Lemma 42.18.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be locally of finite type over $S$. Assume $X$, $Y$ are integral and $n = \dim _\delta (X) = \dim _\delta (Y)$. Let $p : X \to Y$ be a dominant proper morphism. Let $f \in R(X)^*$. Set
Then we have $p_*\text{div}(f) = \text{div}(g)$.
Proof. Let $Z \subset Y$ be an integral closed subscheme of $\delta $-dimension $n - 1$. We want to show that the coefficient of $[Z]$ in $p_*\text{div}(f)$ and $\text{div}(g)$ are equal. We may apply Lemma 42.16.2 to the morphism $p : X \to Y$ and the generic point $\xi \in Z$. Hence we may replace $Y$ by an affine open neighbourhood of $\xi $ and assume that $p : X \to Y$ is finite. Write $Y = \mathop{\mathrm{Spec}}(R)$ and $X = \mathop{\mathrm{Spec}}(A)$ with $p$ induced by a finite homomorphism $R \to A$ of Noetherian domains which induces an finite field extension $L/K$ of fraction fields. Now we have $f \in L$, $g = \text{Nm}(f) \in K$, and a prime $\mathfrak p \subset R$ with $\dim (R_{\mathfrak p}) = 1$. The coefficient of $[Z]$ in $\text{div}_ Y(g)$ is $\text{ord}_{R_\mathfrak p}(g)$. The coefficient of $[Z]$ in $p_*\text{div}_ X(f)$ is
The desired equality therefore follows from Algebra, Lemma 10.121.8. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)