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Hurwitz' theorem states that for a finite separable morphism $f : X \to Y$ of curves of degree $n$ and with ramification divisor $R$, we have

$2 g(X) - 2 = n (2 g(Y) - 2) + \deg(R)$.

Besides, we have $\deg(R)=\sum_{p \in X} (e_p - 1)$ if $f$ has only tame ramification [Hartshorne, IV, Cor. 2.4]. One of the consequences is $g(X) \geq g(Y)$, which also holds when $f$ is not supposed to be separable [loc. cit. 2.5.4].

One purely algebraic application of Hurwitz' theorem is Luroth's theorem, which states every nontrivial intermediate field of $k(t)$ over $k$ is isomorphic to $k(x)$ over $k$. However, it is easy to give a direct algebraic proof of Luroth's theorem, even if $k$ is not supposed to be algebraically closed (which is probably needed for Hurwitz' theorem). Therefore I wonder if there are other algebraic application of Hurwitz' theorem using the correspondence between curves and function fields.

Question: Are there other interesting algebraic applications of Hurwitz' theorem?

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3 Answers

up vote 2 down vote accepted

As far as I remember the Riemann-Hurwitz-formula is used to prove the inequality

$|\mathrm{Aut}(F|K)|\leq 84(g-1)$

for the number of automorphisms of an algebraic function field $F$ of one variable over $K$, where $K$ has characteristic $0$ and $g\geq 2$ holds for the genus of $F|K$.

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Interesting, but is this really purely algebraic? The only definition of the genus of a function field I know is just a disguised version of the geometric definition (mathoverflow.net/questions/152/…). –  Martin Brandenburg Apr 18 '11 at 12:29
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Martin, your question is fine, but the distinction between "algebraic" and "geometric" is a little subjective isn't it? –  Donu Arapura Apr 18 '11 at 12:51
    
Well, is the genus of a function field as it appears in Felipe Voloch's answer a disguised version of the geometric one? One needs an algebraic theory of algebraic functions which is realized in the form of the field theory of extensions of transcendence degree 1 and their valuation/divisor theory. Of course to prove Hurwitz full result one needs Weierstrass points (valuations) and thus the theory of higher derivatives in function fields, which is an algebraization of parts of analysis. Is it pure algebra then? –  Hagen Apr 18 '11 at 13:00
    
@Donu: You're right, perhaps I mean a statement in algebra which becomes a surprising geometric interpretation. –  Martin Brandenburg Apr 18 '11 at 16:55
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Notice that $\deg(R)=0$ and $g(Y)=1$ imply $g(X)=1$.

This means that every unramified cover of an curve of genus $1$ is again a curve of genus $1$. Translating into the algebraic language and using the fact that , if $\textrm{char}(k) \neq 2,3$, any curve of genus $1$ has a birational model of the form $y^2=x^3+px+q$, with $4p^3+27q^2 \neq 0$, we obtain the following result:

Assume $\textrm{char}(k)\neq 2,3$ and let $p, q \in k$ with $4p^3+27q^2 \neq 0$. Then every finite, unramified extension of the quotient field of

$k[x,y]/(y^2-x^3-px-q)$

can be written as the quotient field of

$k[s,t]/(s^2-t^3-at-b)$,

for some $a,b \in k$ with $4a^3+27b^2 \neq 0$.

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On this other thread, about elliptic curves about function fields, the functional analogue of Mazur's theorem was evoked, and described as a simple consequence that the genus of modular curves grows with the conductor. This is indeed an application of the inequality $g(X)\geq g(Y)$ you mention.

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