Showing that $2c_1(f_*\mathscr O_X)=-f_*R_f$ on curves, maybe by local fields

I originally asked this question on Mathematic StackExchange, but it did not seem to be attracting any attention, so now I am trying mathoverflow. I hope it is not too simple or unappropriate a question for this site, and if it is, then please may those in power feel free to delete it.

I would like a way to see that $2c_1(f_*\mathscr O_X)=-f_*R_f,$ where $R_f$ is the ramification divisor of a finite morphism of smooth curves $f\colon X\to Y$ over an algebraically closed field of characteristic $0.$

I am aware that this follows directly from Grothendieck-Riemann-Roch and that it is precisely what is proven in exercise IV. 2.6 of Hartshorne's Algebraic Geometry, where it boils down to the local duality for a finite etale morphism. I would like to have a way of seeing it different from those two.

For instance this paper states it is an easy calculation imitating Chapter 3.6, Proposition 13 of Serre's Local Fields. I do not see how that applies, so I would be very grateful for an explanation of how methods of local fields can be applied to local calculations in algebraic geometry in general, and specifically how to derive the above equality. That is not to say that I would be any lesss happy with a proof of the above equality by some entirely different methods.

Thanks in advance for any help!

• Probably they are referring to a local analysis of the degeneracy locus of the trace pairing on $f_*\mathcal{O}_X$. The trace pairing is equivalent to a homomorphism $f_*\mathcal{O}_X \to \textit{Hom}_{\mathcal{O}_Y}(f_*\mathcal{O}_X,\mathcal{O}_Y)$, and this is injective in the tame case. Applying the Whitney sum formula relates $c_1$ to the cokernel of this injective homomorphism. – Jason Starr Feb 27 '14 at 16:44
• That sounds very interesting. Could you perhaps write it out in more detail as an answer? – A Rock and a Hard Place Feb 27 '14 at 16:55
• @user82454: In general, the degeneracy locus of a generically surjective morphism $E \rightarrow F$ of vector bundles of the same rank is represented by $c_{1}(F)-c_{1}(E)$; all we have to do is note that $f_{\ast}R_f$ the degeneracy locus of Jason's homomorphism. – Yusuf Mustopa Feb 27 '14 at 17:45
• @YusufMustopa: If I understand correctly, you are refering to the Thom-Porteous formula? – A Rock and a Hard Place Feb 27 '14 at 18:18
• @user82454: I was indeed referring to the Thom-Porteous formula. In any event, I'm glad Jason took the time to write the more informative answer below! – Yusuf Mustopa Feb 27 '14 at 20:38

