3
$\begingroup$

I am reading a paper by Darmon and Granville where they made the following claim, with respect to the generalized Fermat curve $Ax^p + By^q = Cz^r$. They said:

"... this (referring to the generalized Fermat curve above) is a curve in an appropriate weighted projective space. However this curve often has genus 0 (for instance, if $p,q$ and $r$ are pairwise coprime),..."

In particular, in such a case Faltings' theorem would not apply.

However I have a hard time seeing why this statement is 'obvious', namely because I don't have a good understanding of what genus means in a weighted projective setting. Further, why should that curve above have genus 0 when $p,q,r$ are pairwise co-prime?

$\endgroup$
1
  • $\begingroup$ Project to the line and use Riemann--Hurwitz. $\endgroup$ Commented Mar 16, 2014 at 0:42

1 Answer 1

7
$\begingroup$

If p, q and r are pairwise coprime the generalized Fermat curve is embedded in the weighted projective space ${\mathbb P}(qr, pr, pq) $ which is isomorphic to a straight ${\mathbb P}^2$ via $(x:y:z)\mapsto (x^p:y^q:z^r) $: the image of the curve is a line, so it is rational.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .