Lemma 27.16.2. In Situation 27.15.1. Let $F_ d$ be the functor associated to $(d, S, \mathcal{A})$ above. If $S$ is affine, then $F_ d$ is representable by the open subscheme $U_ d$ (27.12.0.1) of the scheme $\text{Proj}(\Gamma (S, \mathcal{A}))$.

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 $F_ d = F_ d^{triples}$. $\square$

Comment #4328 by Anu on

In the proof, it is not clear to me why "we may think of $\psi$ as a $\mathcal{O}_{S}$-module map $\mathcal{A}^{(d)} \rightarrow \oplus_{n \geq 0} f_{*}\mathcal{L}^{\otimes n}$". I see that the adjunction $f^{*} \dashv f_{*}$ has been used, but how has the direct image functor commuted past the direct sum?

Comment #4329 by on

@#4328. Because $\psi$ is a graded $\mathcal{O}_T$-algebra map, which requires the summand $f^*\mathcal{A}^{nd}$ to be mapped into the summand $\mathcal{L}^{\otimes n}$. Then apply the adjunction to the map between these summands.

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