$Z(I:J)$ is the Zariski closure of $Z(I)-Z(J)$ Let $(I:J)$ denote the colon ideal (or ideal quotient). It is pretty clear that the Zariski closure of $Z(I)-Z(J)$ is contained in $Z(I:J)$. How can we prove that the the Zariski closure of $Z(I)-Z(J)$ is precisely $Z(I:J)$?
 A: The statement is false. Consider $A=\mathbb{Z}/4$, $I=(0)$, $J=(2)$. Then $Z(I)=SpecA=${$(2)$}, $Z(J)=${$(2)$}, $Z(I)-Z(J)=\emptyset$. On the other hand, $Z(0:J)=(2)$.
A: Re: Last Comment: Yes, it is true then.
If $I$ is radical, then by primary decomposition (valid for any noetherian ring, in particular a polynomial ring), we have $I = \cap_i p_i$ for finitely many primes $p_i$. Then, $Z(I) - Z(J) =\cup_i (Z(p_i)-Z(J))$. Thus, 
$Closure(Z(I)-Z(J))=\cup_i Closure(Z(p_i)-Z(J))$. 
Since ideal quotient commutes with intersection in the first factor, we have 
$Z(I:J)=Z(\cap_i (p_i:J))=\cup_i(Z(p_i:J))$
So it suffices to prove the case where $I=p$ is prime. 
This is easy. We have two cases, (i) $J\subset p$, (ii) there is $x \in J$ not in $p$. For case (i), note that $(p:J)=A$ so $Z(p:J)=\emptyset$. Also, since $J \subset p$, we have $Z(I)-Z(J)=\emptyset$ and that handles (i). For (ii), since $p$ is prime, $yJ \subset p$ only if $yx \in p$ only if $y \in p$. So $(p:J)=p$ and $Z(p:J)=Z(p)$. Moreover, since $J$ does not contain $p$, the set $Z(p)-Z(J)$ is an open subset of the irreducible set $Z(p)$. Thus, its closure is all of $Z(p)$, as desired. 
By the way, I didn't need nullstellensatz or anything like that. All I needed was for the ambient ring to be noetherian and the ideal I to be radical.
Cliffs: The quantity $Closure (Z(I)-Z(J))$ geometrically consists of the components of $Z(I)$ minus the ones contained in $Z(J)$. The ideal quotient $(I:J)$ takes the primary components of $I$ (which, when $I$ is radical, correspond bijectively to the irreducible components of $Z(I)$) and turns all those primary components in common with (or containing those of) $J$ into $1$ and leaves the other ones alone. Thus, taking $Z$, you get the previous set. 
