# Severi Formula for Intersection Multiplicities

I say in advance that I am a novice in Intersection Theory, so forgive me if my question is trivial.

Let $$X\subseteq\mathbb{P}^N$$ be a smooth irreducible projective variety of dimension $$n$$ and $$V, W\subseteq X$$ be two irreducible subvarieties meeting properly in $$X$$. Choose a generic projection $$f\colon X\longrightarrow \mathbb{P}^n$$ and denote by $$i(Z,V\cdot W;X)$$ the intersection multiplicity of $$V$$ and $$W$$ at $$Z$$, in the ambient space $$X$$. As Fulton writes in his Intersection Theory (Example 8.2.6), it holds true the following formula $$i(Z, V\cdot W;X)=i(f(Z),f(V)\cdot f(W);\mathbb{P}^n).$$

This was one of Severi's methods for reducing the intersections on general varieties to intersections on the projective space. I would like to prove the formula above by using the Serre intersection formula.

I am pretty sure that there are results from Homological/Commutative Algebra (that I don't know) that would make the proof of the formula an easy exercise.
Any help is well accepted.

EDIT: If I did everything correctly, the following should be a proof of the claim above (feel free to point out possible mistakes).

Lemma. Let $$f\colon A\longrightarrow B$$ be an injective, flat ring homomorphism and $$\mathfrak p\in\mathsf{Spec}(B), \mathfrak q:=f^{-1}(\mathfrak p)$$. Then $$\mathfrak qB=\mathfrak p.$$

The ring map $$B/\mathfrak qB\longrightarrow B/\mathfrak p$$ is onto. We show that it is injective as well. As $$f$$ is flat and $$A/\mathfrak q\longrightarrow B/\mathfrak p$$ is injective, taking the tensor product by -$$\otimes_AB$$ we get an injective ring homomorphism $$\phi\colon B/\mathfrak qB=B\otimes_A A/\mathfrak q\longrightarrow B\otimes_AB/\mathfrak p.$$ Now notice that $$\phi$$ factorizes as $$B/\mathfrak qB\longrightarrow B/\mathfrak p \longrightarrow B\otimes_AB/\mathfrak p,$$ so the map on the left is injective.

Proof of Claim. We prove the claim for any finite surjective morphism of smooth, irreducible, projective varieties $$f\colon X\longrightarrow Y$$.

First of all, recall that if $$f$$ is as above, then it is flat. We may assume that $$X$$ and $$Y$$ are affine varieties, so we get an injective ring homomorphism $$g\colon A=\mathcal O_{Y,f(Z)}\longrightarrow B=\mathcal O_{X,Z}$$. Denote by $$\mathfrak p,\mathfrak q\subseteq B$$ the ideals corresponding to $$V$$ and $$W$$, respectively. The contractions $$\mathfrak p'$$ of $$\mathfrak p$$ and $$\mathfrak q'$$ of $$\mathfrak q$$ via $$g$$ are the prime ideals corresponding to $$f(V)$$ and $$f(W)$$. Thanks to Lemma we have $$\mathsf{Tor}_\bullet^B(B/\mathfrak p,B/\mathfrak q)=B\otimes_A \mathsf{Tor}_\bullet^A(A/\mathfrak p,A/\mathfrak q),$$ as it is shown here, property (10) at page 2. Thus by using standard properties of the length of modules (see here) we get $$\ell_B\mathsf{Tor}_\bullet^B(B/\mathfrak p,B/\mathfrak q)=\ell_B(B/\mathfrak m_AB)\cdot\ell_A \mathsf{Tor}_\bullet^A(A/\mathfrak p',A/\mathfrak q'),$$ where $$\mathfrak m_A$$ is the maximal ideal of $$A$$. Using our lemma again, and denoting by $$\mathfrak m_B$$ the maximal ideal of $$B$$, we get $$\ell_B(B/\mathfrak m_AB)=\ell_B(B/\mathfrak m_B)=1.$$ So we are done.

First, I assume that $$V$$ and $$W$$ are meeting at $$Z$$, and that's a typo.
1. The Lemma is false. Consider for example $$A = k[x], B=k[y]$$ and the injective flat map $$f:A \rightarrow B$$ such that $$f(x) = y^2$$. Consider the ideal $$\mathfrak{p} = (y-y_0) \in Spec(B)$$ with $$y_0 \ne 0$$. Then $$\mathfrak{q}= f^{-1}(\mathfrak{p})=(x-y_0^2)$$ and $$\mathfrak{q}B = (y^2-y_0^2)=(y-y_0)(y+y_0) \ne \mathfrak{p}$$.
2. The reason it fails is that you assume that $$A/ \mathfrak{q} \rightarrow B/ \mathfrak{p}$$ is injective, which is clearly not the case.
However, the proof is still valid, since your map $$f$$ is a finite morphism between local rings - note that the fibers are finite, and hence locally (in a small enough neighbourhood) there will be only one ideal above $$\mathfrak{p}$$. (That is, you should change the assumptions in your Lemma, but the proof of the claim carries through)