A polynomial that is zero on an open set

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. . .

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.

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.