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Use the following finite-length version of the Hilbert-Serre theorem. Let $S={⨁}_{n\ge 0}{S}_n$ be a graded algebra generated over an Artinian ring $A=S_0$ by finitely many homogeneous elements $x_1,\dots ,x_r$ of positive degrees $e_1,\dots ,e_r$. For a finitely generated nonnegatively graded $S$-module $M$, define its Poincare series of a graded module byEvery component has finite length, and the theorem saysFor 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 isHere $M(-e{)}_n=M_{n-e}$. Additivity of length degree by degree givesThis 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 $k[x_1,x_2,\dots ]$ at $(x_1,x_2,\dots )$, 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 ringis 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, writeDefine $d(R)=d$, the pole order at $t=1$, with $d=0$ when $P_G$ is a polynomial.

Equivalently, the Hilbert–Samuel functionhas generating series $P_G(t)/(1-t)$ and eventually agrees with a polynomial of degree $d$. Indeed, coefficients of $(1-t{)}^{-d-1}$ are $\left(\genfrac{}{}{0ex}{}{n+d}{d}\right)$, and multiplication by $q(t)$ leaves leading term $q(1)n^d/d!$. Consequently $d(R)$ 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 $x^i{y}^{n-i}$ as a basis. ThereforeThe length formula also follows directly by counting monomials of total degree at most $n$ in $R/(x,y{)}^{n+1}$.

True. This is the Artinian commutative ring is Noetherian theorem. One proof uses the nilpotent nilradical $N$ of an Artinian ring $A$. The quotient $A/N$ is a finite product of fields. Each quotient $N^j/N^{j+1}$ is an Artinian module over the semisimple ring $A/N$, hence has finite length and is Noetherian. The finite filtrationthen makes $A$ a Noetherian module over itself, which is exactly the ascending chain condition on its ideals.

Suppose $I$ has $s$ generators. Its associated graded ringis generated in degree one by their initial forms, so there is a graded surjectionBecause $I$ is $\mathfrak{m}$-primary, $R/I$ has finite length of a module. The degree-$n$ piece on the left has lengthso $\mathrm{length}(I^n/I^{n+1})$ grows with degree at most $s-1$. Summing these lengths shows that $\mathrm{length}(R/I^n)$ has polynomial degree at most $s$.