Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

3 Answers 3

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 $f=0$.

share|improve this answer
    
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 $S\subset V(P)$.

share|improve this answer

(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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.