Lemma 27.28.2. In Situation 27.28.1. 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 26.10.4). Moreover, the maps $f^*\mathcal{O}_ Y(n) \to \mathcal{L}^{\otimes n}$ are all isomorphisms.

Proof. By Proposition 27.26.13 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 26.10.4 we see that $f(x) \in W_1$ as desired. By Constructions, Lemma 26.14.1 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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