Let $S$ be the spectrum of $\mathbf{Z}$ or the spectrum of an algebraically closed field. (Actually, one can take $S$ to be any noetherian integral regular scheme.)

Let $f:X\longrightarrow Y$ be a finite morphism of integral normal projective flat $S$-schemes which is etale above the complement of $B$, where $B\subset Y$ is a closed subscheme of codimension $1$. Suppose that $Y$ is regular.

Example. You could take $f$ to be a finite surjective morphism of normal surfaces such that $Y$ is nonsingular.

Since $Y$ is regular, we have a canonical sheaf $\omega_{Y/S}$. Let $s$ be a nonzero rational section of $\omega_{Y/S}$.
Define the cycle $K_{X/S} := \mathrm{div}(s)$. Note that $K_{X/S}$ is a canonical divisor.

Let $f^\ast s$ be the induced nonzero rational section of the line bundle $f^\ast \omega_{Y/S}$ on $X$ and consider the Weil divisor $\mathrm{div}(f^\ast s)$ on $Y$.

Outside $f^{-1}(B)$, we have that $\mathrm{div}(f^\ast s)$ is the pull-back of $K_{Y/S}$. Therefore, there is a Weil divisor $R_f$, supported on $f^{-1}(B)$, such that $\mathrm{div}(f^\ast s) = f^\ast K_{Y/S} + R_f$.

Question 1. How are the coefficients of $R_f$ defined?

Question 2. Is the Weil divisor $\mathrm{div}(f^\ast s)$ a canonical divisor outside the singular locus of $X$?

Question 3. Is $R_f$ independent of $s$?

Note that I work with cycles and not with classes up to linear equivalence.


Your hypothesis imply that $\omega_{Y/S}$ is an invertible sheaf (because $Y\to S$ is locally complete intersection).

(EDIT) As $f$ is flat at points of codimension $1$ ($Y$ is normal) and we are only interested on codimension 1 cycles, we can restrict $Y$ and suppose that $f$ is flat.

Then the dualizing sheaf $\omega_{X/Y}$ is invertible and you have the adjunction formula $$\omega_{X/S}=f^*\omega_{Y/S} \otimes\omega_{X/Y}.$$ The sheaf $\omega_{X/Y}$ is trivial outside of $B$ because $f$ is étale outside of $B$. It can be identified with the sheaf $\mathcal{Hom}_{O_Y}(f_{*}O_{X}, O_{Y})$.

Write $\omega_{X/Y}=O_X(D)$ for some Cartier divisor $D$ on $X$. Its support is contained in $f^{-1}(B)$. For any point $\eta$ of $X$ over a generic point $\xi$ of $B$, the stalk of $\omega_{X/Y}$ at $\eta$ is given by the different ideal of the extension of discrete valuation rings $O_{X,\eta}/O_{Y, \xi}$. The valuation of the different is known to be the ramification index $e_{\eta/\xi}$ minus $1$ when the ramification is tame and bigger or equal to $e_{\eta/\xi}$ otherwise (see Serre: Local fields). So the support of $D$ is equal to $f^{-1}(B)$ and is the ramification locus by definition.

In short, the coefficient of $R_f=D$ at the Zariski closure of $\eta$ is the valuation of the different ideal of $O_{X,\eta}/O_{Y, \xi}$. As for the computation, you can pass to the completions. A finite extension of complete DVR $R'/R$ is monogenous if the residue extension ($k(\eta)/k(\xi)$ in your case) is separable. If $R'=R[\theta]$, and $P(T)\in R[T]$ is the minimal polynomial of $\theta$, then the different ideal is generated by $P'(\theta)$. See Serre's book for more details.

