Lemma 27.12.1 (01N8)—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 (7)

Comment #6488 by Yuto Masamura on August 09, 2021 at 08:14

In the last paragraph of the proof (i.e., the part of uniqueness), it seems that $\alpha^{\otimes n}(\psi^d_\varphi(x))=(\alpha')^{\otimes n}(\psi^d_{\varphi'}(x))$ implies $\psi^d_\varphi(x)=\psi^d_{\varphi'}(x)$ , but does this really hold? How can we prove that?

Comment #6560 by Johan on September 13, 2021 at 13:43

OK, I think you only deduce the uniqueness of $\alpha$ at the very end. In other words, we have already shown that $\varphi = \varphi'$ and we need to show that $\alpha = \alpha'$ . By definition of $U_d$ the sections $\varphi^* x$ for $x \in S_d$ generate $\mathcal{O}_{U_d}(d)$ and hence when you pull them back to $Y$ they generate $\varphi^*\mathcal{O}_{U_d}(d)$ . Thus there can be at most one map $\alpha : \varphi^*\mathcal{O}_{U_d}(d) \to \mathcal{L}$ which sends these sections to their images under $\psi$ . (To see specifically what you asked for: $\psi_\varphi^d$ is the composition of $\psi^d$ and pullback by $\varphi$ , hence if $\varphi = \varphi'$ then of course $\psi_\varphi^d = \psi_{\varphi'}^d$ .)

Comment #6615 by Yuto Masamura on September 24, 2021 at 03:47

Thanks for your comment, but I wanted to ask about the part of uniqueness of $\varphi$ . More precisely, the part 'we have $\alpha^{\otimes n}(\psi_\varphi^d(x))=(\alpha')^{\otimes n}(\psi_{\varphi'}^d(x))$ and $\psi_\varphi^d(x)=\varphi^\sharp(x/f^n)\psi_\varphi^d(f^n)$ and similarly for $\psi_{\varphi'}^d$ , and hence we have $\varphi^\sharp(x/f^n)=(\varphi')^\sharp(x/f^n)$ .'

But it's OK since I solved it by myself. At the time I misunderstood that you deduce $\psi_\varphi^d(x)=\psi_{\varphi'}^d(x)$ from the first equation. The correct way is as follows: first we deduce that $\alpha(\varphi^\sharp(x/f^n))\cdot\psi(f)^n=\alpha'((\varphi')^\sharp(x/f^n))\cdot\psi(f)^n$ over $Y_{\psi(f)}$ and hence $\alpha(\varphi^\sharp(x/f^n))=\alpha'((\varphi')^\sharp(x/f^n))$ . Since $\alpha$ and $\alpha'$ are the identity in degree $0$ (I overlooked this fact), we get the conclusion.

Comment #6856 by Johan on January 12, 2022 at 19:01

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

Comment #10907 by Elías Guisado on October 15, 2025 at 10:24

Typos:

Replace $\mathcal{O}$ by $\mathcal{O}_Y$ in “we define a ring map $\psi_{(f)} : S_{(f)} \to \Gamma(Y_s, \mathcal{O})$ ...,” in “we claim there is a unique ring map $S_{(ff')} \to \Gamma(Y_{ss'}, \mathcal{O})$ ” and in the commutative diagram relating $\psi_{(f)}$ with $\psi_{(f')}$ .

In “because the regular functions $s/s'$ and $\psi_{(f')}(f)$ agree,” replace $\psi_{(f')}(f)$ by $\psi_{(f')}(f/f')$ .

Regarding the “it remains to show that the commutativity of the diagram of the lemma singles out, given $\varphi$ , a unique $\alpha$ . We omit the verification.” I propose we say “restricting $\varphi=\varphi'$ to $Y_{\psi(f)}\to D_+(f)$ , by the commutativity of the diagram we have $\alpha(\varphi^*(\frac{f}{1}))=\alpha(\varphi^*(\psi^d(f)))=\psi(f)=\alpha'(\varphi^*(\frac{f}{1}))$ . That is, $\alpha$ and $\alpha'$ map the trivializing section $\varphi^*(\frac{f}{1})$ of $\varphi^*_f\mathcal{O}_{D_+(f)}(d)$ to the same (trivialing) section of $\mathcal{L}|_{Y_{\psi(f)}}$ . Therefore $\alpha=\alpha'$ .”

Finally (this is aesthetic rather than mathematical), is there some reason why the diagram in the statement is not displayed in this manner or similar?

Comment #10908 by Elías Guisado on October 15, 2025 at 10:26

Remarks to the proof for the vigilant reader:

