Morphism between projective schemes induced by surjection of graded rings Ravi Vakil 9.2.B is "Suppose that $S \rightarrow R$ is a surjection of graded rings. Show that the induced morphism $\text{Proj }R \rightarrow  \text{Proj }S$ is a closed embedding."
I don't even see how to prove that the morphism is affine. The only ways I can think of to do this are to either classify the affine subspaces of Proj S, or to prove that when closed morphisms are glued, one gets a closed morphism.
Are either of those possible, and how can this problem be done?
 A: I think a good strategy could be to verify the statement locally, and then verify that the glueing is successful, as you said. Let us call $\phi:S\to R$ your surjective graded morphism, and $\phi^\ast:\textrm{Proj}\,\,R\to \textrm{Proj}\,\,S$ the corresponding morphism. Note that $$\textrm{Proj}\,\,R=\bigcup_{t\in S_1}D_+(\phi(t))$$
because $S_+$ (the irrelevant ideal of $S$) is generated by $S_1$ (as an ideal), so $\phi(S_+)R$ is generated by $\phi(S_1)$. For any $t\in S_1$ you have a surjective morphism
$S_{(t)}\to R_{\phi(t)}$ (sending $x/t^n\mapsto \phi(x)/\phi(t)^n$, for any $x\in S$), which corresponds to the canonical closed immersion of affine schemes $\phi^\ast_t:D_+(\phi(t))\hookrightarrow D_+(t)$. It remains to glue the $\phi^\ast_t$'s.
A: Almost 7 years late!  Here is my try.  Hallo Thorsten!
I call our maps $f \colon \operatorname{Proj} B \to \operatorname{Proj} A$ and $\varphi \colon A \to B$.  Surjectivity implies that we actually have a well-defined map $\operatorname{Proj} B \to \operatorname{Proj} A$ and a morphism of schemes in this way.
Being a closed immersion is affine-local on the target.  Therefore we can consider some cover of open affines $\bigcup_{j \in J} V_j = \operatorname{Proj} A$ and then check that for each $j \in J$ we have a closed immersion $f \mid_{f^{-1}(V_j)} \colon f^{-1}(V_j) \hookrightarrow V_j$.  This is described in Vakil's notes as an exercise.
We have that the collection over all homogeneous $g \in A$ of $D(g) = \{\,p \in \operatorname{Proj} A \mid g \notin p \,\}$ cover $\operatorname{Proj} A$.  As $\varphi$ is surjective, we have $f^{-1} (D(g)) = D(\varphi(g))$.
We now have
\begin{align*}
f \mid_{D(\varphi(g))} \colon D(\varphi(g)) & \hookrightarrow D(g)  \\
p & \mapsto \varphi^{-1} (p) \, .
\end{align*}
These sets are all open affines!  For any graded ring $R$, we have for any homogeneous $h \in R$ the identification $D(h) = \operatorname{Spec}(R_h)_0 = \operatorname{Spec}\{\, \frac{x}{h^n} \mid n \in \mathbb N, \, \deg x = \deg h \cdot n \,\}$.  (Sometimes, $(R_h)_0$ is confusingly written as $R_{(h)}$.)  Our map can then be seen as
\begin{align*}
f \mid_{\operatorname{Spec} (B_{\varphi(g)})_0} \colon \operatorname{Spec} (B_{\varphi(g)})_0 & \hookrightarrow \operatorname{Spec} (A_g)_0  \\
p & \mapsto \varphi^{-1} (p) \; ,
\end{align*}
which corresponds to the surjective ring homomorphism
\begin{align*}
\varphi (D(g)) \colon (A_g)_0 & \to (B_{\varphi(g)})_0  \\
\frac{x}{g^n} & \mapsto \frac{\varphi(x)}{\varphi(g)^n} \; ,
\end{align*}
which means that $f \mid_{f^{-1}(D(g))} \colon f^{-1}(D(g)) \hookrightarrow D(g)$ is a closed immersion, concluding that $f$ is a closed immersion.
