The twists on the projective line have degree n

Setup. Identify $X={\mathbb{P}}_k^1$ with $\mathrm{Proj}S$ by [F2], and write ${\mathcal{O}}_X(n)$ for the twisting sheaves of this model. The charts $D_{+}(x_0)=\mathrm{Spec}k[t]$ and $D_{+}(x_1)=\mathrm{Spec}k[u]$ cover $X$, meet in $D_{+}(x_0{x}_1)$ with $t=u^{-1}$, and the standard opens form a basis.

$X$ is integral. The point $(0)$ lies in every standard open $D_{+}(f)$ with $f\mathrm{\ne }0$ homogeneous; since these form a basis, $(0)$ lies in every nonempty open subset, so any two nonempty opens meet and $X$ is irreducible. A local ring of $X$ is a localisation of $k[t]$ or $k[u]$ at a prime, by [F4]; if $(a\mathrm{/}s)(b\mathrm{/}t)=0$ in such a localisation, then $uab=0$ for some $u$ in the multiplicative set, so $a=0$ or $b=0$ because $k[t]$ and $k[u]$ are domains: localisations of domains are domains. Hence every local ring is a domain, the nilradical ideal sheaf vanishes, $X$ is reduced, and since $X\mathrm{\ne }\mathrm{\varnothing }$ it is an integral scheme.

$X$ is Noetherian of chain dimension one. The two charts are the spectra of the Noetherian rings $k[t]$, $k[u]$, so $X$ is locally Noetherian and quasi-compact, hence Noetherian, and its underlying space is Noetherian. Since $\dim k[t]=\dim k[u]=1$ and the dimension of a Noetherian space is the supremum of the dimensions of the members of a finite open cover, the chain dimension of $X$ is $\sup \{1,1\}=1$.

$X$ is normal. Each local ring is a prime localisation of $k[t]$ or of $k[u]$, which are integrally closed domains; a prime localisation of an integrally closed domain is integrally closed, and it is Noetherian as a localisation of a Noetherian ring. So every local ring is an integrally closed Noetherian domain, i.e., a normal Noetherian ring, and $X$ is normal.

The section $x_0$ and its effective divisor. On the chart $D_{+}(x_i)$ the module $S(1{)}_{(x_i)}$ is generated by $x_i$ (every element is a sum of terms $a\mathrm{/}{x}_i^m$ with $a\in S_{m+1}$ and $a\mathrm{/}{x}_i^m=(a\mathrm{/}{x}_i^{m+1})x_i$), so the local sections of the global element $x_0$ have coefficients $1$ on $D_{+}(x_0)$ and $u$ on $D_{+}(x_1)$; they agree on the overlap because they are the images of the single element $x_0$, so they glue to a global section of the invertible sheaf ${\mathcal{O}}_X(1)$. The coefficients are regular: $1$ is a unit of $k[t]$ and $u$ is a nonzero element of the domain $k[u]$, so multiplication by their germs is injective on every localisation. By the regular-section lemma there is an effective Cartier divisor $H$ with local-equation datum $\{(D_{+}(x_0),1),(D_{+}(x_1),u)\}$, vanishing subscheme $Z_H$ with ideal sheaf $(1)$ on $D_{+}(x_0)$ and $(u)$ on $D_{+}(x_1)$, and a canonical isomorphism ${\mathcal{O}}_X(H)\cong {\mathcal{O}}_X(1)$.

$X$ is a normal proper curve over $k$ and the degree descends. The structure morphism $X\to \mathrm{Spec}k$ is proper and of finite type, so with steps 1.2 and 1.3 the scheme $X$ is an integral proper $k$-scheme of chain dimension one, a proper curve over $k$; by step 1.4 it is normal. Hence [F8] applies: ${\mathrm{deg}}_k{\mathcal{O}}_X(D):={\mathrm{deg}}_{k}D$ is a well-defined group homomorphism $\mathrm{Pic}(X)\to \mathbb{Z}$, additive in the tensor product, and ${\mathrm{deg}}_k$ of the class of ${\mathcal{O}}_X(D)$ equals ${\mathrm{deg}}_{k}D$ for every divisor $D$ on $X$.

The divisor $H$ is the point at infinity. By step 1.5 the local equation of $H$ on the chart $D_{+}(x_0)$ is the unit $1$, so no point of $D_{+}(x_0)$ lies in $Z_H$; on the chart $D_{+}(x_1)$ the local equation is $u$, and $V(u)$ consists of the maximal ideal $(u)$ alone, with residue field $k[u]\mathrm{/}(u)\cong k$. Hence $Z_H=\{x_{\mathrm{\infty }}\}$ with $\kappa (x_{\mathrm{\infty }})=k$, and $x_{\mathrm{\infty }}\mathrm{\notin }{D}_{+}(x_0)$. Since $x_{\mathrm{\infty }}$ lies in no open subset contained in $D_{+}(x_0)$ while a generic point lies in every nonempty open subset, $x_{\mathrm{\infty }}$ is not the generic point of $X$; its closure is an irreducible closed subset different from $X$, and since $X$ has chain dimension one by step 1.3, every irreducible closed subset other than $X$ is a point, so $\{x_{\mathrm{\infty }}\}$ is closed. Hence $x_{\mathrm{\infty }}$ is a closed point of the curve $X$.