I looked at that section of Local Fields, and I believe this is about the trace pairing. For a finite, flat morphism of Dedekind schemes, $f:X\to Y$, the pushforward sheaf $f_*\mathcal{O}_X$ is a locally free $\mathcal{O}_Y$-module, let's say of (constant) rank $n$. It is also an algebra, i.e., multiplication is a $\mathcal{O}_Y$-module homomorphism, $$\theta_f : f_*\mathcal{O}_X\otimes_{\mathcal{O}_Y} f_*\mathcal{O}_X \to f_*\mathcal{O}_X.$$ By adjunction of tensor and hom, this is equivalent to an $\mathcal{O}_Y$-module homomorphism, $$\widetilde{\theta}_f:f_*\mathcal{O}_X \to \textit{Hom}_{\mathcal{O}_Y}(f_*\mathcal{O}_X,f_*\mathcal{O}_X).$$ But the target module, being End of a locally free sheaf, has a trace map in the usual sense of trace of matrices, $$\text{Tr}_{f_*\mathcal{O}_Y}:\textit{Hom}_{\mathcal{O}_Y}(f_*\mathcal{O}_X,f_*\mathcal{O}_X) \to \mathcal{O}_Y.$$ The composition with $\widetilde{\theta}_f$ is a trace map for $f$, $$\text{Tr}_f:f_*\mathcal{O}_X \to \mathcal{O}_Y.$$ The composition of this trace map with $\theta_f$ is a $\mathcal{O}_Y$-bilinear pairing on the locally free $\mathcal{O}_Y$-module $f_*\mathcal{O}_X$, $$T_f: f_*\mathcal{O}_X\otimes_{\mathcal{O}_Y} f_*\mathcal{O}_X \to \mathcal{O}_Y.$$ This pairing is often called the "trace pairing". Again, by adjunction of tensor and hom, it is equivalent to an $\mathcal{O}_Y$-module homomorphism, $$\widetilde{T}_f:f_*\mathcal{O}_X \to \textit{Hom}_{\mathcal{O}_Y}(f_*\mathcal{O}_X,\mathcal{O}_Y).$$ In particular, taking the associated $n^{\text{th}}$ exterior power of both sides, i.e., taking the determinant, gives a morphism of invertible $\mathcal{O}_Y$-modules, $$\text{det}(\widetilde{T}_f): \text{det}(f_*\mathcal{O}_X) \to \textit{Hom}_{\mathcal{O}_Y}(\text{det}(f_*\mathcal{O}_X),\mathcal{O}_Y).$$ If $f$ is generically étale, then $\text{det}(\widetilde{T}_f)$ is injective. In this case, by definition, the determinant of $f$ is defined to be the unique invertible ideal sheaf $\mathfrak{d}_f$ of $\mathcal{O}_Y$ such that $$\text{Image}(\text{det}(\widetilde{T}_f)) = \textit{Hom}_{\mathcal{O}_Y}(\text{det}(f_*\mathcal{O}_X),\mathcal{O}_Y)\otimes_{\mathcal{O}_Y}\mathfrak{d}_f,$$ as subsheaves of $\textit{Hom}_{\mathcal{O}_Y}(\text{det}(f_*\mathcal{O}_X),\mathcal{O}_Y)$, i.e., $\text{det}(\widetilde{T}_f)$ is an isomorphism, $$\text{det}(\widetilde{T}_f): \text{det}(f_*\mathcal{O}_X) \xrightarrow{\cong} \textit{Hom}_{\mathcal{O}_Y}(\text{det}(f_*\mathcal{O}_X),\mathcal{O}_Y)\otimes_{\mathcal{O}_Y}\mathfrak{d}_f.$$
On the other hand, the different of $f$, $\mathfrak{D}_f$, is defined to be the maximal minimal invertible ideal sheaf on $X$, $$\iota:\mathfrak{D}_f \hookrightarrow \mathcal{O}_X,$$ such that for the associated transpose, $$\iota^\dagger: \mathcal{O}_X \to \textit{Hom}_{\mathcal{O}_X}(\mathfrak{D}_f,\mathcal{O}_X),$$ with its induced pushforward map under $f_*$, $$f_*\iota^\dagger: f_*\mathcal{O}_X \to f_*\textit{Hom}_{\mathcal{O}_X}(\mathfrak{D}_f,\mathcal{O}_X),$$ the $\mathcal{O}_Y$-module homomorphism $\text{Tr}_f$ factors through $f_*\iota^\dagger$, i.e., there exists a (unique) $\mathcal{O}_Y$-module homomorphism, $$\tau_f:f_*\textit{Hom}_{\mathcal{O}_X}(\mathfrak{D}_f,\mathcal{O}_X) \to \mathcal{O}_Y,$$ such that $\text{Tr}_f$ equals $\tau_f\circ f_*\iota^\dagger$. The ramification divisor of $f$, $R_f$, is the unique effective Cartier divisor of $X$ whose associated invertible sheaf equals $\mathfrak{D}_f$. One of the basic computations, Chapter III.3, Proposition 6 on p. 50 of Local Fields, states that the pushforward Cartier divisor $f_*R_f$ equals the effective Cartier divisor on $Y$ associated to the ideal sheaf $\mathfrak{d}_f$, i.e., $$\mathfrak{d}_f = \text{Nm}_f(\mathfrak{D}_f),$$ where $\text{Nm}_f$ is the norm associated to $f$. In particular, the $\mathcal{O}_Y$-module homomorphism $\text{det}(\widetilde{T}_f)$ above now becomes an isomorphism, $$\text{det}(\widetilde{T}_f):\text{det}(f_*\mathcal{O}_X) \xrightarrow{\cong} \textit{Hom}_{\mathcal{O}_Y}(\text{det}(f_*\mathcal{O}_X),\mathcal{O}_Y)(-f_*R_f).$$ Using adjunction of tensor and hom one last time, this is an isomorphism of invertible $\mathcal{O}_Y$-modules, $$\text{det}(f_*\mathcal{O}_X) \otimes_{\mathcal{O}_Y} \text{det}(f_*\mathcal{O}_X) \xrightarrow{\cong} \mathcal{O}_Y(-f_*R_f).$$ This is the identity that you asked about.
• Thank you for the elaborated answer! May I just ask what the morphism $\mu\colon f_*\mathcal O_X\otimes_{\mathcal O_Y} f_*\mathcal O_X\to f_*\mathcal O_X$ is? – A Rock and a Hard Place Feb 27 '14 at 19:55
• @user82454: That is the multiplication map. The sheaf $\mathcal{O}_X$ is a sheaf of rings -- it has a multiplication map. Thus the morphism of ringed spaces, $f$, makes $f_*\mathcal{O}_X$ into a sheaf of $\mathcal{O}_Y$-algebras with an $\mathcal{O}_Y$-bilinear multiplication map. – Jason Starr Feb 27 '14 at 19:57
• @user82454: Now I see what you are asking: $\mu$ is the same as $\theta$. In "Local Fields", they use $\theta$, but I usually call this $\mu$. So I slipped and called it $\mu$ in one place. I fixed it now. – Jason Starr Feb 27 '14 at 20:13
• Actually, there is one detail more that I do not quite understand: where does the minus sign in $-f_*R$ come from? Does the calculation in Local Fields show that $f_*R_f$ is the divisor associated to the sheaf $\mathfrak d_f^{-1}$ or to $\mathfrak d_f$? – A Rock and a Hard Place Feb 27 '14 at 23:18