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Let $X$ be a smooth, projective and geometrically connected curve of genus $g\geq 3$ defined over $\mathbb C$. Suppose the function field $k(X)$ of $X$ takes the form $k(X)=\mathbb C(x)[y]$, where $$y^a=f(x)$$ with $a\geq 3$ an integer and with $f\in \mathbb C[x]$ a separable polynomial of degree $d=\deg(f)\geq 3$.

Question: a) Is it possible to give an upper bound for $d$ and $a$ in terms of $g$?

b) Is there even a precise formula relating $a,d,g$?

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    $\begingroup$ All this follows easily from Riemann-Hurwitz. $\endgroup$
    – Angelo
    Apr 24, 2013 at 18:44

1 Answer 1

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$2g=da+ma_1-2a-d-m+2$, where $a=a_1m,$ and $m$ is the g.c.d ($d,a$).

Edit. Let $\chi(S)=2-2g$ be the Euler characteristic. Hurwitz formula gives $$\chi(S)=2a-r,$$ where $r$ is the ramification: a branch point of order $k$ contributes $k-1$ to $r$. As your polynomial has $d$ simple roots, these roots contribute $(a-1)d$ to the ramification. To find ramification at infinity we write our equation $w^a=P_d(z)$ as $w^a=z^dh(z)=u^d$, where $h$ is a holomorphic function at $\infty$, $h(\infty)\neq 0$, and $u$ a germ of a meromorphic function with a simple pole at $\infty$. The last equation factors into irreducible factors: $$w^a-u^d=\prod_c (w^{a_1}-cu^{d_1}),$$ where $c$ are the roots of unity of degree $m$, $m$ is the greatest common factor of $a$ and $d$, $d=md_1,$ and $a=ma_1$.

From this we conclude that our Riemann surface $S$ has $m$ ramification points of order $a_1$ at $\infty$. So we obtain $$2-2g=2a-(a-1)d-m(a_1-1).$$

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  • $\begingroup$ Thanks a lot for this precise formula. According to Angelo, this formula follows from Riemann-Hurwitz. Can you explain a bit more how you get this formula from R-H? (I have some troubles with ramification at the points at infinity of the induced map $X\to P^1$) $\endgroup$
    – Robert
    Apr 25, 2013 at 10:43
  • $\begingroup$ At the moment I can not vote up your answer, since MO says: Vote Up requires 15 reputation, and I have only 3 reputation. However, as soon as I have enough reputation, I will vote up your answer. To get 15 reputation I have to ask 4 more questions (1 question=3 reputation points)? $\endgroup$
    – Robert
    Apr 27, 2013 at 11:48
  • $\begingroup$ Ramification at $\infty$ is determined by substituting $z= 1/\zeta$, getting rid of the denominator, and then studying what happens at $\zeta=0$. $\endgroup$ Mar 30, 2019 at 11:16

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