Radical ideals and $\operatorname{Proj}$ It is well-known that in $\operatorname{Spec}A$, $V(I)\subset V(J)$ implies $\sqrt{I}\supset\sqrt{J}$. Is it also true in $\operatorname{Proj}A$, where $A$ is an $\mathbb{N}$-graded ring and $I,J\subset A$ are graded ideals? The difficulty in proving this seems to stem from the condition that the elements of $\operatorname{Proj}A$ do not contain  $\sum_{d>0}A_d$.
 A: It is not true: $\mathbb{Z}$ can be considered as an $\mathbb{N}$-graded ring, with $\mathbb{Z}_+=\{0\}$. Then $\operatorname{Proj}\mathbb{Z}=\emptyset$, and if $I=(2)$ and $J=(3)$, then $\emptyset=V_+(I)\subset V_+(J)=\emptyset$, but $\sqrt{I}=I\not\supset J=\sqrt{J}$.
A: Proposition.
Let $J,K\subset S$ be graded ideals. TFAE:


*

*$V_+(J)\subset V_+(K)$

*$(\sqrt{J})_+\supset (\sqrt{K})_+$

*$\sqrt{J}\supset K_+$


In particular, $V_+(J)=V_+(K)$ if and only if $(\sqrt{J})_+=(\sqrt{K})_+$.
Proof. 1$\Rightarrow$2 : Let $\hat{J}:=\bigcap_{\mathfrak{p}\in V_+(J)}\mathfrak{p}$. Using Zorn's Lemma, one can show that $\sqrt{J}$ is the intersection of all graded prime ideals containing $J$. Hence, $\hat{J}_+=(\sqrt{J})_+$.
1$\Rightarrow$2 : If $V_+(J)\subset V_+(K)$, then $\hat{J}\supset\hat{K}$, so $(\sqrt{J})_+=\hat{J}_+\supset\hat{K}_+=(\sqrt{K})_+$.
2$\Rightarrow$3 : $\sqrt{J}\supset(\sqrt{J})_+\supset(\sqrt{K})_+\supset K_+$
3$\Rightarrow$1 : $V_+(J)=V_+(\sqrt{J})\subset V_+(K_+)=V_+(K)$.
Remark. Since $(\sqrt{J_+})_+=(\sqrt{J})_+$, the inclusion $V_+(J)\subset V_+(K)$ depends only on $J_+$ and $K_+$.
