# How to reduce cubics in the plane to a canonical form?

I watched a video from Wildberger in the Differential Geometry series ( first, or third lecture, I don't remember ) where he says the following.

The general format of a cubic curve is $$a x^3 + b y^3 + c x^2 y + d y^2 x + e x^2 + f y^2 + g x y + h x + i y + j = 0$$

and that all of these can be reduced ( transformed, in fact ) to the canonical form $$y^2 = a x^3 + b x + c.$$

He then explains that there are basically three types, one of them being the elliptic curve with no singularities and two others with singularities.

Is this correct? And how do these transformations work?

P.S. Allegedly ( says Wildberger ) Newton studied cubics intensively and categorized them in 80 different types. Is there some book where this is described?

EDIT. I am studying this question which might be a duplicate. Explicit Derivation of Weierstrass Normal Form for Cubic Curve - Couldn't find this before because I did not have the notion of Weierstrass normal form.

• This site uses MathJax, which means you can typset formulas using the same commands you would in TeX. I edited for you. About your questions: 1. Yes, this is correct. 2. See "Weierstrass form" on Wikipedia. 3. No need for "allegedly"; Newton did indeed classify plane cubics into types. I think he described 72 types. I don't know a good source, but googling "newton classification plane cubics" returns lots of relevant things. Commented Jul 7, 2015 at 9:43
• It's perhaps worth saying that this is canonical form only when the characteristic underlying field does not have characteristic $2$ or $3$ (those characteristics have their own forms, however). (Since this is tagged differential-geometry presumably you're only interested in the real and perhaps complex cases anyway.) Commented Jul 7, 2015 at 9:45
• Thanks. I am basically interested in the transformations as such. Can you give an example? Are Newton's categories still relevant? Commented Jul 7, 2015 at 9:46
• Oh, yes, real case only. Commented Jul 7, 2015 at 9:46
• (And actually, the canonical form isn't quite reduced; by applying a the transformation $(x, y) \mapsto (a^{-1 / 3} x, y)$, we can reduce the above form to $$y^2 = x^3 + p x + q .)$$ Commented Jul 7, 2015 at 9:59

Let $C$ be an irreducible cubic curve, and say it's nonsingular. Let $P$ be an inflection point on $C$, which you can find by computing the Hessian and intersecting $C$ with the Hessian curve, and choose a system of coordinates such that $P = [0:0:1]$, and the tangent line at $P$ is given by $T_0 = 0$. The cubic then has equation $$T_2^2T_0 + T_2L_2(T_0,T_1) + L_3(T_0,T_1) = 0,$$ where $L_2$ is a quadratic form and $L_3$ is a cubic form. Since $T_0 = 0$ intersects $C$ at one point, we have that the $T_1^2$ coefficient in $L_2$ is zero, so in affine coordinates $x = T_1/T_0$ and $y = T_2/T_0$, the equation becomes $$y^2 + axy + by + dx^3 + ex^2 + fx + g = 0.$$ Since $d \ne 0$ by assumption, we can assume that $d=1$. Now replacing $y$ with $y + ax/2 + b/2$, we can assume $a = b = 0$. Moreover, by changing variables $x \mapsto x + e/3$, we can assume that $e=0$. We therefore have that the equation is of the form $$y^2 + x^3 + ax + b = 0.$$
On the other hand, if $C$ is singular, and we choose $[0:0:1]$ to be the singular point, the cubic has equation $$T_2L_2(T_0,T_1) + L_3(T_0,T_1) = 0.$$ By linear transformations, we get that either $L_2 = T_0^2$ or $L_2 = T_0T_1$. A similar argument as to the nonsingular case gives that these two cases can be reduced to cuspidal cubics of the form $$y^2 + x^3 = 0,$$ and nodal cubics of the form $$y^2 + x^2(x+1) = 0,$$ respectively. See Dolgachev for details. These give the three cases you stated.