The ideal generated by $X^{2}-Y^{3}$ and $Y^{2}-X^{3}$ is not prime

Question

I tried to show that the ideal generated by $X^{2}-Y^{3}$ and $Y^{2}-X^{3}$ is not prime in $\mathbb{C}[X,Y]$.

I noticed that $X^{2}-Y^{3}-(Y^{2}-X^{3})=(X-Y)(X^{2}+XY+Y^{2}+X+Y)$. Now, if $X-Y$ is in $(X^{2}-Y^{3},Y^{2}-X^{3})$ there exists $f(X,Y),\ g(X,Y)\in \mathbb{C}[X,Y] $ s.t. $$X-Y=f(X,Y)(X^{2}-Y^{3}) + g(X,Y)(Y^{2}-X^{3}).$$ For $Y=0$ it results $X=f(X,0)X^{2} - g(X,0)X^{3}\Rightarrow 1=f(X,0)X-g(X,0)X^{2}$ (False).
A similar argument goes for$X^{2}+XY+Y^{2}+X+Y$. Is this a right solution?

Answer 1

$X(X^2-Y^3)+(Y^2-X^3)=-XY^3+Y^2=Y^2(-XY+1)$. $-XY+1, Y^2$

Suppose that $-XY+1=f(X,Y)(X^2-Y^3)+g(X,Y)(Y^2-X^3)$, we deduce that $1=f(0,Y)(-Y^3)+g(0,Y)Y^2=Y^2(-f(0,Y)Y+g(0,Y))$ impossible.

Suppose that $Y^2=f(X,Y)(X^2-Y^3)+g(X,Y)(Y^2-X^3)$. Set $X=Y=1$, we have $1=f(1,1)(1^2-1^3)+g(1,1)(1^2-1^3)=0$ impossible, so $Y^2$ and $-XY+1$ are not in the ideal $I$ generated by $X^2-Y^3$ and $Y^2-X^3$ but their product is in this ideal so $I$ is not a prime ideal.

Answer 2

Here is a more geometric solution.

To show that $I=(x^2-y^3,y^2-x^3)$ is not prime, one way is to show that their set of common zeroes has cardinality $>1$.

This works because the irreducible components of their common zero set correspond to the radicals of the primary components in the decomposition of $I$.

So suppose $x^2-y^3=y^2-x^3=0$. From the first equality, you get $$x^3-y^2=x^2 x - y^2=y^3x-y^2 = y^2(x-1)=0.$$

But then either $y=0$ or $x=1$. If $x=1$, then we must have $y=1$. If $y=0$, we must have $x=0$. Hence we have found two distinct points solving the equation, hence it cannot be prime.