Hurwitz proved in 1893 that the number of automorphisms of a Riemann surface of genus $g \geq 2$ is bounded by $84(g-1)$. See Wikipedia for some references. I want to understand the proof in the language of algebraic geometry, namely for a complete algebraic curve $X$ over an algebraically closed field of characteristic $0$. It uses Hurwitz' Theorem and it is outlined in this blog entry and also in Hartshorne, Exercise IV.2.5.

The proof starts as follows: It is known that $G := \text{Aut}(X)$ is finite, say of order $n$. Now $G$ acts on the field $K(X)$ and $K(X)^G \subseteq K(X)$ is a finite Galois extension of degree $n$. Thus it corresponds to a finite separable morphism $f : X \to Y$ of degree $n$. Finally Hurwitz' Theorem is applied to $f$.

Now it seems to me that Hartshorne's exercise, Part a, as well as the blog entry ("We’ll think analytically for a moment.[...]") use that $Y$ is the quotient of $X$ by $G$ in the category of ringed spaces, especially in the category of topological spaces. Namely this is needed to show how the fibers and ramification indices of $f$ looks like.

But from abstract nonsense we just get that $Y$ is the quotient of $X$ by $G$ in the category of complete curves, that is $f$ is $G$-invariant and every $G$-invariant dominant morphism $X \to Z$, where $Z$ is a complete curve, factors uniquely through $f$. How does this tell us how $f$ looks like as a map and what are the stalk maps, in order to calculate ramification?

Or do we use the explicitly constructed abstract complete curve associated to $K(X)^G$, thus consider all discrete valuation rings in $K(X)^G$? But then $f$ is, a priori, only defined rationally, right? So to sum up my problems: I don't know how to work with $f$.

  • $\begingroup$ I recommend strongly reading Hurwitz. $\endgroup$
    – roy smith
    Apr 25, 2011 at 3:53
  • $\begingroup$ @roy: Please read my question carefully; Hurwitz' proof is in the language of Riem. surf., where there is no problem. $\endgroup$ Apr 25, 2011 at 8:53

1 Answer 1


Here is an extremely useful fact: if $X$ is a regular variety and $f \colon X \dashrightarrow \mathbf{P}^n$ is a rational map defined on a dense open $U \subset X$, then $\mathcal{f}$ extends uniquely to a maximal open subset $U'$ of $X$ and the complement of $U'$ has codimension at least two. In particular, when $X$ is a smooth curve there is a unique extension to the whole curve. Now apply this to the rational map from $X$ to the complete curve associated to $k(X)^G$, which you know can be embedded in a projective space. I don't have the book on me but this is certainly somewhere in Hartshorne. The proof goes by using that the local ring of a point in codimension one is a DVR, so you can write down an explicit extension of the map $f$ in codim 1 by multiplying through by the lowest possible power of a uniformizer.

On a sidenote, this useful fact is already needed to show that there is a unique smooth complete curve associated to a function field in one variable.

Edit. I think I see now what you are getting at: you want to know why the points of $X/G$ correspond to $G$-orbits on $X$, right? (Knowing this, it immediately follows that if $P$ is a ramification point of order $e_P$, then it has $|G|/e_P$ conjugates under $G$.) There is a general theorem of Emmy Noether on quotients of affine varieties by finite groups which will show this, which can be found in any book on commutative algebra. But for curves I guess you could make a direct proof as follows: pick points $P$ and $Q$ on $X$ in different orbits. Since the function field separates points you can choose a function that vanishes at $P$ but not at $Q$. Do the same thing also for the conjugates of $Q$. Then a linear combination $F$ of these functions is nonzero on all conjugates of $Q$. Take the product of all conjugates of $F$ under the $G$-action: this is a $G$-invariant function which vanishes on the orbit of $P$ but not on the orbit of $Q$. The result follows.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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