Equation of a non-singular cubic curve

Question

The equation of a non-singular cubic curve in affine coordinates is $$y^2+a_1 xy+a_3 y=x^3+a_2x^2+a_4x+a_6 .$$

If $\text{ch } K \neq 2, 3$ then it is written $$y^2=x^3+ax+b .$$

Why do we write it in this form if $\text{ch } K \neq 2, 3 $?

Answer 1

Try a change of variables $X = x +\alpha$ and $Y = y + \beta$ for suitable $\alpha$ and $\beta$, to put it in the required form. This will give equations on $\alpha,\beta$, that you will be able to solve because the characteristic of $k$ is not $2$ not $3$.

Answer 2

For any elliptic curve $C$ over any base field there are affine coordinates $(x, y)$ such $C$ can be written in the form $$y^2+a_1 xy+a_3 y=x^3+a_2x^2+a_4x+a_6. \qquad (\ast)$$ If the characteristic of the underlying field if not $2$ or $3$, then we can choose coordinates $(x, y)$ such that the curve is $$y^2 = x^3 + px + q.$$

As an indication of what goes wrong in the exceptional characteristics, observe that when the characteristic is not $3$, by replacing the coordinate $x$ in $(\ast)$ with the new coordinate $\bar{x} - \frac{a_2}{3}$, $(\ast)$ becomes $$y^2 + a_1 \left(\bar{x} - \frac{a_2}{3}\right) y + a_3 y = \left(\bar{x} - \frac{a_2}{3}\right)^3 + a_2 \left(\bar{x} - \frac{a_2}{3}\right)^2 + a_4 \left(\bar{x} - \frac{a_2}{3}\right) + a_6.$$ In particular, when we expand the terms in $x^2$ cancel out, and by renaming coefficients, we see that the curve's expression in the coordinates $(\bar{x}, y)$ has the form $$y^2 + b_1 xy + b_3 y = x^3 + b_4 x + b_6,$$ that is, for any such curve there are always coordinates $(\bar{x}, y)$ with this form, i.e., without a term in $x^2$. However, this change of coordinates, because it involves the expression $\frac{a_2}{3}$, requires that $3$ be invertible in the underlying field, which requires exactly that the characteristic of that field is not $3$ (by definition, in that characteristic, $0 = 3$).

In characteristic $3$ we can still improve on the general formula; in this case we can write a general nonsingular cubic as $$y^2 = x^3 + c_2 x^2 + c_4 x + c_6,$$ but in characteristic $2$ we are stuck with the form $(\ast)$.