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 26.21.10 and Constructions, Lemma 27.21.4. $\square$

## Comments (2)

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 $\mathcal{O}_S$ in degree 0 and $f_*(\mathcal{O}_X)$ in degrees $\geq 1$, 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 $\mathcal{O}_S$ of $f_*\mathcal{O}_X$. I think this will alleviate your concerns, see here.

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