Lemma 27.4.1 (01LS)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 27: Constructions of Schemes
Section 27.4: Relative spectrum as a functor
Lemma 27.4.1 (cite)

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Lemma 27.4.1. In Situation 27.3.1. Let $F$ be the functor associated to $(S, \mathcal{A})$ above. Let $g : S' \to S$ be a morphism of schemes. Set $\mathcal{A}' = g^*\mathcal{A}$ . Let $F'$ be the functor associated to $(S', \mathcal{A}')$ above. Then there is a canonical isomorphism

$F' \cong h_{S'} \times _{h_ S} F$

of functors.

Proof. A pair $(f' : T \to S', \varphi ' : (f')^*\mathcal{A}' \to \mathcal{O}_ T)$ is the same as a pair $(f, \varphi : f^*\mathcal{A} \to \mathcal{O}_ T)$ together with a factorization of $f$ as $f = g \circ f'$ . Namely with this notation we have $(f')^* \mathcal{A}' = (f')^*g^*\mathcal{A} = f^*\mathcal{A}$ . Hence the lemma. $\square$

Comments (2)

Comment #3837 by anonymous on December 13, 2018 at 06:33

This refers to Tag 01LR just above. It seems that the f on the right hand side should be the morphism making T into an S-scheme. So my suggestion is to write $(f \colon T \rightarrow S) \mapsto \{\text{$ \phi $as above}\}$ . (The right hand side should be in "set brackets" but it did not work in the preview.) In the representability statements below it would be good to say representable by an S-scheme.

Comment #3931 by Johan on January 23, 2019 at 08:28

Dear anonymous, thanks very much for pointing this out. I fixed it instead by replacing the target category with the category of schemes and not the category of schemes over $S$ . See change here.

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