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Let $f:X \rightarrow Y$ be a finite, separable morphism of curves (curve: integral scheme, of dimension 1, proper over an algebraically closed field with all local rings regular). Let $R$ be the ramification divisor on $X$.

Question 1: What is exactly $f_*(R)$? If $R = \sum_{P \in X} r_P \cdot P$ ($r_P$ is the ramification index), then i suppose that $f_*(R) = \sum_{P \in X} r_P \cdot f(P)$. Is that correct?

Now let $B$ be a boundary of $Supp(f_*(R))$, i.e. a divisor on $K_Y$ with nonzero coefficients equal to $1$ on a superset of the support of $f_*(R)$. We know that there is a homomorphism $f^* : Div(Y) \rightarrow Div(X)$ (see Hartshorne bottom of page 137) and so we can talk about $f^* B$.

Question 2: What is $f^{-1} B$?

Motivation: i saw this notation in an exercise where i am asked to prove a Hurwitz-like formula of the form $f^*(K_Y+B)=K_X+f^{-1}B$, where $K_X,K_Y$ are the canonical divisors.

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    If $B$ lives on $X$, then $K_Y+B$ doesn't make sense.2012-11-19
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    @QiL: You are right, i will edit.2012-11-19

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(1) Yes, this is a standard definition. You should have read it before.

(2) This is not a canonical notation. Can you check your exercise ?

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    My exercise is as i have written it.2012-11-19
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    Then it means $f^*B-R$ if the desired equality holds. Strange notation anyway.2012-11-19
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    Maybe if $B=\sum_{Q \in Y} \lambda_Q \cdot Q$ then $f^{-1}B=\sum_{Q \in Y} \sum_{P:f(P)=Q} \lambda_Q \cdot P$?2012-11-19
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    This coincides with $f^*B$ if the support of $B$ is disjoint from the support of $R$. But this is not your case. What is good to know is $f^*K_Y=K_X+R$ (classical Riemann-Hurwitz) and $f^*(K_Y+B)=f^*K_Y+f^*B$. So what must be $f^{-1}B$ ?2012-11-19
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    Yea, i see this, thanks.2012-11-19
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    @Manos: sometimes, when $B$ is viewed as a closed subscheme, $f^{-1}B$ can be $B\times_Y X$. Then as Cartier divisor, it is same as $f^*B$. But definitely different from $f^*B-R$.2012-11-20