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Take a ramified Galois cover $f:X\rightarrow Z$ of Riemann surfaces over $\mathbb{C}$ with Galois group $G$. If $H$ is a non-tirival normal subgroup of the Galois group $G$, this cover factors as $f:X\rightarrow Y\rightarrow Z$ and $X\rightarrow Y$ has Galois group $H$. My question is: How can one determine the number of branch points and the ramification indices of the intermediate cover $X\rightarrow Y$? The example of main interest for me is the following: take the cyclic cover of $\mathbb{P}^{1}(\mathbb{C})$ given by the (affine) equation: $X:=y^{10}=(x-x_{1})^{3}(x-x_{2})^{5}(x-x_{3})^{6}(x-x_{4})^{6}$. The Galois group is $G=\mathbb{Z}/10\mathbb{Z}$ which has a normal subgroup $H\cong\mathbb{Z}/5\mathbb{Z}$. The cover factors as $X\rightarrow Y\rightarrow \mathbb{P}^{1}$ and the Galois group of $X\rightarrow Y$ is $H\cong\mathbb{Z}/5\mathbb{Z}$. By taking the quotient one knows that $Y\cong\mathbb{P}^{1}(\mathbb{C})$ and the last map (i.e. $Y\rightarrow \mathbb{P}^{1}$) has degree $2$ and ramified over $2$ points. How can one compute the number of branch points and ramification indices of the quotient map $X\rightarrow Y$?

Darius Math
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2 Answers2

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(While expecting that this question will be "migrated" to MSE...) Quite analogously to the algebraic number situation, the ramification in this sort of "Kummer extension" $y^n=f(x)$ with $f$ explicitly factored is easy to understand. For one thing, at some stage one proves that ramification degrees are multiplicative in towers. Also, ramification is a local property, so points can be examined individually. Thus, in this example, since $\gcd(10,3)=1$, there is ramification at the point $x_1$ of degree $10$ in the whole, degree $2$ at the bottom, so degree $5$ in $X/Y$. At $x_2$, because $\gcd(10,5)=5$ but $\gcd(2,5)=1$, there is ramification of degree $2$ over $x_2$ at the bottom, but none in $X/Y$. At $x_3,x_4$, since $\gcd(10,6)=2$, there is no ramification at the bottom, but ramification of degree $5$ at the top.

EDIT: ... and, if there were any doubt, if there is no ramification over a point, the number of points lying over it is equal to the degree of the cover, that is, the number of sheets. Thus, here, there is a unique point lying over $x_1$ in $Y$, and still just one at the top, because it is totally ramified, and, similarly one point lying over $x_2$ in $Y$, but five at the top because of the no ramification in $X/Y$. Both $x_3,x_4$ have two points over them in $Y$, because of non-ramification there, and five at the top over each of the two. (Iterated (!?!) correction from earlier garbled version(s)...)

paul garrett
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  • Thank you very much. But I think (in the EDIT) you are making mistakes: above $x_{2}$, there is $1$ point in $Y$ because $(2,5)=1$, and $x_{3}$ and $x_{4}$ both have $2$ points over them in $Y$. This is bacuse $Y\rightarrow \mathbb{P}^{1}$ is in fact the map $\mathbb{P}^{1}\rightarrow \mathbb{P}^{1}$ of degree $2$ ramified over $2$ points (namely, $x_{1}$ and $x_{2}$) so the over is unramified over $x_{3}$ and $x_{4}$ as one expects from the Hurwitz theorem. –  Aug 03 '13 at 18:14
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    Oops, yes, I'll correct it. :) – paul garrett Aug 03 '13 at 18:15
  • something else I don't understand is that you said $x_{2}$ is not a branch point in $X/Y$($gcd(10,5=5)$ but $gcd(2,5)=1)$), but in the EDIT, you mentioned that there is a single point over $x_{2}$ in $X/Y$ ($\Rightarrow x_{2}$ is a branch point in $X/Y$) isn't this in contradiction with what you said earlier? –  Aug 03 '13 at 18:31
  • Sorry, Darius, let me look carefully! :) – paul garrett Aug 03 '13 at 18:31
  • I think the mistake is that the index of ramification is $\frac{10}{gcd(10,a_{i})}$. So the ramification index of $x_{2}$ is $10/5=2$ in the whole and $2$ in the bottom. So if what you said about the multiplicativity of ramification indices is correct, then the ramification index in $X/Y$ is $\frac{2}{2}=1$, i.e. $x_{2}$ is not a branch point of $X/Y$. Above $x_{2}$ in $Y$ there is $1$ point and in $X/Y$ there are $5$ points. –  Aug 03 '13 at 18:37
  • @DariusMath, yes, good to write that expression for the degree of ramification. I think I may finally have the numbers all straight in the second iterate of correction. – paul garrett Aug 03 '13 at 18:59
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In general, consider a cyclic cover of $\mathbb{P}^1$ given by equation $y^N = (x-x_1)^{m_1}\cdots (x-x_k)^{m_k}$. Let $U$ be the complement of the points $x_1,\ldots,x_k$. The fundamental group of $U$ is generated by the loops $\gamma_1,\ldots,\gamma_k,\gamma_\infty$ satisfying $\gamma_1\cdots \gamma_\infty = 1$. The cover defines a homomorphism $\chi:\pi_1(U)\to \mathbb{C}^*$ defined by $\gamma_i\mapsto e^{2\pi i m_i/N}, \gamma_{\infty}\mapsto e^{-2\pi i M/N}$, where $M = \sum m_i$. Thus a nonsingular projective model of the curve is isomorphic to the normalization $X$ of $\mathbb{P^1}$ in the field of rational functions of the quotient of the universal cover of $U$ by the kernel $\Gamma$ of $\chi$. The quotient $\pi_1(U)/\Gamma$ acts on $X$ via the image of $\chi$. Thus, if $m_1,\ldots,m_,M$ generate $\mathbb{Z}/N\mathbb{Z}$, the Galois group is cyclic of order $N$. Its ramification index over $x_i$ is equal to $r_i = (m_i,N)$, and over $\infty$, it is equal to $r_\infty = (M,N)$. The genus of $X$ is equal to $g = 1-N+N(1-1/r_\infty)+ N(\sum (1-1/r_i)$.

user37622
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