The map $\psi_{(f)}$ can be induced from $\psi$ ; one does not need to define it by hand. Let's see how. We have that $n$ -fold multiplication by $s$ induces a map $\mathcal{O}_{Y_s}^{\otimes n}\to\mathcal{L}|_{Y_s}^{\otimes s}$ that is an isomorphism by Modules, Lemma 17.25.10. We denote the inverse of this map by $x\mapsto x/s^{\otimes n}$ , and to the composite $\tag{1}\label{iso} \mathcal{L}|_{Y_s}^{\otimes s}\xrightarrow{(-)/s^{\otimes n}}\mathcal{O}_{Y_s}^{\otimes n}\xrightarrow{\mu_n}\mathcal{O}_{Y_s}$ we denote $x\mapsto x/s^n$ , where $\mu_n$ is the isomorphism $a_1\otimes\cdots \otimes a_n\mapsto a_1\cdots a_n$ . Note that for any $t\in\mathcal{L}^{\otimes n}(V)$ with $V\subset Y_s$ , we have $t=(t/s^n_{|V})s^{\otimes n}_{|V}$ , for these have same image under \eqref{iso}. We have an isomorphism of graded rings $(-)/s^{\otimes *}:\Gamma_*(Y_s,\mathcal{L})\to\Gamma_*(Y_s,\mathcal{O}_Y)$ . The graded ring morphism $\tag{2}\label{comp} S^{(d)} \xrightarrow{\psi}\Gamma_*(Y_s,\mathcal{L}) \xrightarrow{(-)/s^{\otimes *}}\Gamma_*(Y_s,\mathcal{O}_Y) \hookrightarrow\Gamma_*(Y_s,\mathcal{O}_Y,\mathcal{O}_Y)$ (see Modules, Definition 17.25.7) sends $f$ to a unit, and hence it induces a $\mathbf{Z}$ -graded ring homomorphism $\psi_f:(S^{(d)})_f\to\Gamma_*(Y_s,\mathcal{O}_Y,\mathcal{O}_Y)$ (this violates the Stacks Project convention that rings are always $\mathbf{N}$ -graded). The degree $0$ part of \eqref{comp} is a map $(S^{(d)})_{(f)}\to\Gamma_0(Y_s,\mathcal{O}_Y,\mathcal{O}_Y)=\Gamma(Y_s,\mathcal{O}_Y)$ , that along with the identification $(S^{(d)})_{(f)}=S_{(f)}$ , is the map $\psi_{(f)} : S_{(f)} \to \Gamma(Y_s, \mathcal{O}_Y)$ referred in the proof. Indeed, for $a/f^n\in S_{(f)}$ we have $\begin{align*} \psi_{(f)}(a/f^n)&=\psi_{f}(a/f^n)\\ &=\psi_f(a)\psi_f(f)^{-n}\\ &=(\psi(a)/s^{\otimes n})(\psi(f)/s^{\otimes 1})^{-n}\\ &=(\psi(a)/s^{\otimes n})(1_1)^{-n}\\ &=(\psi(a)/s^{\otimes n})1_{-n}, \end{align*}$ where $1_i$ is the “ $i$ -th degree identity section” in $\Gamma_i(Y_s,\mathcal{O}_Y,\mathcal{O}_Y)$ (these satisfy $1_i1_j=1_{i+j}$ ). Therefore, $(\psi(a)/s^{\otimes n})1_{-n}=\mu_n(\psi(a)/s^{\otimes n})=\psi(a)/s^n$ , as desired.

Here's the proof of “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 $\alpha_f$ .” Let $a\in S_{nd}=(S^{(d)})_n$ . Then $\begin{matrix} S^{(d)} &\xrightarrow{\psi^d} &\Gamma_*(D_+(f),\mathcal{O}_{D_+(f)}(d)) &\xrightarrow{\varphi_f^*} &\Gamma_*(Y_s,\varphi_f^*\mathcal{O}_{D_+(f)}(d)) &\xrightarrow{\alpha_f} &\Gamma_*(Y_s,\mathcal{L})\\ a &\mapsto & \dfrac{a}{1}\in S(nd)_{(f)} &\mapsto & \dfrac{a}{1}\otimes 1=\dfrac{a}{f^n}(f^n\otimes 1) &\mapsto & \dfrac{a}{f^n}s^{\otimes n} =\dfrac{\psi(a)}{s^n}s^{\otimes n} =\psi(a) \end{matrix}$ where by $\dfrac{a}{1}\otimes 1$ and $f^n\otimes 1$ we actually mean the homomorphic images under $S(nd)_{(f)}\otimes_{S_{(f)}}\Gamma(Y_s,\mathcal{O}_Y) \to\Gamma_n(Y_s,\varphi^*_f\mathcal{O}_{D_+(f)}(d))$ (see #10899 and take into account 26.7.1). Therefore $\alpha_f\circ\varphi_f^*\circ\psi^d=\psi$ (if you're not satisfied yet, note we may assume $Y_s$ is affine and then pulling back sections is just mapping under the counit $M\to(M\otimes_AB)_A,x\mapsto x\otimes 1$ for the tensors $\dfrac{a}{1}\otimes 1$ and $f^n\otimes 1$ to make sense).

Regarding “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.” One the one hand, it follows that $\varphi^{-1}(U)=(\varphi')^{-1}(U)$ for every open $U\subset X$ ; on the other hand, use Lemma 2 of #10906.

Comment #10909 by Elías Guisado on October 15, 2025 at 10:39

Correction to #10907: Replace “therefore $\alpha=\alpha'$ ” by “therefore $\alpha|_{Y_{\psi(f)}}=\alpha'|_{Y_{\psi(f)}}$ , whence $\alpha=\alpha'$ , for $Y=\bigcup_{f\in S^d}Y_{\psi(f)}$ , since $\mathcal{L}$ is generated by $\psi(S_d)$ .”

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2 comment(s) on Section 27.12: Morphisms into Proj

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