Let $k$ be an algebraically closed field, not necessarily of characteristic $0$. I’m trying to compute the ideal of $C=V(Z-XY,Y^2+XZ-X^2)$ over $k$.
Here is what I have tried so far:
We have $$C=V(Z-XY,Y^2+X^2Y-X^2)=V(Z-XY,Y^2+X^2(Y-1))$$
Let $J=(Z-XY,Y^2+X^2(Y-1))$. According to Macaulay2, at least over $\mathbb{Q}$ we have $$\sqrt{J}=J$$
In fact, it tells me that $J$ is prime over $\mathbb{Q}$, and I imagine this is true in the general case, but I’m struggling to show it. I can prove that both polynomials are irreducible over $k$, but not much else about the ideal.
I’ve also tried showing that $I(C)=J$ directly, and I think I can prove that we can write any $F\in k[X,Y,Z]$ as $$F=A_1(X)+YA_2(X)+[Y^2+X^2(Y-1)]B_1(X,Y)+[Z-XY]B_2(X,Y,Z)$$ But I’m struggling then to show that $F\in I(C)$ implies $A_1(X),A_2(X)=0$.
Any help would be much appreciated.
Note that $k[x,y,z]/(z-xy) \cong k[u,v]$, the polynomial ring in two independent variables, by $x\mapsto u$, $y\mapsto v$ and $z\mapsto uv$. Under this isomorphism, $J$ maps to the ideal $I = (v^2+u^2v-u^2)\subset k[u,v]$. Now, if you show that $v^2+u^2v-u^2$ is irreducible in $k[u,v]$, then your ring is isomorphic to $k[u,v]/(v^2+u^2v-u^2)$, which is an integral domain. This proves that your ideal is prime and thus radical. We are using the Nullstellensatz to say that $I(V(J)) = \sqrt{J} = J$.