Hartshorne's proof of Proposition 2.5, Chapter II of his book Algebraic Geometry

Let $S = \sum_{n\ge 0} S_n$ be a graded commutative ring. Let $f$ be a homogeneous element of $S$ of degree $> 0$. Let $D_+(f) = \{\mathfrak{p} \in$ Proj $S| f \notin \mathfrak{p}\}$. Let $S_{(f)}$ be the degree $0$ part of the graded ring $S_f$, where $S_f$ is the localization with respect to the multiplicative set $\{1, f, f^2,\dots\}$. The proposition states that $D_+(f)$ is isomorphic to Spec $S_{(f)}$ as locally ringed spaces. Part of his proof is as follows. For $\mathfrak{p} \in D_+(f)$, let $\psi(\mathfrak{p}) = (\mathfrak{p}S_f) \cap S_{(f)}$. Then $\psi(\mathfrak{p}) \in$ Spec $S_{(f)}$. He wrote that the properties of localization show that $\psi\colon D_+(f) \rightarrow$ Spec $S_{(f)}$ is bijective. I wonder why $\psi$ is surjective.

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 Given $\mathfrak{p}_0\in\mathrm{Spec}(S_{(f)})$, the ideal $\mathfrak{p}_0S_f$ of $S_f$ is homogeneous, and its radical $\sqrt{\mathfrak{p}_0S_f}$ is homogeneous and prime, so of the form $\mathfrak{q}S_f$ for a homogeneous prime $\mathfrak{q}$ of $S$ with $f\notin\mathfrak{q}$. It can be verified that this $\mathfrak{q}$ maps to $\mathfrak{p}_0$ under $\psi$. – Keenan Kidwell Nov 12 '12 at 0:45

Let $d=\text{deg}(f)$. Given $\mathfrak{p}\in\text{Spec}\left(S_{(f)}\right)$, we define $I_m$ to be $\{b\in S_m|b^d/f^m\in \mathfrak{p}\}$. It can be checked that $I_m$ is an abelian group, that $I:=\oplus I_m$ is a homogeneous prime ideal of $S$ not containing $f$, and that $\psi(I)=\mathfrak{p}$.
 Could you explain why $I_m$ is an abelian group and $I$ is a prime ideal of $S$? – Makoto Kato Dec 22 '12 at 13:01 If $b,c\in S_m$, then $(b+c)^{2d}/f^{2m}\in\mathfrak{p}$ (expand the power of the sum and use that $b,c\in S_m$). Since $\mathfrak{p}$ is prime, this means $(b+c)^d/f^m\in\mathfrak{p}$. With this you prove that $I_m$ is an abelian group. To see that $I$ is prime, you need only check that, for homogeneous elements $f$, $g$ of $S$ such that $fg\in I$, then either $f\in I$ or $g\in I$. But then you use again that $\mathfrak{p}$ is prime, and you're done. – Maximo Carlavilla Jan 4 at 19:06 Sorry, when I wrote $S_m$ both times in the comment, I meant $I_m$. – Maximo Carlavilla Jan 4 at 19:15 Please add the explanation in you answer. – Makoto Kato Jan 4 at 22:07 @YACP The above answer is incomplete. Comments are not answers. The low accept rate is mostly due to (1) I answered some of my questions, (2) there are unanswered questions, (3) I did not accept incomplete answers(like hints). – Makoto Kato Jan 6 at 0:33