The Stacks project

Proof. Consider the diagram

that we constructed in Lemma 42.18.2 starting with $X$ and the rational function $f$ over $S = \mathop{\mathrm{Spec}}(K)$. We will use all the results of this lemma without further mention. We have to show that $c_*\text{div}_ X(f) = c_*p_*\text{div}_ Y(f) = 0$. This is the same as proving that $c'_*q_*\text{div}_ Y(f) = 0$. If $q(Y)$ is a closed point of $\mathbf{P}^1_ K$ then we see that $\text{div}_ X(f) = 0$ and the lemma holds. Thus we may assume that $q$ is dominant. Suppose we can show that $q : Y \to \mathbf{P}^1_ K$ is finite locally free of degree $d$ (see Morphisms, Definition 29.48.1). Since $\text{div}_ Y(f) = [q^{-1}D_0]_0 - [q^{-1}D_\infty ]_0$ we see (by definition of flat pullback) that $\text{div}_ Y(f) = q^*([D_0]_0 - [D_\infty ]_0)$. Then by Lemma 42.15.2 we get $q_*\text{div}_ Y(f) = d([D_0]_0 - [D_\infty ]_0)$. Since clearly $c'_*[D_0]_0 = c'_*[D_\infty ]_0$ we win.

It remains to show that $q$ is finite locally free. (It will automatically have some given degree as $\mathbf{P}^1_ K$ is connected.) Since $\dim (\mathbf{P}^1_ K) = 1$ we see that $q$ is finite for example by Lemma 42.16.2. All local rings of $\mathbf{P}^1_ K$ at closed points are regular local rings of dimension $1$ (in other words discrete valuation rings), since they are localizations of $K[T]$ (see Algebra, Lemma 10.114.1). Hence for $y\in Y$ closed the local ring $\mathcal{O}_{Y, y}$ will be flat over $\mathcal{O}_{\mathbf{P}^1_ K, q(y)}$ as soon as it is torsion free (More on Algebra, Lemma 15.22.11). This is obviously the case as $\mathcal{O}_{Y, y}$ is a domain and $q$ is dominant. Thus $q$ is flat. Hence $q$ is finite locally free by Morphisms, Lemma 29.48.2. $\square$