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Timeline for Genus of Y^3 = X^4 - 1.

Current License: CC BY-SA 3.0

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Feb 28, 2013 at 16:01 comment added Pierre MATSUMI Thank you people. I am Pierre MATSUMI. So for my curve, it will be of genus 3. Very many thanks! Sincerely yours, Pierre MATSUMI
Feb 26, 2013 at 18:37 comment added Alex Suciu Yes, thanks for pointing that out: I was implicitly assuming $f$ is reduced. I added now the more general form of this result.
Feb 26, 2013 at 18:35 history edited Alex Suciu CC BY-SA 3.0
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Feb 26, 2013 at 17:41 comment added Joe Silverman 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.
Feb 26, 2013 at 14:36 history answered Alex Suciu CC BY-SA 3.0