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Motivated by this where it is possible to take certain Fermat curves like $x^3+y^3=1$ into Elliptic curves.

I was wondering if it is always possible to transform any Fermat curve $x^n+y^n=1$ birationally into some hyperelliptic curve?

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1 Answer 1

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There are non-hyperelliptic Fermat curves.

  1. According to "The Group of Automorphisms of the Fermat Curve" (Tzermias 1995), the automorphism group of the Fermat curve with $n \ge 4$ in characteristic $0$ is the semidirect product $\Sigma_3 \ltimes (\Bbb{Z}/n)^2$ which has order $6n^2$.
  2. The genus of the Fermat curve is $g=(n-1)(n-2)/2$.
  3. According to "Automorphism Groups of Hyperelliptic Riemann Surfaces" (Bujulance, Etayo, Martinez 1987), a hyperelliptic Riemann surface of genus $g>15$ has at most $8(g+1)$ automorphisms.
  4. If $n \ge 8$ then $g = (n-1)(n-2)/2 \gt 15$, $8(g+1) = 4(n^2-3n+4) \lt 6n^2$ and so the Fermat curve is not hyperelliptic.
  5. EDIT added: Evidently from Theorem p.175 in the paper cited (3) the bound $8(g+1)$ applies if $g>9$ and so the $n \ge 6$ Fermat curve is not hyperelliptic.
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