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In 5.23., Fulton proves the theorem that if $P$ is an ordinary flex of a plane curve $C=V(F)$, then $C$ and it's Hessian $H$ intersect with multiplicity one, that is $I(P, C \cap H) = 1$.

After a bit of work, we have that it suffices to check the intersection multiplicity of the curves $f = F(X,Y,1) = y + ax^2 + bxy + cy^2+...$ and $g = 2a + 6dx + ...$ in the point $P = [0:0:1]$. Here, $a = 0$ and $d \neq 0$. What I don't see is how this implies the statement. Why do we get $\dim_k \mathcal{O}_0(\mathbb{A}^2) /(f, g) = 1$ from this? I have so far only computed multiplicities when the ideals were just monomials, so I'd appreciate any help.

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I will give you a hint: Use property (5) of the intersection number (cf p.37) – Nils Matthes Jun 21 '12 at 9:10
Very nice, thanks!! Just working with examples where the multiplicities could be computed from the definition made me forget those useful properties (although I shouldn't have omitted the higher order terms of $g$; Fulton has them included, for obvious reasons..) – carlos Jun 22 '12 at 6:38
Glad it helped. Now just post what you did to solve the problem as an answer to your own question and accept your own answer. Then there will be one open problem less in the world (or at least at MSE). – Nils Matthes Jun 22 '12 at 10:59
up vote 2 down vote accepted

@Nils Matthes: Of course.

The property of the intersection number you referred to says that we have an estimate $I_P(C,C') \geq \operatorname{ord}_P(C)\operatorname{ord}_P(C')$, and equality holds iff both curves have distinct tangent lines at $P$. Now we consider $C$ and $V(g)$ as above. Smoothness of $C$ at $P$ holds by assumption, and a straightforward calculation shows that $\frac{\partial g}{\partial X_1}(P)\neq 0$ (again, I omitted some terms in my question, see Fulton's book for details). In particular, $V(g)$ is smooth at $P$ with a different tangent space (after a coordinate transformation, we can assume that $C$ has the tangent line $V(X_2)$ in $P$).

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