topology problem(homemorph)

sam-n

#1

topology problem(homemorph)

**Posted:** Thu Mar 17, 2005 3:00 pm

sam-n

#2

I'm beginner in topology and i found it really exciting but it seems that you're not Mr. Green

**Posted:** Thu Mar 24, 2005 8:24 pm

the game

#3

What is Zariski topology.I think I have heard about it before but I cant remember it right now Blush

.Can you tell me about it.

**Posted:** Thu Mar 24, 2005 8:46 pm

grobber

#4

The problem is I've heard of more than one construction being called the Zariski topology.

First of all, there's this. Another one is number $4$ on this page. They are not the same in $\mathbb R^2$ (although they are the same in $\mathbb R$ ). So, sam-n, could you please give the exact definition of what you're referring to as being the Zariski topology?

**Posted:** Thu Mar 24, 2005 9:54 pm

sam-n

#5

i mean by Zariski topology where a base open set is complement of zero sets of polynomials in two variables

**Posted:** Fri Mar 25, 2005 8:30 am

grobber

#6

Assume they are homeomorphic.

$\mathcal T_2$ has a subbasis of closed sets formed by sets (the straight lines) with the property that each two intersect in at most one point (this means that each closed set can be written as a finite union of arbitrary intersections of lines). The same must then be true for $\mathcal T_1$ .

Consider the curve determined by $xy=1$ (a rectangular hyperbola). This must be a finite union of intersections of sets in ur subbasis for $\mathcal T_1$ . Since it's a finite union, it means that one of these intersections contains infinitely many points $(x,y)$ with the property that $xy=1$ . Since this set is the zero set of a polynomial $P(x,y)=\sum_{i,j=0}^{m,n}a_{ij}x^iy^j$ , it means that $\sum_{i,j=0}a_{ij}x^{i-j}$ is identically null (we get a polynomial with infinitely many roots out of that if we multiply it with a certain power of $x$ ), which, in turn, means that $P(x,y)=0,\ \forall xy=1$ . This means that our set, which is an intersection of elements in the subbasis, contains the set $S=\{(x,y)|xy=1\}$ , so there is at least one set in our subbasis which has $S$ as a subset.

We can do the same for another conic whose intersection with $S$ has at least two points, and we have a contradiction: we have found two elements of the subbase with intersection consisting of at least two points.

Is it correct?

**Posted:** Sat Apr 23, 2005 5:24 pm

sam-n

#7

yes it is what have thought about.

**Posted:** Sun Apr 24, 2005 4:38 pm