Prove that if is infinite then two not empty open sets of (Zariski topology) always intersect.

thanks!

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- November 7th 2008, 06:18 PMroporteZariski in K^n
Prove that if is infinite then two not empty open sets of (Zariski topology) always intersect.

thanks! - November 7th 2008, 11:44 PMMoo
Hello,

An open set of with the Zariski topology means that it is the cartesian product of open sets in , still with the Zariski topology. (It's the product topology)

So . We can define similarly

Where

For and to intersect, we must have :

So we can fix an i and study the intersection.

Since and are open in the Zariski topology (and nonempty), it means that and are**finite**

Hence is finite and**cannot**be equal to K, since K is infinite. So

But , by de Morgan's law.

So

We know that for any sets A and B,

So and hence

It follows that a topological space with the Zariski topology cannot be separate.

Is it clear enough ? (Worried) (do tell me if there is any mistake ><) - November 8th 2008, 12:28 AMNonCommAlg
looking at the complement, you only need to show that the union of two algebraic sets, each not equal to is not equal to in other words we need to show that is an irreducible

algebraic set. this is obvious because if is infinite, then and is a prime ideal of because is an integral domain. Q.E.D. - November 8th 2008, 03:13 PMThePerfectHacker
Hello Moo, but maybe you are using a different definition of Zariski topology?

According to my field theory book it is defined as follows. Let be a commutative ring (assumed to be unitary). Define to be the set of all prime ideals of . Let define . A subset of is closed if and only if it has the form . And this forms a topology on . - November 8th 2008, 10:59 PMMoo
- November 9th 2008, 07:02 AMThePerfectHacker
According to Wikipedia (and my book) what you describe is the finite complement topology and the Zaraski topology is here.