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I am aware of two forms of the Hurwitz formula. The first is more common, and deals only with the degrees. So if $f:X \rightarrow Y$ is a non-constant map of degree $n$ between two projective non-singular curves, with genera $g_X$ and $g_Y$, then $$ 2(g_X-1) = 2n(g_Y-1) + \deg(R), $$ where $R$ is the ramification divisor of $f$. The proof of this was given to me as an exercise when I started my PhD, and I am very happy with it.

However, in some other work that I was doing it appeared that one could strengthen this to say rather that if $K_X={\rm div}(f^*(dx))$ and $K_Y={\rm div}(dx)$ are canonical divisors of $X$ and $Y$, then $$ K_X = n\cdot K_Y + R. $$ I have found this alluded to in a number of places, and even stated in Algebraic Curves Over Finite Fields by Carlos Moreno. However, this was without proof, and every idea of a proof that I have seen is in sheaf theoretic language. I am slowly getting through sheaves and schemes, but I am currently trying to prove this in an elementary manner (fiddling around with orders of $dx$ etc.), in the wildly ramified case (the tamely ramified case is fine).

I would like to know if

  1. It is possible to prove this without using sheaves etc in the wildly ramified case
  2. If so, are there any references that would help with this.

edit: Also, is there a different name for the "more specific" Hurwitz formula?

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There is an IMHO relatively accessible proof in Stichtenoth's Algebraic Function Fields and Codes. It is written without using differential forms (canonical divisor being defined in an adelic way as the divisor for whichs Riemann-Roch holds). Not surprisingly the proof gets a bit technical when dealing with wild ramification. –  Jyrki Lahtonen Jul 23 '12 at 10:54
    
I thought the only proof in there was in terms of degrees, I must not have looked hard enough. Thanks for the recommendation, I will reread it more thoroughly. –  Joe Tait Jul 23 '12 at 11:14
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2 Answers

You might like Griffiths's proof of Riemann-Hurwitz (Introduction to Algebraic Curves page 91, theorem (8.5) ) for a morphism $f:X\to Y$ .
It is very natural: he starts from a meromorphic differential form $\omega \in \Omega(Y)$ (whose existence he provisionally admits ) , lifts it to $f^*\omega \in \Omega(X)$, then computes and compares the divisors $div(f^*\omega), f^*(div(\omega))$ and $R$ locally in coordinates.

Riemann-Hurwitz is not difficult but somewhat explains the slightly confusing result that taking divisors and lifting do not commute for differential forms.

Edit
Stichtenoth gives a proof for fields of any characteristic in Corollary 3.4.13 in the language of function fields, but only under a separability assumption.
The most general formula I'm aware of (but I'm not in the least a specialist) and which does not require separability assumptions is Tate's Genus Formula.
You can find it as Corollary 9.5.20 in Villa's Topics in the Theory of Algebraic Function Fields.

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I'll try to find a reference for the case of an arbitrary field, so as to take your edit into account. –  Georges Elencwajg Jul 23 '12 at 11:31
    
If you do that would be fantastic, but I don't think the same approach works, since this relies on computing the order of the differential of a power of a local representative. But if the power is divisible by the characteristic then this differential is zero. At least this is what has happened in my attempts to extend the result using this style of computation - as I said, if it is possible then I would be very happy, as this is a nice approach. –  Joe Tait Jul 23 '12 at 11:41
    
Dear Joe: I have added an edit with two references which you might find helpful. –  Georges Elencwajg Jul 23 '12 at 12:06
    
Do either of those make statements about the divisors themselves? They both seem to be about the degrees of the divisors, rather than what their values are at specific points/places? I was unaware of the second book though, and it may well be helpful anyway. –  Joe Tait Jul 23 '12 at 12:52
    
Dear Joe, yes you are right. Maybe looking at the proofs of surrounding Tate's result might be of some use. –  Georges Elencwajg Jul 23 '12 at 13:31
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This formula appeared in Miranda's book Algebraic Curves and Riemann Surfaces, page 135. It does not require a proof by shreaves. What you need is just basic definition. Looking at that page and previous pages should give you enough directions to prove this yourself.

Edit: The more "specific" formula you are looking for is just this formula for differential forms.

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Unless I am mistaken, that is for Riemann surfaces, and so will only cover the case of tame ramification? –  Joe Tait Jul 23 '12 at 10:35
    
I realise now that it may not have been very clear that I meant the question for wild ramification. I have edited the question, apologies for the confusion. –  Joe Tait Jul 23 '12 at 10:38
    
yeah, I have no idea how this holds for general algebraic curves with wild ramifications. I assume Silverman's book may have a proof, but I do not know. –  Bombyx mori Jul 23 '12 at 10:38
    
I see. You should ask a real expert (I assume one will appear soon). My knowledge in my Riemann Surface course is not sufficient to answer your question. –  Bombyx mori Jul 23 '12 at 10:41
    
Okay thanks - can I check which of Silverman's books you mean? –  Joe Tait Jul 23 '12 at 10:42
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