Intersection multiplicity

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Intersection multiplicity

Postby Guest » Tue Dec 09, 2014 7:18 pm

Hello!!!

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

That`s what I have tried:

[tex]f[/tex] and [tex]g[/tex] have a common tangent, the [tex]y=0[/tex].

So [tex]I(P, f\cap g) > m_P(f) \cdot m_P(g)=4[/tex]

[tex]f(x, 0)=x^5+x^4 \Rightarrow s=\deg f(x, 0)=5[/tex]

[tex]g(x, 0)=x^6-x^5 \Rightarrow r =\deg g(x, 0)=6[/tex]

[tex]s \leq r[/tex]

So we consider [tex]h(x, y)=g(x, y)-x f(x, y)[/tex]

[tex]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[/tex]

[tex]\deg h(x, 0)=5<r[/tex]

So [tex]I(P, f\cap g)=I(P,f\cap h)[/tex]

[tex]f(x,0)=x^5+x^4\Rightarrow \deg f(x,0)=5=s[/tex]


[tex]h(x,0)=-2x^5\Rightarrow \deg h(x,0)=5=p[/tex]

They have a common tangent, [tex]x=0[/tex], so they don`t intersect traverrsally.

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

[tex]deg h_1(x,0)=4<s,p[/tex]

So, [tex]I(P, f\cap h)=I(P,f\cap h_1)[/tex]

[tex]f(x,0)=x^5+x^4 \Rightarrow \deg f(x,0)=5=s[/tex]

[tex]h_1(x,0)=2x^4\Rightarrow \deg h_1(x,0)=4=t[/tex]

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

[tex]\deg h_2(x,0)=4<s[/tex]

So [tex]I(P, f\cap h_1)=I(P, h_1\cap h_2)[/tex]

[tex]h_1(x,0)=2x^4\Rightarrow \deg h_1(x,0)=4=s[/tex]

[tex]h_2(x,0)=2x^4\Rightarrow \deg h_2(x,0)=4=m[/tex]

They have a common tangent [tex]x=0[/tex], so they don`t intersect traversally.

We consider the polynomial [tex]h_3(x,y)=h_1(x,y)-h_2(x,y)=y^2(1+2x-x^2)[/tex]

[tex]\deg h_3(x,0)=0<s,m[/tex]

So [tex]I(P,h_1\cap h_2)=I(P,h_2\cap h_3)[/tex]

[tex]h_2(x,0)=2x^4\Rightarrow \deg h_2(x,0)=4=m[/tex]

[tex]h_3(x,0)=0\Rightarrow \deg h_3(x,0)=0=n[/tex]

So [tex]I(P,h_2\cap h_3)=I(P,h_2\cap y^2)+I(P,h_2\cap (1+2x-x^2))[/tex]

[tex]I(P,h_2\cap y^2)=8[/tex]

[tex]I(h_2\cap (1+2x-x^2))=0[/tex]

Therefore, [tex]I(P, f \cap g)=8[/tex].


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