A principal divisor of degree zero on the projective line

The two rational points and the divisors of $t-a$, $t-b$. The polynomials $t-a$ and $t-b$ are monic of degree $1$, hence irreducible in $k[t]$: a factorisation into non-units would give two factors of degree at least one, whose degrees add to $1$. So both are monic irreducibles of degree $d=1$, and by [F1] applied to $g=t-a$ and to $g=t-b$, $\mathrm{div}(t-a)=[a]-[\mathrm{\infty }],\mathrm{div}(t-b)=[b]-[\mathrm{\infty }],$ with $[\kappa (a):k]=1=[\kappa (b):k]$. Since $a\mathrm{\ne }b$, the maximal ideals $(t-a)\mathrm{\ne }(t-b)$ are distinct, so $[a]\mathrm{\ne }[b]$ and $[a]-[b]\mathrm{\ne }0$.

The divisor of $f$. The order at a closed point is additive in products and quotients [F2], so for every closed point $x$, ${\mathrm{ord}}_x(f)={\mathrm{ord}}_x(t-a)-{\mathrm{ord}}_x(t-b)$. Summing against $[x]$ as in [F4] and using step 1.1, $\mathrm{div}(f)=\mathrm{div}(t-a)-\mathrm{div}(t-b)=[a]-[\mathrm{\infty }]-[b]+[\mathrm{\infty }]=[a]-[b]\mathrm{.}$ This is a principal divisor by construction — it is the divisor of the nonzero rational function $f$ — and its degree is ${\mathrm{deg}}_k([a]-[b])=[\kappa (a):k]-[\kappa (b):k]=1-1=0$ by [F3] and step 1.1. In particular the zero part $[a]$ and the pole part $[b]$ are individually nontrivial while the total degree vanishes.

A general monic polynomial. Let $g\in k[t]$ be monic of degree $d\ge 0$. By [F6] it factors as $g={\prod }_i{g}_i^{m_i}$ with pairwise distinct monic irreducibles $g_i$ of degree $d_i$ and exponents $m_i\ge 1$, where ${\sum }_i{m}_i{d}_i=d$. Setting $p_i=V(g_i)$ and $Z(g)={\sum }_i{m}_i[p_i]$, additivity [F2] and the divisor formula of [F1] give $\mathrm{div}(g)={\sum }_i{m}_i\mathrm{div}(g_i)={\sum }_i{m}_i([p_i]-d_i[\mathrm{\infty }])=Z(g)-d[\mathrm{\infty }],$ a principal divisor; its degree is ${\mathrm{deg}}_{k}Z(g)-{\mathrm{deg}}_k(d[\mathrm{\infty }])={\sum }_i{m}_i{d}_i-d=0$ by [F3] and the residue-degree formula $[\kappa (p_i):k]=d_i$ of [F1]. For $d\ge 1$ the zero part $Z(g)$ and the pole part $d[\mathrm{\infty }]$ are both nonzero, while $Z(g)\sim d[\mathrm{\infty }]$ because their difference is the principal divisor $\mathrm{div}(g)$; for $d=0$ the polynomial is $g=1$ and $\mathrm{div}(g)=0$, which is the case $Z(g)=0$ and $0[\mathrm{\infty }]=0$ of the formula. Taking $g=t-a$ recovers $\mathrm{div}(t-a)=[a]-[\mathrm{\infty }]$ of step 1.1.

The triviality of the class and the sheaf. Since $\mathrm{div}(f)$ is principal, its class in $\mathrm{CaDiv}({\mathbb{P}}_k^1)\mathrm{/}\mathrm{Prin}({\mathbb{P}}_k^1)$ is zero, and the degree isomorphism of [F1] is injective, so the class of a degree-zero divisor is trivial: $\mathrm{div}(f)\sim 0$ by step 2.1. By the current Cartier/Picard dictionary [F12], the principal-divisor case of the sheaf attachment gives $\mathcal{O}(\mathrm{div}f)\cong {\mathcal{O}}_{{\mathbb{P}}_k^1}(0)\cong {\mathcal{O}}_{{\mathbb{P}}_k^1}$, and the tensor identities give $\mathcal{O}(\mathrm{div}f)\cong \mathcal{O}([a])\otimes \mathcal{O}([b]{)}^{-1}$; the same conclusion via degrees uses [F11], under which a degree-zero invertible sheaf on ${\mathbb{P}}_k^1$ is isomorphic to ${\mathcal{O}}_{{\mathbb{P}}_k^1}$.

