I'm in the process of reading some parts of Fulton's "Introduction to intersection theory" and there's a short part there which I don't quite understand where I think I'm missing something obvious.


Let us assume that we work over $\mathbb{C}$ for simplicity, and let $H_1, \ldots, H_n$ be hypersurfaces in some n-dimensional variety $V$ and let $p$ be an isolated point of $\cap H_i.$ Let $A= \mathcal{O}_{V,p}$ be the local ring of $V$ along $P$ and assume that each $H_i$ is defined by tone element $f_i$ in $A.$ Let $I=(f_1,\ldots,f_n).$ Then $A/I$ is of (Krull) dimension zero. One defines the multiplicity $i(P,H_1\cdots H_n)$ to be the coefficient of $t^n/n!$ in the Hilbert-Samuel polynomial $P(t) = \text{dim }_{\mathbb{C}} (A/I^t) $ (so $P(t)$ is the eventual polynomial for that function).

The projective normal cone and the question

Set $\Lambda = A/I$ and consider the surjection of graded rings $$\Lambda[X_1,\ldots,X_n] \rightarrow \oplus_{t=0}^{\infty} I^t/I^{t+1}$$ taking $X_i$to $f_i \in I/I^2$. The kernel of this homomorphism is a homogenous ideal which defines a subscheme $\mathbf{P}(C)$ of $P^{n-1}_{\Lambda}.$ This is what is known as the projective normal cone. Then, one can see that the degree of $\mathbf{P}(C)$ in $\mathbb{P}^{n-1}_{\Lambda}$, that is, the coefficient of $t^{n-1}/(n-1)!$ in the Hilbert polynomial $P_{\mathbf{P}(C)}(t)$, is precisely the multiplicity defined earlier. This is not hard to show. However:

The claim is that the degree of $P_{\mathbf{P}(C)}(t)$ is the same as the length of $\mathbf{P}(C)$ at the generic point. I can't for the life of me seem to figure out why this should be true, so I would be very grateful for some help.

  • $\begingroup$ Is this claim in Fulton, "Introduction to Intersection Theory in Algebraic Geometry", CBMS Number 54 and if yes where is it stated (I could not find it browsing through the book, only the statements you made above the "claim")? $\endgroup$ – Jürgen Böhm Mar 18 '15 at 16:29
  • $\begingroup$ @JürgenBöhm Yes! It is on pg. 11 "Samuel's intersection multiplicity" at the end of the page, "However, since $\mathbf{P}(C) \subset \mathbf{P}^{n-1}_{\Lambda}$..." and the following explanation. I'm afraid I don't follow his explanation there. $\endgroup$ – user101036 Mar 18 '15 at 17:47
  • $\begingroup$ Did you read my answer? If you find it incomplete or unclear, please notify me, so that I can possibly improve it. $\endgroup$ – Jürgen Böhm Mar 21 '15 at 16:50
  • $\begingroup$ @JürgenBöhm I have looked through it and it seems good! Thanks a lot! I will reread it later tonight or tomorrow, and then I will see if I understand it fully. $\endgroup$ – user101036 Mar 22 '15 at 20:03

$\newcommand{\ideal}[1]{{\mathfrak #1}}$ $\newcommand{\PSP}{{\mathbf P}}$ $\newcommand{\proj}[1]{{\mathrm{proj}}(#1)}$ $\newcommand{\Ass}{\mathrm{Ass}}$ $\newcommand{\supp}{\mathrm{supp}}$ $\newcommand{\length}{\mathrm{length}}$

Consider the sequence $$ 0 \to J \to R = \Lambda[x_1,\ldots,x_n] \to \bigoplus_t I^t/I^{t+1} = U \to 0 $$ so $S = R/J$ is isomorphic to $U$. Note that $\PSP(C) = \proj{R/J}$ and $\dim R/J = n$

Let $\bar{\ideal{m}}$ be the maximal ideal of $A/I$. Then $\ideal{p} = \bar{\ideal{m}}[x_1,\ldots,x_n]=\bar{\ideal{m}}R$ is a prime ideal of $R$ and consists only of nilpotent elements. So it is also the nilradical $\ideal{N}$ of $R$.

First we show, that $J \subseteq \ideal{p} \subseteq R$. It is

$$n = \dim R/J = \dim R/(J+\ideal{N})$$

as every prime $\ideal{q}$ of $R$ that contains $J$ also contains $\ideal{N}$. But if $J + \ideal{N} \supsetneq \ideal{N}$, then as $R/\ideal{N}=R/\ideal{p}$ is the integral affine algebra $A/\ideal{m}[x_1,\ldots,x_n]$, we would necessary have $\dim R/(J+\ideal{N}) < \dim R/\ideal{N} = n$.

So $V(J)$ is an affine algebraic scheme with underlying variety $V(\ideal{p})$.

So it makes sense to consider the length of $(R/J)_{\ideal{p}}$ that you call "the length of $\PSP(C)$ at the generic point" and that we abbreviate now $i(P)'$.

Now set $M_m$ be the $R$-module $R/J$ and consider the typical filtration from the theory of associated primes

$$(*) \quad 0 \to M_{i-1} \to M_i \to R/\ideal{p}_i \to 0$$

where $\ideal{p}_i \in \Ass R/J$ with $\ideal{p} = \min \Ass R/J = \supp R/J$ occuring - and that will be shown next - exactly $i(P)'$ times.

Now simply consider all sequences $(*)$ tensored by $\otimes_R R_\ideal{p}$ to give

$$(**) \quad 0 \to (M_{i-1})_\ideal{p} \to (M_i)_\ideal{p} \to (R/\ideal{p}_i)_\ideal{p} \to 0$$

The $(R/\ideal{p}_i)_\ideal{p}$ are of length $1$ exactly for $\ideal{p} = \ideal{p}_i$ otherwise the localized module itself vanishes.

So as $\length((M_m)_\ideal{p}) = \length (R/J)_\ideal{p} = i(P)'$ we have $\ideal{p}$ occuring $i(P)'$ times in the filtrations $(*)$.

But now consider the Hilbert-polynomials $P_{M_i}$ and $P_{R/\ideal{p}_i} = Q_i$ from the sequences $(*)$. We have the relation

$$(***) \quad P_{M_i} = P_{M_{i-1}} + Q_i$$

The only $Q_i$ that are of degree $n$ are those where $\ideal{p} = \ideal{p}_i$, which occur, as shown above, $i(P)'$ times.

Furthermore for these because of $R/\ideal{p} = A/\ideal{m}[x_1,\ldots,x_n]$:

$$Q_i = t^n/n! + \text{lower terms}$$

So alltogether we get

$$P(t) = P_{R/J} = P_{M_m} = i(P)' t^n/n! + \text{lower terms}$$

As we knew a priori $P(t) = i(P)t^n/n! + \text{lower terms}$ (see Fulton, page 11, (i)) we have the equality $i(P) = i(P')$ which relates the Hilbert-Samuel multiplicity with the one that Fulton introduced earlier on page 9 of the "Introduction".

Note that the key idea of using the $\Ass$-filtration of $R/J$ and localizing by $\ideal{p}$ is exactly used in the same way as in Hartshornes proof of the Bezout-Theorem.


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