The Stacks project

Proof. Write $S = \mathop{\mathrm{Spec}}(R)$ and $A = \Gamma (S, \mathcal{A})$. Then $A$ is a graded $R$-algebra and $\mathcal{A} = \widetilde A$. To prove the lemma we have to identify the functor $F_ d$ with the functor $F_ d^{triples}$ of triples defined in Section 27.12.

Let $(d, f : T \to S, \mathcal{L}, \psi )$ be a quadruple. We may think of $\psi $ as a $\mathcal{O}_ S$-module map $\mathcal{A}^{(d)} \to \bigoplus _{n \geq 0} f_*\mathcal{L}^{\otimes n}$. Since $\mathcal{A}^{(d)}$ is quasi-coherent this is the same thing as an $R$-linear homomorphism of graded rings $A^{(d)} \to \Gamma (S, \bigoplus _{n \geq 0} f_*\mathcal{L}^{\otimes n})$. Clearly, $\Gamma (S, \bigoplus _{n \geq 0} f_*\mathcal{L}^{\otimes n}) = \Gamma _*(T, \mathcal{L})$. Thus we may associate to the quadruple the triple $(d, \mathcal{L}, \psi )$.

Conversely, let $(d, \mathcal{L}, \psi )$ be a triple. The composition $R \to A_0 \to \Gamma (T, \mathcal{O}_ T)$ determines a morphism $f : T \to S = \mathop{\mathrm{Spec}}(R)$, see Schemes, Lemma 26.6.4. With this choice of $f$ the map $A^{(d)} \to \Gamma (S, \bigoplus _{n \geq 0} f_*\mathcal{L}^{\otimes n})$ is $R$-linear, and hence corresponds to a $\psi $ which we can use for a quadruple $(d, f : T \to S, \mathcal{L}, \psi )$. We omit the verification that this establishes an isomorphism of functors $F_ d = F_ d^{triples}$. $\square$