If $I$ is a homogeneous ideal of $A$ contained in $A_+$, then $\sqrt{I} = \bigcap\limits_{I\subset P\in\text{Proj }A} P$?

Question

EDIT:

This is from an exercise of Vakil's Foundations of Algebraic Geometry.

4.5.H: Suppose $I$ is any homogeneous ideal of $S$ contained in $S_+$, and if $f$ is a homogeneous element of positive degree, show that $f$ vanishes on $V(I)$, i.e. $V(I)\subset V(f)$ iff $f^n\in I$ for some $n$.

The definition of $V(I)$ is the set of all homogeneous prime ideals containing $I$ but not $S_+$.

My thoughts:

Now one direct is clear. I want to show the reverse. I think it translates to the following: $$\sqrt{I} = \bigcap_{I\subset P\in\text{Spec }A} P = \bigcap_{I\subset P\in\text{Proj }A} P.$$ But this seems to be false? How should I go about then?

Answer 1

Hint. Reduce the problem modulo $I$ and then consider the localization $S_f$. Recall that any graded ring has a graded prime ideal.