Divisor of a rational function on the projective line

The charts and all points except infinity. 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. The ideal $(u)\subseteq k[u]$ is maximal with residue field $k[u]\mathrm{/}(u)\cong k$, so every prime of $k[u]$ containing $u$ equals $(u)$; hence every point of $U_{\mathrm{\infty }}$ other than $\mathrm{\infty }=(u)$ lies in the basic open $D(u)=U_0\cap U_{\mathrm{\infty }}\subseteq U_0$. Therefore every point of ${\mathbb{P}}_k^1$ other than $\mathrm{\infty }$ lies in $U_0$, and by [F2] it corresponds to a prime $\mathfrak{p}\subseteq k[t]$ with ${\mathcal{O}}_{{\mathbb{P}}_k^1,x}=k[t{]}_{\mathfrak{p}}$.

Integrality, normality and the function field. ${\mathbb{P}}_k^1$ is an integral finite-type $k$-scheme, 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 affine schemes because $k[t]$ and $k[u]$ are domains, and they are glued along the nonempty open subscheme $D(t)\cong D(u)$. The two irreducible open charts have nonempty overlap, which is dense in both, so their union is irreducible, and stalks of the glued scheme are stalks of one of the two charts, so reducedness is inherited; hence ${\mathbb{P}}_k^1$ is integral, and it is finite type over $k$ because its affine charts are. Every local ring of ${\mathbb{P}}_k^1$ is a localisation $k[t{]}_{\mathfrak{p}}$ or $k[u{]}_{\mathfrak{q}}$ of an integrally closed domain [1.1], and 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 $A$, obtained by multiplying the equation of $x$ by $s^n$, whose coefficients lie in $A$; so $sx\in A$ and $x\in S^{-1}A$. The rings $k[t]$, $k[u]$ are Noetherian [F3], so ${\mathbb{P}}_k^1$ is a normal Noetherian scheme [F4]. By [F5] the function field is $\mathrm{Frac}k[t]=k(t)$, and $k(t)=k(u)$ with $t=u^{-1}$ on the overlap; in particular $f=t^2\mathrm{/}(t-1)$ is a global meromorphic unit.

The coordinates are uniformisers. The local ring at $0=(t)$ is $k[t{]}_{(t)}$, with maximal ideal generated by $t$; similarly $t-1$ generates the maximal ideal of $k[t{]}_{(t-1)}$ at $1$, and $u$ generates the maximal ideal of $k[u{]}_{(u)}$ at $\mathrm{\infty }$. Since $\dim k[t]=\dim k[u]=1$ and the ideals $(t)$, $(t-1)$, $(u)$ are nonzero, each of these local rings has dimension one and is therefore a discrete valuation ring [F4]; in each of them the displayed generator is a uniformiser, so $v_{[0]}(t)=v_{[1]}(t-1)=v_{\mathrm{\infty }}(u)=1$ and $v_{[0]}(t-1)=v_{[1]}(t)=v_{\mathrm{\infty }}(1-u)=0$, because $t-1\mathrm{\notin }(t)$, $t\mathrm{\notin }(t-1)$ and $1-u\mathrm{\notin }(u)$ are units.

Orders at the origin and at one. In $k(t)$ one has $f=t^2\cdot (t-1{)}^{-1}$, and ${\mathrm{ord}}_{[0]}(f)=2$, ${\mathrm{ord}}_{[1]}(f)=-1$. By additivity of the order, ${\mathrm{ord}}_{[0]}(f)=2v_{[0]}(t)-v_{[0]}(t-1)=2\cdot 1-0=2$ and ${\mathrm{ord}}_{[1]}(f)=2v_{[1]}(t)-v_{[1]}(t-1)=2\cdot 0-1=-1$.

Order at infinity. On $U_{\mathrm{\infty }}$ the coordinate is $u=1\mathrm{/}t$, and $f=u^{-1}(1-u{)}^{-1}$ in $k(u)=k(t)$, so ${\mathrm{ord}}_{\mathrm{\infty }}(f)=-1$. Indeed $f=\frac{t^2}{t-1}=\frac{u^{-2}}{u^{-1}-1}=\frac{1}{u(1-u)}=u^{-1}(1-u{)}^{-1}$, and with the uniformiser $u$ and the unit $1-u$ of step 1.3, additivity gives ${\mathrm{ord}}_{\mathrm{\infty }}(f)=-v_{\mathrm{\infty }}(u)-v_{\mathrm{\infty }}(1-u)=-1-0=-1$.

Every other prime divisor has order zero. If $Z\mathrm{\ne }[0],[1],[\mathrm{\infty }]$ is a prime divisor with generic point $\xi$, then ${\mathrm{ord}}_Z(f)=0$. By 1.1 the point $\xi$ lies in $U_0$ and corresponds to a prime $\mathfrak{p}\subseteq k[t]$ with ${\mathcal{O}}_{{\mathbb{P}}_k^1,\xi }=k[t{]}_{\mathfrak{p}}$; since $Z$ is a prime divisor, $\dim k[t{]}_{\mathfrak{p}}=1=\mathrm{ht}(\mathfrak{p})$, so $\mathfrak{p}\mathrm{\ne }(0)$ [F4]. The maximal ideals $(t)$ and $(t-1)$ of $k[t]$ are the primes of the points $[0]$ and $[1]$; as they are maximal, $t\in \mathfrak{p}$ forces $\mathfrak{p}=(t)$, and $t-1\in \mathfrak{p}$ forces $\mathfrak{p}=(t-1)$. Hence $t,t-1\mathrm{\notin }\mathfrak{p}$, both elements are units of $k[t{]}_{\mathfrak{p}}$, and $f=t^2(t-1{)}^{-1}$ is a unit of ${\mathcal{O}}_{{\mathbb{P}}_k^1,\xi }$; by [F6] its valuation, and hence ${\mathrm{ord}}_Z(f)$, is $0$.

Conclusion. The principal divisor of $f$ is $\mathrm{div}(f)=2[0]-[1]-[\mathrm{\infty }]$.