The bijection between homogeneous prime ideals of $S_f$ and prime ideals of $(S_f)_0$

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, $\mathfrak{p} \supseteq \Psi(\Phi(\mathfrak{p}))$.

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.