
Lemma 26.4.1. In Situation 26.3.1. 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. $\square$

Comment #3837 by anonymous on

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 on

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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• 3 comment(s) on Section 26.4: Relative spectrum as a functor

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