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?
 A: 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.  
A: 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.
A: 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 $q_i=0$. 
A: 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. 
A: 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. . .
