Yeasterday I asked a question on math stackexchange which simplifies to the following:

Assume that $G$ is a graded ring and $A \subseteq G$ is a homogeneous radical ideal. Is it true that $IV(A) = A $? Here $I(X) $ is the homogeneous ideal generated by homogeneous elements that vanish at every point of $X$ and $V(S)$ is the set of relevent primes that contain $S$.

It is still unanswered. As explained in the math stackexchange question, it suffices to prove the following: If $A \subseteq \mathfrak{p}$ is a prime, then there exists a relevent homogeneous prime $ \mathfrak{q}$ such that $A \subseteq \mathfrak{q} \subseteq \mathfrak{p}$. I understand how to deal with the case when $ \mathfrak{p}$ does not contain the irrelvent ideal, but I am stuck with the case when $ \mathfrak{p}$ does contain the irrelevent ideal. Even in the special case $G = \mathbb{Z}[X,Y] $ I am not sure how to proceed. The primes containing $ (X,Y) $ all look like $(p,X,Y)$ for some prime $ p $, but I don't know how to "choose a line" which passes through this point like you can in the case of $ \mathbb{C}[X,Y]$. I debated whether to ask this question here, but I feel like somthing interesting is going and I just can't see it. I am looking forward to reading some enightened responses!