How to show that $(Y- X^2, Z - X^3) \subseteq k[X,Y,Z]$ is a prime ideal?

Question

I suppose that $k$ is an algebraically closed field (actually, my goal is to show $\mathcal{I}(\mathcal{V}(Y- X^2, Z - X^3)) = (Y- X^2, Z - X^3)$).

(But I think algebraically closed is not necessary to show $(Y- X^2, Z - X^3) \subseteq k[X,Y,Z]$ is a radical ideal...)

My strategy is to show $k[X,Y,Z]/(Y- X^2, Z - X^3)$ has no zero-divisors. I consider the ring homomorphism $\varphi: k[X] \rightarrow k[X,Y,Z]/(Y- X^2, Z - X^3)$; it is clearly surjective. But I have no idea to compute the kernel of $\varphi$.

Does one have solutions or good method? Thank you very much!

Answer 1

1. Use twice that $R[T]/(T-a)\simeq R$ (where $R$ is a commutative ring and $a\in R$): $$K[X,Y,Z]/(Y-X^2,Z-X^3)\simeq k[X,Y]/(Y-X^2)\simeq K[X].$$ This shows that your ideal is prime.

2. You can also use your idea: $f\in\ker\varphi$ iff $f(X)\in(Y-X^2,Z-X^3)$, that is, $f(X)=(Y-X^2)g(X,Y,Z)+(Z-X^3)h(X,Y,Z)$. Now send $Y$ to $X^2$ and $Z$ to $X^3$ and find $f=0$, so $\varphi$ is an isomorphism.

Answer 2

This ideal is generated by irreducible polynomials and there aren't $a(X,Y,Z)$ and $b(X,Y,Z)$ such that $a\cdot (Y-X^2)+b\cdot(Z-X^3)=u$ for any $u\in K$, therefore it is prime. By being prime it is radical.