Severi Formula for Intersection Multiplicities

Question

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.

Answer 1

First, I assume that $V$ and $W$ are meeting at $Z$, and that's a typo.

Regarding your attempted proof:

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)