# Genus of Y^3 = X^4 - 1.

Could you please teach me the genus of Y^3 = X^4 - 1 ?

• "math-philosophy?" :-) – Francesco Polizzi Feb 26 '13 at 9:54
• "the genus" of a planar curve ? – Adrien Hardy Feb 26 '13 at 10:10
• Magma tells me that the projective closure of this curve has no singular points, so it is a smooth quartic curve and has genus 3. – François Brunault Feb 26 '13 at 10:53
• Possibly "teach me" means "explain how to compute", rather than "tell me the answer". One can check that a curve of this sort is nonsingular (as a projective curve) in a minute or two by hand, one really doesn't need Magma. Indeed, it's a nice exercise in a first-year algebraic geometry course to compute the genus of X^n + Y^m = 1 (by hand!). The sequence of blow-ups needed to resolve the singularity mimics the Euclidean algorithm used to compute gcd(m,n). – Joe Silverman Feb 26 '13 at 12:35
• Once you know it is non singular, the genus is the number of lattice points in the interior of the Newton polygon. – Ian Agol Feb 26 '13 at 16:49

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.
• Hmmmm.... I suspect that the OP wanted to know the genus of a smooth projective model, which would be the designularization $\hat C$ of the projective curve given in homogeneous coordinates by $C:X^n+Y^mZ^{n-m}=Z^n$. (For concreteness, I've assumed that $n\ge m$.) The genus of $\hat C$ will not, in general, be $(n-1)(m-1)/2$. I don't recall offhand the formula, but my recollection is that it involves $\gcd(m,n)$, and as I indicated in an earlier comment, is computable by a short exercise in blowing up. – Joe Silverman Feb 26 '13 at 17:41
• Yes, thanks for pointing that out: I was implicitly assuming $f$ is reduced. I added now the more general form of this result. – Alex Suciu Feb 26 '13 at 18:37