In the lecture we had the following two definitions for inflexion point:
- Let $P = (a,b)$, $D = V(f)$ and $C = (a+\lambda t, b+\mu t) = C(t)$, $t \in k$.
$C$ is called inflexion line of $D$ in $P$ if $mult_P(C,D)\geq 3$. Then point $P$ is called inflexion point. - $P\in V(D)$ is called inflexion point, iff $$0 = \begin{Vmatrix} f_{xx}(P) & f_{xy}(P) & f_{x}(P)\\ f_{xy}(P) & f_{yy}(P) & f_{y}(P)\\ f_{x}(P) & f_{y}(P) & 0 \end{Vmatrix}$$
I work with 'Basic Algebraic Geometry' by Shafarevich:
Let $$f(x,y)\overset{Taylor}{=}A(x-a)+B(y-b)+C(x-a)^2+D(x-a)(y-b)+E(y-b)^2+g,$$
where $g$ is a polynomial with terms of degree $\geq 2$.
Then $f(C(t))=(A\lambda + B\mu)t+(C\lambda^2+D\lambda\mu+E\mu^2)t^2+(...)t^3..$
So the intersection multiplicity is $\geq 3$ iff $A\lambda + B\mu=0$ and $C\lambda^2+D\lambda\mu+E\mu^2 =0$.
Then he writes that this condition holds if $C(t)$ is the tangent line to $D$ in $P$ and $C\lambda^2+D\lambda\mu+E\mu^2 = q(\lambda,\mu)$ is divisible by $A\lambda + B\mu = l(\lambda,\mu)$
The basis to the bilinear form is:
$$ \begin{matrix}
& \lambda^2 & \lambda\mu & \mu^2 \\
\hline
q & C & D & E \\
\lambda*l & A & B & 0 \\
\mu*l & 0 & A & B
\end{matrix}$$
So $P\in V(D)$ is inflexion point iff $det\begin{pmatrix}
C & D & E\\
A & B & 0\\
0 & A & B
\end{pmatrix} = 0$.
Then we can also write $det\begin{pmatrix}
f_{xx}(P) & f_{xy}(P) & f_{x}(P)\\
f_{xy}(P) & f_{yy}(P) & f_{y}(P)\\
f_{x}(P)& f_{y}(P)& 0
\end{pmatrix} = det\begin{pmatrix}
2C & D & A\\
D & 2E & B\\
A& B& 0
\end{pmatrix} = -2 * det \begin{pmatrix}
C & D & E\\
A & B & 0\\
0 & A & B
\end{pmatrix}$.
The part I don't understand is why we know that $l(\lambda,\mu)$ must divide $q(\lambda,\mu)$. In the book he writes:
$cu^2+duv+ev^2$ is divisible by $au+bv$ as a homogeneous polynomial in $u,v$.
But why? I guess it should be easy to see, but I didn't see it yet..
