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For an ideal $K\subseteq R[t]$, let $A_n(K)$ consist of the coefficients at $t^n$ of elements of $K$ of polynomial degree at most $n$, including the zero polynomial. Each $A_n(K)$ is an ideal of $R$, and multiplication of representing polynomials by $t$ shows that $A_n(K)\subseteq A_{n+1}(K)$.

If $R$ is a Noetherian ring, this chain stabilizes at some $N$. Choose finite generating sets of an ideal for $A_0(K),\dots ,A_N(K)$, and for each nonzero generator choose a representing polynomial $f_{nj}\in K$ with that leading coefficient and polynomial degree $n$. These finitely many polynomials generate $K$: for any nonzero $f\in K$ of polynomial degree $d$, use the chosen generators at level $\min (d,N)$, multiplied by suitable powers of $t$, to cancel its leading coefficient. Induction on polynomial degree finishes the argument. This proves the Hilbert basis theorem without assuming that $R$ is an integral domain.

A presentation of an algebra over a field with finitely many generators and relations. The Hilbert basis theorem ensures that every finitely generated algebra over a field admits such a presentation. Its module of Kähler differentials is presented by the Jacobian matrix of the relations.

Suppose a compact parameter set carries a locally finite countable multiset of positive quantities, and a fixed injective transform of each labelled quantity is a polynomial in finitely many ambient coordinates. Assume only finitely many labels can have bounded value anywhere in the compact set. Then a uniform finite initial multiset determines the entire multiset. For each source label there are finitely many possible matching labels. Equality for finitely many labels in both directions is a finite union of algebraic sets, obtained from injective matchings and polynomial equalities. These algebraic sets decrease as more labels are included, so they stabilize by the Hilbert basis theorem. A compatible infinite matching in each direction follows by the König infinity lemma, and local finiteness makes multiplicities equal. Squared traces of hyperbolic holonomy representations give the application to the Buser finite length spectrum theorem.

A Laurent polynomial ring is the localization of a ring $k[X_1,\dots ,X_r]$ obtained by inverting all variables. Its elements are finite sums of Laurent polynomials with exponent vectors in ${\mathbb{Z}}^r$. It is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring.

A commutative ring $R$ is a Noetherian ring if every ideal has a finite generating set of an ideal. Equivalently, it satisfies the ascending chain condition on ideals. Indeed, the union of an ascending chain is an ideal, and finitely many generators all belong to one term, after which the chain stabilizes. Conversely, an ideal without a finite generating set of an ideal would allow us to choose successively $a_{n+1}\notin (a_1,\dots ,a_n)$, producing a strictly ascending chain of ideals.

Let $\varphi :R\to S$ be a surjective ring homomorphism, where $R$ is a Noetherian ring. For any ideal $J$ of $S$, its inverse image is an ideal of $R$, say ${\varphi }^{-1}(J)=(a_1,\dots ,a_r)$. Taking images givesThus homomorphic images of Noetherian rings are Noetherian. This also applies to the image of an arbitrary ring homomorphism, by taking its image as the codomain.

The Hilbert basis theorem states that if $R$ is a Noetherian ring, then the polynomial ring $R[t]$ is a Noetherian ring. Repeating this gives the result for finitely many variables.

To prove it, fix an ideal $K\subseteq R[t]$. For $n\ge 0$, define the coefficient ideals of a polynomial idealHere the zero polynomial is allowed. Each $A_n$ is an ideal: addition and multiplication by constants preserve the required degree bound. Multiplying a representing polynomial by $t$ gives $A_n\subseteq A_{n+1}$. The ascending chain condition therefore supplies $N$ such that $A_n=A_N$ for every $n\ge N$.

For each $0\le n\le N$, choose a finite generating set of an ideal $a_{n1},\dots ,a_{n{r}_n}$ for $A_n$, omitting zero generators, and choose $f_{nj}\in K$ of polynomial degree $n$ with leading coefficient $a_{nj}$. If $A_n=0$, choose no such polynomials. We claim that the finitely many $f_{nj}$ generate $K$.

Take a nonzero $f\in K$ of polynomial degree $d$. Put $n=\min (d,N)$. Its leading coefficient belongs to $A_d=A_n$, so write it as ${\sum }_j{b}_j{a}_{nj}$. The polynomialbelongs to $K$ and is either zero or has polynomial degree strictly below $d$. Induction on polynomial degree now expresses $f$ as a combination of the chosen $f_{nj}$; when $d=0$, the displayed difference is zero. Hence $K$ is finitely generated. Since $K$ was arbitrary, $R[t]$ is a Noetherian ring.

