Projective-line curve and divisor basics

Each chart is integral and the overlap is nonempty and dense in each. Thus every nonempty open of either chart meets the overlap, and any two nonempty opens of the glued space meet. The scheme is irreducible and reduced. The same argument over an algebraic closure proves geometric integrality. The chart rings are Noetherian because they are PIDs by [F3], so the finite affine cover makes the scheme Noetherian and licenses the chart-dimension computation in [F2]. The charts are smooth of dimension one; [F1] gives properness, separatedness and finite type. Hence it is a smooth proper geometrically integral curve.

Its genus is zero by the $H^1(\mathcal{O})=0$ calculation in [F5]. Finite points and their residue degrees are those of [F3], and infinity is $u=0$ with residue field $k$.

At $p=V(g)$, $g$ generates the maximal ideal of $k[t{]}_{(g)}$, so its order is one. At every other finite point it is a unit. At infinity, $g(t)=u^{-d}h(u)$ with $h(0)=1$, so its order is $-d$. Thus $\mathrm{div}(g)=[p]-d[\mathrm{\infty }]$.

Write a divisor as $D={\sum }_i{n}_i[p_i]+m[\mathrm{\infty }]$ with $p_i=V(g_i)$ and $d_i=\mathrm{deg}{g}_i$. Then ${\mathrm{deg}}_{k}D={\sum }_i{n}_i{d}_i+m$, and the finite product $f={\prod }_i{g}_i^{n_i}$, allowing negative exponents, satisfies $\mathrm{div}(f)=D-{\mathrm{deg}}_k(D)[\mathrm{\infty }]$. This is the required linear equivalence, also for $D=0$ with empty product $f=1$.

The section $x_0$ has coefficient $1$ on its own chart and coefficient $u$ in the $x_1$-frame on the other chart, so its divisor is exactly $[\mathrm{\infty }]$. The rational-section dictionary gives $\mathcal{O}(1)\cong \mathcal{O}(\mathrm{\infty })$. Steps 1.1–2.2 establish the other assertions. No choices beyond the supplier AC premises are made. ∎