# Thread: Maximal Ideal and Variety

1. ## Maximal Ideal and Variety

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

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

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

However, it doesn't seem that $\displaystyle ker(\phi_a) = I$ is true. Certainly $\displaystyle I \subseteq ker(\phi_a)$. But there may be polynomials in $\displaystyle ker(\phi_a)$ that are not in $\displaystyle I$ it seems. For $\displaystyle ker(\phi_a) = (x_1 - a_1, x_2 - a_2) = \{ q(x_1,x_2) (x_1 - a_1) + r(x_1,x_2)(x_2 - a_2) : q, r \in \mathbb{C}[x_1,x_2] \}$ but there could be some $\displaystyle h(x) \in ker(\phi_a) \backslash I$ if $\displaystyle h(x) = q(x_1,x_2) (x_1 - a_1) + r(x_1,x_2)(x_2 - a_2)$ and there exists $\displaystyle (b_1,b_2) \ne (a_1,a_2)$ such that $\displaystyle q(b_1,b_2) = 0 = r(b_1,b_2)$. 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 $\displaystyle ker(\phi_a) \subseteq I$.

Finally, in working with this problem, I couldn't see a way to get to the structure of $\displaystyle I$ through $\displaystyle V(I)$. I thought maybe a contrapositive or proof by contradiction might work, but to no avail.

2. Originally Posted by huram2215
Suppose that $\displaystyle I$ is a non-zero ideal of $\displaystyle \mathbb{C}[x_1,x_2]$. Define the variety associated with $\displaystyle I$ as $\displaystyle V(I) = \{ (z,w) \in \mathbb{C}^2 : \forall p(x_1,x_2) \in I, p(z,w) = 0 \}$.

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

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

However, it doesn't seem that $\displaystyle ker(\phi_a) = I$ is true. Certainly $\displaystyle I \subseteq ker(\phi_a)$. But there may be polynomials in $\displaystyle ker(\phi_a)$ that are not in $\displaystyle I$ it seems. For $\displaystyle ker(\phi_a) = (x_1 - a_1, x_2 - a_2) = \{ q(x_1,x_2) (x_1 - a_1) + r(x_1,x_2)(x_2 - a_2) : q, r \in \mathbb{C}[x_1,x_2] \}$ but there could be some $\displaystyle h(x) \in ker(\phi_a) \backslash I$ if $\displaystyle h(x) = q(x_1,x_2) (x_1 - a_1) + r(x_1,x_2)(x_2 - a_2)$ and there exists $\displaystyle (b_1,b_2) \ne (a_1,a_2)$ such that $\displaystyle q(b_1,b_2) = 0 = r(b_1,b_2)$. 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 $\displaystyle ker(\phi_a) \subseteq I$.

Finally, in working with this problem, I couldn't see a way to get to the structure of $\displaystyle I$ through $\displaystyle V(I)$. 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 $\displaystyle I=(x_1^2, x_2)$, then V(I)={(0,0)}. In this case I is not a maximal ideal in C[x_1, x_2]. Anyhow, $\displaystyle \text{rad I}=(x_1, x_2)$ is a maximal ideal in C[x_1, x_2].

You might need to modify the claim

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

where I is not necessarily maximal ideal, but satisfying $\displaystyle J(V(I))=\text{rad }I$. See here.

3. yes, the claim is false because if we let $\displaystyle J=\langle x_1-a_1, x_2-a_2 \rangle,$ then $\displaystyle V(J^2)=V(J) \cup V(J)=V(J)=\{(a_1,a_2) \}$ but $\displaystyle J^2$ is not maximal.