Lemma 28.28.2. In Situation 28.28.1. The canonical morphism f : X \to Y maps X into the open subscheme W = W_1 \subset Y where \mathcal{O}_ Y(1) is invertible and where all multiplication maps \mathcal{O}_ Y(n) \otimes _{\mathcal{O}_ Y} \mathcal{O}_ Y(m) \to \mathcal{O}_ Y(n + m) are isomorphisms (see Constructions, Lemma 27.10.4). Moreover, the maps f^*\mathcal{O}_ Y(n) \to \mathcal{L}^{\otimes n} are all isomorphisms.
Proof. By Proposition 28.26.13 there exists an integer n_0 such that \mathcal{L}^{\otimes n} is globally generated for all n \geq n_0. Let x \in X be a point. By the above we can find a \in S_{n_0} and b \in S_{n_0 + 1} such that a and b do not vanish at x. Hence f(x) \in D_{+}(a) \cap D_{+}(b) = D_{+}(ab). By Constructions, Lemma 27.10.4 we see that f(x) \in W_1 as desired. By Constructions, Lemma 27.14.1 which was used in the construction of the map f the maps f^*\mathcal{O}_ Y(n_0) \to \mathcal{L}^{\otimes n_0} and f^*\mathcal{O}_ Y(n_0 + 1) \to \mathcal{L}^{\otimes n_0 + 1} are isomorphisms in a neighbourhood of x. By compatibility with the algebra structure and the fact that f maps into W we conclude all the maps f^*\mathcal{O}_ Y(n) \to \mathcal{L}^{\otimes n} are isomorphisms in a neighbourhood of x. Hence we win. \square
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