Put $A=\mathbb{C}[X,Y]$, $P=(X)$ and $Q=(X^2,Y)$. The quotient ring $A/P\cong \mathbb{C}[Y]$ is an integral domain, so $P$ is a prime ideal and hence a primary ideal. Put $\mathfrak{m}=(X,Y)$. ThenThe polynomial ring $A$ is a Noetherian ring by the Hilbert basis theorem, and $\mathfrak{m}$ is a maximal ideal since $A/\mathfrak{m}\cong \mathbb{C}$. Therefore the maximal-primary ideal criterion from c makes $Q$ $\mathfrak{m}$-primary. In particular its radical of an ideal is $\mathfrak{m}$, distinct from $P$.

Neither primary ideal can be omitted: $X\in P\setminus Q$, while $Y\in Q\setminus P$. Thus the given intersection is already a minimal primary decomposition. To express the ideal by generators, write an element of $P\cap Q$ as $Xh=X^{2}u+Yv$. Reducing modulo the prime ideal $(X)$ gives $Y\bar{v}=0$ in the integral domain $\mathbb{C}[Y]$, so $v=Xw$. Cancelling $X$ in the integral domain $A$ gives $h=Xu+Yw$. HenceBy the first uniqueness theorem for primary decomposition, the associated primes of a module $A/J$ are exactly the two radicals of ideals. Their strict inclusion $P⊊\mathfrak{m}$ givesOne can also see both associated primes of a module directly through annihilators:For the first colon ideal, reducing $hY\in J$ modulo $(X)$ forces $h\in (X)$, and the reverse inclusion follows from $XY\in J$. For the second colon ideal, cancel $X$ in $hX\in X(X,Y)$. Both represented elements of the quotient module $A/J$ are nonzero.

Put $x=I+E_{12}$, $y=I+E_{23}$ and $z=I+E_{13}$. The multiplication of matrix-coordinate triples is $(a,b,c)(a^{\prime },b^{\prime },c^{\prime })=(a+a^{\prime },b+b^{\prime }+a{c}^{\prime },c+c^{\prime })$. In particular $z$ is central, $xy=zyx$, and every element has the unique normal form $y^c{z}^b{x}^a$. ConsequentlyThe distinct normal-form monomials are the group basis, so this is an isomorphism, not merely a surjection from a presented algebra. This is the integral Heisenberg group algebra as a skew Laurent ring.

Here is the needed skew Hilbert basis theorem, including its proof. If $A$ is a left Noetherian ring and $\sigma$ an automorphism, let $J$ be a left ideal of $A[t;\sigma ]$, where $ta=\sigma (a)t$. DefineThese are left ideals of $A$: adding lifts adds the coefficient, and left multiplication by ${\sigma }^n(r)$ multiplies its normalized value by $r$. Multiplication of a lift by $t$ proves $I_n\subseteq I_{n+1}$. The chain stabilizes at some $N$. Choose finite generators $u_{nj}$ for each $I_n$, $0\le n\le N$, and lifts $f_{nj}\in J$ with $t^n$ coefficient ${\sigma }^n(u_{nj})$.

For $f\in J$ of degree $m\ge N$, write its normalized leading coefficient as ${\sigma }^{-m}(\mathrm{lc}f)={\sum }_j{r}_j{u}_{Nj}$. Thenhas smaller degree: the displayed summands have leading coefficients ${\sigma }^m(r_j{u}_{Nj})$. For $m<N$ use the generators for $I_m$ without the shift. Induction on degree expresses every element of $J$ in terms of the finitely many chosen lifts. Thus $A[t;\sigma ]$ is a left Noetherian ring. Passing to the opposite ring replaces $\sigma$ by ${\sigma }^{-1}$ and proves the corresponding right-sided result.

This proof includes the ordinary Hilbert basis theorem when $\sigma$ is the identity. Starting with the field $k$, it makes $k[y,z]$ Noetherian; localization at powers of $y,z$ makes $B$ Noetherian as well. Finally, for a left ideal $L$ in $B[x,x^{-1};\sigma ]$, its intersection $J$ with $B[x;\sigma ]$ has finitely many generators. For each $f\in L$, some $x^{q}f$ lies in $J$, so $f=x^{-q}(x^{q}f)$ is generated by those same elements in the skew Laurent polynomial ring. Right multiplication clears powers for a right ideal. Therefore $kG$ is a left Noetherian ring and a right Noetherian ring.

Whenever the graded pieces $N_i$ are finitely generated over $S_0$, define the Poincare series of a graded module byAdditivity means $\lambda (B)=\lambda (A)+\lambda (C)$ for every short exact sequence $0\to A\to B\to C\to 0$ in the indicated category. Dimension over a field is a basic example; length is another when the modules have finite length.

The Hilbert-Serre theorem needs finite-generation hypotheses: take $S_0$ commutative Noetherian and $S$ generated over it by homogeneous $x_1,\dots ,x_r$ of positive degrees $d_1,\dots ,d_r$. For finitely generated graded $N$, every $N_i$ is then finite over $S_0$, andIf negative grading indices are allowed for the module, the numerator is a Laurent polynomial instead. In standard degree one, the coefficients are eventually polynomial functions of the index.

