Consider the ideal $I=\langle x^4-x^3z-xy^2+y^2z,xy^2-xz-y^2z+z^2 \rangle \subseteq \mathbb{C}[x,y,z]$.
Question : Decompose the radical $\sqrt{I}$ as a finite intersection of prime ideals.
My Attempt : I can see that $I=\langle (x^3-y^2)(x-z),(y^2-z)(x-z) \rangle$. So $V(I)=V(x-z) \cup V(x^3-y^2,y^2-z)$ and hence $I(V(I))=I(V(x-z)) \cap I(V(x^3-y^2,y^2-z))$.
Since $\mathbb{C}$ is algebraically closed, we have $\sqrt{I}=I(V(I))$ by Nullstellensatz. Since $V(x-z)$ is irreducible, we have $I(V(x-z))$ is prime. I want to do the same thing to the ideal $I(V(x^3-y^2,y^2-z))$ but I am not able to show $V(x^3-y^2,y^2-z)$ is irreducible. Or, is this algebraic set irreducible?
Can anyone help me with this step?
$$V(x^3-y^2)=\{(t^2,t^3):t\in\Bbb C\}$$ (to see this let $t=y/x$ on a typical point of the variety). Then $$V(x^3-y^2,y^2-z)=\{(t^2,t^3,t^6):t\in\Bbb C\}.$$ This is the image of $\Bbb A^1(\Bbb C)$ under an algebraic map, so is irreducible.