# 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?

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.

• 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

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.