degree-r Hilbert–Samuel coefficient as Koszul Euler characteristic

If $M=0$, both sides are zero. If $I=R$, the polynomial is zero and the Koszul complex is acyclic. If $r=0$, then $I=0$ and $M$ has finite length; the complex is $M[0]$, its Euler characteristic is ${\mathrm{\ell }}_R(M)$ and $P$ is that constant, so $e_0={\mathrm{\ell }}_R(M)$. The remaining argument concerns $r\ge 1$ and a proper ideal.

Put $c={\mathrm{\ell }}_R(M\mathrm{/}IM)$. For $j\ge 0$, degree-$j$ monomials in the $r$ generators give a surjection $(M\mathrm{/}IM{)}^{a_j}\to I^{j}M\mathrm{/}{I}^{j+1}M$, where $a_j$ is the number of tuples $({\alpha }_1,\dots ,{\alpha }_r)$ of nonnegative integers summing to $j$. Encode a tuple by $j$ marks with $r-1$ separators, including adjacent separators for zero entries. This is a bijection with choices of separator positions, giving $a_j=\left(\genfrac{}{}{0px}{}{j+r-1}{r-1}\right)$. Consequently ${\mathrm{\ell }}_R(I^{j}M\mathrm{/}{I}^{j+1}M)\le c\left(\genfrac{}{}{0px}{}{j+r-1}{r-1}\right)$.

Choose $p>r$ deep enough for the shifted-adic comparison and for $P(p-i-1)={\mathrm{\ell }}_R(M\mathrm{/}{I}^{p-i}M)$ for every $0\le i\le r$. This is possible since only finitely many inequalities are required. The term in cochain degree $-i$ of $K\mathrm{/}{F}^{p}K$ is $(M\mathrm{/}{I}^{p-i}M{)}^{\left(\genfrac{}{}{0px}{}{r}{i}\right)}$. Homology comparison and term cancellation therefore give $\chi (K)={\sum }_{i=0}^r(-1{)}^i\left(\genfrac{}{}{0px}{}{r}{i}\right)P(p-i-1)\mathrm{.}$ The AC hypothesis supplies that in the finite-length and tail lemmas.

Define $\mathrm{\Delta }Q(t)=Q(t)-Q(t-1)$. For $k=0$ the identity ${\mathrm{\Delta }}^{k}Q(t)={\sum }_{i=0}^k(-1{)}^i\left(\genfrac{}{}{0px}{}{k}{i}\right)Q(t-i)$ is the single term $Q(t)$. If it holds at $k$, subtract its value at $t-1$ from its value at $t$. The coefficient of $Q(t-i)$ becomes $(-1{)}^i(\left(\genfrac{}{}{0px}{}{k}{i}\right)+\left(\genfrac{}{}{0px}{}{k}{i-1}\right))=(-1{)}^i\left(\genfrac{}{}{0px}{}{k+1}{i}\right)$, with out-of-range binomials zero; the two endpoint coefficients are $1$ and $(-1{)}^{k+1}$. This proves the identity for every $k$ by induction.

Summing along the $I$-adic filtration gives $0\le {\mathrm{\ell }}_R(M\mathrm{/}{I}^{n+1}M)\le c{\sum }_{j=0}^n\left(\genfrac{}{}{0px}{}{j+r-1}{r-1}\right)=c\left(\genfrac{}{}{0px}{}{n+r}{r}\right)$. For the last identity, tuples in $r$ variables of total at most $n$ correspond bijectively to tuples in $r+1$ variables of total exactly $n$, by adjoining the slack $n-j$; the same separator count applies. This holds for every $n\ge 0$.

The eventual polynomial is nonnegative at all sufficiently large integers. If it is nonzero, its leading coefficient is positive: division by its highest power of $n$ makes the lower terms tend to zero, so the sign is eventually the leading sign. If its degree exceeded $r$, that same division would make the upper bound from the preceding step tend to zero while the polynomial tends to a positive leading coefficient. This is impossible. Hence $P=0$ or $\mathrm{deg}(P)\le r$.

For $d\ge 1$, the binomial expansion gives $t^d-(t-1{)}^d=d{t}^{d-1}$ plus terms of degree at most $d-2$; a constant has difference zero. By linearity, $r$ differences annihilate every monomial of degree less than $r$ and take $t^r$ to $r!$. The degree bound thus gives ${\mathrm{\Delta }}^{r}P=r![T^r]P(T)$, including the zero polynomial. At $t=p-1$ the preceding finite-difference identity is precisely the Euler sum, so it equals $e_r(I,M)$. Together with the initial cases this proves the theorem.