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Thread: Algebraic Varieties and Ideals

  1. #1
    Super Member Bernhard's Avatar
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    Algebraic Varieties and Ideals

    In Chapter 1, Section 4 of Cox et al "Ideals, Varieties and Algorithms, Exercise 3(c) reads as follows:

    Prove the following equality of ideals in $\displaystyle \mathbb{Q}[x,y] $:

    $\displaystyle < 2x^2 + 3y^2 -11, x^2 - y^2 - 3> \ = \ <x^2 - 4, y^2 - 1> $

    Any help with this problem would be appreciated.

    Peter
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  2. #2
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    Re: Algebraic Varieties and Ideals

    I will give you one half:

    $\displaystyle x^2 - 4 = \dfrac{1}{5}(2x^2 + 3y^2 - 11) + \dfrac{3}{5}(x^2 - y^2 - 3)$

    $\displaystyle y^2 - 1 = \dfrac{1}{5}(2x^2 + 3y^2 - 11) - \dfrac{2}{5}(x^2 - y^2 - 3)$

    Your turn.
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Algebraic Varieties and Ideals

    Quote Originally Posted by Deveno View Post
    I will give you one half:

    $\displaystyle x^2 - 4 = \dfrac{1}{5}(2x^2 + 3y^2 - 11) + \dfrac{3}{5}(x^2 - y^2 - 3)$

    $\displaystyle y^2 - 1 = \dfrac{1}{5}(2x^2 + 3y^2 - 11) - \dfrac{2}{5}(x^2 - y^2 - 3)$

    Your turn.
    Thanks for the guidance and help Deveno

    OK so we want to show that in $\displaystyle \mathbb{Q}[x,y] $ we have $\displaystyle <2x^2 + 3y^2 - 11, x^2 - y^2 - 3> \ = \ <x^2 - 4, y^2 - 1> $


    First show that $\displaystyle <x^2 - 4, y^2 - 1> \ \subseteq \ <2x^2 + 3y^2 - 11, x^2 - y^2 - 3> $


    From Deveno's equations above it follows that $\displaystyle x^2 - 4, y^2 - 1 \in \ <2x^2 + 3y^2 - 11, x^2 - y^2 - 3> $

    Thus, $\displaystyle <x^2 - 4, y^2 - 1> \ \subseteq \ <2x^2 + 3y^2 - 11, x^2 - y^2 - 3> $ ... ... ... (1)


    Now show $\displaystyle <2x^2 + 3y^2 - 11, x^2 - y^2 - 3> \ \subseteq \ <x^2 - 4, y^2 - 1> $


    We can write $\displaystyle 2x^2 + 3y^2 - 11 \ = \ 2(x^2 - 4) + 3(y^2 - 1) $

    and $\displaystyle x^2 - y^2 - 3 \ = \ (x^2 - 4) - (y^2 - 1) $

    Thus $\displaystyle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \in <x^2 - 4, y^2 - 1> $

    Thus $\displaystyle <2x^2 + 3y^2 - 11, x^2 - y^2 - 3> \ \subseteq \ <x^2 - 4, y^2 - 1> $ ... ... ... (2)


    Equations (1) (2) $\displaystyle \ \Longrightarrow \ \mathbb{Q}[x,y] $ we have $\displaystyle <2x^2 + 3y^2 - 11, x^2 - y^2 - 3> \ = \ <x^2 - 4, y^2 - 1> $
    Last edited by Bernhard; Aug 2nd 2013 at 04:08 PM.
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    Re: Algebraic Varieties and Ideals

    Note this shows the generators of ideals in Q[x,y] aren't even unique up to units. Also, this is just high-school algebra in fancier clothes.
    Thanks from Bernhard
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