I have a question:
$P =(0,0)$ is a double point on a curve $f(x,y) \in k[x,y]$ where $k$ is algebraically closed field. If degree of $f$ is $n$ and we write $f = f_n + f_{n-1} +......f_m $ where $f_{i}$ is a form of degree $i$ and let $ P $ be a point on $f$ so that $ m> 0$. We say $P$ is a double point on $f$ if $ m =2$. In this case, as $k$ is algebraically closed, we can write $f_2$ as a product of two linear factors, say $L_1$ and $L_2$.
Now we have to show that $P$ is a node if and only if $ (f_{xy}(P)) ^2 \neq f_{xx}(P). f_{yy}(P)$
Here, node = ordinary double point, that is, when powers of all linear factors in factorization of $f_m$ are equal to $1$, in the case here, $P$ is a node iff $f_2 = L_1. L_2$ and $L_1 \neq L_2$.
Now, I don't have a clue how can I get double partial derivatives of $f$ into the picture when the definitions don't include it. I know that to show $P$ is a node it's necessary and sufficient that $L_1$ and $L_2$ are distinct, but after sometime manipulating equations I don't see how I can somehow have them come into the picture.
It would be great if someone could give some hint to this problem. Thanks!
Source: W.Fulton, Algebraic Curves
Let's write $$f_2=ax^2+bxy+xy^2.$$ Then $f_{xx}(P)=(f_2)_{xx}(0,0)=2a$, $f_{xy}(P)=(f_2)_{xy}(0,0)=b$, and $f_{yy}(P)=(f_2)_{yy}(0,0)=2c$. The condition $f_{xy}(P)^2=f_{xx}(P)f_{yy}(P)$ is equivalent to $b^2=4ac$, the condition that $f_2$ has a repeated linear factor.
Note that all second derivatives of $f_3$, $f_4$, etc., vanish at $P=(0,0)$.