# Zariski in K^n

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

thanks!
• November 7th 2008, 11:44 PM
Moo
Hello,
Quote:

Originally Posted by roporte
Prove that if $K$ is infinite then two not empty open sets of $K^n$ (Zariski topology) always intersect.

thanks!

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

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

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

For $\mathcal{P}$ and $\mathcal{O}$ to intersect, we must have :
$\exists i \in [1,n],~ \mathcal{O}_i \cap \mathcal{P}_i \neq \emptyset$

So we can fix an i and study the intersection.
Since $\mathcal{P}_i$ and $\mathcal{O}_i$ are open in the Zariski topology (and nonempty), it means that $\mathcal{P}_i^c$ and $\mathcal{O}_i^c$ are finite

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

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

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

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

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

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 ><)
• November 8th 2008, 12:28 AM
NonCommAlg
Quote:

Originally Posted by roporte
Prove that if $K$ is infinite then two not empty open sets of $K^n$ (Zariski topology) always intersect.

thanks!

looking at the complement, you only need to show that the union of two algebraic sets, each not equal to $K^n,$ is not equal to $K^n.$ in other words we need to show that $K^n$ is an irreducible

algebraic set. this is obvious because if $K$ is infinite, then $\mathcal{I}(K^n)=<0>$ and $<0>$ is a prime ideal of $K[x_1, \cdots , x_n],$ because $K[x_1, \cdots , x_n]$ is an integral domain. Q.E.D.
• November 8th 2008, 03:13 PM
ThePerfectHacker
Quote:

Originally Posted by Moo
Hello,
(do tell me if there is any mistake ><)

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 $R$ be a commutative ring (assumed to be unitary). Define $\text{spec}(R)$ to be the set of all prime ideals of $R$. Let $X\subseteq R$ define $Z(X) = \{ P \in \text{spec}(R) | X\subseteq P\}$. A subset of $R$ is closed if and only if it has the form $Z(X)$. And this forms a topology on $R$.
• November 8th 2008, 10:59 PM
Moo
Quote:

Originally Posted by ThePerfectHacker
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 $R$ be a commutative ring (assumed to be unitary). Define $\text{spec}(R)$ to be the set of all prime ideals of $R$. Let $X\subseteq R$ define $Z(X) = \{ P \in \text{spec}(R) | X\subseteq P\}$. A subset of $R$ is closed if and only if it has the form $Z(X)$. And this forms a topology on $R$.

An open set in the Zariski topology is a subset which has its complement finite.

That's what we were taught in our topology class :/
• November 9th 2008, 07:02 AM
ThePerfectHacker
Quote:

Originally Posted by Moo
An open set in the Zariski topology is a subset which has its complement finite.

That's what we were taught in our topology class :/

According to Wikipedia (and my book) what you describe is the finite complement topology and the Zaraski topology is here.