Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question refers to the Definition in p. 137 from Hartshorne. Let $f: X \rightarrow Y$ be a finite morphism of nonsingular curves. Let $Q$ be a closed point of $Y$ and $P \in X$ such that $f(P)=Q$. Let $t \in \mathcal{O}_{Y,Q}$. Then $\mathcal{O}_{Y,Q}$ is a discrete valuation ring and we can talk about the quantity $v_Q(t)$, where $v_Q$ is the corresponding valuation. Now, $\mathcal{O}_{X,P}$ is also a discrete valuation ring with valuation $v_P$, but Hartshorne is talking about the quantity $v_P(t)$. Does $v_P(t)$ really mean $v_P\left(f_Q(t)\right)$, where $f_Q : \mathcal{O}_{Y,Q} \rightarrow \mathcal{O}_{X,P}$ is the local ring homomorphism induced by the morphism $f$?

The above notation $v_P(t)$ appears in the definition of the group homomorphism $f^*:Div(Y) \rightarrow Div(X)$ by $f^*Q = \sum_{f(P)=Q} v_P(t) \cdot P$

share|cite|improve this question
Dear Manos, you should change your notation for the induced local morphism to $f^*_P : \mathcal{O}_{Y,Q} \rightarrow \mathcal{O}_{X,P}$, because that morphism depends on $P$, not on $Q$. In other words if several $P_i$'s are sent by $f$ to the same $Q$ the induced local morphisms $f^*_{P_i} : \mathcal{O}_{Y,Q} \rightarrow \mathcal{O}_{X,P_i}$ are different and your notation would then be ambiguous.Anyway, you should adopt this modified notation because it is the one universally used since EGA was published :-) – Georges Elencwajg Oct 23 '12 at 20:04
up vote 1 down vote accepted

Yes. And I believe he's taking $t$ to be a uniformizing parameter of the DVR $\mathcal{O}_{Y,Q}$ in the definition of that homomorphism.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.