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Fix a finite-dimensional nonnegative filtration of a ring giving $R$ its almost commutative algebra structure, and choose a good filtration of a module on $M$. When $\mathrm{gr}R$ is a standard graded algebra, the Hilbert-Serre theorem makes the cumulative Hilbert function $h_M(j)={\dim }_k{F}_{j}M$ eventually a polynomial. For $M\ne 0$, define the dimension of a filtered module and multiplicity of a filtered module byUse $d(0)=-\infty$ and $m(0)=0$. With positive weighted generators instead of a standard grading, the cumulative Hilbert function is eventually a quasipolynomial with a common positive leading coefficient across residue classes. Equivalently, $h_M(j)=c{j}^d+O(j^{d-1})$, and the definitions are $d(M)=d$ and $m(M)=d!c$. The common leading coefficient follows from monotonicity: interlacing consecutive residue classes forces their leading coefficients to agree. For a standard grading, multiplicity is a positive integer, since an integer-valued polynomial has an integral expansion in the binomial polynomials; weighted normalization can instead give a positive rational number.

Two good filtrations of a module for this fixed algebra filtration bound each other after fixed index shifts. The corresponding Hilbert functions therefore satisfy $h_F(j-a)\le h_G(j)\le h_F(j+b)$ for fixed $a,b$. Comparing their leading growth shows that the degree and leading coefficient agree. Thus $d(M)$ and $m(M)$ are independent of the module filtration; the fixed algebra filtration is part of the multiplicity convention.

For a submodule $N$, use its subspace filtration and the quotient filtration on $M/N$. These are good because $\mathrm{gr}R$ is Noetherian. The short exact sequence of filtered pieces givesNonzero leading coefficients are positive, soIf $d(N)=d(M/N)$, those leading coefficients add andThis is dimension and multiplicity in a filtered exact sequence.

For the Weyl algebra $A_n(k)$ in characteristic zero, Bernstein inequality for Weyl algebra modules asserts $d(M)\ge n$ for every nonzero finitely generated module $M$. Write its generators as $x_i,D_i$, with $[D_i,x_j]={\delta }_{ij}$ and all other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ give the Bernstein filtrationTake a finite-dimensional generating subspace $V\ne 0$ of $M$ and put $M_j=A_{j}V$.

To prove faithful finite-step action of a Weyl algebra, consider $0\ne a\in A_j$ and choose a nonzero coefficient $c_{\alpha \beta }$ of a monomial of maximal total degree $d\le j$. Applying ${\prod }_i(\mathrm{ad}{D}_i{)}^{{\alpha }_i}(\mathrm{ad}{x}_i{)}^{{\beta }_i}$ to $a$ givesIndeed, these commutators differentiate the ordered coefficients: lower-degree monomials vanish, and another degree-$d$ monomial can survive only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. This is scalar extraction by iterated Weyl commutators.

Expanding these iterated commutators writes $\lambda ={\sum }_{\ell }{u}_{\ell }a{v}_{\ell }$ with $u_{\ell },v_{\ell }\in A_j$. If $a$ annihilated $M_j$, every summand would annihilate $V$, contrary to $\lambda V\ne 0$. Therefore the action map is injective:Taking dimensions and comparing polynomial degrees giveswhich proves Bernstein inequality for Weyl algebra modules. The polynomial representation $k[x_1,\dots ,x_n]$ has dimension $n$, so the bound is sharp.

The associated graded ring of the Weyl algebra is a polynomial ring, so ascending filtered-graded transfer of Noetherianity makes $A_n(k)$ a left Noetherian ring. Thus every submodule of a finitely generated module is finitely generated. If $d(M)=n$, every nonzero subquotient has dimension at most $n$ by the exact-sequence formula, and at least $n$ by Bernstein inequality for Weyl algebra modules. Every such subquotient therefore has dimension $n$ and positive integral multiplicity.

In a strictly descending chain $M=M_0⊋M_1⊋\cdots$, each nonzero quotient $M_i/M_{i+1}$ consumes at least one unit of multiplicity, by dimension and multiplicity in a filtered exact sequence. Hence there are at most $m(M)$ strict steps. This proves the descending chain condition, so $M$ is Artinian. In fact this holonomic Weyl algebra module has module length at most $m(M)$, as in multiplicity bounds the length of a holonomic Weyl module.

