For some positive $a\in\mathbb{R}$ let $$ \dot{x}=ax^2-2xy,\quad \dot{y}=y^2-axy $$ $(0,0)$ is a degenerate equilibrium and thus directional blow-ups are made.
It is said here, p. 4 below:
[...] we use the directional blow ups to desingularize the origin [...] the characteristic polynomial of this system is $\mathcal{F}(x,y)=xy(3y-2ax)$. Therefore we need to do both directional blow ups to get the complete local phase portrait around the origin, as $x=0$ and $y=0$ are both characteristic directions. [...]
A stupid question, but: How to get this characteristic polynomial $\mathcal{F}(x,y)=xy(3y-2ax)$? I do not understand what is meant with "characteristic polynomial" in this cited paper. There are other examples given for which I also do not see how they get what they call "characteristic polynomial".
If I do, for example, the blow-up in the x-direction, $(x,z)\mapsto (x,xz)=(x,y)$, I get the system $$ \dot{x}=ax^2-2x^2z,\qquad \dot{z}=3xz^2-2axz=xz(3z-2a). $$
It looks like it is the equation for $(\dot x,\dot y)\sim(x,y)$, that is, points where the vector field is radial. Then the equation is $$ 0=\det\pmatrix{x&y\\\dot x&\dot y}=xg(x,y)-yf(x,y)=xy((y-ax)-(ax-2y))=xy(3y-2ax). $$