The associated cycle of $H$. Let $Z$ be a prime divisor of $X$. If $Z\mathrm{\ne }\{x_{\mathrm{\infty }}\}$ and $Z$ meets $D_{+}(x_0)$, the local equation $1$ of step 1.5 is a unit at the generic point of $Z$, so its valuation is $0$. If $Z\mathrm{\ne }\{x_{\mathrm{\infty }}\}$ and $Z$ meets $D_{+}(x_1)$ at a prime different from $(u)$, then $u$ is not in that prime, so $u$ is a unit of the corresponding localisation and the valuation is again $0$. Only $Z=\{x_{\mathrm{\infty }}\}$ contributes: by step 2.2 its generic point is $x_{\mathrm{\infty }}$, the local equation on $D_{+}(x_1)$ is $u$, and $k[u{]}_{(u)}$ is a discrete valuation ring with maximal ideal generated by $u$, so $v_{x_{\mathrm{\infty }}}(u)=1$. Therefore $\mathrm{cyc}(H)=1\cdot [x_{\mathrm{\infty }}]$, and ${\mathrm{deg}}_k\mathrm{cyc}(H)=[\kappa (x_{\mathrm{\infty }}):k]=1$.

${\mathrm{deg}}_k{\mathcal{O}}_X(1)=1$. By step 2.1 the descent [F8] applies to $X$, and by step 3.1 the associated cycle of $H$ is $[x_{\mathrm{\infty }}]$; since step 1.5 gives ${\mathcal{O}}_X(H)\cong {\mathcal{O}}_X(1)$, the class $[{\mathcal{O}}_X(1)]=[{\mathcal{O}}_X(H)]$ in $\mathrm{Pic}(X)$ is carried by the canonical class map to $[\mathrm{cyc}(H)]$. As ${\mathrm{deg}}_k$ of the class of ${\mathcal{O}}_X(D)$ equals ${\mathrm{deg}}_{k}D$ for every divisor $D$ and depends only on the class, ${\mathrm{deg}}_k{\mathcal{O}}_X(1)={\mathrm{deg}}_k\mathrm{cyc}(H)=1$.

All nonnegative twists. For $n\ge 1$ the multiplication isomorphisms of [F3] give ${\mathcal{O}}_X(1{)}^{\otimes n}\cong {\mathcal{O}}_X(n)$ by induction, while ${\mathcal{O}}_X(0)={\mathcal{O}}_X$; in $\mathrm{Pic}(X)$ the tensor product is the group law, so additivity of the homomorphism ${\mathrm{deg}}_k$ provided by step 2.1 gives ${\mathrm{deg}}_k{\mathcal{O}}_X(n)=n{\mathrm{deg}}_k{\mathcal{O}}_X(1)=n$ by step 4.1. For $n=0$ the identity class $[{\mathcal{O}}_X]$ has degree $0$ because a homomorphism of groups carries the identity to the identity: ${\mathrm{deg}}_k{\mathcal{O}}_X(0)={\mathrm{deg}}_k{\mathcal{O}}_X=0$.

All negative twists. Let $n<0$ and put $m=-n>0$. The multiplication isomorphism gives ${\mathcal{O}}_X(n){\otimes }_{{\mathcal{O}}_X}{\mathcal{O}}_X(m)\cong {\mathcal{O}}_X(0)={\mathcal{O}}_X$, so in the abelian group $\mathrm{Pic}(X)$ the classes satisfy $[{\mathcal{O}}_X(n)]=[{\mathcal{O}}_X(m){]}^{-1}$. Since ${\mathrm{deg}}_k$ is a group homomorphism and ${\mathrm{deg}}_k{\mathcal{O}}_X(m)=m$ by step 5.1, ${\mathrm{deg}}_k{\mathcal{O}}_X(n)=-{\mathrm{deg}}_k{\mathcal{O}}_X(m)=-m=n$.

Conclusion. Combining steps 5.1 and 6.1, for every integer $n$ the twist satisfies ${\mathrm{deg}}_k{\mathcal{O}}_X(n)=n$. The argument used only ${\mathcal{O}}_X(1)$, its tensor powers and its inverse in $\mathrm{Pic}(X)$; it exhibits no isomorphism of an arbitrary invertible sheaf with a twist, so it claims no classification of line bundles on ${\mathbb{P}}_k^1$. The Axiom of Choice is used exactly through the Proj, properness and twisting-sheaf constructions of [F2], [F3] and [F8] and, via the implication of [F1], through the Dependent Choice hypothesis of the cycle theorem [F10].