Multiplicity and degree of irreducible projective subschemes.

Question

Suppose $X \subset \mathbb{P}^n$ is an irreducible projective scheme. Then its multiplicity $\mu_X$ is defined as the length of the local ring $\mathcal{O}_{X,\eta}$ over itself, where $\eta$ is the generic point of $X$.

The degree $d_X$ of $X$ is defined as $\frac{c}{n!}$, where $c$ is the leading coefficient of the Hilbert polynomial $P_X$ of $X$, and $n$ is the degree of $P_X$.

I would like to show that $d_X = \mu_X \cdot d_{X_{red}}$, where $X_{red} \subset X \subset \mathbb{P}^n$ is the reduced subscheme with the same set of underlying points, but I don't see how to relate the Hilbert polynomial to the multiplicity. So any ideas or hints would be appreciated.

One thought was that maybe even $P_X = \mu_X\cdot P_{X_{red}}$ holds. But on further thought this is inplausible, because $X$ might have embedded components, which do not appear in $Y$, but contribute a part to $P_X$. But doing a (homogeneous) primary decomposition we can decompose $X = X_1 \cup \dots \cup X_r$ scheme theoretically, and $X_1$ corresponds to the maximal component. Because $X_2, \dots X_r$ do not contribute to the leading coefficient of $P_X$, we might assume wlog that $X$ does not have any embedded components. Does this imply $P_X = \mu_X \cdot P_{X_{red}}$?

Answer 1

We will use Proposition I 7.4 from Hartshorne's Algebraic geometry (p. 50), that is:

Let $M$ be a finitely generated graded module over a noetherian graded ring $S$. Then there exists a filtration $0 = M^0 \subset M^1 \subset \dots \subset M^r = M$ by graded submodules, such that for each $i$, $M^i / M^{i+1} \cong (S/\mathfrak{p}_i)(l_i)$, where $\mathfrak{p}_i$ is a graded homogeneous prime ideal of $S$, and $l_i \in \mathbb{Z}$. The filtration is not unique, but for any such filtration we have:

1. if $\mathfrak{p}$ is a homogeneous prime ideal of $S$, then $\mathfrak{p} \supset \text{Ann }M \Leftrightarrow \mathfrak{p} \supset \mathfrak{p_i}$ for some $i$. In particular, the minimal elements of the set $\{\mathfrak{p}_1, \dots,\mathfrak{p}_r\}$ are just the minimal primes of $M$, i.e. the primes which are minimal containing $\text{Ann }M$.
2. for each minimal prime of $M$, the number of times which $\mathfrak{p}$ occurs in the set $\{\mathfrak{p_1},\dots,\mathfrak{p}_r\}$ is equal to the length of $M_\mathfrak{p}$ over the local ring $S_\mathfrak{p}$ (and hence is independent of the filtration).

To compare the Hilbert polynomial of $X$ and $X_{red}$, we consider the homomorphism on homogeneous coordinate rings $$ S(X) = k[x_0,\dots,x_n]/I \rightarrow k[x_0,\dots,x_n]/\sqrt{I} = S(X_{red}).$$

Applying the proposition, we see that $P_X(l) = \sum_i \dim(S(X)/\mathfrak{p}_i)_{l_i + l}$ for $l \gg 0$. Only the minimal primes of $S(X)$ contribute to the leading coefficient of $P_X$, because the degree of the Hilbert polynomial equals the dimension of the scheme. The only minimal prime is the nilradical, so we see that the leading coefficient is a multiple of the leading coefficient of $S(X_{red})$, noting that the shift $l \mapsto l + l_i$ does not change the leading coefficient of a polynomial.

According to part 2. of the proposition, the factor of the degree equals $\text{length}_{S_\mathfrak{p}} M_\mathfrak{p}$. But following the proof of the proposition, one can check that this also holds if we make to homogeneous localisation, i.e. take $S_{(\mathfrak{p})}$ and $M_{(\mathfrak{p})}$ instead.

I think this works without mentioning primary decompositions, though the connection would be interesting.