Let $C$ be a nonsingular algebraic curve in $ℂℙ^2$ of degree $d$. Define a point of inflection of $C$. What is the maximum number of points of inflection that $C$ can have, as a function of $d$? Justify your answer briefly. [You may assume that $C$ and its Hessian curve have no common component.]

Let $C$ be defined by $P(x,y, z)=0$, $P$ homogeneous of degree $d$. The Hessian $H_p$ of $p$ is \[ H_p(x, y, z)=\det\begin{bmatrix} P_{x x} & P_{x y} & P_{x z} \\ P_{y x} & P_{y y} & P_{y z} \\ P_{z x} & P_{z y} & P_{z z} \end{bmatrix} \] A point of inflection of $C$ is a (nonsingular) point $[x, y, z]$ of $C$ with $H_p(x, y, z)=0$.
Case $d=1$
$H_p=0$, every point of $C$ is a point of inflection (so, infinitely many points).
Case $d=2$
$H_p(x, y, z)$ is constant and nonzero (as $C$ is nonsingular), no inflection points.
Case $d>2$
{inflection points} is intersection of $C$, degree $d$, and $H_p=0$, degree $3(d-2)$. As can assume no common component by hint, by weak Bézout there are at most $3 d(d-2)$ inflection points.

Let $C$ be the nonsingular cubic curve in $ℂℙ^2$ defined by the equation \[ x^3+y^3+3 x z^2=0 . \] Find all the points of inflection of $C$.

\begin{align*}H_p&=\begin{vmatrix} 6 x & 0 & 6 z \\ 0 & 6 y & 0 \\ 6 z & 0 & 6 x \end{vmatrix}\\ & =216\left(x^2 y-y z^2\right) \\ & =216 y(x-z)(x+z) \end{align*} So $3d(d-2)=9$ inflection points divide into:
$y=0$ and $x^3+y^3+3 x z^2=0$:
$[0,0,1],[\sqrt{3} i, 0,1],[-\sqrt{3} i, 0,1]$
$x=z$ and $x^3+y^3+3 x z^2=0$:
$[1, \sqrt[3]{-4}, 1]$ for 3 cube roots of $-4$.
$x=-z$ and $x^3+y^3+3 x z^2=0$:
$[1, \sqrt[3]{-4},-1]$ for 3 cube roots of $-4$.
Sanity check: 9 points of inflection, consistent with (a). In fact, every nonsingular cubic has 9 points of inflection, which are order-3 points in the group law of cubic.

Show that $C$ in part (b) can be taken by a projective transformation to a cubic $C_λ$ of the form \[ y^2 z-x(x-z)(x-λ z)=0, \] for some $λ ∈ ℂ ∖\{0,1\}$ which you should determine.

Follow the algorithm in the proof in the notes for normal form of cubic.
Step 1: Choose a point of inflection, and apply a projective transformation so that the point of inflection is $[0,1,0]$ with tangent line $z=0$.
Choose point $[0,0,1]$, which has tangent line $x=0$.
So: apply projective transformation $(x, y, z)↦(\tilde{x}, \tilde{y}, \tilde{z})$ with \begin{align*} x&=\tilde{z}\\ y&=\tilde{x}\\ z&=\frac{i}{\sqrt3}\tilde{y} &\text{factor } \frac{i}{\sqrt3} \text{ gives nicer constant} \\ \end{align*} Then \begin{align*} \tilde{P}(\tilde{x}, \tilde{y}, \tilde{z})&=\tilde{z}^3+\tilde{x}^3-\tilde{y}^2 \tilde{z} \\ &=-\tilde{y}^2 \tilde{z}+(\tilde{x}+\tilde{z})\left(\tilde{x}+e^{2 π i / 3} \tilde{z}\right)\left(\tilde{x}+e^{-2 π i / 3} \tilde{z}\right) \end{align*} Change sign of $\tilde{P}$ without changing curve: \begin{aligned} \tilde{y}^2 \tilde{z} & -(\tilde{x}-a \tilde{z})(\tilde{x}-b \tilde{z})(\tilde{x}-c \tilde{z}), \\ a & =-1,   b=-e^{2 π i /3},   c=-e^{-2 π i /3} . \end{aligned}
Step 2: Apply projective transformation
$(\tilde{x}, \tilde{y}, \tilde{z}) ↦(\tilde{\tilde{x}}, \tilde{\tilde{y}}, \tilde{\tilde{z}})$ with \begin{aligned} & \tilde{\tilde{x}}=\frac{\tilde{x}-a \tilde{z}}{b-a} \\ & \tilde{\tilde{y}}=(b-a)^{-3 / 2} \tilde{y} \\ & \tilde{\tilde{z}}=\tilde{z} \end{aligned} gives \[\tilde{\tilde{P}}(\tilde{x}, \tilde{y}, \tilde{z})=(b-a)^{-3} ⋅\left(\tilde{\tilde{y}}^2 \tilde{\tilde{z}}-\tilde{\tilde{x}}(\tilde{\tilde{x}}-\tilde{\tilde{z}})(\tilde{\tilde{x}}-λ \tilde{\tilde{z}})\right) \] for $\begin{aligned}[t]λ&=\frac{c-a}{b-a}\\ & =\frac{e^{-2 π i / 3}-1}{e^{2 π i / 3}-1} \\ & =\frac{-\frac{3}{2}-i\frac{\sqrt{3}}{2}}{-\frac{3}{2}+i\frac{\sqrt{3}}{2}}=\frac{\left(-\frac{3}{2}-i \frac{\sqrt{3}}{2}\right)^2}{3}=\frac{1}{2}+\frac{i \sqrt{3}}{2} . \end{aligned}$

