Intersection multiplicity of the curves

Question

I want to find the intersection multiplicity of the curves $f(x,y)=x^5+x^4+y^2$ and $g(x,y)=x^6-x^5+y^2$ at the point $P=(0,0)$.

That`s what I have tried:

$f$ and $g$ have a common tangent, the $y=0$.

So $I(P, f\cap g) > m_P(f) \cdot m_P(g)=4$

$$f(x, 0)=x^5+x^4 \Rightarrow s=\deg f(x, 0)=5$$

$$g(x, 0)=x^6-x^5 \Rightarrow r =\deg g(x, 0)=6$$

$$s \leq r$$

So we consider $h(x, y)=g(x, y)-x f(x, y)$

$$h(x, y)=x^6-x^5+y^2-x(x^5+x^4+y^2)=x^6-x^5+y^2-x^6-x^5-xy^2 \\ \Rightarrow h(x, y)=-2x^5+y^2-xy^2$$

$\deg h(x, 0)=5<r$

So $I(P, f\cap g)=I(P,f\cap h)$

$$f(x,0)=x^5+x^4\Rightarrow \deg f(x,0)=5=s$$

$$h(x,0)=-2x^5\Rightarrow \deg h(x,0)=5=p$$

They have a common tangent, $x=0$, so they don`t intersect traverrsally.

We consider the polynomial $$h_1(x,y)=h(x,y)+2f(x,y)=3y^2-xy^2+2x^4$$

$deg h_1(x,0)=4<s,p$

So, $I(P, f\cap h)=I(P,f\cap h_1)$

$f(x,0)=x^5+x^4 \Rightarrow \deg f(x,0)=5=s$

$h_1(x,0)=2x^4\Rightarrow \deg h_1(x,0)=4=t$

They have a common tangent $x=0$,so they don`t intersect traversally. We consider the polynomial $h_2(x,y)=2f(x,y)-xh_1(x,y)=2x^4+2y^2-3xy^2+x^2y^2$

$\deg h_2(x,0)=4<s$

So $I(P, f\cap h_1)=I(P, h_1\cap h_2)$

$h_1(x,0)=2x^4\Rightarrow \deg h_1(x,0)=4=s$

$h_2(x,0)=2x^4\Rightarrow \deg h_2(x,0)=4=m$

They have a common tangent $x=0$, so they don`t intersect traversally.

We consider the polynomial $h_3(x,y)=h_1(x,y)-h_2(x,y)=y^2(1+2x-x^2)$

$\deg h_3(x,0)=0<s,m$

So $I(P,h_1\cap h_2)=I(P,h_2\cap h_3)$

$h_2(x,0)=2x^4\Rightarrow \deg h_2(x,0)=4=m$

$h_3(x,0)=0\Rightarrow \deg h_3(x,9)=9=n$

So $I(P,h_2\cap h_3)=I(P,h_2\cap y^2)+I(P,h_2\cap (1+2x-x^2))$

$I(P,h_2\cap y^2)=8$

$I(h_2\cap (1+2x-x^2))=0$

Therefore, $I(P, f \cap g)=8$.

Is it right? Do we find that $f$ and $h$ have a common tangent from $f(x,0)$ and $h(x,0)$ ?

Answer 1

See here for the definining properties of the intersection property. These allow us to compute:

$~~~I_P(x^5+x^4+y^2,x^6-x^5+y^2)\\ =I_P(x^5+x^4+y^2,(x^6-x^5+y^2)-(x^5+x^4+y^2))\\ =I_P(x^5+x^4+y^2,x^4 (x^2-2x-1))\\ =I_P(x^5+x^4+y^2,x^4)+I_P(x^5+x^4+y^2,x^2-2x-1)\\ =I_P(x^5+x^4+y^2,x^4)\\ =I_P(y^2,x^4)\\ =2 \cdot 4 \cdot I_P(y,x)\\ =8$

Answer 2

From the system of equations,

$$x^5+x^4+y^2=x^6-x^5+y^2=0,$$

you can eliminate $y$, and

$$-x^6+2x^5+x^4=0.$$

This polynomial has a quadruple root at $0$.

Answer 3

$ f(x,y) =0 $ (red curve) has no real common tangent with $ g(x,y) =0 $ (blue curve) at $ P(0,0) $.