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Math Help - Zariski in K^n

  1. #1
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    Zariski in K^n

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

    thanks!
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    Moo
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    Hello,
    Quote Originally Posted by roporte View Post
    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 ? (do tell me if there is any mistake ><)
    Last edited by Moo; November 8th 2008 at 12:27 AM.
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    Quote Originally Posted by roporte View Post
    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.
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    Quote Originally Posted by Moo View Post
    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.
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    Moo
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    Quote Originally Posted by ThePerfectHacker View Post
    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 :/
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    Quote Originally Posted by Moo View Post
    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.
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