An almost commutative algebra over $\mathbb{C}$ has an exhaustive nonnegative increasing multiplicative ascending filtrationfor which the associated graded ringis a commutative finitely generated complex algebra. This includes the customary degree filtration with $F_{0}R=\mathbb{C}$, finitely many degree-one generators, and commutators lowering degree. The finite-generation clause is part of the convention used here and is necessary for the conclusion.

A finitely generated commutative complex algebra is a quotient of a polynomial ring $\mathbb{C}[X_1,\dots ,X_d]$. Repeated application of the central-variable case of the Hilbert basis theorem makes that polynomial ring Noetherian; quotients preserve the ascending chain condition by lifting ideal chains. Thus $\mathrm{gr}R$ is Noetherian.

We prove ascending filtered-graded transfer of Noetherianity directly. Give a left ideal $I\subseteq R$ the induced ascending filtration $I\cap F_{n}R$. Its associated graded module embeds into $\mathrm{gr}R$ as a homogeneous ideal, denoted $\mathrm{gr}I$. Choose finitely many homogeneous generators ${\overline{u}}_1,\dots ,{\overline{u}}_m$ of this ideal, of degrees $d_1,\dots ,d_m$, and lifts $u_j\in I\cap F_{d_j}R$. Homogeneous generators may be chosen by taking homogeneous components of any finite generating set.

For $u\in I\cap F_{n}R$, its degree-$n$ symbol can be expressed aswith terms of negative coefficient degree omitted. Lift each coefficient to $a_j\in F_{n-d_j}R$. ThenInduction on $n$, ending at $F_{-1}R=0$, expresses $u$ in the left ideal generated by the $u_j$. Hence every left ideal is finitely generated. Applying the same argument to right ideals, or to the opposite ring, provesThis degree-lowering proof terminates in finitely many steps and requires no completeness assumption on the filtered ring. Without finite generation of the graded algebra, a counterexample would be the commutative polynomial ring in infinitely many variables, whose ideals $(X_1)⊊(X_1,X_2)⊊\cdots$ do not stabilize.

For the universal enveloping algebra $U(\mathfrak{g})$, use the degree ascending filtration inherited from the tensor algebra of the finite-dimensional Lie algebra $\mathfrak{g}$. For $x,y\in \mathfrak{g}$, the defining relation is $xy-yx=[x,y]\in F_{1}U(\mathfrak{g})$, whereas the product $xy$ has filtration degree at most $2$. Thus the degree-one symbols commute in $\mathrm{gr}U(\mathfrak{g})$. Those symbols generate the graded algebra, so it is a quotient ofThis already proves the required almost-commutative property. The Poincaré-Birkhoff-Witt theorem gives the stronger identification $\mathrm{gr}U(\mathfrak{g})\cong \mathrm{Sym}(\mathfrak{g})$, but injectivity of this map is not needed for the chain-condition argument. Therefore

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 $R/(x,y{)}^{n+1}$.

A local ring is a nonzero ring with one maximal ideal $\mathfrak{m}$. Write $\kappa =R/\mathfrak{m}$ for its residue field. Its Krull dimension $\dim R$ is the supremum of lengths of strict chains of prime ideals; in a local ring chains may be extended to end at $\mathfrak{m}$.

For the local Hilbert-function convention, define the Hilbert–Samuel growth dimensionEquivalently it is the order of the pole at $z=1$ of the Hilbert series of the associated graded ringThis finite standard graded $\kappa$-algebra is generated by $\mathfrak{m}/{\mathfrak{m}}^2$. The Hilbert-Serre theorem makes the cumulative Hilbert function eventually polynomial, which establishes the definition. For an Artinian local ring, the polynomial is constant and nonzero, so $d(R)=0$.

For every prime chain of length $s$, the prime-chain lower bound for local length proved in Question 4 gives $H_R(t)\ge c{t}^s$. A polynomial of degree $d(R)$ cannot satisfy this when $s>d(R)$. Thus the requested inequality isFor clarity, the embedding dimension is the different invariantThe equality follows from the Nakayama lemma. Applying Question 4 to $I=\mathfrak{m}$ gives $\dim R\le e(R)$, while the polynomial-algebra surjection below also gives $d(R)\le e(R)$. In particular, growth dimension and embedding dimension should be kept distinct.

