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**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.