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Thread: Varieties

  1. #1
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    Varieties

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

    In lecture, my professor said $\displaystyle 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 $\displaystyle I $ is prime then $\displaystyle 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 $\displaystyle V $ be the $\displaystyle y $-axis contained in $\displaystyle \mathbb{R}^2 $ ($\displaystyle V=\mathbb{V}(x) $).

    In lecture, my professor said $\displaystyle 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 $\displaystyle I $ is prime then $\displaystyle 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 $\displaystyle V$ is $\displaystyle \langle x \rangle$ and clearly $\displaystyle \mathbb{R}[x,y]/\langle x \rangle \cong \mathbb{R}[y]$ is an integral domain. thus $\displaystyle \langle x \rangle$ 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 $\displaystyle V$ is $\displaystyle \langle x \rangle$ and clearly $\displaystyle \mathbb{R}[x,y]/\langle x \rangle \cong \mathbb{R}[y]$ is an integral domain. thus $\displaystyle \langle x \rangle$ is a prime ideal.
    Expanding on this, what if the field wasn't $\displaystyle \mathbb{R} $ but arbitrary, call it $\displaystyle F $.

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

    Thanks!
    Last edited by mathman88; Apr 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 $\displaystyle F $ is an algebraically closed field.

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

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

    I'm stuck after this. What is $\displaystyle \text{Rad}(x) $ explicitly? Also I'm able to show if $\displaystyle A $ is a prime ideal then $\displaystyle 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 $\displaystyle F $ is an algebraically closed field.

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

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

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

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

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

    Now what about that field where $\displaystyle 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 $\displaystyle \mathbb{F}_2 $ will do the trick.
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