A regular local ring is a Noetherian local ring with $e(R)=\dim R$. Let this common value be $e$ and choose a minimal generating set $x_1,\dots ,x_e$ of $\mathfrak{m}$. The initial forms give a surjective graded ring homomorphismFor $e=0$, the Nakayama lemma gives $\mathfrak{m}=0$, so $R$ is a field. Assume $e>0$. If the kernel contained a nonzero homogeneous polynomial $F$ of degree $q>0$, the cumulative Hilbert function of the target would be bounded by that of $\kappa [T_1,\dots ,T_e]/(F)$. Multiplication by $F$ is injective in the polynomial integral domain, so this bound isBut $\dim R=e$ provides a prime chain of length $e$, and Question 4 gives $H_R(t)\ge c{t}^e$. This is a contradiction. Hence the associated graded ring of a regular local ring isan integral domain.

Finally the Krull intersection theorem gives ${\bigcap }_j{\mathfrak{m}}^j=0$. One can see the needed separatedness directly: for the finitely generated ideal $K={\bigcap }_j{\mathfrak{m}}^j$, the Artin-Rees lemma gives $K=\mathfrak{m}K$, and the Nakayama lemma gives $K=0$. Therefore each nonzero $a\in R$ has finite $\mathfrak{m}$-adic order $r$, with nonzero initial form in ${\mathfrak{m}}^r/{\mathfrak{m}}^{r+1}$. For nonzero $a,b$, their initial forms have nonzero product in the graded integral domain. It follows that $ab\ne 0$, and indeed its order is the sum of their orders. Thus $R$ is an integral domain.

For $a\in R$, write $m_a$ for multiplication by $a$ in ${\mathrm{End}}_k(R)$. The algebraic differential operator ring is defined by the filtrationThe order of an algebraic differential operator is the least $j$ for which $\theta \in D_k^j(R)$; the zero operator may be assigned order $-\infty$. In degree zero, commuting with all multiplication maps means $\theta (a)=a\theta (1)$, so $D_k^0(R)=\{m_a:a\in R\}\cong R$. The commutator identity $[UV,m_a]=U[V,m_a]+[U,m_a]V$ proves by induction that $D_k^r{D}_k^s\subseteq D_k^{r+s}$, so the union is a ring under addition and composition.

The derivations of an algebra formEvery such derivation has $\delta (1)=0$ and $[\delta ,m_a]=m_{\delta (a)}$, so has differential order at most one. Conversely, if $\theta \in D_k^1(R)$, set $c=\theta (1)$ and $\delta =\theta -m_c$. Then $\delta (1)=0$ and $[\delta ,m_a]$ is a multiplication map. Evaluating at $1$ shows it is $m_{\delta (a)}$; evaluating at $b$ gives the Leibniz rule. Hence the first-order algebraic differential operator decomposition isIt is a direct sum as $k$-vector spaces: a multiplication map that is a derivation vanishes at $1$, and is zero.

The complex Weyl algebra has generators $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ and relationsTake $n\ge 1$. If a nonzero unital module $M$ had finite dimension of a vector space over $\mathbb{C}$ $m$, its representing matrices would satisfy $[{\partial }_1,x_1]=I_m$. Taking the matrix trace gives $0=m$, impossible in characteristic zero. The Weyl algebra has no nonzero finite-dimensional modules. For $n=0$ the algebra is $\mathbb{C}$ and this assertion has the obvious exception.

For a nonzero finitely generated Weyl algebra module, use the Bernstein filtrationThe ordered monomials form a vector space basis, soTo justify the ordered basis, the relations move all ${\partial }_i$ to the right and give spanning. Independence follows in the action on $\mathbb{C}[x_1,\dots ,x_n]$: apply a putative zero operator ${\sum }_{\mathbf{b}}{p}_{\mathbf{b}}(x){\partial }^{\mathbf{b}}$ coefficientwise to the formal exponential $e^{t\cdot x}$, obtaining ${\sum }_{\mathbf{b}}{p}_{\mathbf{b}}(x)t^{\mathbf{b}}=0$. Thus every coefficient vanishes.