The sections of $\mathrm{div}(f)$, directly. Let $g\in k(t{)}^{\times }$. By [F4], $g\in L(\mathrm{div}f)$ if and only if ${\mathrm{ord}}_x(g)+{\mathrm{ord}}_x(f)\ge 0$ at every closed point $x$, which by the additivity of [F2] is the same as ${\mathrm{ord}}_x(gf)\ge 0$ at every closed point $x$, that is $\mathrm{div}(gf)\ge 0$ by [F4]. So $g\in L(\mathrm{div}f)$ if and only if $gf\in L(0)$, and the space $L(0)$ is the constant field $k$: a nonzero rational function $h\in k(t{)}^{\times }$ factors by [F6] as $h=c{\prod }_i{g}_i^{n_i}$ with $c\in k^{\times }$, pairwise distinct monic irreducibles $g_i$ of degree $d_i$ and exponents $n_i\in \mathbb{Z}$, so that $\mathrm{div}(h)={\sum }_i{n}_i([p_i]-d_i[\mathrm{\infty }])$ by [F1] and the additivity of the orders [F2]; effectivity forces $n_i\ge 0$ for every $i$ (from the coefficient at $p_i$) and $-{\sum }_i{n}_i{d}_i\ge 0$ (from the coefficient at $\mathrm{\infty }$), and since $d_i\ge 1$ while $n_i\ge 0$ this gives $n_i=0$ for all $i$ and $h=c$. Hence $g\in L(\mathrm{div}f)$ if and only if $gf=c$ for some $c\in k$, i.e. $L(\mathrm{div}f)=k\cdot (1\mathrm{/}f)$, a one-dimensional space spanned by $1\mathrm{/}f=(t-b)\mathrm{/}(t-a)$, and $l(\mathrm{div}f)=1$ by [F5]. In particular $f\mathrm{\notin }L(\mathrm{div}f)$: the function $f$ corresponds to $c=gf=f^2\mathrm{\notin }k$, and directly $\mathrm{div}(f)+\mathrm{div}(f)=2[a]-2[b]$ has coefficient $-2<0$ at $[b]$, so it is not effective.

The cohomological reading. By step 3.1, $\mathcal{O}(\mathrm{div}f)\cong {\mathcal{O}}_{{\mathbb{P}}_k^1}$; by [F7] the structure sheaf has $h^0({\mathcal{O}}_{{\mathbb{P}}_k^1})=1$, and by [F8] it has $h^1({\mathcal{O}}_{{\mathbb{P}}_k^1})=0$. Therefore $l(\mathrm{div}f)=h^0(\mathrm{div}f)=1$ and $i(\mathrm{div}f)=h^1(\mathrm{div}f)=0$, in agreement with the direct computation of step 3.2; here $l(D)=h^0(D)$ and $i(D)=h^1(D)$ are the dimensions attached to $D$ and its sheaf [F5], the sheaf-theoretic equality being the current dictionary [F12].

Riemann-Roch and the Euler characteristic. The Euler characteristics are $\chi ({\mathcal{O}}_{{\mathbb{P}}_k^1})=h^0(\mathcal{O})-h^1(\mathcal{O})=1-0=1$ and, by step 4.1, $\chi (\mathcal{O}(\mathrm{div}f))=h^0(\mathrm{div}f)-h^1(\mathrm{div}f)=1-0=1$, so $\chi (\mathcal{O}(\mathrm{div}f))=\chi (\mathcal{O})$ as the degree shift [F10] requires for the degree-zero divisor $\mathrm{div}(f)$: $\chi (\mathcal{O}(\mathrm{div}f))-\chi (\mathcal{O})=0={\mathrm{deg}}_k\mathrm{div}(f)$ by step 2.1. Riemann-Roch [F9] reads $l(\mathrm{div}f)-i(\mathrm{div}f)=1-0=1={\mathrm{deg}}_k\mathrm{div}(f)+1-g=0+1-0,$ the genus of the projective line being $g=g({\mathbb{P}}_k^1)=0$ by [F1] and the degree being computed in step 2.1.

Assembly and the current supplier route. Step 2.1 computes $\mathrm{div}(f)=[a]-[b]$ of degree zero, step 3.1 identifies the divisor class as trivial and the attached sheaf as ${\mathcal{O}}_{{\mathbb{P}}_k^1}$, steps 3.2 and 4.1 compute $L(\mathrm{div}f)=k\cdot (1\mathrm{/}f)$ and $(l,i)=(1,0)$, step 5.1 reads the degree shift and Riemann-Roch as $1-0=0+1-0$, and step 2.2 gives $\mathrm{div}(g)=Z(g)-d[\mathrm{\infty }]$ for every monic polynomial of degree $d$. The sheaf identifications of steps 3.1 and 4.1 use the current interfaces [F12]; the degree-zero claim is computed explicitly in step 2.1, and the direct divisor calculations of steps 2.1, 2.2 and 3.2 use the current order route. The Axiom of Choice enters only through the suppliers recorded in [F13]: the functions $f$, $1\mathrm{/}f$ and the factorisations are exhibited by formulas, and no family of objects is selected.