$p$ is an inflection point if and only if the Hessian $\mathcal{H}_{P}$ vanishes at $p$

Question

I'm trying to comprehend the proof of the following proposition.

Let $C$ be a projective plane curve over $\mathbb{C}$ and let $p$ be a smooth point of $C .$ Let $P$ be a homogeneous polynomial in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right]$ with no repeated factors such that $C$ is the zero set of $P$ in $\mathbb{P}_{\mathbb{C}}^{2}$. Then $p$ is an inflection point of $C$ if and only if the Hessian $\mathcal{H}_{P}$ vanishes at $p$.

The presented proof: we may assume that $p=[0,0,1]$ and that the tangent line to $C$ at $p$ is defined by $x_{0}=0 .$ Multiplying $P$ with a non-zero constant, we may also assume that $\partial_{x_{0}} P(0,0,1)=1$. We set $Q(x, y)=P(x, y, 1)$. By a previous result, we have that $$ \mathcal{H}_{P}(0,0,1)=(d-1)^{2} \cdot \operatorname{det}\left(\begin{array}{ccc} \partial_{x}^{2} Q(0,0) & \partial_{x} \partial_{y} Q(0,0) & 1 \\ \partial_{x} \partial_{y} Q(0,0) & \partial_{y}^{2} Q(0,0) & 0 \\ 1 & 0 & 0 \end{array}\right)=-(d-1)^{2} \partial_{y}^{2} Q(0,0). $$ (Here's the part that I don't understand) By Example $11.21, p$ is an inflection point of $C$ if and only if the polynomial $Q(0, y)$ has a zero of multiplicity at least 3 at $y=0 .$ Taking the Taylor expansion of $Q$ around $(0,0),$ we see that this is equivalent to saying that $\partial_{y}^{2} Q(0,0)=0$.

For additional information, "Example 11.21" is given:

Let $C$ be an irreducible projective plane curve of degree $d$ over $\mathbb{C},$ and let $p$ be a point of $C .$ Set $m=\operatorname{mult}_{p} C .$ Show that for all but finitely many lines $L$ through $p$ in $\mathbb{P}_{\mathbb{C}}^{2},$ the line $L$ meets $C$ in exactly $d-m+1$ points. Deduce that there exists a line in $\mathbb{P}_{C}^{2}$ that meets $C$ in precisely $d$ points.

I am very confused. Can someone help me understand the proof? (I would also welcome explanations that don't use example 11.21 too)

Answer 1

If $p = [0,0,1]$ is an inflection point of $C$, then $i(T_PC,C) ≥ 3$ which means $$\newcommand\mb{\mathbb} 3 ≤ i(T_PC,C) = \dim_{\mb C} \mb C[x,y]_{(x,y)}/(x,Q(x,y)) = \dim_{\mb C}\mb C[y]_{(y)}/(Q(0,y)) = v_y(Q(0,y)).$$ Thus we can write $Q(0,y)= y^3g(y)$. Hence $\partial_y^2 Q(0,0) = 0$ and $\newcommand\mc{\mathcal} \mc H_P(0,0,1)=0$.

Conversely if $\mc H_P(0,0,1)=0$ then $\partial_y^2 Q(0,0) = 0$.

As $P \in C$ we have $Q(0,0)=0$ and as $T_PC = \{x_0=0\}$ we have $\partial_yQ(0,0)=0$.

So we can factor $Q(0,y) = y^3g(y)$ hence $i(T_PC,C) = v_y(Q(0,y)) \geq3 $.