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I want to compute the intersection multiplicity of $YZ=X^2$ and $YZ=(X+Z)^2$ at $P=(0:1:0)$ In an affine nbd of $P$, let $(X:Y:Z)=(x:1:z)$ $$I_P=\dim \mathcal{O}_{\mathbb{A}_k^2,(0,0)}/(x^2-z,x^2+2xz+z^2-z)$$

In my previous question, I learned that if $(0,0)$ is the only zero, I can change the local ring to $k[x,z]$. But the problem occurs since there are intersection points not only $(0,0)$, but also $(-2,4)$. I saw the Fulton's book(section 2.9 prop 6), which says that $$k[x,z]/I \simeq \mathcal{O}_{(0,0)}/I\mathcal{O}_{(0,0)} \times \mathcal{O}_{(-2,4)}/I\mathcal{O}_{(-2,4)}$$ So now how can I compute the intersection multiplicity?

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up vote 6 down vote accepted

That there are multiple points of intersection does not really affect your calculation. In order to compute the intersection multiplicity at $P,$ we must pass to the local ring, which forgets other intersection points. Thus, you can compute $$I_P=\dim_k k[x,z]_{(x,z)}/(z-x^2,z-(x+z)^2)_{(x,z)},$$

i.e. we can localize $k[x,z]$ at the prime $(x,z)$ and then take the quotient. Of course, localization commutes with quotients, so we have

$$I_P=\dim_k (k[x,z]/(z-x^2,z-(x+z)^2))_{(x,z)}=\dim_k (k[x]/(x^2-(x+x^2)^2))_{(x)}=\dim_k (k[x]/(x^3(-2-x)))_{(x)}=\dim_k k[x]/(x^3)$$

since $-2-x$ is a unit in the localization $k[x]_{(x)}$. This shows that the multiplicity is $I_P=3.$

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Minor quibble: there really should be $\dim_k$ in front of every step in the last string of equalities. –  Michael Joyce Nov 20 '12 at 17:52
    
Thanks to good answer. All the things are now much clear to me. But I have a question. Isn't it $(k[x]/(x^3))_{(x)}$ in the last? If so, then why is $\dim_k (k[x]/(x^3))_{(x)}=3$? –  Gobi Nov 20 '12 at 18:03
    
@MichaelJoyce, Whoops, thanks for editing, and pointing that out! –  Andrew Nov 20 '12 at 19:38
    
@Gobi, yes, you are correct, but it's not hard to show that $(k[x]/(x^3))_{(x)}\cong k[x]/(x^3).$ (I.e., the latter is already local.) –  Andrew Nov 20 '12 at 19:42
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