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 (9)

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.

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