The Stacks project

Lemma 42.18.3. Let $K$ be any field. Let $X$ be a $1$-dimensional integral scheme endowed with a proper morphism $c : X \to \mathop{\mathrm{Spec}}(K)$. Let $f \in K(X)^*$ be an invertible rational function. Then

\[ \sum \nolimits _{x \in X \text{ closed}} [\kappa (x) : K] \text{ord}_{\mathcal{O}_{X, x}}(f) = 0 \]

where $\text{ord}$ is as in Algebra, Definition 10.121.2. In other words, $c_*\text{div}(f) = 0$.

Proof. Consider the diagram

\[ \xymatrix{ Y \ar[r]_ p \ar[d]_ q & X \ar[d]^ c \\ \mathbf{P}^1_ K \ar[r]^-{c'} & \mathop{\mathrm{Spec}}(K) } \]

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. $\square$

Comments (4)

Comment #176 by Adeel on

After the diagram, p and c should be switched in .

Comment #6610 by WhatJiaranEatsTonight on

Do I miss some details? I think that by lemma 18.1, we know that . And obviously g is a unit on . Thus .

But the proof seems much longer than I expected. I don't know if I miss some crucial details or misunderstand something.

Comment #6611 by WhatJiaranEatsTonight on

I see. The two schemes are not of the same dimension. Sorry for my ignorance.

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