# Algebraic applications of Hurwitz' theorem

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?

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 $F|K$.

• 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/…). Apr 18 '11 at 12:29
• Martin, your question is fine, but the distinction between "algebraic" and "geometric" is a little subjective isn't it? 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? Apr 18 '11 at 13:00
• @Donu: You're right, perhaps I mean a statement in algebra which becomes a surprising geometric interpretation. Apr 18 '11 at 16:55

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 $4a^3+27b^2 \neq 0$.

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.

Ree gave an application of Riemann-Hurwitz to permutation groups. If $$g \in S_n$$ (the symmetric group), then let $$v(g)$$ be $$n$$ minus the number of cycles of $$G$$. Ree observed that if $$g_1,...,g_k \in S_n$$ with the subgroup generated by them acting transitively on the set of $$n$$ elements, and with $$v(g_1) + \cdots + v(g_k) = n-1$$, then any product $$g_{i_1} \cdots g_{i_k}$$ is an $$n$$-cycle (where $$i_j$$ are the result of permuting the indices).

To prove this we form a branched cover over the sphere, branched over $$k+1$$ points. The first $$k$$ points we call $$z_j$$ and each has monodromy $$g_{i_j}$$. The last point $$z_*$$ has monodromy $$(g_{i_1} \cdots g_{i_k})^{-1}$$. Applying Riemann-Hurwitz to the resulting covering yields $$v(g_1) + \cdots + v(g_k) + v((g_{i_1} \cdots g_{i_k})^{-1}) - 2n - 2 = 2g \geq 0$$ Therefore, there is only one cycle in $$(g_{i_1} \cdots g_{i_k})^{-1}$$ and so there is only one cycle in $$g_{i_1} \cdots g_{i_k}$$. Ree mentions that he could not find a purely algebraic proof. I don't know if there is still no algebraic proof known.