The Stacks project

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

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 \phi. (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 . See change here.

There are also:

  • 9 comment(s) on Section 27.4: Relative spectrum as a functor

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01LS. Beware of the difference between the letter 'O' and the digit '0'.