The Stacks project

Lemma 27.12.1. Let $S$ be a graded ring, and $X = \text{Proj}(S)$. Let $d \geq 1$ and $U_ d \subset X$ as above. Let $Y$ be a scheme. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Let $\psi : S^{(d)} \to \Gamma _*(Y, \mathcal{L})$ be a graded ring homomorphism such that $\mathcal{L}$ is generated by the sections in the image of $\psi |_{S_ d} : S_ d \to \Gamma (Y, \mathcal{L})$. Then there exist a morphism $\varphi : Y \to X$ such that $\varphi (Y) \subset U_ d$ and an isomorphism $\alpha : \varphi ^*\mathcal{O}_{U_ d}(d) \to \mathcal{L}$ such that $\psi _\varphi ^ d$ agrees with $\psi $ via $\alpha $:

\[ \xymatrix{ \Gamma _*(Y, \mathcal{L}) & \Gamma _*(Y, \varphi ^*\mathcal{O}_{U_ d}(d)) \ar[l]^-\alpha & \Gamma _*(U_ d, \mathcal{O}_{U_ d}(d)) \ar[l]^-{\varphi ^*} \\ S^{(d)} \ar[u]^\psi & & S^{(d)} \ar[u]^{\psi ^ d} \ar[ul]^{\psi ^ d_\varphi } \ar[ll]_{\text{id}} } \]

commutes. Moreover, the pair $(\varphi , \alpha )$ is unique.

Proof. Pick $f \in S_ d$. Denote $s = \psi (f) \in \Gamma (Y, \mathcal{L})$. On the open set $Y_ s$ where $s$ does not vanish multiplication by $s$ induces an isomorphism $\mathcal{O}_{Y_ s} \to \mathcal{L}|_{Y_ s}$, see Modules, Lemma 17.25.10. We will denote the inverse of this map $x \mapsto x/s$, and similarly for powers of $\mathcal{L}$. Using this we define a ring map $\psi _{(f)} : S_{(f)} \to \Gamma (Y_ s, \mathcal{O})$ by mapping the fraction $a/f^ n$ to $\psi (a)/s^ n$. By Schemes, Lemma 26.6.4 this corresponds to a morphism $\varphi _ f : Y_ s \to \mathop{\mathrm{Spec}}(S_{(f)}) = D_{+}(f)$. We also introduce the isomorphism $\alpha _ f : \varphi _ f^*\mathcal{O}_{D_{+}(f)}(d) \to \mathcal{L}|_{Y_ s}$ which maps the pullback of the trivializing section $f$ over $D_{+}(f)$ to the trivializing section $s$ over $Y_ s$. With this choice the commutativity of the diagram in the lemma holds with $Y$ replaced by $Y_ s$, $\varphi $ replaced by $\varphi _ f$, and $\alpha $ replaced by $\alpha _ f$; verification omitted.

Suppose that $f' \in S_ d$ is a second element, and denote $s' = \psi (f') \in \Gamma (Y, \mathcal{L})$. Then $Y_ s \cap Y_{s'} = Y_{ss'}$ and similarly $D_{+}(f) \cap D_{+}(f') = D_{+}(ff')$. In Lemma 27.10.6 we saw that $D_{+}(f') \cap D_{+}(f)$ is the same as the set of points of $D_{+}(f)$ where the section of $\mathcal{O}_ X(d)$ defined by $f'$ does not vanish. Hence $\varphi _ f^{-1}(D_{+}(f') \cap D_{+}(f)) = Y_ s \cap Y_{s'} = \varphi _{f'}^{-1}(D_{+}(f') \cap D_{+}(f))$. On $D_{+}(f) \cap D_{+}(f')$ the fraction $f/f'$ is an invertible section of the structure sheaf with inverse $f'/f$. Note that $\psi _{(f')}(f/f') = \psi (f)/s' = s/s'$ and $\psi _{(f)}(f'/f) = \psi (f')/s = s'/s$. We claim there is a unique ring map $S_{(ff')} \to \Gamma (Y_{ss'}, \mathcal{O})$ making the following diagram commute

\[ \xymatrix{ \Gamma (Y_ s, \mathcal{O}) \ar[r] & \Gamma (Y_{ss'}, \mathcal{O}) & \Gamma (Y_{s, '} \mathcal{O}) \ar[l]\\ S_{(f)} \ar[r] \ar[u]^{\psi _{(f)}} & S_{(ff')} \ar[u] & S_{(f')} \ar[l] \ar[u]^{\psi _{(f')}} } \]

