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Use unital rings, and distinguish left and right Noetherianity. A useful form of the noncommutative Hilbert basis theorem is the following: if a ring $B$ is generated by a subring $A$ and an element $x$, if $A+xA=A+Ax$, and if $A$ is a right Noetherian ring, then $B$ is a right Noetherian ring. The corresponding left-handed statement holds as well. In particular, a central polynomial ring over a Noetherian ring is Noetherian, and a skew polynomial ring of an automorphism preserves either side's Noetherianity.

Here is a proof of the general right-handed statement. Put $F_{-1}=0$ andThe equality follows inductively from $A+xA=A+Ax$. These groups form an exhaustive ascending filtration, stable under multiplication by $A$ on either side, with $F_j{F}_k\subseteq F_{j+k}$. For any $b\in A$ and $n\ge 0$, moving coefficients one step at a time also givesFix a right ideal $I$ of $B$ and define coefficient setsEach $L_n$ is a right ideal of $A$. Additive closure is immediate. If $a\in L_n$ and $b\in A$, use the preceding transport identity to obtainHere $(a{x}^n)c$ has this membership because $I$ is a right ideal and $F_{n-1}$ is stable under right multiplication by $A$. Also $L_n\subseteq L_{n+1}$, by multiplying $a{x}^n\in I+F_{n-1}$ on the right by $x$. The ascending chain condition in $A$ yields $N$ with $L_n=L_N$ for $n\ge N$.

For each $0\le k\le N$, choose finite right generators $a_{kj}$ of $L_k$ and lifts $f_{kj}\in I$ with $f_{kj}-a_{kj}{x}^k\in F_{k-1}$. These finitely many lifts generate $I$ as a right ideal of $B$. To prove this, let $f\in I\cap F_d$ and write $f=a{x}^d+v$, $v\in F_{d-1}$. Then $a\in L_d$. Put $k=\min (d,N)$ and write $a={\sum }_j{a}_{kj}{b}_j$. Transport each coefficient as $b_j{x}^d=x^d{c}_j+v_j$ with $v_j\in F_{d-1}$. ConsequentlyInduction on $d$, with $F_{-1}=0$, expresses $f$ in the right ideal generated by the lifts. Every right ideal is therefore finitely generated, proving the theorem. Passing to the opposite ring proves the left-handed version. This proof does not require unique coefficient normal forms.

A poly-(cyclic or finite) group has a finite subnormal serieswhose factors $G_i/G_{i-1}$ are cyclic or finite. Normality is required between consecutive terms. We show inductively that each group ring $\mathbb{Z}[G_i]$ is both a left Noetherian ring and a right Noetherian ring.

The starting ring $\mathbb{Z}$ is Noetherian, since every ideal is principal. The other module fact we use is that a finitely generated module over a Noetherian ring on the relevant side is a Noetherian module: it is a quotient of a finite direct sum of regular modules, and the ascending chain condition passes to finite sums and quotients. For a finite sum this follows by induction, comparing the intersections of a chain with one summand and its images in the quotient by that summand.

If $H◃K$ has finite index, choosing coset representatives shows that $\mathbb{Z}[K]$ is finite free as a left and as a right $\mathbb{Z}[H]$-module. By finite-module extension preserves Noetherianity, every one-sided ideal of $\mathbb{Z}[K]$ is finitely generated already as a $\mathbb{Z}[H]$-module, and those same generators generate it as a $\mathbb{Z}[K]$-ideal. Thus either chain condition passes through a finite factor.

If $K/H$ is infinite cyclic, choose $t\in K$ lifting its generator. Every element of $K$ has a unique form $h{t}^n$, and conjugation by $t$ gives an automorphism $\sigma$ of $\mathbb{Z}[H]$. HenceFor the polynomial subring $C=\mathbb{Z}[H][t;\sigma ]$, the condition $A+tA=A+At$ follows from invertibility of $\sigma$, so the proved noncommutative Hilbert basis theorem makes $C$ Noetherian on both sides. A right ideal $J$ of the skew Laurent polynomial ring is generated by $J\cap C$: for any $f\in J$, choose $m$ large enough that $f{t}^m\in C$, and recover $f=(f{t}^m)t^{-m}$. Finite generators of $J\cap C$ over $C$ therefore generate $J$ over the Laurent ring. Left multiplication by a large power of $t$ proves the left-handed statement. These are the Laurent-extension facts used in the induction.

Every factor of the subnormal series is covered by one of these two cases. ThereforeThe argument proves the more general result that group rings of poly-(cyclic or finite) groups are Noetherian over any coefficient ring satisfying both chain conditions.

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