To prove this, regard $N$ as finite over the surjective polynomial presentation $S_0[X_1,\dots ,X_r]\to S$. Induct on $r$. At $r=0$ only finitely many homogeneous generator degrees occur, so the series is a polynomial. For the last variable $x$ of degree $d$, put $K=(0{:}_{N}x)$ and $Q=N/xN$. The polynomial ring is Noetherian by the Hilbert basis theorem, hence $K$ is finite; both $K$ and $Q$ are finite modules over the ring with that variable removed. The degree-$i$ exact sequence isAdditivity, followed by summing over $i$, givesThe induction hypothesis supplies the remaining denominator factors. This proves Hilbert-Serre theorem for an additive coefficient function. If all $d_i=1$, expanding $(1-t{)}^{-r}$ proves eventual polynomial coefficients directly.

An arbitrary positively graded ring in the introductory wording is not enough by itself. For example $S=k[X_1,X_2,\dots ]$ with every variable of degree one and $N=S$ is a cyclic graded module, but $N_1$ is not finite-dimensional over $k$, so the dimension-valued series is not even defined. The theorem above states the standard hypotheses that make the requested definition and rationality valid.

Let $I=(a_1,\dots ,a_s)$ and form the Rees algebra $\mathcal{R}={⨁}_{j\ge 0}{I}^j{t}^j$. It is generated over $R$ by $a_{1}t,\dots ,a_{s}t$, so the Hilbert basis theorem makes it Noetherian. The Rees module $\mathcal{M}={⨁}_{j\ge 0}{I}^{j}M{t}^j$ is finite over it, generated by finitely many generators of the module $M$ in degree zero. Hence its graded submodulehas finitely many homogeneous generators, all in degrees at most some $k$.

For $i\ge k$, each degree-$d$ generator contributes a multiple from $I^{i-d}(N\cap I^{d}M)$. This is contained in $I^{i-k}(N\cap I^{k}M)$, because multiplication by $I^{k-d}$ puts its coefficient both in $N$ and in $I^{k}M$. Thus $N\cap I^{i}M\subseteq I^{i-k}(N\cap I^{k}M)$. The reverse inclusion follows directly by multiplication: it stays in $N$ and in $I^{i}M$. ConsequentlyThis is the Artin-Rees lemma, proved through finite graded generation rather than an assumption that the intersection filtration initially has that form.

A precise version of the Hilbert-Serre theorem is as follows. Let $R={⨁}_{j\ge 0}{R}_j$ be a commutative graded ring, generated over an Artinian degree-zero ring $R_0$ by homogeneous elements $y_1,\dots ,y_s$ of positive degrees $d_1,\dots ,d_s$. Let $M$ be a finitely generated graded module, bounded below. Each component has finite module length over $R_0$. Its Poincare series of a graded module, or Hilbert series, isWhen $R_0$ is a field, these lengths are dimensions. The theorem assertsFor a nonnegatively graded module the numerator can be taken in $\mathbb{Z}[t]$. The finite-generation and finite-length hypotheses specify the meaning of the series; they are part of the theorem's statement.

We prove it by induction on $s$. Regard $M$ as a finite graded module over $S=R_0[Y_1,\dots ,Y_s]$, with $\mathrm{deg}{Y}_i=d_i$, acting through the specified generators of $R$. This polynomial ring is Noetherian by the Hilbert basis theorem and Question 1. When $s=0$, finitely many homogeneous generators over $R_0$ mean that only finitely many components are nonzero, so the series is a Laurent polynomial. For the induction step, put $y=Y_s$, $d=d_s$, $K=(0{:}_{M}y)$ and $C=M/yM$. Both $K$ and $C$ are finite graded modules; $y$ annihilates them, so they are modules over $R_0[Y_1,\dots ,Y_{s-1}]$. The finite generation of $K$ uses Noetherianity of $S$.

The exact sequence of graded modulesgives, by additivity of component lengths,Here the shift is defined so that $P_{M(-d)}=t^d{P}_M$. By induction both series on the right have denominator ${\prod }_{i<s}(1-t^{d_i})$. Dividing by $1-t^d$ proves the result. The annihilator term cannot be omitted unless multiplication by $y$ is injective.

In the standard grading $d_i=1$, the coefficients of $(1-t{)}^{-s}$ are $\left(\genfrac{}{}{0ex}{}{j+s-1}{s-1}\right)$. Multiplying by the finite numerator shows that the component dimensions agree with a polynomial for sufficiently large $j$. Cumulative dimensions have generating function $P_M(t)/(1-t)$ and are also eventually polynomial. This consequence supplies the dimension used for Weyl algebra modules.

