# Multiplicity and degree of irreducible projective subschemes.

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 $$P_X = \mu_X \cdot P_{X_{red}}$$?

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.