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Math Help - Varieties

  1. #1
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    Varieties

    Let  V be the  y -axis contained in  \mathbb{R}^2 (  V=\mathbb{V}(x) ).

    In lecture, my professor said  V is irreducible (which makes sense), but I don't know how to show it. One hint he gave us is that "a prime ideal is radical" i.e. if  I is prime then  I = \text{Rad}(I) .

    Any insight is much appreciated.
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  2. #2
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    Quote Originally Posted by mathman88 View Post
    Let  V be the  y -axis contained in  \mathbb{R}^2 (  V=\mathbb{V}(x) ).

    In lecture, my professor said  V is irreducible (which makes sense), but I don't know how to show it. One hint he gave us is that "a prime ideal is radical" i.e. if  I is prime then  I = \text{Rad}(I) .

    Any insight is much appreciated.
    well, a variety is irreducible iff its ideal is prime. now it's easy to see that in your example the ideal of V is \langle x \rangle and clearly \mathbb{R}[x,y]/\langle x \rangle \cong \mathbb{R}[y] is an integral domain. thus \langle x \rangle is a prime ideal.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    well, a variety is irreducible iff its ideal is prime. now it's easy to see that in your example the ideal of V is \langle x \rangle and clearly \mathbb{R}[x,y]/\langle x \rangle \cong \mathbb{R}[y] is an integral domain. thus \langle x \rangle is a prime ideal.
    Expanding on this, what if the field wasn't  \mathbb{R} but arbitrary, call it  F .

    My professor said  V is irreducible when  F is algebraically closed. How would that be shown? Also what would be an example of a non algebraically closed field where  V is reducible?

    Thanks!
    Last edited by mathman88; April 19th 2010 at 02:29 PM.
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  4. #4
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    So in general, here's what I'm thinking:

    We have that  F is an algebraically closed field.

    Now  V is irreducible  \iff \mathbb{I}(V) is prime. But by Hilbert's Nullstellensatz we see that  \mathbb{I}(V) = \text{Rad}(x) .

    So  V is irreducible  \iff \text{Rad}(x) is prime.

    I'm stuck after this. What is  \text{Rad}(x) explicitly? Also I'm able to show if  A is a prime ideal then  A = \text{Rad}(A) , but does the other direction work too?

    Thanks!
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  5. #5
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    Quote Originally Posted by mathman88 View Post
    So in general, here's what I'm thinking:

    We have that  F is an algebraically closed field.

    Now  V is irreducible  \iff \mathbb{I}(V) is prime. But by Hilbert's Nullstellensatz we see that  \mathbb{I}(V) = \text{Rad}(x) .

    So  V is irreducible  \iff \text{Rad}(x) is prime.

    I'm stuck after this. What is  \text{Rad}(x) explicitly? Also I'm able to show if  A is a prime ideal then  A = \text{Rad}(A) , but does the other direction work too?

    Thanks!
     I=\langle x \rangle is prime since  F[x,y]/\langle x \rangle \cong F[y] is an integral domain, so  \text{Rad}(x) = \langle x \rangle by the theorem you said you knew how to prove.

    Thus  V is irreducible  \iff \langle x \rangle is prime...
    Last edited by chiph588@; April 20th 2010 at 05:46 AM.
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  6. #6
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    Thanks!

    Now what about that field where  V is reducible? I'm having trouble picturing what that would be.
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    I think  \mathbb{F}_2 will do the trick.
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