The following is a question from Rosen's "Number Theory in Function Fields"-
Let $l$ be a prime not equal to $char(F)$ and $K=F(x,y)$ a function field where $x$ and $y^l=(x-a_1)^{n_1}\cdots (x-a_m)^{n_m}$, where all $a_i$ are different and $l$ does not divide $n_i$. Compute the genus of $K$.
My strategy was to find all ramified primes, and then use Riemman-Hurwitz formula. It is quite straightforward to see that, when we denote $P_i$ the prime in $F(x)$ which correspond to $x-a_i$, that $P_i$ is ramified in $K$ with ramification index $e_i=l$. Indeed, if $\mathfrak{P}_i$ is any prime over $P_i$, we can take the valuation at $\mathfrak{P}_i$ from both sides of the defining equation of the function field to get (we use the fact that $e(\mathfrak{P}_i|P)$ is the integer such that for elements of $F(x)$, $v_{\mathfrak{P}_i}(s)=e(\mathfrak{P}_i|P)v_{P_i}(f)$)-
$$l\cdot v_{\mathfrak{P}_i}(y)=n_i\cdot e(\mathfrak{P}_i|P)$$
Hence, $l$ must divide $e(\mathfrak{P}_i|P)$, and as this number is at most $l$ (as $l=[K:F]$), we have an equality. Now from the equation which connects the ramification indices and relative degrees of primes over $P_i$ in $K$, we get that there is only one such prime, and with relative degree 1, and at said, $e(\mathfrak{P}_i|P)=1$.
I'm having a hard time computing the ramification index at other primes. Geometrically it seems that primes $P_\alpha$ in $F(x)$ which corresponds to $x-\alpha$ for $\alpha\neq a_1,...,a_m$ should not ramify in $K$- because if we enlarge $F$ enough to contain the $l$-th roots of $(\alpha-a_1)^{n_1}\cdots (\alpha-a_m)^{n_m}$, we'll end up with $l$ different points above our point $\alpha$, and locally the situation is of a homeomorphism if we think of it over $\mathbb{C}$. I have not managed however to formalize this argument. I tried to come up with an algebraic argument- maybe see how the local discriminant behave or see how the ideal $(x-\alpha)$ splits. However, it seems that we must calculate the integral closure of $F[x]_{(x-\alpha)}$ in $K$ in order to proceed, and I do not see how to do this, as we don't even know if the mentiond $l$-th roots are in our field or not. I also need to find if there is a ramification at infinity, and this seems to be a more delicate point- it seems to depend on wehther $l$ divides $n_1+...+n_m$ or not, but I don't see eacatly how to formulate the relation; moreover it's possible that $char(F)$ will divide $n_1+...+n_m$, and make the situation more complicated.
So I guess my question is how can we find the ramification at all these primes? Do we must find the integral closure?
Thanks in advance.
EDIT: After some time I got a solution, I post it here to anyone who might be interested in the future. By computing the discriminant of the basis $1,y,...,y^{l-1}$ we find that the only primes which might be ramified are those that correspond to $x-a_i$. As claimed above, each of this is totally ramified. Also, they're all tamely ramified- since we have the assumption $p\nmid l$. Finally, by similar considerations to what done in the question, if $l\nmid n_1+...+n_m$- the prime at infinity is totally ramified (and also tamely ramified). And if $l|n_1+...+n_m$, we make the change of variables $s=1/x$, $t=ys^{(n_1+...+n_m)/l}$, which gives us $g(s,t)=t^l=(1-a_1s)^{n_1}\cdots(1-a_ms)^{n_m}$. Thus the elements $1,t,...,t^{l-1}$ spans the field $k(x,y)=k(s,t)/k(s)$ and are integral. Computing the discriminant of that base implies that $s=1/x$ is unramified. Having all this ramification data enable us to use Riemman-Hurwitz and find the genus.