# Degeneracy divisor of the “trace” morphism

Let $f\colon X\to Y$ be a finite morphism of smooth curves over an alg. closed field of characteristic zero. I recently asked how methods reminiscent of basic algebraic number theory can be used to see that $2c_1(\mathscr O_X)=-f_*R_f$ where $R_f$ is the raminification divisor of $f,$ and I got a great answer explaining just that. A comment of Yusuf Mustopa caught my interest though.

The trace pairing gives rise to an injection $f_*\mathscr O_X\to \mathscr{H}om_{\mathscr O_Y}(f_*\mathscr O_X,\mathscr O_Y)$ (see Jason Starr's answer to the previous question for details). Via the Thom-Porteous formula, my previous question boils down to showing that the degeneracy divisor of this morphism of vector bundles is linearly equivalent to the divisor $f_*R_f.$

Can that be seen explicitly?

By that I mainly mean: not by translating it to statements about norms, discriminants and differents as implicitly done in Jason Starr's answer mentioned above.

• You can reduce to the case of a local ring, or better yet, complete local rings. Then the only finite morphism of complete local rings $k[[x]]$ is the raising to a power map $x \to x^n$, and you can check that both conditions occur if and only if $n>1$. I'm not sure if you'd like that proof. – Will Sawin Mar 5 '14 at 4:52
• In fact, not only is the effective Cartier divisor $f_*R_r$ "linearly equivalent" to the effective Cartier divisor $\text{Zero}(\widetilde{T}_f)$, they are even equal as effective Cartier divisors. As Will says, this stronger claim is a local statement that can even be checked after base change; after 'etale base change, you can arrange that the target is a DVR and the domain is a product of DVRs. In characteristic $0$, you only need to consider $x\mapsto x^n$, just as Will suggests. However, in char $p>0$, there are many more cases (moduli of non-tame covers). Best advice: read Serre. – Jason Starr Mar 5 '14 at 10:53
• @Will Sawin: Actually I quite like passing locally to completion as a roof. Thanks also to Jason Starr for the advice; I am following it and am indeed reading Serre. However, I thought having a more direct proof for characteristic $0$ (this is not just a technical assumption to make things easier, but really the case that interests me most, since I have some background in complex analytic geometry) would be very nice. – A Rock and a Hard Place Mar 6 '14 at 23:18