# Is a polynomial that vanishes a nonempty open subset of $\mathbb{K}^n$, $\mathbb{K} \subseteq \mathbb{C}$, necessarily zero?

Let be $\mathbb K$ a subfield of $\mathbb C$ and consider $\mathbb K^n$ with the Euclidean topology.

If $p \in \mathbb K[x_{1},...,x_{n}]$ vanishes on a nonempty open subset on $\mathbb K^n$, is it the null polynomial?

I know that it is true for $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$, but I don't now if it is true in the general case.

• I edited the question to specify that the open subset ought to be nonempty, which I assume you wanted anyway. – Travis Willse Feb 28 '15 at 14:52

Without loss of generality assume that $\mathbb{Q} = \mathbb{K}$. Your nonempty open set will contain a set of the form $U=((a_1,b_1) \times \cdots \times (a_n, b_n)) \cap \mathbb{Q}^n$ where $a_i<b_i$. We want to show that $U$ lies dense in $\mathbb{C}^n$ with respect to the Zariski topology. The closure of $U$ with respect to the Euclidean topology is $U'=([a_1,b_1] \times \cdots \times [a_n, b_n]) \cap \mathbb{R}^n$ which contains an open subset of $\mathbb{R}^n$. Since the Zariski closure of $U$ contains $U'$ we see that a polynomial vanishing on $U$ has to vanish on $U'$. Since you know that the claim is true for $\mathbb{K}=\mathbb{R}$ we are done.