# How does one get from smooth, projective, algebraic curve of genus one to the usual equation for elliptic curves?

I am used to understanding elliptic curves as a non-singular curve over some field given by the equation $$y^2 = x^3 +ax + b.$$ However, I have also seen that elliptic curves can be characterized as a smooth, projective, algebraic curves of genus one. Clearly the former definition satisfies this characterization, but I am curious why this definition captures all such curves.

Worded another way, how might one take the set of smooth, projective, algebraic curve of genus one and determine that these are precisely those of the form $$y^2 = x^3 +ax + b$$? Why couldn't some be of the form $$y^3 = x^3 + ax + b$$, for example?

• This is a standard (and fun!) exercise in using the Riemann-Roch theorem, which you should be able to find in many textbooks discussing algebraic curves. Commented Sep 26, 2020 at 17:27
• Probably here on math.SE also. Here it is in some slides: cr.yp.to/talks/2009.03.26/slides.pdf Commented Sep 26, 2020 at 17:28
• Your second definition is not quite right. You need a $k$-point also Commented Sep 26, 2020 at 17:29
• Here it is on math.SE: math.stackexchange.com/questions/2234229/… Commented Sep 26, 2020 at 17:31

2. The usual proof that every smooth genus one curve (edit: with a rational point!) has a Weierstrass equation goes through the Riemann-Roch theorem. That proof just tells you that it's of the form $$y^2 + axy = bx^3 + cx^2 + dx + e,$$ but you can eliminate most of the constants in characteristic other than $$2$$ and $$3$$ by changes of variables. It's a little long to reproduce as a math stackexchange answer but should be in any book on algebraic curves.
3. In fact in characteristics $$2$$ and $$3$$ it's not true that you have an equation of the form you wrote for every elliptic curve.