Choose a finite-dimensional nonzero generating subspace $M_0\subseteq M$ and set $M_j=F_j{A}_n{M}_0$. The associated graded module is finitely generated over the polynomial associated graded ring, so the Hilbert-Serre theorem implies ${\dim }_{\mathbb{C}}{M}_j$ is eventually a polynomial. DefineThis Bernstein growth dimension of a Weyl algebra module equals the Gelfand–Kirillov dimension of a module, not its ordinary vector-space dimension. If a different finite generating subspace is used, each lies in some fixed filtration step of the other. The corresponding $M_j$ are sandwiched between shifted filtrations, so the degree is unchanged.

We prove Bernstein inequality for Weyl algebra modules using an injectivity estimate. The faithful finite-step action of a Weyl algebra claim is that the mapis injective for every $j$. Induct on $j$. For $j=0$, a scalar annihilating $M_0\ne 0$ is zero. Suppose $a\in F_j{A}_n$ annihilates $M_j$. For any generator $z=x_i$ or ${\partial }_i$, the commutator $[a,z]$ belongs to $F_{j-1}{A}_n$ and annihilates $M_{j-1}$: both $a(zv)$ and $z(av)$ vanish. The induction hypothesis gives $[a,z]=0$ for every generator. In the ordered basis,In characteristic zero these identities force $a$ to be a scalar. Since it annihilates $M_0$, it is zero. This proves injectivity.

ConsequentlyThe left side grows as a positive constant times $j^{2n}$; the right side has growth degree $2d(M)$. Comparing powers provesThe polynomial module $\mathbb{C}[x_1,\dots ,x_n]$, with the usual multiplication and differentiation actions, has $d(M)=n$, showing the bound is sharp.

Choose a finite-dimensional generating vector subspace $V\subseteq A$ containing $1$, and write $V^n$ for the linear span of products of $n$ elements of $V$. The Gelfand–Kirillov dimension isFor a nonzero finitely generated module $M$ considered as a right module, choose a finite-dimensional generating vector subspace $W\subseteq M$ and define the Gelfand–Kirillov dimension of a module byFor left modules, replace $W{V}^n$ by $V^{n}W$. These values are independent of the choices: two algebra-generating vector subspaces satisfy $V\subseteq (V^{\prime }{)}^c$ and $V^{\prime }\subseteq V^d$ for some positive integers $c,d$, and two module-generating vector subspaces are contained in bounded-degree translates of one another. The corresponding growth bounds differ only by a constant rescaling and shift of $n$, which do not change the limit superior.

Let $S$ be a nonzero standard graded algebra, generated by $r$ elements of degree one. It is a homogeneous quotient ring of the polynomial ring $k[z_1,\dots ,z_r]$. By the Hilbert-Serre theorem, its Hilbert series is a rational function whose only possible pole is at $t=1$. After canceling, write it as $P(t)/(1-t{)}^d$ with $P(1)\ne 0$, where $0\le d\le r$. The cumulative dimensions are the coefficients of $P(t)/(1-t{)}^{d+1}$ and are eventually a polynomial in $n$ of degree $d$, with positive leading coefficient. With $V=S_0\oplus S_1$, these cumulative dimensions equal $\dim V^n$. ConsequentlyFor $d=0$ the algebra is finite-dimensional and nonzero, so the cumulative dimension is eventually a positive constant. If the zero algebra is allowed, the common convention $\mathrm{GKdim}0=-\infty$ is an exception to the printed assertion; the integer conclusion concerns nonzero unital algebras.

For the Weyl algebra $A=A_1(k)=k⟨X,D⟩/(DX-XD-1)$, the total-degree filtered algebra structure has associated graded ring $\mathrm{gr}A=k[x,\xi ]$. Its ordered monomials $X^i{D}^j$ form a basis. Every finitely generated module has a good filtration of a module, and its associated graded module is finitely generated over $k[x,\xi ]$. The Hilbert-Serre theorem therefore shows that its Gelfand–Kirillov dimension is an integer in $\{0,1,2\}$.

