# Radical of the ideal generated by $x^3 - y^6$ and $xy-y^3$

I am following Andreas Gathmann's notes on algebraic geometry.

He asks, right after he shows that $$V$$ and $$I$$ are bijection between algebraic varieties and radical ideals (so I suppose I must use this) to find the radical of $$\langle x^3 - y^6, xy-y^3 \rangle$$. I have tried something, but it is wrong, and I don't know why.

I have consider the variety generated by this ideal, who is the set of points such that

\begin{align} x^3 - y^6 = 0 \\ xy - y^3 = 0 \end{align}

By factorizing, we get:

\begin{align} (x - y^2)(x^2 + xy^2 + y^4) = 0 \\ (x - y^2)y = 0 \end{align}

For the second equation to be true, either $$x-y^2 = 0$$ or $$y=0$$. If the former is true, then the first equation must be true. If $$y=0$$, the first equation become $$x^3 = 0$$, so $$x=0$$. But the point $$(0,0)$$ is in the zero locus of $$x-y^2$$, so $$V(x^3 - y^6, xy-y^3) = V(x - y^2)$$, and as this is radical, then the radical of $$\langle x^3 - y^6, xy-y^3 \rangle$$ is $$\langle x - y^2 \rangle$$.

I know all of this is wrong only because I have computed with Maple that the radical of this ideal is $$\langle x^3 - xy^2, xy - y^3 \rangle$$. What mistake did I do?

• The point $(0,0)$ is not the zero locus of $x-y^2$. It's part of the zero locus, sure. – Arthur May 23 at 19:55
• Your Maple is wrong! It's obvious that $(x-y^2)^3$ belongs to the given ideal which at its turn is contained in the ideal generated by $x-y^2$. – user26857 May 23 at 20:44