There exists a non-empty open set $U ⊆ \Bbb R^2$ such that $f(x, y) = 0$ for every $(x, y) ∈ U$. Show that $f = 0$, i.e. $f$ is identically zero. Let $f ∈ \Bbb R[x, y]$ be such that there exists a non-empty open set $U ⊆ \Bbb R^2$ such that $f(x, y) = 0$ for every $(x, y) ∈ U$. Show that $f = 0$, i.e. $f$ is identically zero.
My try: Since $f ∈ \Bbb R[x, y]$, where $\Bbb R[x,y]$ is the polynomial ring in two variables, $f$ can be expressed as $a_0x^n+a_1x^{n-1}y+...+a_ny^n$ for some $n$ and $a_0,a_1,...,a_n$.
If we can show that each of the coefficients $a_0,a_1,...,a_n$ are zero then we are done and for that we have to find out $n^{th}$ order partial derivatives at $(0,0)$.
Partial derivatives of the form $$\frac{\delta^n}{\delta x^i \delta y^{n-i}} f$$
I can't figure out what to do next. Help Needed!!
Also other methods of solution greatly accepted.
 A: Let $(x_0,y_0)$ be an inner point of this set $U$.
Your function can be written as $$f(x,y) = \sum_{i=0}^n\sum_{j=0}^m f_{ij} (x-x_0)^i (y-y_0)^j$$ with constants $f_{ij}$.
Now, obviously, $$\frac{\partial^{s+p}f }{\partial x^s\partial y^p}(x_0,y_0)=0$$  by the definition of the set $U$. On the other hand, by a direct calculations you can show that $$\frac{\partial^{s+p}f }{\partial x^s\partial y^p}(x_0,y_0)=s!p!f_{sp}$$under the hypothesis that $s\le n$, $p\le m$.
This implies that all coefficients $f_{ij}$ are zero and hence $f$ is identically zero.
A: An algebraic approach. Because $U$ is open, there is a rectangle $[x_0,x_1]\times [y_0,y_1]\subseteq U$. Think of $f$ as an element of $(\mathbb{R}[x])[y]$, so we have $f_0,\dotsc,f_n\in \mathbb{R}[X]$ with
$$f(x,y)=f_x(y)=\sum_{i=1}^n f_i(x)y^i$$
For every $x=[x_0,x_1]$. We know that $f_{x}=0\in \mathbb{R}[y]$ because there are infinte many zeros namely $[y_0,y_1]$. So $f_i(x)=0$ for each $x=[x_0,x_1]$. So  $f_i\in \mathbb{R}[x]$ has infinete many zeros namely $[x_0,x_1]$ and thus $f_i=0\in \mathbb{R}[x]$. So $f=0$.
A: My initial instinct is that if $f$ is zero on an open subset of the domain, $f$ is zero on a lot of points, infinitely many in fact. You can use this to your advantage with Lagrange Interpolation. It takes only a finite number of points to determine a polynomial function of a certain degree. Call this degree $d$. Call the number of zero points needed to determine this polynomial $n$. Then pick another zero point (so that we now have $n+1$ points) at which our interpolated polynomial is nonzero. Having to incorporate this point now make the function identically zero. The trick here is to ensure that you are able to pick the right points, but I believe something along these lines could work for even some non-open sets.
Utilizing the open set specifically though, you will need the fact that if a function of one variable is zero on an interval, then the function is zero. Since $U$ is an open set, there is an open disc within $U$, call it $D$. Take any $x$ coordinate that is in $D$ and restrict $f$ to just this coordinate. This restricted $f$ is zero on an interval, so it must also be zero. This works for any $x$ coordinate in $D$, so that now for every $y$, we have an interval at which we know $f$ restricted to that $y$ is zero. The rest of the proof then follows.
