Lemma 27.4.2. In Situation 27.3.1. Let $F$ be the functor associated to $(S, \mathcal{A})$ above. If $S$ is affine, then $F$ is representable by the affine scheme $\mathop{\mathrm{Spec}}(\Gamma (S, \mathcal{A}))$.

Proof. Write $S = \mathop{\mathrm{Spec}}(R)$ and $A = \Gamma (S, \mathcal{A})$. Then $A$ is an $R$-algebra and $\mathcal{A} = \widetilde A$. The ring map $R \to A$ gives rise to a canonical map

$f_{univ} : \mathop{\mathrm{Spec}}(A) \longrightarrow S = \mathop{\mathrm{Spec}}(R).$

We have $f_{univ}^*\mathcal{A} = \widetilde{A \otimes _ R A}$ by Schemes, Lemma 26.7.3. Hence there is a canonical map

$\varphi _{univ} : f_{univ}^*\mathcal{A} = \widetilde{A \otimes _ R A} \longrightarrow \widetilde A = \mathcal{O}_{\mathop{\mathrm{Spec}}(A)}$

coming from the $A$-module map $A \otimes _ R A \to A$, $a \otimes a' \mapsto aa'$. We claim that the pair $(f_{univ}, \varphi _{univ})$ represents $F$ in this case. In other words we claim that for any scheme $T$ the map

$\mathop{\mathrm{Mor}}\nolimits (T, \mathop{\mathrm{Spec}}(A)) \longrightarrow \{ \text{pairs } (f, \varphi )\} ,\quad a \longmapsto (f_{univ} \circ a, a^*\varphi )$

is bijective.

Let us construct the inverse map. For any pair $(f : T \to S, \varphi )$ we get the induced ring map

$\xymatrix{ A = \Gamma (S, \mathcal{A}) \ar[r]^{f^*} & \Gamma (T, f^*\mathcal{A}) \ar[r]^{\varphi } & \Gamma (T, \mathcal{O}_ T) }$

This induces a morphism of schemes $T \to \mathop{\mathrm{Spec}}(A)$ by Schemes, Lemma 26.6.4.

The verification that this map is inverse to the map displayed above is omitted. $\square$

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