A nonzero module of Gelfand–Kirillov dimension zero here would have eventually constant cumulative dimension, hence be finite-dimensional. This is impossible in characteristic zero: taking the trace of the endomorphisms representing $DX-XD=1$ gives $0={\dim }_{k}M$. For right modules the right-action operators reverse composition and give the negative identity instead, with the same contradiction. ThusBoth occur. The regular module $A$ has $\dim V^n=\left(\genfrac{}{}{0ex}{}{n+2}{2}\right)$, hence Gelfand–Kirillov dimension two. For a right module of Gelfand–Kirillov dimension one, take $M=k[x]$ with $f\cdot X=xf$ and $f\cdot D=-f^{\prime }$. These actions satisfy the defining relation because $(f\cdot D)\cdot X-(f\cdot X)\cdot D=f$. The module is a cyclic module generated by $1$, and $\dim (1\cdot V^n)=n+1$. This also gives the lower bound for this particular Weyl algebra without invoking a general inequality for higher Weyl algebras.

Choose homogeneous algebra generators $x_1,\dots ,x_r$ of positive degrees $d_1,\dots ,d_r$ for $S$; they exist because $S$ is Noetherian. For a finitely generated graded module $V={⨁}_j{V}_j$, its Poincare series of a graded module isThe Hilbert-Serre theorem states thatfor a Laurent polynomial $Q(t)\in \mathbb{Z}[t,t^{-1}]$.

We prove this by induction on $r$. For $r=0$, $S=k$ and $V$ is finite-dimensional, so its series is a Laurent polynomial. For $x=x_r$ of degree $d=d_r$, multiplication gives the exact sequence of graded modulesAdditivity of the Hilbert series yieldsBoth modules on the right are finitely generated over $k[x_1,\dots ,x_{r-1}]$, so the induction hypothesis proves the formula.

For the finitely generated commutative algebra $R$, let $R_j$ be the stated degree filtration. Its associated graded ringis a standard graded algebra generated by the initial forms of $x_1,\dots ,x_n$. SinceHilbert–Serre shows that the quantity ${\dim }_k{R}_j$ agrees with a polynomial for all sufficiently large $j$. The growth of a finitely generated commutative algebra has polynomial degree $\dim R$, independently of the chosen finite generating set.

Forwith the four displayed generators, a basis of $R_j$ consists of the Laurent monomials $Y_1^a{Y}_2^b$ satisfying $|a|+|b|\le j$. There are $4r$ points with $|a|+|b|=r>0$, and one point at the origin. Hence the growth of the two-variable Laurent polynomial algebra is

Assume first that $f\ne 0$ and letbe the least total degree of a nonzero homogeneous part of $f$. The associated graded ring iswhere $f_d$ is the initial homogeneous form. Multiplication by the nonzero polynomial $f_d$ is injective in $k[x,y]$, so the degree-$j$ component has dimensionSumming the components of degrees below $n$ givesThus the Hilbert polynomial isIts leading coefficient is the order of vanishing, or multiplicity, of the plane curve $f=0$ at the origin. If $f=0$, no relation is imposed and $\chi =n(n+1)/2$ for every $n$.

For the maximal ideal $\mathfrak{m}$ of $(A,\mathfrak{m})$, the associated graded ring isIf $x+{\mathfrak{m}}^{r+1}$ and $y+{\mathfrak{m}}^{s+1}$ are homogeneous classes, their product isIt is a graded algebra over the residue field $k=A/\mathfrak{m}$.

The Hilbert series isThe Hilbert-Serre theorem makes this rational. The number $d(G_{\mathfrak{m}}(A))$ is the order of its pole at $t=1$, as recorded by the pole dimension of an associated graded ring.

The Dimension theorem for Noetherian local rings states

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 $s$.

For an ascending multiplicative filtration with $1\in F_{0}A$, its Rees ring contains the central element $h$. The quotient by $h$ is the associated graded ring, and the quotient by $h-1$ is the original algebra. If $[F_{i}A,F_{j}A]\subseteq F_{i+j-1}A$, commutators of lifts are divisible by $h$. Their first-order term defines the Poisson bracket from a central deformation parameter on the associated graded ring.