Use the Bernstein filtration $B_m{A}_n$, in which both coordinate generators and differentiation generators have degree one. The ordered monomial basis of a Weyl algebra givesChoose a finite-dimensional generating subspace $M_0$ of a nonzero finitely generated left module $M$ and set $M_m=B_m{A}_n{M}_0$. This is a good filtration of a module: the symbols of the generators generate its associated graded module over the displayed polynomial ring. The Hilbert-Serre theorem makes ${\dim }_{\mathbb{C}}{M}_m$ an eventual polynomial. Define the Bernstein growth dimension of a Weyl algebra module byThe dimension is independent of the chosen finite generating subspace. Each of two finite generating sets is contained in a fixed filtration step for the other, giving $M_m\subseteq M_{m+c}^{\prime }$ and $M_m^{\prime }\subseteq M_{m+c^{\prime }}$. Such fixed shifts preserve the degree of growth. The same argument applies to any good Bernstein filtration. This dimension agrees with the Gelfand–Kirillov dimension of a module; one may set $d(0)=-\infty$.

Let the nonconstant element $D$ have Bernstein degree $d\ge 1$. Since ${\mathrm{gr}}_B{A}_n$ is a domain, the product of two nonzero symbols cannot vanish. HenceRight multiplication by $D$ is injective, using the domain property of $A_n$. Thus the quotient filtration on $M=A_n/A_{n}D$ has cumulative dimensionsThe leading terms of degree $2n$ cancel, and the leading surviving term isIt is nonzero, because $d\ge 1$. ThereforeEquivalently the Hilbert series of the associated graded quotient is $(1-t^d)/(1-t{)}^{2n}$. Here $d$ is Bernstein degree, not differential order: multiplication by $x_1$ has differential order zero but Bernstein degree one. Using the order filtration in this dimension argument would lose the finite-dimensional filtration pieces.

The Hilbert basis theorem says that if $B$ is a commutative Noetherian ring, then $B[t]$ is Noetherian; iteration treats any finite number of variables. Here is a complete leading-coefficient proof. For an ideal $I\subseteq B[t]$, let $L_d$ be the set of degree-$d$ coefficients of its polynomials of degree at most $d$, including zero coefficients. These are ideals of $B$, and multiplication by $t$ gives $L_d\subseteq L_{d+1}$. The ascending chain condition makes the chain stationary from some $N$ onward. Choose finite generators for $L_0,\dots ,L_N$ and lift them to polynomials $f_{dj}\in I$ of degree at most $d$. If $f\in I$ has degree $m$, set $d=\min (m,N)$ and express its leading coefficient as ${\sum }_j{a}_{dj}{b}_j$, where $a_{dj}$ is the chosen degree-$d$ coefficient. Then $f-{\sum }_j{b}_j{t}^{m-d}{f}_{dj}$ has smaller degree. Induction proves that the finitely many lifts generate $I$. This proves the theorem, including coefficient rings with zero divisors.

Put $A=\mathbb{C}G$ and $L=\mathbb{C}[X,X^{-1}]$. The infinite dihedral group has normal forms $X^m$ and $X^{m}Y$. The relations move every $Y$ to the right and reduce its exponent modulo two; the action $X:n\mapsto n+1$, $Y:n\mapsto -n$ on $\mathbb{Z}$ shows that these forms are distinct. ThusThe Hilbert basis theorem makes $\mathbb{C}[X]$ Noetherian, and its localization $L$ is Noetherian: an ideal of a localization is generated by the images of the finite generators of its contraction. Since $A$ is free of rank two over $L$ on the left, it is a Noetherian module over $L$. Every left ideal of $A$ is an $L$-submodule, so all its ascending chains stabilize. Therefore $\mathbb{C}G$ is left Noetherian.

For the center of the infinite dihedral group algebra, write an element as $f(X)+g(X)Y$. Commuting it with $X$ gives $g(X)(X^{-1}-X)=0$ in the Laurent integral domain, forcing $g=0$. Commuting $f(X)$ with $Y$ then gives $f(X)=f(X^{-1})$. A symmetric Laurent polynomial is a linear combination of $1$ and $X^j+X^{-j}$, $j\ge 1$. With $z=X+X^{-1}$, the recursionexpresses $S_j=X^j+X^{-j}$ as a polynomial in $z$. Conversely every polynomial in $z$ is central. The highest Laurent exponent of a nonzero polynomial in $z$ shows that $z$ is transcendental. HenceThe identities $X^{-1}=z-X$ and $X^2-zX+1=0$ give $L=R+RX$. Independence follows by applying $X\mapsto X^{-1}$ to a relation $f(z)+Xg(z)=0$ and subtracting: $(X-X^{-1})g(z)=0$, so $g=f=0$. Consequentlyis a free $R$-module of rank four.

