# Question on why short Weierstrass can't be used for curves with char=2

An elliptic curve given by $E: y^2=x^3+ax+b$ with $a,b \in K$ and $Δ(E)=-16(4a^3+27b^2) \neq 0$ is adequate for elliptic curves with $char\neq2,3$ Because of the factor -16 in the definition of $Δ(E)$, according to this definition there are no elliptic curves in characteristic 2.

Ok I understand the short Weierstrass equation, and I understand the discriminant equation. So my questions are:

1. A long Weierstrass equation is needed only for curves with char=2 or 3?
2. Why does the factor of -16 imply that curves of characteristic 2 do not exist?
3. I read in another paper that "$E: y^2=x^3+ax+b$ does not apply in char 2 because $E: y^2=x^3+ax+b$ is always singular in char 2" but I'm not sure what this means. Singular means the discriminant =0 right? I guess this is related to #2, how does is the value of the characteristic used in the discriminant equation?

Thanks!

You can check that $E: f(x,y)=0$ is singular in characteristic 2, using the definition of singular point:
A point $P=(x_0,y_0)$ on $E$ is singular if $\partial f/\partial x = \partial f/\partial y = 0$ at $P$.
Now take $f(x,y) = y^2 - (x^3+ax+b)$, and suppose for simplicity $K=\mathbb{F}_2$. Then, $P=(a,b)$ is a point on the curve, and you can check using the definition above that the point $P$ is singular.
Elliptic curves in characteristic $2$ do in fact exist, but they cannot be written in short Weierstrass form with a non-singular model. For instance $$y^2+xy=x^3+1$$ is an elliptic curve with non-zero discriminant.
• My favorite example in char. $2$ is $y^2+y=x^3$. There, $\partial f/\partial y=1$ (!) May 10, 2015 at 4:40