$\overline{V(I)-V(J)}=V(\bigcup_{n=1}^{\infty}I\colon J^n)$

Question

Is it true that $\overline{V(I)-V(J)}=V\left(\bigcup_{n=1}^\infty I\colon J^n\right)$? If not, is it true for noetherian rings?

Answer 1

I assume that $I,J$ are ideals of a commutative ring $R$?

After replacing $R$ by $R/I$ we may clearly assume that $I=0$, so that $0:J^n=\mathrm{Ann}(J^n)$.

Let $I \subseteq R$ be an ideal. Then the following are equivalent:

$\overline{\mathrm{Spec}(R) \setminus V(J)} \subseteq V(I)\\ \mathrm{Spec}(R) \setminus V(J) \subseteq V(I)\\ V(J) \cup V(I)=\mathrm{Spec}(R)\\ IJ \subseteq \sqrt{0}\\I \subseteq (\sqrt{0}:J) \\ V(\sqrt{0}:J) \subseteq V(I)$

We have used here that $(\sqrt{0}:J)$ is a radical ideal, which is easy to prove. This proves $$\overline{\mathrm{Spec}(R) \setminus V(J)} = V\bigl((\sqrt{0}:J)\bigr)$$

We always have $(\sqrt{0}:J) \supseteq \sqrt{\bigcup_{n \geq 1} \mathrm{Ann}(J^n)}$, and the converse inclusion holds when $J$ is finitely generated (this is easy to check). In general we won't have equality.