Can we find an ideal $I$ of $\mathbb{C}[X,Y]$ such that $\lbrace(x,y)\in \mathbb{C}^2 : \vert x\vert = \vert y\vert\rbrace = V(I)$, the variety of $I$?
My attempt: Such points $(x,y)$ would lie on a circle of the same radius in the complex plane with center the origin, so each $(x,y)$ solves $X^2 + Y^2 - a^2 = 0$ for some real $a$. But then surely $a$ can range over all the reals, so the ideal would contain infinitely polynomials of the form $X^2 + Y^2 - a^2$. That would contradict Hilbert's Basis Theorem, which states that all the ideals of the given polynomial ring are finitely generated.
Can anyone suggest a correct method for this question?
Suppose $p \in \mathbb{C}[x, y]$ is a polynomial in the ideal of $S = \{ (x, y) \in \mathbb{C}^2 : |x| = |y| \}$, i.e. $p(x, y) = 0$ whenever $(x, y) \in S$. Then for any $a \ne 0$, $p(a, y)$ is a polynomial in $y$ which has infinitely many roots (since there are infinitely many points on the complex circle $|y| = |a| > 0$). That implies that $p(a, y)$ is the zero polynomial, which further implies that $x - a \mid p(x, y)$.
Now, since this holds for all $a \ne 0$, the only possibility is $p(x, y) \equiv 0$. Therefore, $I(S) = \langle 0 \rangle$, and in particular the Zariski closure of $S$ is all of $\mathbb{C}^2$.