It exists because we may use the rule $x/(ff')^ n \mapsto \psi (x)/(ss')^ n$, which “works” by the formulas above. Uniqueness follows as $\text{Proj}(S)$ is separated, see Lemma 27.8.8 and its proof. This shows that the morphisms $\varphi _ f$ and $\varphi _{f'}$ agree over $Y_ s \cap Y_{s'}$. The restrictions of $\alpha _ f$ and $\alpha _{f'}$ agree over $Y_ s \cap Y_{s'}$ because the regular functions $s/s'$ and $\psi _{(f')}(f)$ agree. This proves that the morphisms $\psi _ f$ glue to a global morphism from $Y$ into $U_ d \subset X$, and that the maps $\alpha _ f$ glue to an isomorphism satisfying the conditions of the lemma.

We still have to show the pair $(\varphi , \alpha )$ is unique. Suppose $(\varphi ', \alpha ')$ is a second such pair. Let $f \in S_ d$. By the commutativity of the diagrams in the lemma we have that the inverse images of $D_{+}(f)$ under both $\varphi $ and $\varphi '$ are equal to $Y_{\psi (f)}$. Since the opens $D_{+}(f)$ are a basis for the topology on $X$, and since $X$ is a sober topological space (see Schemes, Lemma 26.11.1) this means the maps $\varphi $ and $\varphi '$ are the same on underlying topological spaces. Let us use $s = \psi (f)$ to trivialize the invertible sheaf $\mathcal{L}$ over $Y_{\psi (f)}$. By the commutativity of the diagrams we have that $\alpha ^{\otimes n}(\psi ^ d_{\varphi }(x)) = \psi (x) = (\alpha ')^{\otimes n}(\psi ^ d_{\varphi '}(x))$ for all $x \in S_{nd}$. By construction of $\psi ^ d_{\varphi }$ and $\psi ^ d_{\varphi '}$ we have $\psi ^ d_{\varphi }(x) = \varphi ^\sharp (x/f^ n) \psi ^ d_{\varphi }(f^ n)$ over $Y_{\psi (f)}$, and similarly for $\psi ^ d_{\varphi '}$. By the commutativity of the diagrams of the lemma we deduce that $\varphi ^\sharp (x/f^ n) = (\varphi ')^\sharp (x/f^ n)$. This proves that $\varphi $ and $\varphi '$ induce the same morphism from $Y_{\psi (f)}$ into the affine scheme $D_{+}(f) = \mathop{\mathrm{Spec}}(S_{(f)})$. Hence $\varphi $ and $\varphi '$ are the same as morphisms. Finally, it remains to show that the commutativity of the diagram of the lemma singles out, given $\varphi $, a unique $\alpha $. We omit the verification. $\square$

Comments (4)

Comment #6488 by Yuto Masamura on

In the last paragraph of the proof (i.e., the part of uniqueness), it seems that implies , but does this really hold? How can we prove that?

Comment #6560 by on

OK, I think you only deduce the uniqueness of at the very end. In other words, we have already shown that and we need to show that . By definition of the sections for generate and hence when you pull them back to they generate . Thus there can be at most one map which sends these sections to their images under . (To see specifically what you asked for: is the composition of and pullback by , hence if then of course .)

Comment #6615 by Yuto Masamura on

Thanks for your comment, but I wanted to ask about the part of uniqueness of . More precisely, the part 'we have and and similarly for , and hence we have .'

But it's OK since I solved it by myself. At the time I misunderstood that you deduce from the first equation. The correct way is as follows: first we deduce that over and hence . Since and are the identity in degree (I overlooked this fact), we get the conclusion.

Comment #6856 by on

I've decided to leave this as is for now.

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