The module-relative Hilbert–Samuel polynomial exists without a dimension theorem

If $I=R$ or $M=0$, every quotient in the statement is zero, giving the polynomial zero. More generally put $N=M\mathrm{/}IM$. If $N=0$, then $M=IM$ implies $M=I^{a}M$ by repeated multiplication for every $a\ge 1$, so again all quotients are zero. Uniqueness in these cases follows from the polynomial argument at the end. Henceforth $I\mathrm{\ne }R$ and $c={\mathrm{\ell }}_R(N)>0$.

For a simple $R$-module $S$ and $0\mathrm{\ne }s\in S$, $Rs=S$; the map $R\to S$ taking $a$ to $as$ is onto and its kernel is maximal, since a nontrivial intermediate ideal would give a nontrivial submodule. Thus the kernel is $\mathfrak{m}$ and $S\cong k$. A composition series $0=N_0\subset \cdots \subset N_c=N$ therefore satisfies $\mathfrak{m}{N}_j\subseteq N_{j-1}$. Iteration gives ${\mathfrak{m}}^{c}N=0$.

Set $A=R\mathrm{/}{\mathfrak{m}}^c$, a Noetherian ring. Each ${\mathfrak{m}}^j\mathrm{/}{\mathfrak{m}}^{j+1}$ for $0\le j<c$ is generated by finitely many elements and killed by $\mathfrak{m}$, hence is a finite-dimensional $k$-space. Indeed a finite spanning list can be reduced, by deleting any member in the span of the others, to a basis; the flag of initial basis spans has simple factors $k$. Its $R$-length is its dimension. Applying length additivity to the finite $\mathfrak{m}$-adic filtration gives ${\mathrm{\ell }}_R(A)<\mathrm{\infty }$.

Choose generators $f_1,\dots ,f_s$ of $I$. For each $j\ge 0$, multiplication by the finitely many monomials $f^{\alpha }$ with $\mathrm{\mid }\alpha \mathrm{\mid }=j$ gives a surjection from copies of $N$ onto $G_j=I^{j}M\mathrm{/}{I}^{j+1}M$: coefficients modulo $IM$ suffice because $f^{\alpha }IM\subseteq I^{j+1}M$. In particular every $G_j$ has finite length and is killed by ${\mathfrak{m}}^c$. The actions $X_i\bar{m}=\overline{f_{i}m}$ commute, and define $G={⨁}_{j\ge 0}{G}_j$ as a finite graded module over $A[X_1,\dots ,X_s]$, generated by a finite generating list of $N$ in degree zero.

For any finite graded $A[X_1,\dots ,X_s]$-module $V$, replacing a finite generating list by all its homogeneous components gives homogeneous generators. Their degrees have a lower bound. Each $V_j$ is a quotient of finitely many copies of $A$, indexed by monomials of the required degree times these generators. Thus $H_V(t)={\sum }_j{\mathrm{\ell }}_R(V_j)t^j$ is a well-defined formal Laurent series. When $s=0$, only the finitely many generator degrees can occur, so $H_V$ is a Laurent polynomial.

Suppose $s>0$ and rationality with denominator $(1-t{)}^{s-1}$ has been proved for every finite graded module over $A[X_1,\dots ,X_{s-1}]$. Put $V(-1{)}_j=V_{j-1}$, $U=\ker (X_s:V(-1)\to V)$ and $W=V\mathrm{/}{X}_{s}V$. Iterating polynomial extension from the Noetherian ring $A$ makes $A[X_1,\dots ,X_s]$ Noetherian. Consequently $U$ is finite as a submodule of finite $V(-1)$; $W$ is finite as a quotient. Both are killed by $X_s$, so their same finite generators generate them over the ring with $s-1$ variables.

In degree $j$, the exact sequence $0\to U_j\to V_{j-1}\stackrel{X_s}{\to }{V}_j\to W_j\to 0$, split at its image, yields ${\mathrm{\ell }}_R(V_j)-{\mathrm{\ell }}_R(V_{j-1})={\mathrm{\ell }}_R(W_j)-{\mathrm{\ell }}_R(U_j)$. Hence $(1-t)H_V=H_W-H_U$. The induction hypothesis gives a Laurent polynomial numerator divided by $(1-t{)}^s$. Together with the base case this proves that form for all $s$.

The filtration of $M\mathrm{/}{I}^{n+1}M$ has successive factors $G_0,\dots ,G_n$, so its length is $h(n)={\sum }_{j=0}^n{\mathrm{\ell }}_R(G_j)$. Its generating series is $H_G(t)\mathrm{/}(1-t)=Q(t)\mathrm{/}(1-t{)}^{s+1}$ for a Laurent polynomial $Q(t)={\sum }_a{q}_a{t}^a$. The geometric series and repeated convolution give the coefficient $\left(\genfrac{}{}{0px}{}{n-a+s}{s}\right)$ of $t^n$ in $t^a\mathrm{/}(1-t{)}^{s+1}$ when $n\ge a$. One can count the convolution terms as nonnegative $(s+1)$-tuples with sum $n-a$, separated by $s$ dividers. Thus for $n$ beyond the finite set of exponents of $Q$, $h(n)={\sum }_a{q}_a\left(\genfrac{}{}{0px}{}{n-a+s}{s}\right)$, a rational polynomial in $n$. For $s=0$ the binomial is $1$, as required for $I=0$.

If two rational polynomials eventually equal $h(n)$, their difference vanishes at infinitely many distinct integers. Division by $T-a$ at a root lowers the degree by one, so a nonzero polynomial of degree $d$ has at most $d$ distinct roots. The difference is therefore zero. This also proves the uniqueness left open in the zero cases.