Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

References

$[1]$ 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.

share|improve this question

2 Answers 2

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.

share|improve this answer
    
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, $\mathfrak{p} \supseteq \Psi(\Phi(\mathfrak{p}))$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.