The twist index on the projective line is an isomorphism invariant

By [F1] the two charts $U_0,U_1$ cover ${\mathbb{P}}_k^1$ and their overlap is $U_0\cap U_1=D_1^{(0)}$, with coordinate ring the localisation $k[u,u^{-1}]=k[u{]}_u$ in which $v=u^{-1}$. By [F2] and [F5] the prescription $e_1=u^n{e}_0$ glues the two trivial invertible sheaves $k[u]$ on $U_0$ and $k[v]$ on $U_1$ to an invertible sheaf $\mathcal{O}(n)$ on ${\mathbb{P}}_k^1$, trivialized on $U_i$ by the frame $e_i$, which is a global frame of $\mathcal{O}(n){\mathrm{\mid }}_{U_i}$; the same holds with $m$ in place of $n$, with frames $e_i^{(m)}$, and on the overlap $e_1^{(n)}=u^n{e}_0^{(n)}$, $e_1^{(m)}=u^m{e}_0^{(m)}$.

Let $\mathrm{\Phi }:\mathcal{O}(n)\to \mathcal{O}(m)$ be a morphism of $\mathcal{O}$-modules. By [F4] it is determined by its restrictions ${\mathrm{\Phi }}_i$ to the two charts, and on the chart $U_i$ the source and target are free of rank one with frames $e_i^{(n)}$ and $e_i^{(m)}$, so ${\mathrm{\Phi }}_i$ is given by a section $a_i\in \mathrm{\Gamma }(U_i,\mathcal{O})$, namely ${\mathrm{\Phi }}_i(e_i^{(n)})=a_i{e}_i^{(m)}$, with $a_0\in k[u]$ and $a_1\in k[v]$. Conversely, two such sections define a morphism exactly when the two local morphisms agree on the overlap, and since $e_1^{(n)}=u^n{e}_0^{(n)}$ and $e_1^{(m)}=u^m{e}_0^{(m)}$ there, [F3] and [F4] turn this into the single compatibility relation $a_1{u}^m=u^n{a}_0\text{in }k[u,u^{-1}]\mathrm{.}$ Moreover $\mathrm{\Phi }$ is an isomorphism if and only if both $a_0$ and $a_1$ are units: if $\mathrm{\Phi }$ is invertible, its inverse has components $b_i$ with $a_i{b}_i=1$, and if $a_0,a_1$ are units, the components $a_i^{-1}$ satisfy the same relation and give an inverse.

Suppose now that $\mathrm{\Phi }$ is an isomorphism. By step 2.1 there are units $a_0\in k[u{]}^{\times }$ and $a_1\in k[v{]}^{\times }$ with $a_1{u}^m=u^n{a}_0$ in $k[u,u^{-1}]$; multiplying by $u^{-n}$ gives $a_0=a_1{u}^{m-n}$. By [F6], applied to the domain $k$, units of $k[u]$ and of $k[v]$ are nonzero constants, so $a_0=c_0$ and $a_1=c_1$ with $c_0,c_1\in k^{\times }$. Put $N=m-n$ and $c=c_0\mathrm{/}{c}_1\in k^{\times }$; then $u^N=c$ in $k[u,u^{-1}]$. If $N>0$, the polynomial $u^N-c\in k[u]$ becomes zero in the localisation, so [F7] gives an $M\ge 0$ with $u^{M+N}=c{u}^M$ in $k[u]$. This is an equality of polynomials of degrees $M+N$ and $M$ by [F8], impossible since $M+N>M$. If $N<0$, then $K=-N>0$ and $c^{-1}=u^K$, so the same argument applied to $c{u}^K-1$ gives $c{u}^{M+K}=u^M$ in $k[u]$ for some $M\ge 0$, again an equality of polynomials of degrees $M+K$ and $M$, impossible. Hence $N=0$ and $n=m$.

Conversely, if $n=m$ then the two gluing prescriptions coincide and $\mathcal{O}(n)=\mathcal{O}(m)$. This proves the equivalence: distinct indices give nonisomorphic twists. Finally, if invertible sheaves $\mathcal{M},{\mathcal{M}}^{\mathrm{\prime }}$ satisfy $\mathcal{M}\cong \mathcal{O}(n)$, ${\mathcal{M}}^{\mathrm{\prime }}\cong \mathcal{O}(m)$ and $\mathcal{M}\cong {\mathcal{M}}^{\mathrm{\prime }}$, composing isomorphisms gives $\mathcal{O}(n)\cong \mathcal{O}(m)$ and hence $n=m$ by step 3.1, so the twist index of an invertible sheaf on ${\mathbb{P}}_k^1$ isomorphic to a twist is well defined. No choice principle is used: the two frames, the two components $a_0,a_1$ and the integer $N$ are finite data.