Prove that if $\displaystyle K$ is infinite then two not empty open sets of $\displaystyle K^n$ (Zariski topology) always intersect.

thanks!

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

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

An open set $\displaystyle \mathcal{O}$ of $\displaystyle K^n$ with the Zariski topology means that it is the cartesian product of open sets in $\displaystyle K$, still with the Zariski topology. (It's the product topology)

So $\displaystyle \mathcal{O}=\mathcal{O}_1 \times \mathcal{O}_2 \times \dots \times \mathcal{O}_n=\underset{i \in [1,n]}{\LARGE{\rm{X}}} \mathcal{O}_i$. We can define similarly $\displaystyle \mathcal{P}=\underset{i \in [1,n]}{\LARGE{\rm{X}}} \mathcal{P}_i$

Where $\displaystyle \forall i \in [1,n],~ \mathcal{O}_i \text{ and } \mathcal{P}_i \text{ are ope}\text{n sets in the Zariski topology.}$

For $\displaystyle \mathcal{P}$ and $\displaystyle \mathcal{O}$ to intersect, we must have :

$\displaystyle \exists i \in [1,n],~ \mathcal{O}_i \cap \mathcal{P}_i \neq \emptyset$

So we can fix an i and study the intersection.

Since $\displaystyle \mathcal{P}_i$ and $\displaystyle \mathcal{O}_i$ are open in the Zariski topology (and nonempty), it means that $\displaystyle \mathcal{P}_i^c$ and $\displaystyle \mathcal{O}_i^c$ are**finite**

Hence $\displaystyle \mathcal{P}_i^c \cup \mathcal{O}_i^c$ is finite and**cannot**be equal to K, since K is infinite. So $\displaystyle \mathcal{P}_i^c \cup \mathcal{O}_i^c \neq K$

But $\displaystyle \mathcal{P}_i^c \cup \mathcal{O}_i^c=\left(\mathcal{P}_i \cap \mathcal{O}_i\right)^c$, by de Morgan's law.

So $\displaystyle \left(\mathcal{P}_i \cap \mathcal{O}_i\right)^c \neq K=\emptyset^c$

We know that for any sets A and B, $\displaystyle A=B \Leftrightarrow A^c=B^c$

So $\displaystyle \boxed{\mathcal{P}_i \cap \mathcal{O}_i \neq \emptyset}$ and hence $\displaystyle \mathcal{P} \cap \mathcal{O} \neq \emptyset$

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 ><) - Nov 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 $\displaystyle K^n,$ is not equal to $\displaystyle K^n.$ in other words we need to show that $\displaystyle K^n$ is an irreducible

algebraic set. this is obvious because if $\displaystyle K$ is infinite, then $\displaystyle \mathcal{I}(K^n)=<0>$ and $\displaystyle <0>$ is a prime ideal of $\displaystyle K[x_1, \cdots , x_n],$ because $\displaystyle K[x_1, \cdots , x_n]$ is an integral domain. Q.E.D. - Nov 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 $\displaystyle R$ be a commutative ring (assumed to be unitary). Define $\displaystyle \text{spec}(R)$ to be the set of all prime ideals of $\displaystyle R$. Let $\displaystyle X\subseteq R$ define $\displaystyle Z(X) = \{ P \in \text{spec}(R) | X\subseteq P\}$. A subset of $\displaystyle R$ is closed if and only if it has the form $\displaystyle Z(X)$. And this forms a topology on $\displaystyle R$. - Nov 8th 2008, 10:59 PMMoo
- Nov 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.