= Solution
Use the following finite-length version of the <Hilbert-Serre theorem>. Let $S=\bigoplus_{n\geq0}S_n$ be a graded algebra generated over an <Artinian ring> $A=S_0$ by finitely many homogeneous elements $x_1,\ldots,x_r$ of positive degrees $e_1,\ldots,e_r$. For a finitely generated nonnegatively graded $S$-module $M$, define its <Poincare series of a graded module> by
$$
P_M(t)=\sum_{n\geq0}\ell_A(M_n)t^n.
$$
Every component has finite length, and the theorem says
$$
\boxed{P_M(t)=\frac{f(t)}{\prod_{i=1}^r(1-t^{e_i})},\qquad f(t)\in\mathbb Z[t].}
$$
For a <module> whose grading is merely bounded below, the same statement holds with a Laurent-polynomial numerator. When the generators all have degree one, the denominator is $(1-t)^r$.
Here is an induction proof. An <Artinian ring> is <Noetherian> by question 1, so the <Hilbert basis theorem> makes $S$ <Noetherian>. If there are no positive-degree generators, $S=A$ and a finite homogeneous generating set for $M$ occupies only finitely many degrees. Each component is a finite <module> over the <Artinian ring> $A$, hence has finite <length of a module>, and $P_M$ is a polynomial.
For $r>0$, put $x=x_r$, $e=e_r$, $N=\{m\in M:xm=0\}$ and $C=M/xM$. Both are finitely generated graded <modules> annihilated by $x$, hence <modules> over $S/(x)$, which is generated by the first $r-1$ homogeneous elements. Their multiplication exact sequence is
$$
0\longrightarrow N(-e)\longrightarrow M(-e)\mathrel{\mathop{\longrightarrow}^{x}}M\longrightarrow C\longrightarrow0.
$$
Here $M(-e)_n=M_{n-e}$. Additivity of length degree by degree gives
$$
\boxed{(1-t^e)P_M(t)=P_C(t)-t^eP_N(t).}
$$
This is the <Hilbert series multiplication exact sequence>. By induction, the right-hand side has denominator $\prod_{i<r}(1-t^{e_i})$. Division by $1-t^e$ completes the proof of the <Hilbert-Serre theorem>. The finite-component and finite-generation claims also follow from the finite set of positive-degree algebra and <module> generators; no analytic convergence of a series is involved.
For the local invariant, use the usual <Noetherian local ring> hypothesis of <Hilbert–Samuel growth dimension>. Locality alone does not guarantee finite lengths or polynomial growth; the printed question leaves this finiteness assumption implicit. For instance, in the <localization at a prime ideal> of the <polynomial ring> $k[x_1,x_2,\ldots]$ at $(x_1,x_2,\ldots)$, the <vector space> $\mathfrak m/\mathfrak m^2$ has the infinitely many independent classes of the variables, so its length is not finite. Write $\mathfrak m=P$ and $\kappa=R/\mathfrak m$. Its <associated graded ring>
$$
G=\operatorname{gr}_{\mathfrak m}R=\bigoplus_{n\geq0}\mathfrak m^n/\mathfrak m^{n+1}
$$
is generated over $\kappa$ in degree one, because $\mathfrak m$ is finitely generated. Thus its <Hilbert series> is rational with a denominator that is a power of $1-t$. After cancelling factors, write
$$
P_G(t)=\frac{q(t)}{(1-t)^d},\qquad q(1)\neq0.
$$
\b[Define $d(R)=d$, the pole order at $t=1$], with $d=0$ when $P_G$ is a polynomial.
Equivalently, the <Hilbert–Samuel function>
$$
H_R(n)=\ell_R(R/\mathfrak m^{n+1})=\sum_{i=0}^n\dim_\kappa(\mathfrak m^i/\mathfrak m^{i+1})
$$
has generating series $P_G(t)/(1-t)$ and eventually agrees with a polynomial of degree $d$. Indeed, coefficients of $(1-t)^{-d-1}$ are $\binom{n+d}{d}$, and multiplication by $q(t)$ leaves leading term $q(1)n^d/d!$. Consequently \b[$d(R)$ is the degree of the cumulative <Hilbert-Samuel polynomial>], rather than the degree of the individual graded-component function; the latter has degree $d-1$ when $d>0$. This pole/growth invariant also equals <Krull dimension> by the local dimension theorem, although that theorem is not required to define it here.
For an example, take $R=k[x,y]_{(x,y)}$. Its <associated graded ring> is $k[x,y]$ with the ordinary degree grading: degree $n$ has the $n+1$ monomials $x^iy^{n-i}$ as a basis. Therefore
$$
P_G(t)=\frac1{(1-t)^2},\qquad H_R(n)=\binom{n+2}{2},\qquad\boxed{d\bigl(k[x,y]_{(x,y)}\bigr)=2.}
$$
The length formula also follows directly by counting monomials of total degree at most $n$ in $R/(x,y)^{n+1}$.
Back to article page
