On verifying Proj S is a scheme In Hartshorne II Prop 2.5, it says $D_+{(f)}$ is homeomorphic to $\text{Spec}(S_{(f)})$, but I cannot prove it. Since $D_+{(f)}$ homeomorphic to $S_f$, I have to show $\text{Spec}(S_{(f)})$ homeomorphic to $\text{Spec}(S_f)$.
 Take $p \in \text{Spec} S$. I cannot show $S_f(pS_f \cap pS_(f))=pS_f$.
 A: You already defined the (homeo)morphism $\phi:D_+(f)\to\textrm{Spec}\,S_{(f)}$ by $\mathfrak p\mapsto \mathfrak pS_f\cap S_{(f)}$. Its inverse $\psi$ sends $\mathfrak q\mapsto \ell^{-1}(\mathfrak qS_f)$, where $\ell:S\to S_f$ is localization at $f$. (I write $\ell^{-1}(-)$ instead of $S\cap -$ as I do not have Hartshorne's book here and cannot check if $S$ is integral.)
Hint: It should be clear that $\mathfrak q\subset \phi(\psi(\mathfrak q))$ for every $\mathfrak q\in \textrm{Spec}\,S_{(f)}$. For the reverse inclusion, show that given an element $x\in \phi(\psi(\mathfrak q))$, a sufficiently high power of $x$ lies in $\mathfrak q$.
To show that $Y=\textrm{Proj}\,S$ is a scheme, you may set $\mathscr O_Y(D_+(f))=S_{(f)}$ and notice that for every open covering $D_+(f)=\bigcup_iD_+(f_i)$ the sequence
$$0\to S_{(f)}\to\prod_iS_{(f_i)}\rightrightarrows\prod_{i,j}S_{(f_if_j)}$$ is an equalizer diagram. As $f$ varies in $S_+$, the $D_+(f)$ generate the topology of $Y$, so we get a uniquely determined sheaf $\mathscr O_Y$ on the whole $Y$. And finally, by construction, $(D_+(f),\mathscr O_Y|_{D_+(f)})\cong (\textrm{Spec}\,S_{(f)},\mathscr O_{\textrm{Spec}\,S_{(f)}})$.
