TITLE: Zariski topology in the complex plane: an example
QUESTION [9 upvotes]: I want to find the closure under the zariski topology, of this set $
\left\{ {\left( {x,y} \right) \in {\Bbb C}^2 ;\left| x \right| + \left| y \right| = 1} \right\}
$
I have no idea what I can do

REPLY [2 votes]: Let $S$ be your set, and let $f\in\mathbb C[x,y]$ a polynomial vanishing on $S$. 
We prove $f=0$ as follows.
Let $(a,b)$ be a nonzero vector in $\mathbb C^2$. 
The restriction of $f$ to the line 
$$
L:=\{(ta,tb)\ |\ t\in\mathbb C\}
$$
is a polynomial in $t$ vanishing on $S\cap L$, which is the set of those $(ta,tb)$ with 
$$
|t|=\frac{1}{|a|+|b|}\quad.
$$
This set being infinite, $f$ vanishes on $L$. As $\mathbb C^2$ is covered by such lines, we have indeed $f=0$.
Variation: $f(ru,(1-r)v)$ is a polynomial in $r,u,v$ vanishing on 
$$
0\le r\le1,\quad |u|=1=|v|.
$$
From this it is easy to deduce $f=0$.