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For $H\le G$ of finite subgroup index, let $H^{\prime }=[H,H]$ and choose a right-coset transversal $T$ for the cosets $Ht$. For each $g\in G$ writeThe transfer homomorphism isThe product is taken in the abelianization, so its order is irrelevant. If another transversal has representative $a_{t}t$ for $Ht$, with $a_t\in H$, its factor is $a_{t}h(t,g)a_{t_g}^{-1}$. The map $t\mapsto t_g$ permutes $T$, so the extra factors cancel in $H/H^{\prime }$. This proves independence of the transversal. AlsoMultiplying over $t$ and again permuting the second indices proves $V(g_1{g}_2)=V(g_1)V(g_2)$. Thus it is indeed a group homomorphism.

The Burnside transfer theorem states that if a Sylow $p$-subgroup $P$ of a finite group satisfies $P\subseteq Z(N_G(P))$, then $G$ has a normal $p$-complement: a normal subgroup of order $|G|/|P|$. The hypothesis first makes $P$ abelian. It also makes fusion inside $P$ trivial. To prove this, suppose $u,v\in P$ and $u^g=v$, using $u^g=g^{-1}ug$. Both $P$ and $P^g$ lie in $C_G(v)$: the former because $P$ is abelian, and the latter because $v\in P^g$. They are Sylow subgroups of that centralizer. Choose $c\in C_G(v)$ with $(P^g{)}^c=P$. Then $gc\in N_G(P)$ and $u^{gc}=v^c=v$, while the central-normalizer hypothesis gives $u^{gc}=u$. Thus $u=v$.

Evaluate transfer to the abelian group $P$ on $u\in P$. In the action of $u$ on right cosets, a cycle of length $r$ contributeswith $t_r=t_0$. This is a conjugate of $u^r$ inside $P$ and hence equals $u^r$ by the fusion argument. Summing the cycle lengths therefore givesThe exponent is coprime to $p$, so this power map is an automorphism of the finite abelian $p$-group $P$. Transfer is consequently onto $P$, and its kernel is normal of subgroup index $|P|$ and of order prime to $p$. This proves the theorem. Notice also that a normal $p$-complement consists exactly of the elements of prime-to-$p$ order: they map trivially to the $p$-group quotient, while all elements of the complement have such order. It is therefore unique and a characteristic subgroup.

Now let $G\le A_p$ be primitive of degree $p=2q+1$, with $q$ prime. Every nontrivial normal subgroup of a faithful primitive group action is transitive: its orbits form a $G$-invariant block partition, and singleton orbits would make it trivial. Such a subgroup has order divisible by $p$. A Sylow $p$-subgroup $P$ has order $p$, since $p!$ contains only one factor $p$. The normalizer of a regular $p$-cycle in $S_p$ is the affine group $x\mapsto ax+b$ on ${\mathbb{F}}_p$, of order $p(p-1)$. Translations are even, whereas a generating multiplier gives a $(p-1)$-cycle and is odd. Its normalizer in $A_p$ therefore has order $p(p-1)/2=pq$.

Suppose $G$ is not simple and choose $1<N◃G$ proper. It is transitive, so we can choose $P\le N$. The Frattini argument gives $G=N{N}_G(P)$: for each $g$, the subgroup $P^g$ is a Sylow subgroup of $N$ and can be conjugated back to $P$ within $N$. Since $N_G(P)$ contains $P$ and lies in the order-$pq$ normalizer, it has order $p$ or $pq$. If its order were $p$, the Burnside theorem applied to $G$ would give a normal $p$-complement. A nontrivial such complement cannot be transitive because its order is prime to $p$, contradicting primitivity. It must be trivial, giving $G=P$, contrary to the choice of $N$. Thus $|N_G(P)|=pq$.