Let $M$ be a nonzero simple module on the left. For any $m\ne 0$, $M=Am$, so it is finite over $R$. Every central $r\notin P={\mathrm{Ann}}_R(M)$ acts bijectively: its kernel and image are $A$-submodules, and simplicity forces kernel zero and image $M$. To prove that $P$ is maximal, not merely prime, put $B=R/P$; then $M$ is a faithful finite $B$-module. For $0\ne b\in B$, $bM=M$. If $m_1,\dots ,m_d$ generate $M$, write $m_i=b{\sum }_j{c}_{ij}{m}_j$. The adjugate matrix identity for $I-bC$ shows that $\det (I-bC)$ annihilates every generator and hence $M$. Faithfulness makes this determinant zero in $B$. Its constant term is one, so it has form $1-bq$, proving $bq=1$. Thus $B$ is a field, and $P$ is a maximal ideal of $R$. This proves the general principle for simple modules over an algebra finite over its center in this instance.

Every maximal ideal of $\mathbb{C}[z]$ is $(z-\lambda )$ for some $\lambda \in \mathbb{C}$, by factorization of complex polynomials. The module $M=Am$ is therefore a quotient of the four-dimensional algebra $A/(z-\lambda )A$. In particular ${\dim }_{\mathbb{C}}M\le 4<\infty$, as required.

The actual simple dimensions are at most two. If $\lambda \ne \pm 2$, choose $u\ne \pm 1$ with $u+u^{-1}=\lambda$. The matrices $X=\mathrm{diag}(u,u^{-1})$ and $Y=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ give a representation of $A/(z-\lambda )A$ onto $M_2(\mathbb{C})$: polynomial functions of $X$ give the two diagonal matrix units, and multiplication by $Y$ gives the off-diagonal ones. Both algebras have dimension four, so this is an isomorphism. If $\lambda =\pm 2$, the element $X\mp 1$ has square zero and is conjugated to its negative by $Y$; its generated two-sided ideal is square zero. It annihilates every simple module, since its action on a simple module is either zero or the whole module, and the latter contradicts its square being zero. The remaining quotient is $\mathbb{C}[C_2]$, whose two simple modules have $Y=\pm 1$. Thus the generic simple modules have dimension two, and the four exceptional one-dimensional characters have $X=\pm 1$, $Y=\pm 1$.

Use generators $x,D$ with $Dx-xD=1$. The ordered monomials $x^i{D}^j$ form a basis: construct the Weyl algebra as the differential Ore extension of $\mathbb{C}[x]$ with $Df=fD+f^{\prime }$, which has unique ordered polynomial expressions. Giving both generators degree one produces the Bernstein filtration, whose associated graded ring is $\mathbb{C}[x,\xi ]$. If $a,b\ne 0$, their nonzero leading symbols have nonzero product in this polynomial integral domain. Thus $ab$ has nonzero leading symbol and cannot vanish. Hence $A_1(\mathbb{C})$ has no nontrivial zero divisors.

Here is the left-handed Ore theorem. Let $S$ be a multiplicative set of regular elements of a ring, containing $1$ and excluding $0$. There is an injective Ore localization whose elements are left fractions $s^{-1}a$ if and only ifThis is the left Ore condition. For a noncommutative domain, taking $S=A\setminus \{0\}$ gives a division ring exactly when this condition holds. With zero-divisor denominators a further left reversibility condition is needed; the regular-denominator statement is the one used here.

Necessity follows by writing $a{s}^{-1}=t^{-1}b$ in the quotient and multiplying by $t$ and $s$, giving $ta=bs$ in the embedded original ring. For a construction proving sufficiency, use the left Ore fraction construction: pairs $(s,a)$ represent $s^{-1}a$, and $(s,a)$ and $(t,b)$ are equivalent whenfor suitable $u,v\in A$. Reflexivity and symmetry are immediate. For transitivity, combine $us=vt=w$ and $u^{\prime }t=v^{\prime }q=w^{\prime }$ by a common left multiple $hw=k{w}^{\prime }\in S$. Then $(hv-k{u}^{\prime })t=0$, so regularity gives $hv=k{u}^{\prime }$, which combines the numerator equalities and proves transitivity.

Addition uses the common denominator $w$, with numerator $ua+vb$. To multiply $s^{-1}a$ by $t^{-1}b$, choose $v\in S,c\in A$ with $va=ct$; setThe Ore condition supplies a common refinement whenever a representative or an Ore relation is changed, and regularity cancels the shared denominator, proving independence of choices. The same refinements verify the ring axioms; the formulas are simply addition and multiplication after clearing left denominators. The map $a\mapsto (1,a)$ is injective, since equivalence to zero would give $ua=0$ with $u\in S$. Each $s\in S$ becomes invertible. A homomorphism inverting $S$ extends uniquely by $(s,a)\mapsto f(s{)}^{-1}f(a)$, giving the universal property. For all nonzero denominators in a noncommutative domain, a nonzero fraction has inverse $(s^{-1}a{)}^{-1}=a^{-1}s$, so the quotient is a division ring. This is the requested proof sketch, with the fraction equivalence and multiplication made explicit.

It remains to check the left Ore condition for $A_1(\mathbb{C})$. Its graded polynomial ring is Noetherian by the Hilbert basis theorem. Lifting homogeneous generators of a graded left ideal and successively cancelling leading symbols proves that the filtered algebra is a left Noetherian ring: each subtraction lowers the nonnegative degree, so it terminates.

