The degree of the Hilbert-Samuel polynomial equals the dimension of the support

Put $A=R\mathrm{/}{\mathrm{Ann}}_R(M)$ and let $J=IA$. Then $A$ is Noetherian local, $M$ is a faithful finite $A$-module, and $\mathrm{Spec}(A)={\mathrm{Supp}}_R(M)$. If $m_1,\dots ,m_s$ generate $M$, there is a surjection $A^s\twoheadrightarrow M$ and a faithful injection $A\hookrightarrow M^s$, $a\mapsto (a{m}_1,\dots ,a{m}_s)$. The surjection gives ${\chi }_{J,M}(n)\le s{\chi }_{J,A}(n)$. Applying [L3] to the injection gives a bounded shift $c$ and the reverse estimate ${\chi }_{J,A}(n-c)\le s{\chi }_{J,M}(n)$ for large $n$. Hence $P_{J,A}$ and $P_{I,M}$ have the same degree. It remains to prove the theorem for the ring $A$.

Let $r=\mathrm{deg}{P}_{J,A}$ and let $d$ be the least number of generators of an ideal of definition of $A$; [L1] identifies $d$ with $\dim A$. Choose an ideal of definition $Q=(x_1,\dots ,x_d)$. The ideals $J$ and $Q$ are finite by [L6] and have the same radical ${\mathfrak{m}}_A$. Raising finite generating sets to suitable powers and expanding products therefore gives positive integers $u,v$ with $J^u\subseteq Q$ and $Q^v\subseteq J$. The resulting linear reindexing inequalities between ${\chi }_{J,A}$ and ${\chi }_{Q,A}$ show that their eventual polynomials have the same degree. If $d=0$, then $Q=0$ is an ideal of definition, so $A$ has finite length and $r=\mathrm{deg}{P}_{Q,A}=0$. If $d>0$, every $Q^n\mathrm{/}{Q}^{n+1}$ is a quotient of a direct sum of $\left(\genfrac{}{}{0px}{}{n+d-1}{d-1}\right)$ copies of $A\mathrm{/}Q$, indexed by the degree-$n$ monomials in the $x_i$. Its length is therefore bounded by a polynomial of degree $d-1$, and summing the graded-piece lengths gives $r=\mathrm{deg}{P}_{Q,A}\le d$.

We prove $\dim A\le r$ by induction on $r$. If $r=0$, the increasing integer sequence ${\mathrm{\ell }}_A(A\mathrm{/}{J}^{n+1})$ is eventually constant, so $J^n\mathrm{/}{J}^{n+1}=0$ for all large $n$. Thus the finite ideal $J^n$ from [L6] satisfies $J(J^n)=J^n$. By [L5], some $a\in J$ has $(1-a)J^n=0$; since $a\in J\subseteq {\mathfrak{m}}_A$, [L7] makes $1-a$ a unit, and hence $J^n=0$. Because $J$ is an ideal of definition, $\sqrt{J}={\mathfrak{m}}_A$; nilpotence of $J$ then makes every prime equal to ${\mathfrak{m}}_A$, so $\dim A=0$.

Assume $r>0$ and take a strict chain ${\mathfrak{p}}_0\subset {\mathfrak{p}}_1\subset \cdots \subset {\mathfrak{p}}_e={\mathfrak{m}}_A$. If $e=0$, there is nothing to prove, so assume $e>0$. Put $B=A\mathrm{/}{\mathfrak{p}}_0$. The quotient maps $A\mathrm{/}{J}^{n+1}A\twoheadrightarrow B\mathrm{/}{J}^{n+1}B$ give ${\chi }_{J,B}(n)\le {\chi }_{J,A}(n)$, hence $s:=\mathrm{deg}{P}_{J,B}\le r$. Choose $x\in {\mathfrak{p}}_1\setminus {\mathfrak{p}}_0$ and write $\bar{x}$ for its nonzero image in the domain $B$. Multiplication by $\bar{x}$ is injective, so $0\to B\stackrel{\bar{x}}{\to }B\to C:=B\mathrm{/}\bar{x}B\to 0$ is short exact; the image of $J$ is an ideal of definition in both $B$ and $C$. Applying [L2] at the maximum of $s$ and $\mathrm{deg}{P}_{J,C}$ cancels the two degree-$s$ contributions from $B$; if $\mathrm{deg}{P}_{J,C}\ge s$, it would force the nonzero leading coefficient of $P_{J,C}$ to vanish. Hence $\mathrm{deg}{P}_{J,C}<s\le r$. The images of ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_e$ give a strict prime chain of length $e-1$ in $C$. The induction hypothesis applied to $C$ gives $e-1\le \dim C\le \mathrm{deg}{P}_{J,C}\le r-1,$ and therefore $e\le r$. Since the chain was arbitrary, $\dim A\le r$.

Steps 1.2 and 1.4 give $r=d=\dim A$, and step 1.1 transfers this equality to $M$. Therefore $\mathrm{deg}{P}_{I,M}=\dim \mathrm{Supp}(M)$.