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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?

user26857
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Hint. Reduce the problem modulo $I$ and then consider the localization $S_f$. Recall that any graded ring has a graded prime ideal.

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user26857
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  • So the statement regarding the radical is correct? – mez Feb 18 '15 at 00:58
  • @mez Localizing at $f$ throw away the irrelevant maximal ideal $S_+$, this is why the hint works. For the radical you also want to lose it, but I'm afraid this is not possible: if $I=(X^2,Y)$ in $S=K[X,Y]$, then its radical is exactly $S_+$. (There is no other homogeneous prime containing $I$ excepting $S_+$.) But the radical of $I$ is the intersection of all homogeneous primes containing $I$. – user26857 Feb 18 '15 at 01:03
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    I see, so the equality should be $\sqrt{I} = A_+\cap \bigcap\limits_{I\subset P\in \text{Proj } A} P = \bigcap\limits_{I\subset P} P$ where all ideals considered are homogeneous. – mez Feb 18 '15 at 05:41