A morphism of curves is said to be Galois if the corresponding extension of function fields is Galois.

Let $f:X\longrightarrow Y$ be a finite morphism of smooth projective connected curves over $\mathbf{C}$ (or Riemann surfaces if you prefer) with branch locus $B\subset Y$ and ramification locus $R\subset X$. There exists a finite morphism $g:W\longrightarrow X$ of smooth projective curves such that the composition $f\circ g:W\longrightarrow Y$ is Galois. Take the Galois closure for example.

Can we say something about the branch locus of $g$? For example, when does it lie in $R$?

In my problem I would like to be able to choose $g$ such that the branch locus of $g$ lies in $R$. I have a feeling this is not always possible unfortunately.

The following might help actually. In my problem, $Y$ is the projective line and $f$ is a Belyi morphism. Thus, what I would like to see is that I can choose $g:W\longrightarrow X$ such that the composition $f\circ g$ is Belyi and Galois. Again, this seems unlikely to be possible.

  • 1
    $\begingroup$ The composition $f \circ g$ should be etale over the the complement $u$ of the branch locus of $f$ in $X$. This would imply that the branch locus of $g$ lies in the ramification locus of $f$. To prove the you could form the Galois closure of the restriction of $f$ to the preimage of $U$ inside the Galois category of etale coverings (SGA1... ) $\endgroup$ Commented Mar 5, 2011 at 12:22
  • $\begingroup$ Ok. I'll look it up. $\endgroup$ Commented Mar 6, 2011 at 12:51
  • $\begingroup$ I guess Dmitri's answer is a special case of yours, right? $\endgroup$ Commented Mar 6, 2011 at 16:29

3 Answers 3


This is to address the last part of your question -- about Belyi morphism. The answer is YES: you can always get $g: W\to X$ such that $f\circ g: W\to \mathbb CP^1$ is Belyi.

Indeed, $X\to \mathbb CP^1$ is Belyi if it ramifies only over $0,1,\infty$. Let us now conisder $\hat X=f^{-1}(\mathbb CP^1\setminus (0,1,\infty))$. Clearly, $f_*\pi_1(\hat X)$ is a finite index subgroup of $\pi_1(\mathbb CP^1\setminus (0,1,\infty))=F_2$. The map $f:X\to \mathbb CP^1$ is Galois iff $f_*\pi_1(\hat X)$ is normal in $F_2$.

Now, take a normal subgroup $N$ in $F_2$ of finite index, contained in $f_*(\pi_1(\hat X))$ and take the corresponding cover $\mathbb CP^1\setminus (0,1,\infty)$, call it $\hat W$. The completion $W$ of $\hat W$ is what you are looking for.

Added. In order to find a normal subgroup of finite index we use the following standard lemma (I forgot the name):

Lemma. Let $G$ be any group and $G_1$ be any finite index subgroup. Then there exists $N\subset G_1$ of finite index, and such that $N$ is normal in $G$.

Proof. $G$ is acting on the left on the finite set of cosets $G/G_1$. Let $N$ be the subgroup of $G$ that acts trivially on $G/G_1$. Clearly $N$ is normal in $G$ and the index of $N$ is at most $\#(G/G_1)!$

EDIT (by SK): fixed LaTeX

  • $\begingroup$ Why does there exist a normal subgroup N in F_2 of finite index, contained in f_*(pi_1(X))? $\endgroup$ Commented Mar 6, 2011 at 12:51
  • $\begingroup$ I mean f_*(pi_1(\hat{X})). $\endgroup$ Commented Mar 6, 2011 at 12:52

Your feeling is correct, in fact the branch locus of $\mathcal{g}$ usually does not lie in $R$.

For example, take a general projection $f \colon X \to L$, where $X \subset \mathbb{P}^2$ is a smooth cubic and $L$ is a general line. Since the class of $X$ is $3(3-1)=6$, it follows that there are six distinct tangent lines to $X$ through a general point of $\mathbb{P}^2$, so $f$ is branched in six points, with branching number $2$ (this means that around any of the six branching points the local monodromy is a transposition).

Scheme theoretically we have


where $R$ is the ramification divisor and $R'$ is a degree six divisor on $X$ where $f$ is not ramified.

Now take the double cover $g \colon W \to X$ such that $h=f \circ g \colon W \to L$ is the Galois closure of $f$. Since points lying in the same fiber of $h$ must have conjugate (hence isomorphic) stabilizers, it follows that $g$ is ramified precisely over $R'$, in particular it is unramified over $R$.

More generally, the same argument shows that if $g \colon W \to X$ is any finite cover such that $f \circ g \colon W \to L$ is Galois, then the branch locus of $g$ must contain $R'$.


As correctly awnsered by Dimitri, given a Belyi map $f:X\to Y$, there exists a finite cover $g:Z\to X$ such that the composition $h=f\circ g:Z\to Y$ is Belyi and Galois (this is true in the more general setting of finite covers of curves, namely, the Galois closure of a cover $X\to Y$ does not change the branch locus in $Y$). Working in characteristic $0$, the description of the branch locus $B'$ of $g$ essentially follows from Abhyankar's lemma: first of all, $B'$ is contained in the set $f^{-1}(\lbrace\infty,0,1\rbrace)\supset R$. Let $e_\infty$ be the lcm of the ramification indices of the points of $R$ lying in the fiber $f^{-1}(\infty)$. Then a point point $P\in f^{-1}(\infty)$ belongs to $B'$ if and only if its ramifiaction index is strictly less than $e_\infty$. The same applies also for the points in $f^{-1}(0)$ and $f^{-1}(1)$ and leads to a complete description of $B'$.

  • $\begingroup$ Thnx! So if I understand correctly, we know that the branch locus of g lies in f^{-1}(0,1,infty). Then, there is a simple criterion (which you describe) to decide which points in f^{-1}(0,1,infty) are branch points. And this is not in contradiction to Francesco's answer because his R' is contained in f^{-1}(0,1,infty), right? $\endgroup$ Commented May 3, 2011 at 8:12
  • $\begingroup$ You're welcome! And yes, you're right, no contradiction with the nice example of Francesco: the Galois closure of his degree 3 cover is ramified over 6 points (and not 3, but the criterion works in full generality), with ramification index 2 (the gcd of the ramification indices over each branch point). Therefore, g is unramified over R and ramifies exaclty over R'. $\endgroup$
    – Leonardo
    Commented May 3, 2011 at 12:01

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