TITLE: 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.
QUESTION [1 upvotes]: 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.

REPLY [1 votes]: 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 $f=0$.
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