Let $f : X \to Y$ be a finite morphism between compact complex manifolds of the same dimension $n$. We denote by $R_f \subset X$ the ramification divisor of $f$ and by $J_f \subset X$ the set of points where $\operatorname{Jac} f = 0$.

It is easy to see that $R_f = J_f$ as sets ($J_f \subset R_f$ is trivial; the other inclusion follows from the inverse function theorem). Do the two also have the same scheme structure?

This is true in the case of curves, pretty much by local inspection. I have a vague feeling one can prove this holds in the general case too by first reducing to affine spaces and then inducting on the dimension: given a point of interest we'd first pick a hyperplane $X_1$ in $X$ that intersects the sets nicely, then project away from a point in $Y$ to an affine space $Y_1$ and the composition $f_1: X_1 \to X \to Y \to Y_1$ would be finite of the right order at the corresponding point of interest. Eventually we'd be dealing with curves where things are true and try to bootstrap our way up again. This feels messy (and maybe doesn't work) so if there's a simpler argument I'd love to see it.

  • 3
    $\begingroup$ What is your definition of $R_f$? One of the definitions in the above situation is to use the natural inclusion $f^*K_Y\to K_X$ where $K$ stands for canonical bundles. Then $R_f$ is the scheme-theoretic support of $K_X/f^*K_Y$. With this definition, it is a tautology that $R_f=J_f$ scheme-theoretically. $\endgroup$ – Mohan Aug 3 '13 at 23:08
  • $\begingroup$ Ah bon? That might be my problem, I've had trouble finding a good definition of $R_f$ in higher dimensions. The ones I've seen essentially copy the curve one, mumbling something about local multiplicity. You don't happen to know a good reference for these things? $\endgroup$ – Gunnar Þór Magnússon Aug 4 '13 at 0:01
  • $\begingroup$ If everything is smooth, the definition I give above should be the standard one. A point $x\in X$ is ramified if the map is not etale there and then the definition I give is the correct one. I will look for a standard reference. $\endgroup$ – Mohan Aug 4 '13 at 2:29

I don't know it is exactly what you want or not. But a reference not in the context of complex manifolds but smooth schemes is for example "Neron models" by Bosch-Lütkebomert, setion 2.2: If $f:X\rightarrow Y$ is a morphism of smooth schemes, then $f$ is etale at $x\in X$ if and only if the canonical homomorphism $(f^{*}\Omega_{Y}^{1})_{x}\rightarrow (\Omega_{X}^{1})_{x}$ is isomorphism. Now this last condition is equivalent to smoothness of $f$ at $x$ (proposition 8) or the fact that $Jac(f)\neq0$ at $x$. This shows that in fact $J_{f}$ and $R_{f}$ are the same at least as sets. But it is also proved there that the map $(f^{*}\Omega_{Y}^{1})_{x}\rightarrow (\Omega_{X}^{1})_{x}$ is isomorphism if and only if the canonical homomorphism $(f^{*}\Omega_{Y}^{1})\otimes k(x)\rightarrow (\Omega_{X}^{1})\otimes k(x)$ is injective. This in turn shows that $R_{f}$ and $J_{f}$ can both be thought of as the kernel of this restricted homomorphism at each point. So $R_{f}$ and $J_{f}$ are the same as schemes. In that book, they also mention (p.40) that this is the counterpart of Jacobi criterion in differential geometry.


Given an effective divisor $D$, how do we associate a subscheme to it. If it is an irreducible divisor, then it corresponds to an ideal $I$ that is locally principal, and the subscheme corresponding to the divisor is the subscheme corresponding to the ideal. For an effective divisor, presumably we just take the product of the ideals. I can't see another effective way to take the local multiplicity into account scheme-theoretically.

I think it's clear that the ramification divisor corresponds in this way to the Jacobian ideal $(f^* K_Y) \otimes K_X^{-1}$. This is a locally principal ideal, so we check that they have the same multiplicity on codimension $1$ points, which we can do by the same argument as for curves.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.