Lemma 27.4.4. In Situation 27.3.1. The scheme $\pi : \underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A}) \to S$ constructed in Lemma 27.3.4 and the scheme representing the functor $F$ are canonically isomorphic as schemes over $S$.

Proof. Let $X \to S$ be the scheme representing the functor $F$. Consider the sheaf of $\mathcal{O}_ S$-algebras $\mathcal{R} = \pi _*\mathcal{O}_{\underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A})}$. By construction of $\underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A})$ we have isomorphisms $\mathcal{A}(U) \to \mathcal{R}(U)$ for every affine open $U \subset S$; this follows from Lemma 27.3.4 part (1). For $U \subset U' \subset S$ open these isomorphisms are compatible with the restriction mappings; this follows from Lemma 27.3.4 part (2). Hence by Sheaves, Lemma 6.30.13 these isomorphisms result from an isomorphism of $\mathcal{O}_ S$-algebras $\varphi : \mathcal{A} \to \mathcal{R}$. Hence this gives an element $(\pi , \varphi ) \in F(\underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A}))$. Since $X$ represents the functor $F$ we get a corresponding morphism of schemes $can : \underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A}) \to X$ over $S$.

Let $U \subset S$ be any affine open. Let $F_ U \subset F$ be the subfunctor of $F$ corresponding to pairs $(f, \varphi )$ over schemes $T$ with $f(T) \subset U$. Clearly the base change $X_ U$ represents $F_ U$. Moreover, $F_ U$ is represented by $\mathop{\mathrm{Spec}}(\mathcal{A}(U)) = \pi ^{-1}(U)$ according to Lemma 27.4.2. In other words $X_ U \cong \pi ^{-1}(U)$. We omit the verification that this identification is brought about by the base change of the morphism $can$ to $U$. $\square$

Comment #8433 by on

I think the expression $(\underline{\text{Spec}}_S(\mathcal{A}), \varphi) \in F(\underline{\text{Spec}}_S(\mathcal{A}))$ should be $(\pi, \varphi)\in F(\underline{\text{Spec}}_S(\mathcal{A}))$ instead. Here's an alternative way to phrase the proof that doesn't require invoking $X$: "Let $U\subset S$ be open affine. Since the functor $F_U$ is represented by $\pi^{-1}(U)$, let $(\pi^U,\varphi^U)\in F_U(\pi^{-1}(U))$ be the universal element. By the proof of 26.15.4 (we verified the hypotheses in the proof of 27.4.3), it suffices to see that $(\pi,\varphi)|_U=(\pi^U,\varphi^U)$. This follows from the construction of $\pi$ and $\varphi$."

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