Radical of $I = (xy-x^3, y^2-yx^2)$

Question

Quick question about finding $\sqrt I$, for $I = (xy-x^3, y^2-yx^2) \subset k[x,y]$.

Attempt:

Using for $I, J$ ideals we have $\sqrt{IJ} = \sqrt I \cap \sqrt J$.

$\sqrt{(xy-x^3, y^2-yx^2)} = \sqrt{(x(y-x^2), y(y-x^2))} =$

$= \sqrt{x, y(y-x^2)} \cap \sqrt{(y-x^2), y(y-x^2)} =$

$= \sqrt{x, y} \cap \sqrt{x,(y-x^2)} \cap \sqrt{(y-x^2), y} \cap \sqrt{(y-x^2)} =$

$= (x, y) \cap (x,y-x^2) \cap (y-x^2, y) \cap (y-x^2) = $

$= (y-x^2)$

Is there a quicker way to do this? E.g. can one conclude something from the fact that $V(I) = V(\sqrt I)$?

Answer 1

Slightly shorter: the radical of an ideal is the intersection of the prime ideals which contain it. So, determine the prime ideals which contain $xy-x^3$ and $y^2-yx^2$. It is not hard to see there are exactly $2$: the prime ideal $\;(y-x^2)$ and the maximal ideal $\;(x,y)$, which contains the latter. So the radical is $(y-x^2)$.