Show that, for general $λ ∈ ℂ ∖\{0,1\}$, the cubic $C_λ$ in (c) may be taken to a different curve $C_{\tilde{λ}}$ by a projective transformation with matrix of the form \begin{pmatrix} a & 0 & b \\ 0 & c & 0 \\ 0 & 0 & d \end{pmatrix} and find the possibilities for $\tilde{λ}$ as a function of $λ$ (there are six, including $\tilde{λ}=λ)$.

Try projective transformation \[ \left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{lll} a & 0 & b \\ 0 & c & 0 \\ 0 & 0 & d \end{array}\right)\left(\begin{array}{c} \tilde{x} \\ \tilde{y} \\ \tilde{z} \end{array}\right)=\left(\begin{array}{c} a \tilde{x}+b \tilde{z} \\ c \tilde{y} \\ d \tilde{z} \end{array}\right) \] This takes $y^2 z-x(x-z)(x-λ z)$ to \[ c^2 d \tilde{y}^2 \tilde{z}-a^3\left(\tilde{x}+\frac{b}{a} \tilde{z}\right)\left(\tilde{x}+\left(\frac{b-d}{a}\right) \tilde{z}\right)\left(\tilde{x}+\left(\frac{b-λ d}{a}\right) \tilde{z}\right) \] To make this of the form \[ \tilde{y}^2\tilde{z}-\tilde{x}(\tilde{x}-\tilde{z})(\tilde{x}-λ \tilde{z}), \] let’s choose $a, b, c, d$ so that: \begin{aligned} c^2 d&=1 \\a^3&=1\\ \left\{\frac{b}{a}, \frac{b-d}{a}, \frac{b-λ d}{a}\right\}&=\{0,-1,-\tilde{λ}\} \end{aligned} as multiply $a, b, c, d$ by $\sqrt[3]{1}$ makes no difference, can fix $a=1$.
Also fix $c=d^{-1 / 2}$. The first two equations hold. There are 3! possibilities depending on the order in which we identify the sets: Note: these acts as a group of Mobius transformations on $λ↦\tilde{λ}$, isomorphic to $S_3$. Can check your calculations by composing Mobius transformations and getting back one of the same 6.
The Arithmetic of Elliptic Curves, Joseph H. Silverman page 54
the Legendre form of the equation of the elliptic curve is $y^2=x(x-1)(x-λ)$, where $λ ≠ 0,1$. $j$ can be expressed in terms of $λ$ as \[ j=256 \frac{(λ^2-λ+1)^3}{λ^2(λ-1)^2} . \] $λ$ is not uniquely defined, depending on which two roots of the cubic are chosen to be mapped to 0 and 1 , it can take 6 values, $λ, 1-λ, 1 / λ, 1 /(1-λ)$, $λ /(λ-1)$ and $(λ-1) / λ$, but they all give the same value of $j$.