The intersection $N_N(P)$ again has order $p$ or $pq$. If it had order $pq$, the whole normalizer would lie in $N$ and the Frattini equality would give $G=N$. Hence $N_N(P)=P$. Applying Burnside transfer within $N$ gives its normal $p$-complement $K$. By the uniqueness argument, $K$ is a characteristic subgroup of $N$, and so normal in $G$. Primitivity again forces this prime-to-$p$ subgroup to be trivial. Thus $N=P$ is normal in $G$, and $G=N_G(P)$ has order $pq$. The even primitive groups of safe-prime degree conclusion is

Write $\hat{X}=X_{/Y}$. The formal completion of a scheme restricts on each chart toThus $\hat{X}$ is a formal scheme over $\mathrm{Spf}\mathbb{Z}[[q]]$. The ordinary torus $D(q)$ has disappeared from its underlying space: every point of $\hat{X}$ lies on $Y$. In particular, only adjacent completed charts intersect, and their two possible branch overlaps inside any one chart are disjoint.

Construct the action using the integral lattice automorphismIt carries $v_i$ to $v_{i+3}$ and ${\sigma }_i$ to ${\sigma }_{i+3}$, so induces a toric morphism $g:X\to X$ that is an automorphism. It preserves the lattice projection to the second coordinate, hence preserves $q$ and $Y$. On the dense algebraic torus it isIn the local coordinates this sends $U_i$ to $U_{i+3}$ withThese identities show regularity on every chart without requiring $q$ to be invertible there. They also make the action continuous in the $q$-adic topology, so it extends to $\hat{X}$. Its powers shift $D_i$ to $D_{i+3k}$; no nonzero power is the identity or fixes a point of the formal support. Therefore $G=⟨g⟩\cong \mathbb{Z}$ is a nontrivial infinite cyclic group acting freely.

Construct the quotient explicitly, using three representative chartswith $q=X_i{Y}_i$. Glue their branch opens cyclically, taking subscripts modulo three, byThese are genuine isomorphisms of formal schemes. For example, the completed localization of an adic ring on the first open iswhere $X_i=q/Y_i$. On the second it is $\mathbb{Z}[X_{i+1},X_{i+1}^{-1}][[q]]$, and the gluing sends $X_{i+1}$ to $Y_i^{-1}$ and $Y_{i+1}$ to $q{Y}_i$. Thus it preserves $q$ and is continuous, with a continuous inverse.

Inside ${\mathfrak{V}}_i$ we haveConsequently there are no triple-overlap conditions left to check: the two branch identifications in a chart have disjoint domains. By gluing of schemes along open subschemes in its formal version, the three charts produce a formal scheme $\mathfrak{C}$ over $\mathrm{Spf}\mathbb{Z}[[q]]$. Equivalently, reduce the same gluing modulo $q^{n+1}$ for every $n$; this constructs compatible schemes $C_n$, all on the same underlying space, with $C_n=C_{n+1}\times \mathrm{Spec}\mathbb{Z}[q]/(q^{n+1})$. Their inverse system has precisely the displayed formal charts.

For every integer $i$, map ${\hat{U}}_i$ isomorphically to ${\mathfrak{V}}_{\bar{i}}$ by identifying its coordinates with the corresponding capital-letter coordinates. The adjacent-chart formulas agree with the cyclic identifications, so these maps glue toThey are invariant under $g$, and ${\pi }^{-1}({\mathfrak{V}}_i)$ is the disjoint union of the translated opens ${\hat{U}}_{i+3k}$, each mapping isomorphically onto ${\mathfrak{V}}_i$. The group permutes these copies transitively. Thus the fibres of $\pi$ are exactly the $G$-orbits, and invariant sections on this inverse image are exactly the sections on ${\mathfrak{V}}_i$.

