Genus of Y^3 = X^4 - 1.  Could you please teach me the genus of Y^3 = X^4 - 1 ? 
 A: I don't know which definition of genus you are using, but you can deform this smooth projective quartic curve to four projective lines, any two of which intersect at one point. There are three holes in this configuration, and since genus is preserved under the deformation, the genus is 3.
A: The complex curve $X^n + Y^m = 1$ is the Milnor fiber (at the origin) of the weighted-homogeneous polynomial $f(X,Y)=X^n + Y^m$.  Suppose $\gcd(n,m)=1$.  Then the Milnor fiber deform-retracts onto a (minimal) Seifert fiber for the singularity link, which is an $(n,m)$-torus link. This Seifert fiber consists of $n$ stacked disks, each one joined to the one above by $m$ once-twisted bands. It is now a simple exercise to see that the genus of this surface (equal to the Milnor number of $f$) is $(n-1)(m-1)/2$. 
More generally, if $f=f(z_1,\dots,z_m)$ is a weighted-homogeneous polynomial with weights $(w_1,\dots, w_m)$, then the Milnor fiber $f=1$ has the homotopy type of a wedge of $(m-1)$-dimensional spheres, and the (Milnor) number of these spheres is given by $\mu=(w_1−1)(w_2−1)\cdots (w_m−1)$, according to John Milnor and Peter Orlik, Isolated singularities defined by weighted homogeneous polynomials, Topology 9 (1970), 385-393. 
