Lemma 28.26.10. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Set $S = \Gamma _*(X, \mathcal{L})$. Assume (a) every point of $X$ is contained in one of the open subschemes $X_ s$, for some $s \in S_{+}$ homogeneous, and (b) $X$ is quasi-compact. Then the canonical morphism of schemes $f : X \longrightarrow \text{Proj}(S)$ of Lemma 28.26.9 above is quasi-compact with dense image.

**Proof.**
To prove $f$ is quasi-compact it suffices to show that $f^{-1}(D_{+}(s))$ is quasi-compact for any $s \in S_{+}$ homogeneous. Write $X = \bigcup _{i = 1, \ldots , n} X_ i$ as a finite union of affine opens. By Lemma 28.26.4 each intersection $X_ s \cap X_ i$ is affine. Hence $X_ s = \bigcup _{i = 1, \ldots , n} X_ s \cap X_ i$ is quasi-compact. Assume that the image of $f$ is not dense to get a contradiction. Then, since the opens $D_+(s)$ with $s \in S_+$ homogeneous form a basis for the topology on $\text{Proj}(S)$, we can find such an $s$ with $D_+(s) \not= \emptyset $ and $f(X) \cap D_+(s) = \emptyset $. By Lemma 28.26.9 this means $X_ s = \emptyset $. By Lemma 28.17.2 this means that a power $s^ n$ is the zero section of $\mathcal{L}^{\otimes n\deg (s)}$. This in turn means that $D_+(s) = \emptyset $ which is the desired contradiction.
$\square$

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