Principal divisors on the projective line have degree zero

Charts and points. By [F1], ${\mathbb{P}}_k^1=U_0\cup U_{\mathrm{\infty }}$, the overlap $D(t)\subseteq U_0$ is identified with $D(u)\subseteq U_{\mathrm{\infty }}$, and $tu=1$ on it; every point of ${\mathbb{P}}_k^1$ other than $\mathrm{\infty }=(u)$ lies in $U_0$ and corresponds to a prime $\mathfrak{p}\subseteq k[t]$ with ${\mathcal{O}}_{{\mathbb{P}}_k^1,x}=k[t{]}_{\mathfrak{p}}$. Since $(u)\subseteq k[u]$ is maximal with $k[u]\mathrm{/}(u)\cong k$, every prime of $k[u]$ containing $u$ equals $(u)$; hence $U_{\mathrm{\infty }}\setminus \{\mathrm{\infty }\}=D(u)=U_0\cap U_{\mathrm{\infty }}$.

Integrality, normality, function field, dimension. ${\mathbb{P}}_k^1$ is an integral finite-type $k$-scheme of chain dimension one, every local ring of it is an integrally closed domain, and its function field is $k(t)=k(u)$ with $t=u^{-1}$. $U_0$ and $U_{\mathrm{\infty }}$ are integral and glued along the nonempty open $D(t)\cong D(u)$, so ${\mathbb{P}}_k^1$ is integral, and finite type over $k$ because its affine charts are. Localisations of integrally closed domains are integrally closed: if $x\in \mathrm{Frac}(A)$ is integral over $S^{-1}A$ with monic equation $x^n+{\sum }_{i<n}(a_i\mathrm{/}{s}_i)x^i=0$, then with $s={\prod }_{i<n}{s}_i$ the element $sx$ satisfies the monic equation $(sx{)}^n+{\sum }_{i<n}{a}_i(s^{n-i}\mathrm{/}{s}_i)(sx{)}^i=0$ over $A$, so $sx\in A$ and $x\in S^{-1}A$. Hence every local ring is an integrally closed domain and ${\mathbb{P}}_k^1$ is a normal Noetherian scheme [F4]; by [F5] its function field is $k(t)=k(u)$. For the dimension, $U_0=\mathrm{Spec}k[t]$ is a Noetherian space of chain dimension $\dim k[t]=1$ whose proper closed subsets are finite unions of points; since $U_0$ is dense in ${\mathbb{P}}_k^1$, the only proper irreducible closed subsets of ${\mathbb{P}}_k^1$ are the points, so its chain dimension is one (Chain dimension and the empty-space convention).

Factorisations and the degree count. $p={\prod }_i{r}_i^{a_i}$ and $q={\prod }_j{s}_j^{b_j}$ with pairwise distinct monic irreducibles $r_i$ and $s_j$, no $r_i$ equal to any $s_j$, and $m={\sum }_i{a}_i\mathrm{deg}{r}_i$, $n={\sum }_j{b}_j\mathrm{deg}{s}_j$. The monic polynomial $p$ factors as a product of monic irreducibles: a factorisation $p=c\prod r_i^{a_i}$ has leading coefficient $c\prod (\text{leading coefficients})=c$ if each $r_i$ is monic, so $c=1$; the same holds for $q$. Coprimality of $p$ and $q$ says no monic irreducible divides both, so the two families are disjoint. Degrees add: $m=\mathrm{deg}p={\sum }_i{a}_i\mathrm{deg}{r}_i$ and $n=\mathrm{deg}q={\sum }_j{b}_j\mathrm{deg}{s}_j$.

Orders at the finite points. For every monic irreducible $r\in k[t]$, ${\mathrm{ord}}_{[r]}(f)$ equals $a_i$ if $r=r_i$, equals $-b_j$ if $r=s_j$, and is $0$ otherwise. Let $r$ be monic irreducible and $\mathfrak{p}=(r)$. The local ring ${\mathcal{O}}_{{\mathbb{P}}_k^1,[r]}=k[t{]}_{\mathfrak{p}}$ is a one-dimensional local domain [F4] and hence a discrete valuation ring whose maximal ideal is generated by the uniformiser $r$ [F7]. If $r=r_i$ then $r_i\mid p$ and $r_i\nmid q$, so $q$ is a unit of $k[t{]}_{\mathfrak{p}}$ and additivity gives ${\mathrm{ord}}_{[r]}(f)=a_i\cdot 1-0=a_i$; if $r=s_j$ then $p$ is a unit and $q=r^{b_j}\cdot (\text{unit})$, giving ${\mathrm{ord}}_{[r]}(f)=-b_j$; and if $r$ divides neither $p$ nor $q$, both are units and the order is $0$.

Order at infinity. ${\mathrm{ord}}_{\mathrm{\infty }}(p)=-m$, ${\mathrm{ord}}_{\mathrm{\infty }}(q)=-n$, and hence ${\mathrm{ord}}_{\mathrm{\infty }}(f)=n-m$. On $U_{\mathrm{\infty }}$ one has $t=u^{-1}$, so for a monic polynomial $g(t)=t^d+c_{d-1}{t}^{d-1}+\cdots +c_0$ of degree $d$, $g=u^{-d}(1+c_{d-1}u+\cdots +c_0{u}^d)$ with the second factor equal to $1$ at $u=0$, hence a unit of $k[u{]}_{(u)}=O_{{\mathbb{P}}_k^1,\mathrm{\infty }}$; with the uniformiser $u$ this gives ${\mathrm{ord}}_{\mathrm{\infty }}(g)=-d$. Applying this to $p$ and $q$ and using additivity yields ${\mathrm{ord}}_{\mathrm{\infty }}(f)=(-m)-(-n)=n-m$.

The divisor. $\mathrm{div}(f)={\sum }_i{a}_i[r_i]-{\sum }_j{b}_j[s_j]+(n-m)[\mathrm{\infty }]$, a finite Weil sum, and all its terms are closed points, so it is an element of $\mathrm{Div}({\mathbb{P}}_k^1)$. By steps 1.4 and 1.5 these are exactly the nonzero orders of $f$ at prime divisors; the remaining prime divisors have order zero. The support is finite, so the locally finite sum of [F7] is this finite sum, and since ${\mathbb{P}}_k^1$ has chain dimension one its prime divisors are closed points [F6].

Degree. ${\mathrm{deg}}_k\mathrm{div}(f)={\sum }_i{a}_i\mathrm{deg}{r}_i-{\sum }_j{b}_j\mathrm{deg}{s}_j+(n-m)=m-n+(n-m)=0$. By step 1.3 the finite terms have total degree ${\sum }_i{a}_i\mathrm{deg}{r}_i-{\sum }_j{b}_j\mathrm{deg}{s}_j=m-n$, by [F9] the residue degree of $[r]$ is $[\kappa ([r]):k]=\mathrm{deg}r$ and $[\kappa (\mathrm{\infty }):k]=1$, and by [F6] the degree is additive over the coefficients; adding the coefficient ${\mathrm{ord}}_{\mathrm{\infty }}(f)=n-m$ of step 1.5 gives $m-n+(n-m)=0$.

Conclusion. On ${\mathbb{P}}_k^1$ the principal divisor of $f=p\mathrm{/}q$ is ${\sum }_i{a}_i[r_i]-{\sum }_j{b}_j[s_j]+(n-m)[\mathrm{\infty }]$ and has degree zero; the Axiom of Choice is inherited from the two-affine projective-line construction [F1] and the properness theorem [F6]. The finite factorisation and valuation computation make no further choice.