In any left Noetherian noncommutative domain, nonzero principal left ideals intersect. Otherwise $Ab\cap Aa=0$ with $a,b\ne 0$ would makean infinite direct sum of nonzero left ideals. To verify directness, reduce a finite relation modulo $Aa$ to make its first summand zero, then cancel $a$ on the right and repeat. Its increasing finite partial sums contradict the ascending chain condition. A nonzero element of $Aa\cap As$ therefore has the form $ta=bs\ne 0$, with $t\ne 0$, proving the required condition. Consequently $A_1(\mathbb{C})$ has a classical ring of left quotients, and it is a division ring.

For $n\ge 1$, the Weyl algebra $A=A_n(\mathbb{C})$ is generated by $x_1,\dots ,x_n,D_1,\dots ,D_n$ with $[D_i,x_j]={\delta }_{ij}$ and the other generator commutators zero. Iterating the differential Ore extension construction gives the ordered basis $x^{\alpha }{D}^{\beta }$. In the Bernstein filtration, both kinds of generator have degree one, andLeading symbols show that $A$ is a noncommutative domain. The Hilbert basis theorem and homogeneous-generator lifting by decreasing degree make it both left and right Noetherian.

It is also a simple ring. In a nonzero two-sided ideal, choose a nonzero element and a monomial $x^{{\alpha }_0}{D}^{{\beta }_0}$ of maximal total degree with nonzero coefficient $c$. Applying $(\mathrm{ad}D{)}^{{\alpha }_0}(\mathrm{ad}x{)}^{{\beta }_0}$ eliminates every lower-degree term and every different term of that maximal degree. The result is the nonzero scalar $(-1{)}^{|{\beta }_0|}{\alpha }_0!{\beta }_0!c$. Thus the ideal contains $1$. Here the scalar extraction by iterated Weyl commutators uses characteristic zero. Despite simplicity, $A$ has no nonzero finite-dimensional modules: in such a representation $[D_1,x_1]=I$ would have trace $0=\dim M$, impossible over $\mathbb{C}$. Its interesting module theory is necessarily infinite-dimensional.

For a finitely generated left module $M$, choose a finite-dimensional generating subspace $M_0$ and set $M_j=F_{j}A{M}_0$. This is a good filtration of a module. Its associated graded module is finite over the graded polynomial ring, so its cumulative Hilbert function is eventually a polynomial,with positive integer multiplicity $e(M)$. These are the Bernstein growth dimension of a Weyl algebra module and its multiplicity of a filtered module. Two good filtrations bound each other after fixed shifts: lift finite homogeneous generators in one filtration, bound their degrees in the other, and use compatibility with $F_{j}A$; then reverse the roles. The growth degree and leading coefficient therefore do not depend on the good filtration of a module. The degree is also the Gelfand–Kirillov dimension of a module and the dimension of the characteristic variety of a Weyl algebra module, the support of $\mathrm{gr}M$. Since $M$ is a quotient of a finite sum of copies of $A$, $d(M)\le 2n$.

The essential lower bound is the Bernstein inequality for Weyl algebra modules:Here is a full proof. Multiplication gives a linear map $F_{j}A\to {\mathrm{Hom}}_{\mathbb{C}}(M_j,M_{2j})$. Prove it injective by induction on $j$. At $j=0$ a scalar killing $M_0\ne 0$ must vanish. If $a\in F_{j}A$ kills $M_j$, then for every generator $z$, its commutator $[a,z]\in F_{j-1}A$ kills $M_{j-1}$: both $azm$ and $zam$ are zero for $m\in M_{j-1}$. Induction gives $[a,z]=0$ for all generators. In the ordered basis, $[D_i,a]$ differentiates the $x_i$ exponents and $[x_i,a]$ differentiates the $D_i$ exponents with a minus sign. Vanishing of all these commutators in characteristic zero forces $a$ to be a scalar. It then vanishes because it kills $M_0$. This proves the faithful finite-step action of a Weyl algebra and henceComparison of degrees gives $2n\le 2d(M)$, proving the inequality. Equally, without invoking eventual polynomiality, this estimate directly bounds the Gelfand–Kirillov dimension of a module below by $n$. Right modules satisfy the same inequality through the filtration-preserving isomorphism $A\cong A^{\mathrm{op}}$, $x_i\mapsto x_i$, $D_i\mapsto -D_i$.

A nonzero finitely generated module with $d(M)=n$ is a holonomic Weyl algebra module; the zero module is conventionally included. This minimal-growth class is stable under submodules, quotient modules and extensions. For $0\to N\to M\to Q\to 0$, take the induced and quotient filtrations. Their graded sequence is exact; Noetherianity of $\mathrm{gr}A$ makes the induced filtrations good. Thus the Hilbert functions add andif both nonzero terms have the top dimension, their multiplicities add. For a holonomic Weyl algebra module, every nonzero subquotient has dimension $n$, by the upper bound inherited from $M$ and Bernstein's lower bound. Each nonzero factor in a strict chain therefore consumes positive integral multiplicity. This proves the multiplicity bounds the length of a holonomic Weyl module:In particular a holonomic module is both Noetherian and Artinian and has finite length, although its vector-space dimension is infinite. This distinction separates finite-dimensional representation theory from finite-length Weyl module theory.

