Results 1 to 6 of 6

Thread: Zariski in K^n

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    48

    Zariski 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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by roporte View Post
    Prove that if $\displaystyle K$ is infinite then two not empty open sets of $\displaystyle K^n$ (Zariski topology) always intersect.

    thanks!
    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 ? (do tell me if there is any mistake ><)
    Last edited by Moo; Nov 8th 2008 at 12:27 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by roporte View Post
    Prove that if $\displaystyle K$ is infinite then two not empty open sets of $\displaystyle K^n$ (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 $\displaystyle K^n,$ 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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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 $\displaystyle R$ 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 $\displaystyle R$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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 $\displaystyle R$ 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 $\displaystyle R$.
    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 :/
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The zariski topology
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 30th 2011, 10:26 AM
  2. Zariski Topology
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Nov 29th 2010, 08:15 PM
  3. The zariski topology
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Feb 10th 2009, 01:43 PM
  4. Zariski closure
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Nov 7th 2008, 07:49 PM

Search Tags


/mathhelpforum @mathhelpforum