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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