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](https://www.math.leidenuniv.nl/scripties/MasterJavanpeykar.pdf) 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!



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