  • $\begingroup$ 1. The morphism $f:X\longrightarrow Y$ need not be a local complete intersection. Again, how is $\omega_{X/Y}$ defined? 2. Similarly, the morphism $X\longrightarrow S$ need not be a local complete intersection. The scheme $X$ is only normal. How is $\omega_{X/S}$ defined? (I am looking at Definition 6.4.7 of your book.) 3. Thus, $R_f$ is given by the divisor of any section of $\omega_{X/Y}$? 4. Looking at Serre's book, there is also an upper bound for the valuation of the different. Namely, e-1+v(e). In the unequal characteristic case this is bounded from above by 2e-1. continued below... $\endgroup$
    – Tamed
    Jul 21 '11 at 14:18
  • $\begingroup$ To define $\omega_{X/S}$ I guess one could extend the line bundle $\omega_{\mathrm{Reg}(X)/S}$ on the regular part of $X$ to $X$. This extension is unique because the the singular locus of $X$ is of codimension greater or equal to 2. But what about $\omega_{X/Y}$? $\endgroup$
    – Tamed
    Jul 21 '11 at 14:22
  • $\begingroup$ Look also at 6.4.25 and 6.4.26. To avoid complications, you can restrict to $X\setminus f^{-1}(f(X_{sing}))$ following your remark. $\endgroup$
    – Qing Liu
    Jul 21 '11 at 14:40
  • $\begingroup$ Many thanks! I think I will be able to figure it out now. $\endgroup$
    – Tamed
    Jul 21 '11 at 14:43
  • $\begingroup$ I got it. One combines Theorem 6.4.32 and Lemma 6.4.26. Thanks again! $\endgroup$
    – Tamed
    Jul 21 '11 at 14:47

For the sake of simplicity, let $S=\operatorname{Spec}(k)$. I also suppose that there is no wild ramification, for instance requiring that $\textrm{char}(k) > \deg(f)$.

Answer to Question 1. It depends on the local behaviour of the cover around $R_f$. For instance, if $X$ is also smooth and $f$ is a Galois cover with group $G$, the multiplicity of each component $R_i$ of $R_f$ is $|\textrm{Stab}(R_i)|-1$, where $\textrm{Stab}(R_i)$ is the stabilizer subgroup of $R_i$.

Therefore, if $f$ is a double cover each component of $R_f$ appears with multiplicity $1$, if $f$ is a cyclic triple cover each component appears with multiplicity $2$ and so on.

Answer to Question 2. Yes, essentially by definition of canonical divisor. In particular, when $X$ is also smooth we have the identity $$K_X = f^*K_Y +R_f,$$ which is known as Hurwitz formula.

Answer to Question 3. Yes, $R_f$ is independent of $s$. In fact, its support coincides with the locus of points $x \in X$ where the differential $$df_x \colon T_xX \longrightarrow T_{f(x)}Y$$ is not an isomorphism, and this clearly does not depend on $s$.

  • $\begingroup$ 1. Thank you for the illuminating examples. Is there a general definition of the multiplicity of $B_{ij}$, where $B_{ij}$ is an irreducible component of $B_i$ and $B_i$ is an irreducible component of $B$? 2. If I simply define $K_{X/S}$ to be the divisor of $f^\ast s$, I get that $K_{X/S} = f^\ast K_{Y/S} + R_f$. When $X$ is regular, this implies, I believe, the Hurwitz formula. Note again that here we have an equality of cycles. 3. It's clear that the support of $R_f$ is independent of $s$. But why are the coefficients of $R_f$ independent of $s$? I mean, if I would have started with......... $\endgroup$
    – Tamed
    Jul 21 '11 at 12:10
  • $\begingroup$ ....another section $t$, I could, a priori, get another divisor $R_f^\prime$ with the same support as $R_f$ but different coefficients. And for my last question, I can kind of see that $S$ being Spec k doesn't really matter for the second and third question. But how does it matter for the first question? $\endgroup$
    – Tamed
    Jul 21 '11 at 12:11
  • $\begingroup$ 1) One can give a general definition of branching order $e_j$ around a component $R_j$ of the ramification divisor $R_f$, so that $R_f= \sum (e_j-1)R_j$. You can find it in Barth-Peters-Van de Ven book "Complex Algebraic Surfaces", Chapter I, Lemma (16.1). In the case of Galois covers, it coincides with the definition I gave in the answer. 3) Not only the support, but also the coefficients of $R_f$ are independent of $s$; in fact, the coefficient of the component $R_j$ equals $\dim X - \textrm{rank}(df_x)$, where $x \in R_j$ is a general point. $\endgroup$ Jul 21 '11 at 13:04
  • $\begingroup$ 4) Maybe the definition of branching order should be slightly changed in the case where $S$ is not $\textrm{Spec}(k)$, but I do not think this is particularly difficult. I'm always supposing there is no wild ramification here. $\endgroup$ Jul 21 '11 at 13:07

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