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I heard $E:y^2＝1＋x^4$ is Elliptic curve because it is birational to $y^2＝x^3-4x$. But I have a question.

The projective closure of $E$ is $Y^2Z^2＝Z^4＋X^4$, and this is smooth curve（if it is not smooth, then the curve is not elliptic curve）, so its genus is $（4-1）（4-2）/2＝3$. But elliptic curve has genus 1. Where am I missing?

Poitou-Tate
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    See also this post, with the comments about the genus. "As I said, it's not a non-singular quartic." – Dietrich Burde Apr 16 '21 at 13:02
  • Thank you for your comment, but then, this is hyper elliptic curve but not elliptic curve? what is the genus of $E$? – Poitou-Tate Apr 16 '21 at 13:24
  • An alternative approach to using that birational model explicitly: The curve $E$ is NOT smooth at its point at infinity $(0:1:0)$. If one blows it up there, one will get that the genus (of the smooth model) is $1$ (char 0) - e.g., use Riemann-Hurwitz, or calculate the space of regular differentials - exercises well worth doing... – peter a g Apr 16 '21 at 14:11
  • Before blowing up, the curve's genus is 3, right? – Poitou-Tate Apr 16 '21 at 15:15
  • Sorry I just deleted a comment, as I misread yours. – peter a g Apr 16 '21 at 18:08
  • According to the (the inadequate) https://en.wikipedia.org/wiki/Genus%E2%80%93degree_formula the arithmetic genus is indeed $3$. After blowing up so that one a smooth curve, the arithmetic and geometric genus (genera!) coincide. If we can end up with an ell. curve (ok, after choosing a base point), that common number has to be one. – peter a g Apr 16 '21 at 18:13
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    @peterag It is maybe not completely clear from your comments that the for the OPs curve the geometric and arithmetic genera do not conincide since it is not smooth – Mummy the turkey Apr 16 '21 at 20:04
  • @Mummytheturkey I certainly had been trying to convey that idea... – peter a g Apr 16 '21 at 20:51
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    @Mummytheturkey Yes, I also wanted to write "elliptic curve". So I have deleted the old comment. Still, the answer by Angina Seng certainly is useful. – Dietrich Burde Apr 17 '21 at 08:12

1 Answers1

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The confusion comes from the fact that you've heard the definition of an elliptic curve require $E$ to be smooth. In fact, it's often convenient to allow $E$ to be singular, but asserting that the base point $O$ is nonsingular.

From the genus degree formula you are getting the wrong answer because $C$ is singular - and the formula only holds for smooth plane curves. You assert your curve is smooth but this is not the case - on the affine patch $y = 1$ we have $f(x,z) = z^2 - x^4 - z^4$ which has vanishing first partial deriviatives at $(0,0)$, hence $C$ is not smooth at the point at infinity.

To put $E$ in Weierstrass form it is suggested in the comments that you can blowup the singular point (at infinity). However in the spirit of Silverman Chapter III, let's just think about what $x$ and $y$ "are" in a Weierstrass equation, and follow the proof that one must exist.

You have your special point $O$, and $\{1,x'\}$ is a basis for $L(2O)$ and $\{1,x',y'\}$ is a basis for $L(3O)$ - then $x'$ and $y'$ will satisfy some Weierstrass equation. So one should just go out and find $x'$ and $y'$ in the wild.

In our case, if we take $O = (0,1)$ we can just take $x' = \frac{y + 1}{x^2}$ and $y' = \frac{y + 1}{x^3}$. Now $L(6O)$ is overloaded with stuff - namely $1, x', y', (x')^2, (y')^2, x'y',(x')^3$, so we just do linear algebra looking for a relation amongst these, one sees that $$2(y')^2 + x'^3 - x' = 0$$ modulo the equation for your curve. Replacing $x'' = -2x'$ and $y'' = 2y'$ we see that $$(y'')^2 = (x'')^3 - 4x''$$ and we have our Weierstrass equation.

Mummy the turkey
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  • Actually, I had not been suggesting blowing up the orig. curve to get a model in Weierstrass form. Rather, I was suggesting doing so as a means to see that the (geometric) genus of a/the desingularization was 1. Likewise, it is "a" way of getting the genus of an hyp. ell. curve, and to see what the reg. diff'ls are. – peter a g Apr 16 '21 at 21:02