The polynomial module $\mathbb{C}[x_1,\dots ,x_n]$, with $x_i$ acting by multiplication and $D_i$ by differentiation, is generated by $1$. Its $M_j$ consists of polynomials of degree at most $j$, with dimension $\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)$, so $d=n$ and $e=1$. It is simple: differentiating a maximal-degree monomial of a nonzero polynomial isolates a nonzero constant, and multiplication by polynomials then generates the whole module. A second basic example is the delta moduleUsing the reverse ordered basis $D^{\beta }{x}^{\alpha }$, its vector space is $\mathbb{C}[D_1,\dots ,D_n]$; $D_i$ acts by multiplication and $x_i$ by negative differentiation. The same argument proves simplicity and holonomicity. These two examples are exchanged by the automorphism $x_i\mapsto D_i$, $D_i\mapsto -x_i$. Free modules, in contrast, have growth dimension $2n$ and are not holonomic. Thus the central organizing invariants are good filtration of a module, growth dimension, characteristic support and multiplicity, with Bernstein's estimate explaining why dimension $n$ is the natural smallest nonzero size.

Put $A=\mathbb{C}[V]={⨁}_{d\ge 0}{A}_d$ and $R=A^G$. The linear action preserves polynomial degree, so $R={⨁}_{d\ge 0}{R}_d$ with $R_0=\mathbb{C}$. We prove Hilbert finite generation for reductive invariants by combining finite ideal generation in $A$ with a degree-lowering argument in $R$.

Over $\mathbb{C}$, representations of a reductive algebraic group are completely reducible. On each finite-dimensional $A_d$, project its isotypic component for the trivial representation onto itself and its nontrivial summands onto zero. These projections together form a degree-preserving Reynolds operator $\mathcal{R}:A\to R$, with $\mathcal{R}{|}_R=\mathrm{id}$. The crucial additional property isHere is why it holds rather than being an arbitrary projection property. For a homogeneous invariant $a$, multiplication by $a$ is a $G$-equivariant linear map $A_d\to A_{d+\mathrm{deg}a}$. It sends invariant vectors to invariant vectors. On a nontrivial irreducible summand, its composition with projection onto the target's trivial summand is zero, since there is no nonzero equivariant map from a nontrivial irreducible representation to the trivial representation. It therefore commutes with the Reynolds operators. Summing the homogeneous components of $a$ and $h$ proves the displayed module property in general.

Let $R_{+}={⨁}_{d>0}{R}_d$ and form the ideal $I=A{R}_{+}$ in the ambient polynomial ring. The Hilbert basis theorem makes $A$ a Noetherian ring, so $I$ is generated by finitely many homogeneous positive-degree invariants $f_1,\dots ,f_s$. To justify choosing the generators from this particular set, start with any finite ideal generating set; each of its members is a finite sum of multiples of homogeneous members of $R_{+}$, and the union of those finitely many invariants then generates $I$. If $I=0$, there are no positive-degree invariants and the answer is simply $R=\mathbb{C}$.

Otherwise consider a homogeneous invariant $f$ of degree $d>0$. Since $f\in I$, writewith $h_i=0$ when that degree is negative. Taking the degree-$d$ component of any ideal expression gives exactly such an expression. Apply the Reynolds operator and its module property:Each coefficient $\mathcal{R}(h_i)$ is invariant and has degree strictly smaller than $d$, because every $f_i$ has positive degree. Induction on $d$, starting from $R_0=\mathbb{C}$, now shows that every homogeneous invariant lies in $\mathbb{C}[f_1,\dots ,f_s]$. An arbitrary invariant is a finite sum of invariant homogeneous components, soThis proves finite generation as an algebra. No Noetherian property of the polynomial invariant ring was assumed in obtaining the generators.

Give $U(\mathfrak{g})$ its degree filtration. The Poincaré-Birkhoff-Witt theorem identifies its associated graded ring with ${\mathrm{Sym}}_k\mathfrak{g}$, a polynomial ring in ${\dim }_k\mathfrak{g}$ variables. The Hilbert basis theorem makes this graded algebra Noetherian on both sides.

Here is the required transfer, including why a finite list of symbols gives actual generators. For a right ideal $I$ of a nonnegatively filtered ring $A$, its induced graded ideal $\mathrm{gr}I$ is a homogeneous right ideal of $\mathrm{gr}A$. Choose finite homogeneous generators and lift them to $u_1,\dots ,u_m\in I$. For $u\in I$ of degree $d$, express its leading symbol as a graded right linear combination of these symbols, using homogeneous coefficients of complementary degrees. Lift the coefficients to $A$ and subtract the corresponding ${\sum }_i{u}_i{a}_i$. The remainder has degree strictly below $d$. Induction, terminating because degrees are nonnegative, proves $I={\sum }_i{u}_{i}A$. The identical argument with coefficients on the left treats left ideals.

