronger, but still reasonably easy to prove, statement follows from the combinatorial Nullstellensatz. It's actually enough to require that $p$ is identically zero on a lattice with sufficiently many points.

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