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It is well-known that if $S$ is a graded ring, and $f$ is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localization $S_f$ and the prime ideals of $S_{(f)}$, the subring of $S_f$ comprising the homogeneous elements of degree $0$.

This is proposition II.2.5b in Hartshorne, exercise 5.5B in Ravi Vakil's notes $[1]$ (p. $130$ of the February $24$, $2012$ version), and proposition 8.1.21 of Akhil Mathew's notes $[2]$ (p. $136$). Unfortunately I cannot follow any of those proofs to my own satisfaction, perhaps because I'm not well-versed in commutative algebra.

The crux of the proof appears to be to show that, given a homogeneous prime ideal $\mathfrak{p}$ of $S$ not containing $f$, the construction used to obtain a prime ideal $\Psi (\mathfrak{q})$ of $S$ from a prime ideal $\mathfrak{q}$ of $S_{(f)}$ will recover $\mathfrak{p}$ when $\mathfrak{q} = S_f \mathfrak{p} \cap S_{(f)}$. To be precise, let $\Psi (\mathfrak{q})$ be the homogeneous ideal of $S$ generated by $$\bigcup_{d \in \mathbb{N}} \{ s \in S_d : s / f^d \in \mathfrak{q} \}$$ and let $\Phi (\mathfrak{p}) = S_f \mathfrak{p} \cap S_{(f)}$. It's easy to see that $\Phi \circ \Psi$ acts as the identity on $\operatorname{Spec} A$ (or, for that matter, the set of all ideals of $A$), but I cannot see any obvious reason why $\Psi \circ \Phi$ should act as the identity on the set of prime ideals of $S$ not containing $f$. A detailed proof of this point would be much appreciated.


$[1]$ Foundations of Algebraic Geometry.

$[2]$ Algebraic geometry notes (covers material at the level of the first and second volume of EGA): html page, and pdf file.

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Let me try. I follow the hints from Ravi's notes.

For each prime ideal $\mathfrak{p}$ of $S_{(f)}$, define $\alpha(\mathfrak{p})=\mathfrak{q}$ a prime homogeneous ideal of $S_f$ as $\oplus Q_i$, where $Q_i\subset (S_f)_i$ and $a\in Q_i$ iff $a^{\deg f}/f^i\in \mathfrak{p}$. Then show $\mathfrak{q}$ is a homogeneous prime ideal. (I think this is the hard part of the proof, but there are wonderful hints in Ravi's notes.)

Define $\beta:\mathrm{Proj}(S_f) \to \mathrm{Spec}(S_{(f)})$, $\beta(\mathfrak{q})=\mathfrak{q}\cap S_{(f)}$.

Of course $\beta\alpha=1$ (notice that $Q_0=\mathfrak{p}$).

Another direction: Let $\mathfrak{q}=\oplus Q_i\in \mathrm{Proj}(S_f)$, $a\in Q_i$, then the degree of $a^{\deg f}/f^i$ is zero, so $a^{\deg f}/f^i\in \mathfrak{q}\cap S_{(f)}$, that is to say $\alpha\beta(\mathfrak{q})\supset\mathfrak{q}$, but if $a\in (S_f)_i$ and $a^{\deg f}/f^i\in Q_0\subset \mathfrak{q}$, since $\mathfrak{q}$ is prime, then $a\in \mathfrak{q}$. Hence $\alpha\beta=1$.

Sorry, maybe this is not what you want.

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It is not hard to show that $\mathfrak{q}$ (your notation) is a homogeneous prime ideal, unless I'm missing a subtle point. (The hints given in Ravi Vakil's notes seem superfluous to me.) I'm not entirely convinced by your argument to show that $\mathfrak{q} \supseteq \alpha \beta (\mathfrak{q})$, but I think I've found a fix I'm happy with... – Zhen Lin Jul 30 '11 at 3:37
up vote 1 down vote accepted

I guess I should post an answer and resolve this question...

Let $D = \{ \mathfrak{p} \in \operatorname{Proj} S : f \notin \mathfrak{p} \}$. Suppose $\mathfrak{p} \in D$. If $s \in \mathfrak{p} \cap S_d$, i.e. if $s$ is an element of $\mathfrak{p}$ of degree $d$, then $s / f^d \in \Phi (\mathfrak{p})$, so certainly $s \in \Psi (\Phi (\mathfrak{p}))$, and thus $\mathfrak{p} \subseteq \Psi (\Phi(\mathfrak{p}))$.

Conversely, if $s \in \Psi (\Phi (\mathfrak{p})) \cap S_d$, then $s / f^d \in \Phi (\mathfrak{p})$, so there is some $s' \in \mathfrak{p}$ such that $s' / f^{d'} = s / f^d$. This implies, for some $e$, $f^e ( f^d s' - f^{d'} s ) = 0$. Observe that $s' \in \mathfrak{p}$ but $f^{d' + e} \notin \mathfrak{p}$, so $s \in \mathfrak{p}$ since $\mathfrak{p}$ is a homogeneous prime ideal. Hence, $\mathfrak{p} \supseteq \Psi(\Phi(\mathfrak{p}))$.

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