Invertible twists for degree-one generated rings

The degree-one charts cover. Let $\mathfrak{p}\in X=\mathrm{Proj}S$. If $S_1\subseteq \mathfrak{p}$ then $S_{+}\subseteq \mathfrak{p}$ by [F1], contradicting $\mathfrak{p}\in \mathrm{Proj}S$; hence there is $x\in S_1$ with $x\mathrm{\notin }\mathfrak{p}$. Therefore every point of $X$ lies in some $D_{+}(x)$ with $x\in S_1$, and the family $\{D_{+}(x):x\in S_1\}$ is an open cover of $X$.

A frame on each degree-one chart. Fix $x\in S_1$ and an integer $n$. In the localised ring $S[x^{-1}]$ the element $x$ is a unit, so $x^n$ is defined for every integer $n$ and is a unit; as an element of the graded module $S(n)[x^{-1}]$ it has degree $0$. Every class of $S(n{)}_{(x)}$ has the form $a\mathrm{/}{x}^k$ with $a\in S(n{)}_k=S_{n+k}$ homogeneous of degree $n+k$, and $a\mathrm{/}{x}^k=(a\mathrm{/}{x}^{k+n})x^n$ with $a\mathrm{/}{x}^{k+n}\in S_{(x)}$; hence multiplication by $x^n$ maps $S_{(x)}$ onto $S(n{)}_{(x)}$, and it is injective because $x^n$ is a unit of $S[x^{-1}]$, so $S(n{)}_{(x)}$ is free of rank one with basis $x^n$. By [F3] this basis is a frame of ${\mathcal{O}}_X(n)$ on $D_{+}(x)$.

Invertibility. By step 1.1 the degree-one charts $D_{+}(x)$ with $x\in S_1$ cover $X$, and by step 1.2 the restriction of ${\mathcal{O}}_X(n)$ to each of them is free of rank one; hence ${\mathcal{O}}_X(n)$ is locally free of rank one, that is invertible, by [F4]. This is claim (1).

Multiplication on a chart is an isomorphism. Fix $x\in S_1$ and integers $m,n$. On $D_{+}(x)$ the multiplication map of [F3] sends $S(m{)}_{(x)}{\otimes }_{S_{(x)}}S(n{)}_{(x)}\to S(m+n{)}_{(x)}$ and, in terms of the frames of step 1.2, $x^m\otimes x^n\mapsto x^{m+n}$, which is a unit of $S(m+n{)}_{(x)}$; being a map of free rank-one modules carrying a generator to a generator, it is an isomorphism. The identifications are compatible with restrictions to smaller standard opens, since both sides are given by the same localisation maps.

Global isomorphism. The maps of step 2.2 on the charts $D_{+}(x)$, $x\in S_1$, glue: on overlaps $D_{+}(x)\cap D_{+}(y)=D_{+}(xy)$ both restrictions are the multiplication maps computed in the localisation at $xy$, so they agree by the canonicity of the localisation maps in [F3]; the cover of step 1.1 and the local isomorphisms of step 2.2 therefore yield a global isomorphism ${\mathcal{O}}_X(m)\otimes {\mathcal{O}}_X(n)\to {\mathcal{O}}_X(m+n)$, which is claim (2).

Conclusion. Step 2.1 proves invertibility of every ${\mathcal{O}}_X(n)$ and step 3.1 proves that all multiplication maps are isomorphisms. If $S_{+}=0$ then $S=S_0$ is generated by $S_1=0$ over $S_0$, and $X=\mathrm{Proj}S=\mathrm{\varnothing }$ because no homogeneous prime satisfies $S_{+}\subseteq ̸\mathfrak{p}$; the statements are then vacuous, and the empty scheme is covered by the empty family of charts. The Axiom of Choice [A1] is inherited from the Proj construction [F2], [F3]; no choice is made here. [A1, F2, step 2.1, step 3.1, cases: empty X] \qed