Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

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 invetible and you have the adjunction formula $$\omega_{X/S}=f^*\omega_{Y/S} \otimes\omega_{X/Y}.$$ The sheaf $\omega_{X/Y}$ is trivial out side of $B$ because $f$ is étale out side 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.

share|improve this answer
    
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... –  Tamed Jul 21 '11 at 14:18
    
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 $\omega_{X/Y}$? –  Tamed Jul 21 '11 at 14:22
    
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. –  Qing Liu Jul 21 '11 at 14:40
    
Many thanks! I think I will be able to figure it out now. –  Tamed Jul 21 '11 at 14:43
    
I got it. One combines Theorem 6.4.32 and Lemma 6.4.26. Thanks again! –  Tamed Jul 21 '11 at 14:47
add comment

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 $s$.

share|improve this answer
    
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......... –  Tamed Jul 21 '11 at 12:10
    
....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? –  Tamed Jul 21 '11 at 12:11
    
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. –  Francesco Polizzi Jul 21 '11 at 13:04
    
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. –  Francesco Polizzi Jul 21 '11 at 13:07
add comment

Your Answer

 
discard

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.