The Stacks project

Proof. Let $f : X \to S$ be a finite morphism. Then $f_*\mathcal{O}_ X$ is a quasi-coherent $\mathcal{O}_ S$-module (Lemma 29.11.5) of finite type (by our definition of finite morphisms and Properties, Lemma 28.16.1). We claim there is a closed immersion

\[ \sigma : X \longrightarrow \mathbf{P}(f_*\mathcal{O}_ X) = \underline{\text{Proj}}_ S(\text{Sym}^*_{\mathcal{O}_ S}(f_*\mathcal{O}_ X)) \]

over $S$, which finishes the proof. Namely, we let $\sigma $ be the morphism which corresponds (via Constructions, Lemma 27.16.11) to the surjection

\[ f^*f_*\mathcal{O}_ X \longrightarrow \mathcal{O}_ X \]

coming from the adjunction map $f^*f_* \to \text{id}$. Then $\sigma $ is a closed immersion by Schemes, Lemma 26.21.11 and Constructions, Lemma 27.21.4. $\square$


Comments (10)

Comment #1835 by on

In order to apply to tag 27.16.11 shouldn't the graded algebra structure be made more explicit? In particular: one puts in degree 0 and in degrees , and then one can appeal to 27.16.11. I might be overlooking something, but right now I don't see how we can use the referenced lemma.

Comment #1872 by on

Yes, I have now added the explicit form of the projective bundle as the relative Proj of the symmetric algebra over of . I think this will alleviate your concerns, see here.

Comment #7118 by Arnab Kundu on

Maybe a silly point: but why is the map a closed immersion? I understand that the graph of is a closed immersion by Lemma O1KS and Lemma 01OD. Am I missing something?

Comment #7119 by on

Not silly! The reference should have been to Lemma 26.21.11. Will fix this later.

Comment #8444 by on

Argh, the reference to Lemma 26.21.11 is still wrong because the Proj is over and not over . The best thing would be for people to work out for themselves why is a closed immersion (by doing a local computation). But we can also see it using theory. Write as where . (Namely, we have and there is a relative version of this.) Then with there is a map of graded -algebras. (Namely, given a ring map there is a graded -algebra map , and there is a relative version of this.) The map is surjective in all sufficiently high degrees. (Namely, in degrees .) Whence the corresponding morphism is everywhere defined and a closed immersion, see Lemma 27.18.3.

Comment #8447 by on

By the discussion in "N'eron models" around finite flat morphisms, after a faithfully flat base change there is a filtration of f_*O_X whose associated graded pieces are f-pushforwards of pushforwards of invertible O_S-modules along sections of f (i.e., multisections are "scheme-theoretic unions" of sections after faithfully flat base change). The composition of those sections with sigma give a corresponding filtration of the structure sheaf of the relative Proj (by ideal sheaves) compatible with the filtration on the pushforward of O_X. You can use this to prove that the morphism from the structure sheaf of the relative Proj to the pushforward of O_X is surjective.

Comment #8451 by on

I think my comment was wrong. There is a filtration after base change, and you can use it to prove the result. However, that is not the shortest proof, and I do not know if the filtration comes from a sequence of sections. I need to double-check what is in "N'eron models" -- they have effectively evicted us from our offices while they repair the HVAC system.


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 0B3I. Beware of the difference between the letter 'O' and the digit '0'.