Lemma 29.44.16. A finite morphism is projective.

**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

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

coming from the adjunction map $f^*f_* \to \text{id}$. Then $\sigma $ is a closed immersion. Namely, affine locally on $S$ we can write $X = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$. Since $X$ is finite over $S$ we may choose $a_1, \ldots , a_ n \in A$ generating $A$ as an $R$-module. Then the $R$-algebra map $R[T_0, T_1, \ldots , T_ n] \to \text{Sym}_ R^*(A)$ sending $T_0$ to $1$ and $T_ i$ to $a_ i$ for $1 \leq i \leq n$ is surjective. Whence $\mathbf{P}(f_*\mathcal{O}_ X)$ is a closed subscheme of $\mathbf{P}^ n_ S$ by Constructions, Lemma 27.11.3. Thus it suffices to prove that the induced morphism $X \to \mathbf{P}^ n_ S$ is a closed immersion (see for example, Lemma 29.2.2). The reader checks that $X \to \mathbf{P}^ n_ S$ has image contained in the open $D_+(T_0) \cong \mathbf{A}^ n_ S$ and that $X \to \mathbf{A}^ n_ S$ corresponds to the surjective $R$-algebra map $R[x_1, \ldots , x_ n] \to \text{Sym}_ R^*(A)$ sending $x_ i$ to $a_ i$. Whence $X \to \mathbf{P}^ n_ S$ is an immersion (as a composition of a closed immersion and an open immersion). Since $X$ is finite over $S$ the image of this immersion is closed (this uses Lemma 29.44.11, Lemma 29.41.7, and Constructions, Lemma 27.13.4). We conclude by Schemes, Lemma 26.10.4. $\square$

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