Lemma 27.16.6. In Situation 27.15.1. The scheme $\pi : \underline{\text{Proj}}_ S(\mathcal{A}) \to S$ constructed in Lemma 27.15.4 and the scheme representing the functor $F$ are canonically isomorphic as schemes over $S$.

**Proof.**
Let $X$ be the scheme representing the functor $F$. Note that $X$ is a scheme over $S$ since the functor $F$ comes equipped with a natural transformation $F \to h_ S$. Write $Y = \underline{\text{Proj}}_ S(\mathcal{A})$. We have to show that $X \cong Y$ as $S$-schemes. We give two arguments.

The first argument uses the construction of $X$ as the union of the schemes $U_ d$ representing $F_ d$ in the proof of Lemma 27.16.5. Over each affine open of $S$ we can identify $X$ with the homogeneous spectrum of the sections of $\mathcal{A}$ over that open, since this was true for the opens $U_ d$. Moreover, these identifications are compatible with further restrictions to smaller affine opens. On the other hand, $Y$ was constructed by glueing these homogeneous spectra. Hence we can glue these isomorphisms to an isomorphism between $X$ and $\underline{\text{Proj}}_ S(\mathcal{A})$ as desired. Details omitted.

Here is the second argument. Lemma 27.15.5 shows that there exists a morphism of graded algebras

over $Y$ which on sections over affine opens of $S$ agrees with (27.10.1.3). Hence for every $y \in Y$ there exists an open neighbourhood $V \subset Y$ of $y$ and an integer $d \geq 1$ such that for $d | n$ the sheaf $\mathcal{O}_ Y(n)|_ V$ is invertible and the multiplication maps $\mathcal{O}_ Y(n)|_ V \otimes _{\mathcal{O}_ V} \mathcal{O}_ Y(m)|_ V \to \mathcal{O}_ Y(n + m)|_ V$ are isomorphisms. Thus $\psi $ restricted to the sheaf $\pi ^*\mathcal{A}^{(d)}|_ V$ gives an element of $F_ d(V)$. Since the opens $V$ cover $Y$ we see “$\psi $” gives rise to an element of $F(Y)$. Hence a canonical morphism $Y \to X$ over $S$. Because this construction is completely canonical to see that it is an isomorphism we may work locally on $S$. Hence we reduce to the case $S$ affine where the result is clear. $\square$

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