A genus-zero curve with a degree-one divisor is the projective line

From a degree-one divisor to a rational point. Let $D$ be a divisor on $C$ with ${\mathrm{deg}}_k(D)=1$. By [F1] one has $l(D)\ge {\mathrm{deg}}_k(D)+1-g=2$, so ${\dim }_{k}L(D)\ge 2$ and there is a nonzero $f\in L(D)$. By [F3] the divisor $E:=\mathrm{div}(f)+D$ is effective and linearly equivalent to $D$, and by [F4] ${\mathrm{deg}}_k(\mathrm{div}f)=0$, so ${\mathrm{deg}}_k(E)={\mathrm{deg}}_k(D)=1$. Write $E={\sum }_x{n}_x[x]$ with $n_x\ge 0$ by [F2]; then $1={\mathrm{deg}}_k(E)={\sum }_x{n}_x[\kappa (x):k]$ is a sum of nonnegative terms, so exactly one closed point $q$ has $n_q\ge 1$, with $n_q[\kappa (q):k]=1$ and $n_x=0$ for $x\mathrm{\ne }q$; hence $n_q=1$ and $[\kappa (q):k]=1$, that is, $q$ is a $k$-rational closed point by [F2]. Conversely, if $p$ is a closed point with $[\kappa (p):k]=1$ then ${\mathrm{deg}}_k([p])=1$ by [F2], so $C$ admits a divisor of degree $1$; this proves the equivalence of the two hypotheses of the statement.

A nonconstant function with poles at most at $q$. With $q$ as in step 1.1 one has ${\mathrm{deg}}_k([q])=[\kappa (q):k]=1$ by [F2], so [F1] gives $l([q])\ge 1+1-g=2$. The divisor $[q]$ is effective, so every constant $c\in k^{\times }$ satisfies $\mathrm{div}(c)+[q]=[q]\ge 0$ and lies in $L([q])$ by [F3]; thus $k\cdot 1\subseteq L([q])$ is a subspace of dimension one, and since ${\dim }_{k}L([q])\ge 2$ it is proper, so by [F7] there is $h\in L([q])\setminus k\cdot 1$. The function $h$ is nonconstant, and $h\in L([q])$ means $\mathrm{div}(h)+[q]\ge 0$ by [F3].

The pole divisor of $h$ is $[q]$. From $\mathrm{div}(h)+[q]\ge 0$ at every closed point $x\mathrm{\ne }q$ the order ${\mathrm{ord}}_x(h)$ is $\ge 0$, and at $q$ it is $\ge -1$; hence the pole divisor $(h{)}_{\mathrm{\infty }}={\sum }_{{\mathrm{ord}}_x(h)<0}(-{\mathrm{ord}}_x(h))[x]$ of [F5] satisfies $(h{)}_{\mathrm{\infty }}\le [q]$. Since $h$ is nonconstant, [F5] exhibits the finite locally free morphism ${\phi }_h$ whose fibre over infinity is $(h{)}_{\mathrm{\infty }}$, of degree $[k(C):k(h)]\ge 1$; in particular $(h{)}_{\mathrm{\infty }}\mathrm{\ne }0$. As $(h{)}_{\mathrm{\infty }}\le [q]$ and $[q]$ has coefficient one at $q$ and zero elsewhere, the only nonzero effective divisor dominated by $[q]$ that is nonzero is $[q]$ itself, so $(h{)}_{\mathrm{\infty }}=[q]$.

Degree one and birationality. By [F5] the degree of ${\phi }_h$ is ${\mathrm{deg}}_k(h{)}_{\mathrm{\infty }}={\mathrm{deg}}_k[q]=1$ by step 3.1 and [F2], that is $[k(C):k(h)]=1$; the pullback of the coordinate function of ${\mathbb{P}}_k^1$ is $h$, so the image of $k({\mathbb{P}}_k^1)\to k(C)$ is $k(h)$ and the extension $k(C)\mathrm{/}k(h)$ is trivial, $k(C)=k(h)$. Hence ${\phi }_h$ is dominant with pullback an isomorphism of function fields, that is, ${\phi }_h$ is birational.

Conclusion. By [F6] the birational map ${\phi }_h:C⇢{\mathbb{P}}_k^1$ of step 4.1 is represented by a $k$-isomorphism $C\to {\mathbb{P}}_k^1$; hence $C\cong {\mathbb{P}}_k^1$, which is the claim. The proof used one nonzero $f\in L(D)$ in step 1.1, one nonconstant $h\in L([q])$ in step 2.1, and the morphism ${\phi }_h$ supplied by [F5]; the Axiom of Choice is inherited only through the suppliers recorded in [F8].