1

Why is a genus 1 curve smooth and is it still true for a non-zero genus one in general?

According to this answer, Elliptic curve（which is defined as genus 1 curve with base point） need not to be smooth. They are just binational to the smooth 3-dimmensional curve.

Are there any example of elliptic curve which is not smooth? Thank you.

Poitou-Tate
  • 6,813

1 Answers1

2

It depends where you are coming from. Every non-smooth curve is birational to a smooth curve of the same genus. Hence if a curve $C$ has genus $1$ and has a smooth rational point, you can just blow up the singular points. A concrete example of such a family are (the projective closure of) $$y^2 = f(x)$$ where $f(x)$ is of degree $4$, has no repeated roots, and $O = (a, 0)$ is a rational point (where $f(a) = 0$). In this example the point at infinity $(0:1:0)$ is singular.

Mummy the turkey
  • 4,456
  • Thank you very much. I 'm not familiar with blow up, so I may not understand your answer, but are there concrete example of genus 1 singular（not smooth）curve with base point ? – Poitou-Tate Sep 26 '20 at 10:03
  • 1
    @bellow the example I gave has the point at infinity as a singular point, I'll edit my answer to make this clear. The blow up is just an example of a way to find a birational equivalence from a singular curve to a smooth curve – Mummy the turkey Sep 26 '20 at 10:59
  • Thank you very much! y^2＝x（x-1）（x＋１）（Xー２） is example? How to show [0:1:0] is singular point? – Poitou-Tate Sep 26 '20 at 11:15
  • 1
    @bellow Yes that is an example. Write $F(X,Y,Z) = Z^2Y^2 - Z^4f(X/Z)$ take the affine piece where $Y = 1$. Then compute the partial derivatives wrt $X$ and $Z$. – Mummy the turkey Sep 26 '20 at 12:13