<s>[INST] Determine the radical of an ideal Determine the radical of the ideal $(x^3-y^6,xy-y^3)$ in $C[x,y]$.
I used Nullstellensatz theorem $\sqrt{I}=I(V(I))$.
Factorization gives:
$$x^3-y^6=(x-y^2)(x+(\frac{1}{2}+\frac{\sqrt{3}}{2}i) y^2)(x+(\frac{1}{2}-\frac{\sqrt{3}}{2}i) y^2)$$
$$xy-y^3=y(x-y^2)$$
so $V(I)={(x,y)|x-y^2=0}$
Hence $\sqrt{I}=(x-y^2)$. Is this right?
 [/INST] The result is right.
It is a good idea to factorise your polynomials. The correct factorisation is the following:
$x^3-y^6=(x-y^2)(y^4+xy^2+x^2)$ and $xy-y^3=y(x-y^2)$
Then, a point $(x,y)$ lies in $V(I)$ if and only if one of the following hold:
$x=y^2$ or 
$y=0$ and $y^4+xy^2+x^2=0$
As this latter only gives $(0,0)$, already in $V(x-y^2)$, you have indeed $\sqrt{I}=I(V(I))=(x-y^2)$.
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