To verify the quotient property also for morphisms, take a $G$-invariant morphism $h:\hat{X}\to \mathfrak{T}$. Its restrictions to ${\hat{U}}_0,{\hat{U}}_1,{\hat{U}}_2$ agree on the adjacent branch opens. On the closing overlap they agree because the translated chart ${\hat{U}}_3$ has the same restriction as ${\hat{U}}_0$ under $g$. Hence they descend to a unique morphism $\mathfrak{C}\to \mathfrak{T}$. This proves that the gluing is the cyclic formal quotient of the infinite toric chain:Its fibre modulo $q$ is a triangle of three projective lines over $\mathbb{Z}$, each meeting its two neighbours at its two boundary sections. More generally, a shift by any $m\ge 3$ gives an $m$-gon. Completion is important here: translates of a formal chart are disjoint, whereas all ordinary toric charts still share the generic torus.

Throughout, the braces denote a multiset, so repeated splitting degrees retain their multiplicities. If the multisets agree after a common shift, permuting summands gives $E_a\cong E_b\otimes \mathcal{O}(c)$. Tensoring all the fibres of a vector bundle by the same line bundle leaves their one-dimensional quotient spaces unchanged. HenceThis proves one implication in the isomorphism classification of abstract rational scrolls.

For the converse first take $n\ge 3$. Every divisor class on $F_a$ has the form $xL+yM$. If its square is zero, restriction to a fibre ${\mathbb{P}}^{n-1}$ giveswhere $h$ is the hyperplane class and $h^2\ne 0$ because $n-1\ge 2$. Thus $y=0$. It follows that $L$ is intrinsically characterized as the primitive nonzero nef class whose square is zero. An isomorphism $u:F_a\to F_b$ therefore carries $L_b$ to $L_a$; the sign is positive because both classes are nef and effective. The complete pencil $|L|$ determines the ruling, sofor an automorphism $\sigma$ of ${\mathbb{P}}^1$.

On a fibre, the isomorphism preserves the positive generator ${\mathcal{O}}_{{\mathbb{P}}^{n-1}}(1)$ of its Picard group. Consequentlyfor some integer $c$. Push forward along the ruling, using ${\pi }_{\ast }\mathcal{O}(M)=E$ and the projection formula for sheaves:Every automorphism of ${\mathbb{P}}^1$ preserves each $\mathcal{O}(d)$. The uniqueness of the splitting degrees in the Birkhoff–Grothendieck theorem now implies equality of the two multisets up to a shift. Replacing this $c$ by $-c$ gives the sign convention in the final formula. One can see that uniqueness numerically: for a split bundle,The first difference counts the degrees with $a_i+t\ge 0$, and the second difference counts the multiplicity of $a_i=-t$. Hence these dimensions determine every splitting degree, with multiplicity.

For $n=2$, subtract the smaller degree and write the normalized bundle as $\mathcal{O}\oplus \mathcal{O}(a)$, $a\ge 0$. Its surface is the Hirzebruch surface ${\mathbb{F}}_a$. Let $B$ be its negative section and $L$ its fibre, so $B^2=-a$, $BL=1$, $L^2=0$. The nef cone is generated by $L$ and $B+aL$: a nef divisor $xL+yB$ must have $y\ge 0$ from intersection with a fibre and $x-ay\ge 0$ from intersection with $B$, and these two inequalities also express it as a nonnegative combination of those generators. If $a>0$, a class $uL+v(B+aL)$ with $u,v\ge 0$ has squarewhich vanishes only when $v=0$. Thus the ruling is again intrinsic, and the same bundle argument applies. If $a=0$, the surface is ${\mathbb{P}}^1\times {\mathbb{P}}^1$ and has two nef square-zero rays, its two rulings. This distinguishes it from every $a>0$ surface. Isomorphisms may exchange its rulings, but both normalized splitting degrees are zero, so the stated classification is unchanged.

For $n=1$, every scroll is ${\mathbb{P}}^1$, and any two single splitting degrees differ by a common shift. Thus all ranks are covered:Reading the braces as ordinary sets would lose information and make the assertion false when degrees repeat. For example, $(0,0,1,2)$ and $(0,1,1,2)$ have the same underlying set but not the same splitting multiset; their minima are both zero, so no nonzero common shift can reconcile them.

