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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

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up vote 2 down vote accepted

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 $f=0$.

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I proved in a different way, thanks for all! All the answers was useful – Arkj Mar 28 '12 at 2:28

Your set $S$ has closure $\bar S=\mathbb C^2$.
To prove that it is enough to prove that no non-zero polynomial $P(x,y)$ vanishes on $S$.
Here is a proof using real manifolds:

The set $S$ is a real submanifold of dimension $3$ of $\mathbb C^2=\mathbb R^4$ since its equation is $f(x,y)=(x_1^2+x_2^2)^{1/2}+(y_1^2+y_2^2)^{1/2}=1$ and $f$ has non-zero gradient on $S$.
But except at finitely many points $V(P)$, the variety defined by $P$, is smooth and thus a real manifold of dimension $2$, so that you can't have $S\subset V(P)$.

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(expanding my comment to an answer)

Let $S$ be your set $S=\{(x,y)\in\mathbb{C}^2\mid |x|+|y|=1\}$. If we can show that the only bivariate polynomial that vanishes on all the points of $S$ is the constant zero, then the ideal $\cal{I}(S)$ consists of the zero polynomial alone, and hence $\overline{S}=\cal{V}(\cal{I}(S))=\mathbb{C}^2$.

Let $f(x,y)$ be a polynomial that vanishes at all the point of $S$. Then all the complex numbers $x$ on the circle $|x|=1/2$ must be zeros of the univariate polynomial $g(x):=f(x,1/2)$. There are infinitely many numbers $x$ with this property, so we must have $g(x)=0$. Therefore $y-\frac12$ must be a factor of $f(x,y)$. But we can repeat the same argument for any complex number $y_0$ such that $|y_0|<1$. So all such polynomials $y-y_0$ must be factors of $f(x,y)$. But the polynomial $f(x,y)$ has a finite degree, so this is impossible by unique factorization.

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