This ascending filtered-graded transfer of Noetherianity proves that $U(\mathfrak{g})$ is right and left Noetherian. A quotient ring inherits either property: pull an ascending ideal chain back to the original ring, or take images of ideal generators. Part (a) therefore gives $R$ right and left Noetherian. Alternatively, under the broader almost commutative algebra convention, its commutative finitely generated associated graded ring is directly a quotient of a finite-variable polynomial ring, and the same leading-symbol argument proves the conclusion.

Use the usual commutative unital-ring convention for Noetherian rings. A ring is Noetherian exactly when every ideal is finitely generated, equivalently when ascending chains of ideals stabilize. Indeed, the union of an ascending chain is an ideal; finitely many generators all lie in a single chain member. Conversely, if an ideal cannot be finitely generated, adjoining a new element at each stage constructs a strictly ascending chain.

First assume $R[X]$ is Noetherian. Evaluation at zero gives a surjective ring homomorphism $R[X]\to R$ with kernel $(X)$. A quotient of a Noetherian ring is Noetherian: lift an ideal to the source, take finitely many generators there, and map those generators down. Thus $R\cong R[X]/(X)$ is Noetherian.

For the converse, assume $R$ is Noetherian and take any ideal $I\subseteq R[X]$. For $d\ge 0$, let $J_d$ be the set of coefficients of $X^d$ in elements of $I$ of degree at most $d$. It is an ideal of $R$, since sums and scalar multiples of such polynomials stay in $I$. Multiplication by $X$ shows $J_d\subseteq J_{d+1}$. The ascending chain stabilizes, say at $N$.

For each $0\le d\le N$, choose finitely many generators $a_{d,j}$ of $J_d$ and corresponding $f_{d,j}\in I$ of degree at most $d$ having coefficient $a_{d,j}$ at $X^d$. Discard zero generators. We claim these finitely many polynomials generate $I$. If a nonzero $f\in I$ has degree $m\le N$, express its leading coefficient as ${\sum }_j{r}_j{a}_{m,j}$ and subtract ${\sum }_j{r}_j{f}_{m,j}$. The degree decreases. If $m>N$, use $J_m=J_N$, express the leading coefficient using the $a_{N,j}$, and subtractAgain the leading term cancels and the degree decreases. Induction on degree, including degree zero, expresses every $f$ in the claimed ideal. Therefore every ideal of $R[X]$ is finitely generated, proving the Hilbert basis theorem andThe argument is the bounded-degree coefficient proof of Hilbert basis theorem; it never divides by a leading coefficient and therefore does not require $R$ to be an integral domain. With the analogous left-ideal convention the same proof applies to left-Noetherian rings with a central indeterminate.

Hilbert's basis theorem: if $R$ is a commutative Noetherian ring with identity, then the polynomial ring $R[x]$ is Noetherian. Iteration gives the same assertion for every finite number of variables.

Here is a bounded-degree coefficient proof of Hilbert basis theorem. Let $I$ be any ideal of $R[x]$. For $n\ge 0$, define $J_n\subseteq R$ to be the coefficients of $x^n$ in the polynomials of $I$ of degree at most $n$, including the zero coefficient. Closure under addition and scalar multiplication makes $J_n$ an ideal of $R$. Multiplication of a polynomial by $x$ shows $J_n\subseteq J_{n+1}$. By the ascending chain condition, there is $N$ such that $J_N=J_{N+1}=\cdots$.

Each $J_n$ for $0\le n\le N$ is finitely generated. Choose generators $a_{n,1},\dots ,a_{n,r_n}$ and corresponding polynomials $f_{n,j}\in I$ of degree at most $n$ whose coefficient of $x^n$ is $a_{n,j}$. There are only finitely many chosen polynomials. We prove they generate $I$ by induction on degree.

Take a nonzero $f\in I$ of degree $d$ and leading coefficient $a$. If $d\le N$, express $a={\sum }_j{c}_j{a}_{d,j}$ and subtract ${\sum }_j{c}_j{f}_{d,j}$. If $d>N$, use $a\in J_d=J_N$, express $a={\sum }_j{c}_j{a}_{N,j}$, and subtract ${\sum }_j{c}_j{x}^{d-N}{f}_{N,j}$. In both cases the leading coefficient cancels, leaving a polynomial in $I$ of smaller degree. For degree zero the remainder is zero, and the induction completes the claim.

Thus every ideal of $R[x]$ is finitely generated, proving the Hilbert basis theorem. The proof uses no cancellation of nonzero elements of $R$, so it remains valid when $R$ has zero divisors.

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