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Vakil's 4.5.H reads as follows

Suppose $I$ is any homogeneous ideal of $S_•$ contained in $S_+$, and $f$ is a homogeneous element of positive degree. Show that $f$ vanishes on $V(I)$ (i.e., $V(I) ⊂ V(f)$) if and only if $f^n ∈ I$ for some $n$. (Hint: Mimic the affine case; see Exercise 3.4.J.)

It seems like one cannot exactly mimic the affine case without knowing something like "for a homogenus ideal $I$, $\sqrt{I}$ is the intersection of all homogeneous primes that contain it." (EDIT: Maybe a better way of saying this is that minimal primes of a homogeneous ideal are homogeneous. )But this statement has not been mentioned so far in Vakil, so I am wondering if there is an easier way to do it.

A similar question has been asked here, but that seemed to use a lot of results that had not been developed in Vakil (yet).

William C. Newman
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    One minor typographical thing: your quoted section from Vakil seems to have lost its math formatting. Secondly, in Vakil's introduction, he says that "[t]he reader should be familiar with some basic notions in commutative ring theory, in particular the notion of ideals ... and localization." I think the answer to the linked question is that it exactly uses these ideas, which Vakil says one should already be somewhat familiar with (and thus he doesn't develop these ideas in the text). If this is unsatisfying, what level of technology would suffice for "easier"? – KReiser May 23 '20 at 22:09
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    Whatever you are saying is in fact true. If you have a homogeneous ideal $I$, then $I$ is the intersection of homogeneous prime ideals containing $I$. To see this, show that $p\in \text{Spec }S$, then $p^h=\langle x\in p | x \text{ is homogeneous} \rangle$ is in fact a homogeneous prime ideal. – user6 May 23 '20 at 23:31
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    @Soumik: I think you meant to say that $\color{red}{\sqrt I}$ (not $I$) is the intersection of the prime ideals containing $I$. – Aryaman Maithani Jun 05 '21 at 19:07
  • Oh yes of course! – user6 Jun 08 '21 at 01:02

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$\DeclareMathOperator{\Proj}{Proj}\newcommand{\fp}{\mathfrak{p}}$Let $I$ and $f$ be as mentioned in the post.

First, assume that $V(I) \subset V(f)$. We wish to show that $f \in \sqrt{I}$.
Suppose not. Then, $f \notin \fp$ for some ordinary prime $\fp$. Now, consider its homogenisation $\fp^{\ast}$. Since $\fp^{\ast}$ is a subset of $\fp$, it follows that $f \notin \fp^{\ast}$.
On the other hand, since $I$ is homogeneous, it follows that $I \subset \fp^{\ast}$. Thus, $\fp^{\ast} \in V(I) \setminus V(f)$, a contradiction.

Conversely, assume that $f \in \sqrt{I}$. Let $\fp$ be a homogeneous prime containing $I$. Then, $f \in \sqrt{I} \subset \sqrt{\fp} = \fp$ showing that $\fp$ contains $f$. This proves $V(I) \subset V(f)$.

The only "extra bit" which looks like Vakil had not mentioned is the boldened part. That is not particularly difficult to show.
By definition, $\fp^{\ast} = \bigoplus_{n \geqslant 0} (\fp \cap S_n)$. It is clear that $\mathfrak{p}^\ast$ is an ideal and is homogeneous, almost by construction. Since $\fp^{\ast} \subset \fp$, it also follows that $\fp^{\ast}$ does not contain $S_+$.
To see that it is prime, it suffices to only check for homogeneous elements. (Exercise 4.5.C. (c))

Suppose $a, b$ are homogeneous such that $ab \in \mathfrak{p}^\ast$. Then, $ab \in \mathfrak{p} \cap S_n$, where $n = \deg(ab).$ Thus, $ab \in \mathfrak{p}$ and hence, $a \in \mathfrak{p}$ or $b \in \mathfrak{p}.$ In either case, $a \in \mathfrak{p}^\ast$ or $b \in \mathfrak{p}^\ast.$

Aryaman Maithani
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