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Math Help - Cubic surface

  1. #1
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    Cubic surface

    We define the cubic V in \mathbb{R}^3 by the equations: y-x^2=0, z-x^3=0.

    Prove that \mathcal{I}(V)=<y-x^2,z-x^3>

    thanks!
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  2. #2
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    Quote Originally Posted by roporte View Post
    We define the cubic V in \mathbb{R}^3 by the equations: y-x^2=0, z-x^3=0.

    Prove that \mathcal{I}(V)=<y-x^2,z-x^3>

    thanks!
    let J=<y-x^2,z-x^3>. clearly J \subseteq \mathcal{I}(V). now suppose f(x,y,z) \in \mathcal{I}(V). then f(x,x^2,x^3)=0.

    since f \in \mathbb{R}[x,y,z], we have: f(x,y,z)=\sum {c_{ijk}}x^i y^j z^k, for some c_{ijk} \in \mathbb{R}. thus:

    f(x,y,z)=\sum c_{ijk}x^i(y-x^2 + x^2)^j (z-x^3 + x^3)^k<br />

    =p(x,y,z)(y-x^2) + q(x,y,z)(z-x^3) + f(x,x^2,x^3)

    =p(x,y,z)(y-x^2) + q(x,y,z)(z-x^3) \in J.

    thus \mathcal{I}(V) \subseteq J and we're done.
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