Homogeneous ideal of height $2$ in $\mathbb C[X,Y]$

Question

If $J$ is a homogeneous ideal of height $2$ in $\mathbb C[X,Y]$ such that $J\subseteq (X,Y)$, then does there necessarily exist an integer $n\ge 1$ such that $X^n,Y^n \in J$ ?

Answer 1

The minimal primes over an ideal $J = (f_1,\ldots,f_n)$ correspond to intersection points $V(f_1) \cap \cdots \cap V(f_n)$. The radical of $J$ is the intersection of the minimal primes, and the condition $x,y \in \sqrt{J}$ thus means the only intersection point is the origin.

If $f_1,\ldots,f_n$ have a common zero at another point $(a,b) \neq 0$, then by homogeneity they have common zeros along the line $\lambda\cdot(a,b)$. But this implies they are divisible by the equation $\ell$ of the line, hence $J \subset (\ell)$ has height 1.

So if $J$ has height two, their only common zero is the origin, and $x,y \in \sqrt{J}$.

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First (wrong) attempt: No. The minimal primes over $J = (f,g)$ correspond to common zeros where $f=g=0$. The radical of $J$ is the intersection of the minimal primes, and the condition $x,y \in \sqrt{J}$ thus means the only common zero is at the origin. This is true for monomial ideals but not for more general homogeneous ideals.

For instance take $f=x + y$ and $g=x^3+y^3$, which intersect at $(1,-1)$.