Prove that $\sqrt{I}=I$

Question

Let $f(x,y)=y^2-x^2-x^3$ and consider the ideal $I=(f)$ in $\mathbb C[x,y]$. I want to show that the radical ideal $\sqrt{I}$ is equal to $I$.

Any help? I've tried an approach using the division algorithm in 2 variables but became too messy. Is there any general approach one can follow to prove that $\sqrt{(g)}=(g)$ for $g \in k[x_1,\cdots,x_n]$?

Thanks!

Answer 1

An expansion of my comment:

$I$ is a radical ideal of $R$ if and only if $R/I$ has no nilpotents. But in this case, $\Bbb C[x,y]/(y^2-x^3-x^2)$ is better than that, it’s an integral domain. You’re just using the congruence $y^2\equiv x^3+x^2\pmod I$, and you should think of this as looking at the extension $\Bbb C(x)\bigl(\sqrt{x^3+x^2}\,\bigr)\supset\Bbb C(x)$.