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Math Help - Maximal Ideal and Variety

  1. #1
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    Maximal Ideal and Variety

    Suppose that I is a non-zero ideal of \mathbb{C}[x_1,x_2]. Define the variety associated with I as V(I) = \{ (z,w) \in \mathbb{C}^2 : \forall p(x_1,x_2) \in I, p(z,w) = 0 \}.

    Claim: If there exists  a = (a_1,a_2) \in \mathbb{C}^2 such that V(I) = \{ (a_1,a_2) \} then I is a maximal ideal.

    Failed attempt: For all p(x_1,x_2) \in I, p(a_1,a_2) = 0 from the definition of V(I). Define the substitution homomorphism \phi_a: \mathbb{C}[x_1,x_2] \rightarrow \mathbb{C} as \phi_a(p(x_1,x_2)) = p(a_1,a_2). It suffices to prove that ker(\phi_a) = I, for then, by Hilbert's Nullstellensatz, I is a maximal ideal.

    However, it doesn't seem that ker(\phi_a) = I is true. Certainly I \subseteq ker(\phi_a). But there may be polynomials in ker(\phi_a) that are not in I it seems. For ker(\phi_a) = (x_1 - a_1, x_2 - a_2) = \{ q(x_1,x_2) (x_1 - a_1) + r(x_1,x_2)(x_2 - a_2) : q, r \in \mathbb{C}[x_1,x_2] \} but there could be some h(x) \in ker(\phi_a) \backslash  I if h(x) = q(x_1,x_2) (x_1 - a_1) + r(x_1,x_2)(x_2 - a_2) and there exists (b_1,b_2) \ne (a_1,a_2) such that q(b_1,b_2) = 0 = r(b_1,b_2). It seems this could plausibly occur, but I haven't been able to construct such a counter example. So I am unable to prove that ker(\phi_a) \subseteq I.

    Finally, in working with this problem, I couldn't see a way to get to the structure of I through V(I). I thought maybe a contrapositive or proof by contradiction might work, but to no avail.
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  2. #2
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    Quote Originally Posted by huram2215 View Post
    Suppose that I is a non-zero ideal of \mathbb{C}[x_1,x_2]. Define the variety associated with I as V(I) = \{ (z,w) \in \mathbb{C}^2 : \forall p(x_1,x_2) \in I, p(z,w) = 0 \}.

    Claim: If there exists  a = (a_1,a_2) \in \mathbb{C}^2 such that V(I) = \{ (a_1,a_2) \} then I is a maximal ideal.

    Failed attempt: For all p(x_1,x_2) \in I, p(a_1,a_2) = 0 from the definition of V(I). Define the substitution homomorphism \phi_a: \mathbb{C}[x_1,x_2] \rightarrow \mathbb{C} as \phi_a(p(x_1,x_2)) = p(a_1,a_2). It suffices to prove that ker(\phi_a) = I, for then, by Hilbert's Nullstellensatz, I is a maximal ideal.

    However, it doesn't seem that ker(\phi_a) = I is true. Certainly I \subseteq ker(\phi_a). But there may be polynomials in ker(\phi_a) that are not in I it seems. For ker(\phi_a) = (x_1 - a_1, x_2 - a_2) = \{ q(x_1,x_2) (x_1 - a_1) + r(x_1,x_2)(x_2 - a_2) : q, r \in \mathbb{C}[x_1,x_2] \} but there could be some h(x) \in ker(\phi_a) \backslash I if h(x) = q(x_1,x_2) (x_1 - a_1) + r(x_1,x_2)(x_2 - a_2) and there exists (b_1,b_2) \ne (a_1,a_2) such that q(b_1,b_2) = 0 = r(b_1,b_2). It seems this could plausibly occur, but I haven't been able to construct such a counter example. So I am unable to prove that ker(\phi_a) \subseteq I.

    Finally, in working with this problem, I couldn't see a way to get to the structure of I through V(I). I thought maybe a contrapositive or proof by contradiction might work, but to no avail.
    I have a question on your claim. If we choose I=(x_1^2, x_2), then V(I)={(0,0)}. In this case I is not a maximal ideal in C[x_1, x_2]. Anyhow, \text{rad I}=(x_1, x_2) is a maximal ideal in C[x_1, x_2].

    You might need to modify the claim

    ker (\phi_a)= \text{rad} I,

    where I is not necessarily maximal ideal, but satisfying  J(V(I))=\text{rad }I. See here.
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  3. #3
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    yes, the claim is false because if we let J=\langle x_1-a_1, x_2-a_2 \rangle, then V(J^2)=V(J) \cup V(J)=V(J)=\{(a_1,a_2) \} but J^2 is not maximal.
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