Lemma 27.4.3. In Situation 27.3.1. The functor $F$ is representable by a scheme.

**Proof.**
We are going to use Schemes, Lemma 26.15.4.

First we check that $F$ satisfies the sheaf property for the Zariski topology. Namely, suppose that $T$ is a scheme, that $T = \bigcup _{i \in I} U_ i$ is an open covering, and that $(f_ i, \varphi _ i) \in F(U_ i)$ such that $(f_ i, \varphi _ i)|_{U_ i \cap U_ j} = (f_ j, \varphi _ j)|_{U_ i \cap U_ j}$. This implies that the morphisms $f_ i : U_ i \to S$ glue to a morphism of schemes $f : T \to S$ such that $f|_{U_ i} = f_ i$, see Schemes, Section 26.14. Thus $f_ i^*\mathcal{A} = f^*\mathcal{A}|_{U_ i}$ and by assumption the morphisms $\varphi _ i$ agree on $U_ i \cap U_ j$. Hence by Sheaves, Section 6.33 these glue to a morphism of $\mathcal{O}_ T$-algebras $f^*\mathcal{A} \to \mathcal{O}_ T$. This proves that $F$ satisfies the sheaf condition with respect to the Zariski topology.

Let $S = \bigcup _{i \in I} U_ i$ be an affine open covering. Let $F_ i \subset F$ be the subfunctor consisting of those pairs $(f : T \to S, \varphi )$ such that $f(T) \subset U_ i$.

We have to show each $F_ i$ is representable. This is the case because $F_ i$ is identified with the functor associated to $U_ i$ equipped with the quasi-coherent $\mathcal{O}_{U_ i}$-algebra $\mathcal{A}|_{U_ i}$, by Lemma 27.4.1. Thus the result follows from Lemma 27.4.2.

Next we show that $F_ i \subset F$ is representable by open immersions. Let $(f : T \to S, \varphi ) \in F(T)$. Consider $V_ i = f^{-1}(U_ i)$. It follows from the definition of $F_ i$ that given $a : T' \to T$ we gave $a^*(f, \varphi ) \in F_ i(T')$ if and only if $a(T') \subset V_ i$. This is what we were required to show.

Finally, we have to show that the collection $(F_ i)_{i \in I}$ covers $F$. Let $(f : T \to S, \varphi ) \in F(T)$. Consider $V_ i = f^{-1}(U_ i)$. Since $S = \bigcup _{i \in I} U_ i$ is an open covering of $S$ we see that $T = \bigcup _{i \in I} V_ i$ is an open covering of $T$. Moreover $(f, \varphi )|_{V_ i} \in F_ i(V_ i)$. This finishes the proof of the lemma. $\square$

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