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

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

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"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

## 2 Answers

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.

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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.

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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
Thank you people. I am Pierre MATSUMI. So for my curve, it will be of genus 3. Very many thanks! Sincerely yours, Pierre MATSUMI – Pierre MATSUMI Feb 28 '13 at 16:01