Proof. Let $U_ d \to S$ be the scheme representing the functor $F_ d$ defined above. Let $\mathcal{L}_ d$, $\psi ^ d : \pi _ d^*\mathcal{A}^{(d)} \to \bigoplus _{n \geq 0} \mathcal{L}_ d^{\otimes n}$ be the universal object. If $d | d'$, then we may consider the quadruple $(d', \pi _ d, \mathcal{L}_ d^{\otimes d'/d}, \psi ^ d|_{\mathcal{A}^{(d')}})$ which determines a canonical morphism $U_ d \to U_{d'}$ over $S$. By construction this morphism corresponds to the transformation of functors $F_ d \to F_{d'}$ defined above.

For every affine open $\mathop{\mathrm{Spec}}(R) = V \subset S$ setting $A = \Gamma (V, \mathcal{A})$ we have a canonical identification of the base change $U_{d, V}$ with the corresponding open subscheme of $\text{Proj}(A)$, see Lemma 27.16.2. Moreover, the morphisms $U_{d, V} \to U_{d', V}$ constructed above correspond to the inclusions of opens in $\text{Proj}(A)$. Thus we conclude that $U_ d \to U_{d'}$ is an open immersion.

This allows us to construct $X$ by glueing the schemes $U_ d$ along the open immersions $U_ d \to U_{d'}$. Technically, it is convenient to choose a sequence $d_1 | d_2 | d_3 | \ldots$ such that every positive integer divides one of the $d_ i$ and to simply take $X = \bigcup U_{d_ i}$ using the open immersions above. It is then a simple matter to prove that $X$ represents the functor $F$. $\square$

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