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Let us work over a field $\Bbbk$ of characteristic not equal to two. Let $d\in\Bbbk\setminus\{0,1\}$. It is said in the wikipedia article about Edwards curves that the plane quartic defined by the equation $$x^2+y^2 = 1 + d\cdot x^2y^2$$ is birationally equivalent to a curve in Weierstraß form, i.e. a plane cubic. However, the genus of a plane curve of degree $d$ is equal to $$g(d)=\frac{(d-1)(d-2)}{2}.$$ Furthermore, the genus is a birational invariant and $g(4)=3\ne 1=g(3)$. How can it be that the Edwards curve is birationally equivalent to a cubic?

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The expression you have written down is the genus of a nonsingular plane projective curve. The projective closure of the Edwards curve in $\mathbf P^2$ is not smooth!

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  • $\begingroup$ d'uh. Indeed. Thanks. $\endgroup$ Oct 4, 2013 at 20:00
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    $\begingroup$ Dear Marie, do you think I should change my user name to Georgette? $\endgroup$ Oct 4, 2013 at 20:54
  • $\begingroup$ @GeorgesElencwajg I think Georges is a beautiful name, but sometimes we all need a bit of extra spice in our lives! $\endgroup$ Oct 4, 2013 at 21:17
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    $\begingroup$ Dear Marie, thanks for your flattering answer! Needless to say, my question was tongue-in-cheek but the feminine version Georgette of Georges really existed in France and was discontinued, I would guess, about 60 years ago. Since first names have a sort of cyclic existence, I can't exclude the possibility that Georgette will be all the rage in trendy families in a few years ... That'll be the day! [By the way,+1 for your answer]. $\endgroup$ Oct 4, 2013 at 22:01
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    $\begingroup$ Thank you @GeorgesElencwajg! I know exactly what you are talking about: one of my grandmothers was named Albertine! $\endgroup$ Oct 4, 2013 at 22:47

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