TITLE: Zariski topology, non-empty intersection of open sets
QUESTION [5 upvotes]: Let $k$ be a field and $X=\mathbb k^n$. A subset $Y \subset X$ is closed if there are $f_1,\ldots,f_m \in k[x_1,.\ldots,x_n]$ such that $Y=\{a \in k^n :f_i(a)=0 \space \forall \space i\}$, we write $Y=V(f_1,\ldots,f_m)$. The Zariski topology is $T=\{U \subset X : X \setminus U=V(f_1,\ldots,f_m) \}$. 
I am trying to show that given the Zariski topology on $k^n$, if $U$ and $V$ are non-empty open subsets, then $U \cap V$ is non-empty. I don't know how to prove this so any hints would be greatly appreciated.

REPLY [2 votes]: Let $U,V$ be two open sets of $k^n$ under Zariski topology. Thus they have associated two families of nonzero polynomials $f_1,...f_m;g_1,...,g_l\in k[x_1,...,x_n]$ such that $$f_i(a)=0\,\forall a\not\in U,\forall i\qquad g_j(b)=0\,\forall b\not\in V,\forall j.$$ Note that the intersection $U\cap V$ has associated the products $f_ig_j$, varying $i,j$. Since $k[x_1,...,x_n]$ is an integer domain, all these products are not zero.
Now, suppose that $U\cap V=\varnothing$. This means that all points of $k^n$ are roots of all $f_ig_j$. But this is a contradiction because this would imply $f_ig_j=0$ for all $i,j$.