Take the equation $y^{d}=\Pi_{1}^{n}(x-t_{i})^{m_{i}}$ over $\mathbb{C}$. This affine equation gives a cyclic cover of $\mathbb{P}^{1}$. Now it is usually said without explanation that if the sum $\sum m_{i}$ is not divisible by $d$, then the family is ramified over $\infty$. My question is how to see explicitly--in terms of equations maybe--that this is the case.
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2 Answers
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One way to see this is the following. Let $U$ be the complement of the points $t_1,\ldots,t_n$ in $\mathbb{P}^1(\mathbb{C})$. The fundamental group $\pi_1(U)$ is generated by elements $\gamma_1,\ldots,\gamma_n,\gamma_\infty$ with relation $\gamma_\infty^{-1} = \gamma_1\cdots \gamma_n$. Consider the character $\chi:\pi_1(U)\to \mathbb{C}^*$ defined by $\chi(\gamma_i) = e^{2\pi m_i/d}, i = 1,\ldots,n$, $\chi(\gamma_\infty) = e^{-2\pi M/d}$, where $M = \sum m_i$. Let $V$ be the quotient of the universal cover of $U$ by the kernel of $\chi$. The normalization of your curve is isomorphic to the normalization of $\mathbb{P}^1$ in $V$. It is a smooth complete curve on which the cyclic group of order $d$ acts as the Galois group with quotient $\mathbb{P}^1$. The ramification over $\infty$ is defined by the value of $\chi$ at $\gamma_\infty$, i.e. $e^{2\pi M/d}$. It is trivial, and hence the cover is unramified over $\infty$ if and only if $d|M$. The ramification index is equal M\mod d.$
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$\begingroup$ Thanks for the argument. But isn't it that if you remove $n$ points of $\mathbb{P}^{1}$, the fundamental group is generated by $\gamma_{1}$,...,$\gamma_{n}$ with $\gamma_{1}...\gamma_{n}=1$? and of course we can't assume from the beginning that $\infty$ is a removed (i.e. branch point). So does $\gamma_{\infty}$ always appear? $\endgroup$ Commented Aug 3, 2013 at 7:17
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1$\begingroup$ @DariusMath I believe user37622 meant to remove $t_1,\ldots,t_n$ from $\mathbb{C}$ itself, instead of the projective line. We assume from the beginning that $\gamma_\infty$ appears. $\endgroup$– S. Carnahan ♦Commented Aug 3, 2013 at 7:52
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$\begingroup$ Ah, OK. If we remove the points form $\mathbb{C}$ (i.e. remove the points together with $\infty$ from $\mathbb{P}^{1}$) then the argument will be correct. Thanks! $\endgroup$ Commented Aug 3, 2013 at 8:14
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Another way to do it is to compute explicitly the curve above the point at infinity, just by looking at the equation. The argument works over any field, we can of course assume that this one is algebraically closed.
The curve $y^d=\prod_{i=1}^n (x-t_i)^{m_i}$ is affine, and covers the affine line which coordinate is $x$. If you want to see what happens at infinity, you just change coordinates on $\mathbb{P}^1$, which corresponds to take affine coordinate $X=1/x$.
Assume that each $t_i$ is not equal to zero (otherwise, make an affine change of coordinates at begin). We then have
$\prod_{i=1}^n (x-t_i)^{m_i}=\frac{\prod_{i=1}^n (1/t_i-1/x)^{m_i}}{\prod_{i=1}^n (xt_i)^{m_i}}$.
(a) If $\sum m_i$ is a multiple of $d$, you write $\sum m_i=ad$ and make a change of variables $Y=yx^a$ and get
$(\prod_{i=1}^n t_i^{m_i} )Y^d=\prod_{i=1}^n (1/t_i-X)^{m_i}$
In this case, the curve is not ramified at infinity, which is $X=0$.
(b) If $\sum m_i$ is not a multiple of $d$, you write $\sum m_i=ad-k$ where $1\le k< d$ and make a change of variables $Y=yx^a$ and get
$(\prod_{i=1}^n t_i^{m_i} )Y^d=X^k\prod_{i=1}^n (1/t_i-X)^{m_i}$
In this case, the curve is ramified at infinity.
answered Aug 3, 2013 at 15:58
$\sqrt[d]\prod_1^n(z-t_i)^{m_i}$along the circle $\{|z|=R\}$, $R\gg0$. How many different branches do you obtain? $\endgroup$