Let $S=k[x_1,\ldots,x_r]$ with positive degrees $d_i=\deg x_i$, and let $M$ be a finitely generated graded $S$-module whose graded pieces are finite-dimensional over $k$. The <Hilbert-Serre theorem> states that
$$
\boxed{H_M(t)=\sum_n\dim_k(M_n)t^n
=\frac{P(t)}{\prod_{i=1}^r(1-t^{d_i})}}
$$
for some Laurent polynomial $P(t)\in\mathbb Z[t,t^{-1}]$. For the standard grading, the <Hilbert function> $n\mapsto\dim_kM_n$ consequently agrees for all sufficiently large $n$ with a polynomial in $n$.

For the proof, induct on $r$. When $r=0$, $M$ is finite-dimensional and its <Hilbert series> is a Laurent polynomial. For $r>0$, multiplication by $x_r$ gives an exact sequence of graded modules
$$
0\longrightarrow K\longrightarrow M(-d_r)
\xrightarrow{x_r}M\longrightarrow C\longrightarrow0.
$$
Both $K$ and $C$ are annihilated by $x_r$, so they are finitely generated graded modules over $k[x_1,\ldots,x_{r-1}]$. Additivity of the Hilbert series yields
$$
(1-t^{d_r})H_M(t)=H_C(t)-H_K(t).
$$
The induction hypothesis supplies the required denominator for the right side and proves the rational formula. When all $d_i=1$, expanding $(1-t)^{-r}$ shows that its coefficients are binomial polynomials in $n$, which proves eventual polynomiality.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 5 / ii / Solution

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Let $S=k[x_1,\ldots,x_r]$ be graded with positive degrees $d_i=\deg x_i$, and let $M$ be a finitely generated graded $S$-module with finite-dimensional graded pieces. Then
$$
H_M(t)=\frac{P(t)}{\prod_{i=1}^r(1-t^{d_i})}
$$
for a Laurent polynomial $P(t)\in\mathbb Z[t,t^{-1}]$. For a standard grading, $\dim_kM_n$ consequently agrees with a polynomial for all sufficiently large $n$.


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 Incoming links: Hilbert-Serre theorem