Why weierstrass equation should not be used for elliptic curves over the field char=2?

Question

According to my source, this is because 'Weierstrass equation is singular', and it shows singularity by the point $(x_0,y_0)$, where $x_0$ is the root of $-3x^2-A$(maybe on a field extension), and $y_0$ satisfies $y_0^2=x_0^3+Ax+B$. He claims the derivatives $f_x(x_0,y_0)=f_y(x_0,y_0)=0$ by using char=2. \
But this still does not make sense to me, why we care about a derivate of point on a field extension? \
For derivates, my guess is that zero derivatives implies multiple roots, which weierstrass equation not allowed(but I also dont know why not). But as I said above, $(x_0,y_0)$ may be a point out of our initial field, why we care?

Answer 1

Actually, the problem is the discriminant. By definition, an elliptic curve must have non-zero discriminant, otherwise it is degenerate. A clear explanation of this with examples of finite fields by John D. Cook is at All elliptic curves over fields of order 2 and 3. With the short Weierstrass form $\,y^2 = x^3 + Ax + B\,$ there are no curves with non-zero discriminant in a field of characteristic $2$.

The article by Cook makes an important point:

The two curves in the example of Lendstra and Pila are the only ones over GF(2) with five points. So the two curves of order 5 over GF(2) are

$$y^2 + y = x^3 + x^2$$ $$y^2 + y = x^3 + x.$$

They determine the same set of points but are algebraically different.

He explains what this means with

Both determine the same set of points, but the two curves are algebraically different because $(0,0) + (0,0)$ equals $(1,1)$ on the first curve and $(1,0)$ on the second.

If the curve is degenerate, then you can't even define the addition of points on the curve.