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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?

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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.

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Hmmm, is it true that gluing together closed morphisms always gives a closed morphism? – only Apr 21 '12 at 22:00
Oh, I think I can prove it works if there are a finite number of closed morphisms, just by gluing the images and then using finite union of closed subschemes are closed. I don't see how to do it here in general though, because I need to glue together infinitely many... – only Apr 21 '12 at 22:07
There is no problem in the infinite case too. If you have an arbitrary open covering $(X_t)$ of a scheme $X$ and a closed subscheme $U_t\subset X_t$ for every $t$, such that $U_s\cap X_t$ and $U_t\cap X_s$ are the same closed subscheme of $U_t\cap U_s$, then you have a unique closed subscheme $U\subset X$ s.t. $U\cap X_t=U_t$ for any $t$. You may take this property as a synonymous of having a closed immersion. Apply it to $X=\textrm{Proj}\,\,S$ and $X_t=D_+(t)$, $U_t=D_+(\phi(t))$. – Brenin Apr 21 '12 at 22:29
Note that for every $t,s\in S_1$, the morphisms of rings $\phi_t$ and $\phi_s$ coincide on $S_{(ts)}$, the localization corresponding to $D_+(t)\cap D_+(s)$. So the condition in the assumption is satisfied. – Brenin Apr 21 '12 at 22:31

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