How to pass from general weierstrass equation of elliptic curves to shorter ones

I found a page on wikipedia about weierstrass equations of elliptic curves.

https://fr.wikipedia.org/wiki/%C3%89quation_de_Weierstrass

The page says that one can put weierstrass equations in shorter forms depending on the characteristic of the field. However there is an interesting separation between supersingular and ordianire curves. Unfortunetly there is no detail what changes I have to apply in order to pass from the general weierstrass equation to one of the 4 provided. Does anyone know how to do this?

Let's start with the long Weierstrass equation for an elliptic curve over a field $K$: $$y^2+a_1xy+a_3y = x^3+a_2x^2+a_4x+a_6$$

If the characteristic of the $K$ is not 2 then we can make the substitution $y_1=\dfrac{a_1}{2}(y-a_1x-a_3)$ which leads us to a medium equation:

$$y_1^2 = 4x^3 + b_2x^2 + 2b_4x+b_6$$

If the characteristic is not 3 then we can make another substitution $x_1=\dfrac{1}{36}(x-3b_2), y_2=\dfrac{y_1}{108}$ to get a short equation:

$$y^2=x^3-27c_4x-54c_6$$

If however, the characteristic of $K$ equals $2$ or $3$ then we can not do this, but you can instead show that it can be reduced to one of the four forms in your link.

Assume the characteristic is $3$. Then we can eliminate the $a_1$ and $a_3$ as above to get to a medium form. (Note we can take $y=2y'$ then divide through by $4$ to get rid of the coefficient in front of $x^3$. Now if $b_2=0$ (happens if and only if the j-invariant is $0$) then we are in the supersingular case so we are done.

Otherwise take $x=x'+\dfrac{2b_4}{b_2}$ to get something in the regular case.

Now assume the characteristic is $2$ and we start with the long equation. If $a_1=0$ (iff the j-invariant is 0 so we are in the supersingular case again), then take $x=x'+a_2$.

Otherwise, take $x=a_1^2x'+\dfrac{a_3}{a_1}$ and $y=a_1^3y' + \dfrac{a_1^2a_4+a_3^2}{a_1^3}$ to get something in the regular case.