Maximal Ideal and Variety

**huram2215** · May 26th 2010, 07:02 PM

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.

**TheArtofSymmetry** · May 27th 2010, 02:10 AM

Originally Posted by huram2215

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.

**NonCommAlg** · May 27th 2010, 02:22 AM

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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