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If $A$ is a left Noetherian ring and a right Noetherian ring, and $G$ is a poly-(cyclic or finite) group, then $A[G]$ satisfies both chain conditions. Induct along the subnormal series. A finite factor gives a finite free extension on each side, handled by finite-module extension preserves Noetherianity. An infinite cyclic factor gives a skew Laurent polynomial ring, with the twisting automorphism induced by conjugation by a lift of the cyclic generator. The noncommutative Hilbert basis theorem and clearing negative powers prove the required chain conditions at that step. In particular $\mathbb{Z}[G]$ is Noetherian.

Suppose first that $M$ is a Noetherian module. An ascending chain of submodules of $N$ is also a chain in $M$, so it stabilizes. A chain in the quotient module $M/N$ lifts to a chain in $M$, so it also stabilizes. Thus both $N$ and $M/N$ are Noetherian modules.

Conversely, let $M_1\subseteq M_2\subseteq \cdots$ be a chain of submodules of $M$. If $N$ and $M/N$ are Noetherian modules, both $M_i\cap N$ and $(M_i+N)/N$ stabilize, say for $i\ge i_0$. For $x\in M_{i+1}$, equality of the images gives $y\in M_i$ with $x-y\in N$. Then $x-y\in M_{i+1}\cap N=M_i\cap N$, and hence $x\in M_i$. This proves the extension direction of Noetherian modules in a short exact sequence.

The left regular module $R$ is Noetherian because $R$ is a left Noetherian ring. Applying the extension result inductively makes each finite direct sum $R^s$ Noetherian. Every finitely generated module on the left is a quotient module of such a direct sum, so it is Noetherian.

A poly-(cyclic or finite) group has a finite subnormal serieswhose factors are cyclic groups or finite groups. We prove that the group ring $R[G_i]$ is a left Noetherian ring by induction. Put $A=R[G_{i-1}]$. A finite factor makes $R[G_i]$ a finite free left $A$-module, using a coset transversal. Its left ideals are $A$-submodules, so finite-module extension preserves Noetherianity handles this case.

If the factor is infinite cyclic, choose a lift $t$ of a generator and let $\sigma (a)=ta{t}^{-1}$. The group ring is the skew Laurent polynomial ring $A[t,t^{-1};\sigma ]$. Here is the required left-sided skew Hilbert basis theorem. For a left ideal $L\subseteq A[t;\sigma ]$, defineEach $I_n$ is a left ideal of $A$. Left multiplication by $t$ gives $I_n\subseteq I_{n+1}$. The ascending chain condition makes these ideals constant for $n\ge n_0$; choose finitely many generators for each $I_n$, $0\le n\le n_0$, and lift them to polynomials $f_{nj}\in L$.

If $f\in L$ has degree $d\ge n_0$ and leading coefficient $a_d$, writeSubtracting ${\sum }_j{\sigma }^d(r_j)t^{d-n_0}{f}_{n_{0}j}$ cancels the leading coefficient. For $d<n_0$, use the lifts for $I_d$ instead. Induction on polynomial degree shows that the finitely many $f_{nj}$ generate $L$. Thus the skew polynomial ring of an automorphism is a left Noetherian ring.

Finally, for a left ideal $K$ of $A[t,t^{-1};\sigma ]$, its intersection with $A[t;\sigma ]$ has finitely many generators. Every $f\in K$ satisfies $t^{m}f\in A[t;\sigma ]$ for some $m\ge 0$, so those same generators generate $K$ after multiplication by $t^{-m}$. This completes the induction and proves that $R[G]$ is left Noetherian, the left-sided form of group rings of poly-(cyclic or finite) groups are Noetherian.

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.