The Stacks project

Proof. The (not graded) $S$-module maps $S \to S(nd)$, and $M \to M(nd)$, given by $x \mapsto f^ n x$ become isomorphisms after inverting $f$. The first shows that $S_{(f)} \cong S(nd)_{(f)}$ which gives an isomorphism $\mathcal{O}_{D_{+}(f)} \cong \mathcal{O}_ X(nd)|_{D_{+}(f)}$. The second shows that the map $S(nd)_{(f)} \otimes _{S_{(f)}} M_{(f)} \to M(nd)_{(f)}$ is an isomorphism. The case of the map (27.10.1.2) is a consequence of the case of the map (27.10.1.1). $\square$

Proof. Under the assumptions of the lemma $X$ is covered by the open subsets $D_{+}(f)$ with $f \in S_1$ and the lemma is a consequence of Lemma 27.10.2 above. $\square$

Proof. If $x \in D_{+}(fg)$ with $\deg (f) = \deg (g) + d$ then on $D_{+}(fg)$ the sheaves $\mathcal{O}_ X(dn)$ are generated by the element $(f/g)^ n = f^{2n}/(fg)^ n$. This implies $x$ is in the open subset $W$ defined in (1) by arguing as in the proof of Lemma 27.10.2.

Conversely, suppose that $\mathcal{O}_ X(d)$ is free of rank 1 in an open neighbourhood $V$ of $x \in X$ and all the multiplication maps $\mathcal{O}_ X(nd)|_ V \otimes _{\mathcal{O}_ V} \mathcal{O}_ X(md)|_ V \to \mathcal{O}_ X(nd + md)|_ V$ are isomorphisms. We may choose $h \in S_{+}$ homogeneous such that $x \in D_{+}(h) \subset V$. By the definition of the twists of the structure sheaf we conclude there exists an element $s$ of $(S_ h)_ d$ such that $s^ n$ is a basis of $(S_ h)_{nd}$ as a module over $S_{(h)}$ for all $n \in \mathbf{Z}$. We may write $s = f/h^ m$ for some $m \geq 1$ and $f \in S_{d + m \deg (h)}$. Set $g = h^ m$ so $s = f/g$. Note that $x \in D_{+}(g)$ by construction. Note that $g^ d \in (S_ h)_{-d\deg (g)}$. By assumption we can write this as a multiple of $s^{\deg (g)} = f^{\deg (g)}/g^{\deg (g)}$, say $g^ d = a/g^ e \cdot f^{\deg (g)}/g^{\deg (g)}$. Then we conclude that $g^{d + e + \deg (g)} = a f^{\deg (g)}$ and hence also $x \in D_{+}(f)$. So $x$ is an element of the set defined in (2).

The existence of the generating section $s = f/g$ over the affine open $D_{+}(fg)$ whose powers freely generate the sheaves of modules $\mathcal{O}_ X(nd)$ easily implies that the multiplication maps $\widetilde M(nd)|_ W = \widetilde M|_ W \otimes _{\mathcal{O}_ W} \mathcal{O}_ X(nd)|_ W \to \widetilde{M(nd)}|_ W$ (see 27.10.1.5) are isomorphisms. Compare with the proof of Lemma 27.10.2. $\square$

Proof. Let $D_{+}(ab) \subset W$ be a standard affine open with $a, b \in S$ homogeneous and $\deg (a) = \deg (b) + d$. Note that $D_{+}(ab) \cap D_{+}(f) = D_{+}(abf)$. On the other hand the restriction of $s$ to $D_{+}(ab)$ corresponds to the element $f/1 = b^ nf/a^ n (a/b)^ n \in (S_{ab})_{nd}$. We have seen in the proof of Lemma 27.10.4 that $(a/b)^ n$ is a generator for $\mathcal{O}_ W(nd)$ over $D_{+}(ab)$. We conclude that $W_ s \cap D_{+}(ab)$ is the principal open associated to $b^ nf/a^ n \in \mathcal{O}_ X(D_{+}(ab))$. Thus the result of the lemma is clear. $\square$

Proof. Since $Y$ is quasi-compact there exist finitely many homogeneous $f_ i \in S_{+}$, $i = 1, \ldots , n$ such that the standard opens $D_{+}(f_ i)$ give an open covering of $Y$. Let $d_ i = \deg (f_ i)$ and set $d = d_1 \ldots d_ n$. Note that $D_{+}(f_ i) = D_{+}(f_ i^{d/d_ i})$ and hence we see immediately that $Y \subset W_ d$, by characterization (2) in Lemma 27.10.4 or by (1) using Lemma 27.10.2. Note that (1) implies (2), (3) and (4) by Lemma 27.10.4. (Note that (3) is a special case of (4).) Assertion (5) follows from Lemma 27.10.5. Assertions (6) and (7) follow because the open subsets $D_{+}(f)$ form a basis for the topology of $X$ and are affine. $\square$

Proof. Let $f \in S$ be homogeneous of degree $d > 0$. Recall that $\widetilde{M}|_{D_{+}(f)}$ corresponds to the $S_{(f)}$-module $M_{(f)}$ by Lemma 27.8.4. Thus we can define a canonical map

which makes sense because $f|_{D_+(f)}$ is a trivializing section of the invertible sheaf $\mathcal{O}_ X(d)|_{D_+(f)}$, see Lemma 27.10.2 and its proof. Since $\widetilde{M}$ is quasi-coherent, this leads to a canonical map

via Schemes, Lemma 26.7.1. We obtain a global map if we prove that the displayed maps glue on overlaps. Proof of this is omitted. We also omit the proof of the final statement. $\square$