TITLE: How to explicitly see the ramification over infinity
QUESTION [7 upvotes]: 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.

REPLY [7 votes]: 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.