I want to calculate the order of the point $M$ at the curve $f(x,y) \in \mathbb{R}[x,y]$, when:
$$M(0,1), f=(x^2-1)^2-y^2(3-2y)$$
That's what I have tried:
$$f(x,y)=x^4-2x^2+1-3y^2+2y^3$$
$$f_4(x,y)=x^4$$
$$f_3(x,y)=2y^3$$
$$f_2(x,y)=-2x^2-3y^2$$
$$f_0(x,y)=1$$
We check if $M$ is singular.
$$f(0,1)=0$$ $$f_x(x,y)=4x^3-4x$$ $$f_x(0,1)=0$$ $$f_y(x,y)=-6y+6y^2$$ $$f_y(0,1)=0$$
Taylor expansion of $f(x,y)$ at the point $M(0,1)$
$$f(x,y)=f(0,1)+[f_x(0,1)(x-0)+f_y(0,1)(y-1)]+ \dots$$
$$f(x,y)=-2x^2+3(y-1)^2+2(y-1)^3+x^4$$
So, the order of $M$ is $2$.
Could you tell me if it is right or if I have done something wrong?
I think you did right. You could use other methods also: use the linear change of co-ordinates so that the point $(0, 1)$ becomes the point $(0, 0)$. Then write $f(x, y)$ as $f = f_0 + f_1 + \cdots,$ where $f_d$ is a homogeneous polynomial of degree $d$ in $x$ and $y$. Then the order of the point $(0, 1)$ will be the least $r$ such that $f_r\neq 0.$(R. Hartshorne, Algebraic Geometry, Chapter I, Ex. 5.3)
In this case, use the co-ordinate transformation: $X = x, Y = y-1.$ Then the resulting polynomial will be $f(X, Y) = X^4 + 2Y^3 + - 2X^2 + 3Y^2$ (this is the one you got at the end). So the order of $M$ is 2.