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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?

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  • $\begingroup$ The point $(0,0)$ is not the zero locus of $x-y^2$. It's part of the zero locus, sure. $\endgroup$ – Arthur May 23 at 19:55
  • $\begingroup$ 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 $x-y^2$. $\endgroup$ – user26857 May 23 at 20:44

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