The Stacks project

Proof. Observe that $\text{div}_ T(q^*(s)) = \text{div}_{\mathcal{O}_ T}(q^*(s))$ by the compatibility between the constructions given in Spaces over Fields, Sections 72.6 and 72.7. We will show the agreement with $\text{div}_{\mathcal{O}_ T}(q^*(s))$ in this proof. We will use all the properties of $q : T \to X$ stated in Lemma 82.17.3 without further mention. Let $Z \subset T$ be a prime divisor. Then either $Z \to X$ is dominant or $Z = q^{-1}(Z')$ for some prime divisor $Z' \subset X$. If $Z \to X$ is dominant, then the coefficient of $Z$ in either side of the equality of the lemma is zero. Thus we may assume $Z = q^{-1}(Z')$ where $Z' \subset X$ is a prime divisor. Let $\xi ' \in |Z'|$ and $\xi \in |Z|$ be the generic points. Then we obtain a commutative diagram

see Decent Spaces, Remark 68.11.11. Choose a trivialization $s_{\xi '}$ of $\mathcal{L}_{\xi '} = c_{\xi '}^*\mathcal{L}$. Then we can use the pullback $s_\xi $ of $s_{\xi '}$ via $h$ as our trivialization of $\mathcal{L}_\xi = c_\xi ^* q^*\mathcal{L}$. Write $s/s_{\xi '} = a/b$ for $a, b \in \mathcal{O}_{X, \xi '}$ nonzerodivisors. By definition the coefficient of $Z'$ in $\text{div}_\mathcal {L}(s)$ is

Since $Z = q^{-1}(Z')$, this is also the coefficient of $Z$ in $q^*\text{div}_\mathcal {L}(s)$. Since $\mathcal{O}_{X, \xi '}^ h \to \mathcal{O}_{T, \xi }^ h$ is flat the elements $a, b$ map to nonzerodivisors in $\mathcal{O}_{T, \xi }^ h$. Thus $q^*(s)/s_\xi = a/b$ in the total quotient ring of $\mathcal{O}_{T, \xi }^ h$. By definition the coefficient of $Z$ in $\text{div}_ T(q^*(s))$ is

The proof is finished because these lengths are the same as before by Algebra, Lemma 10.52.13 and the fact that $\mathfrak m_\xi ^ h = \mathfrak m_{\xi '}^ h\mathcal{O}_{T, \xi }^ h$ shown in Lemma 82.17.3. $\square$