One useful version is the effective global Torelli theorem for K3 surfaces. Let $X,Y$ be complex K3 surfaces and letpreserve the integral cup-product pairing and the Hodge decomposition. Suppose that $\varphi$ carries some Kähler class of $Y$ to a Kähler class of $X$. ThenConversely every isomorphism has these properties. In particular, for projective K3 surfaces an integral Hodge isometry carrying an ample polarization to an ample polarization is induced by a unique isomorphism of the polarized surfaces. This is the effective global Torelli statement.

The Kähler-cone condition is necessary if the prescribed isometry itself is to be induced by an isomorphism. If $\delta$ is the class of an effective smooth rational $(-2)$-curve, the reflectionis integral, preserves the pairing, and preserves the Hodge decomposition because $\delta$ has type $(1,1)$. It sends $\delta$ to $-\delta$. An automorphism preserves effectivity, whereas $-\delta$ is not effective: it has negative intersection with an ample class. Thus this Hodge isometry is not an automorphism pullback. Requiring an isometry to respect the Kähler chamber removes precisely this kind of obstruction.

The assertions are relative independence results. We construct countermodels in a background model of ZFC; if ZF is consistent, its constructible universe gives a model of ZFC as well. The original model satisfies foundation and choice, supplying the positive sides.

For failure of foundation, use a Rieger-Bernays permutation model. Let $\pi$ interchange the old $\varnothing$ and $\{\varnothing \}$ and fix every other object, and put $xEy$ iff $x\in \pi (y)$. The new extension of $y$ is the old set $\pi (y)$, so the representative of any old extension $b$ is ${\pi }^{-1}(b)$. Bijectivity gives extensionality. Translate formulas recursively by this definition of $E$. Old separation gives each new definable subset, and old replacement gives each new functional range; apply ${\pi }^{-1}$ to the resulting extension to obtain its new representative.

Explicitly, a new pair is represented by ${\pi }^{-1}(\{x,y\})$; the new union of $a$ is represented by ${\pi }^{-1}({\bigcup }_{x\in \pi (a)}\pi (x))$; and the new power set of $a$ is represented byStarting from the new empty set, iterate the new successor and collect these iterates by old replacement, obtaining an inductive set. The old axiom of choice selects a member from each nonempty extension and thereby gives new choice functions. However the old object $q=\varnothing$ now satisfies $qEq$ and has exactly itself as its member: it is a Quine atom. Its new singleton has no $E$-minimal member, so foundation fails.

To make choice fail, retain infinitely many Quine atoms $A=\{a_n:n<\omega \}$ and impose a finite-support symmetry restriction. A precise background construction uses tagged set codes. Each atom $a$ has extension $\{a\}$. For every other set of available objects $S$, introduce a unique ordinary set code with extension $S$, except that $S=\{a\}$ is already represented by $a$ itself. Iterate this construction through the ordinals. Omitting the duplicate singleton codes makes the resulting relation extensional. Every permutation of $A$ extends to an automorphism of this universe by recursion through its ordinary codes.

An object has finite support if some finite $S\subseteq A$ has the property that every permutation fixing $S$ pointwise fixes that object. Retain the hereditarily finite-supported Quine-atom model: the object and every object reached through its membership relation must have finite support. The atomic self-loops are permitted. The atom set $A$ and all pure sets are retained.

The union of supports for a domain and finitely many parameters supports any definable subset or functional image: every permutation fixing those supports preserves the defining formula. Members of the subset or image are already hereditary, so this verifies separation and replacement. The collection of all retained subsets of a retained set is invariant under any permutation fixing that set's support, and each of its members is hereditary; this verifies the Axiom of power set. Pairs and unions have the union of the requisite finite supports, and the pure natural-number hierarchy verifies infinity. The coding still verifies extensionality.

If a choice function on all two-element subsets of $A$ had finite support $S$, choose distinct $a,b\notin S$. The transposition of $a,b$ fixes $S$ and the unordered pair $\{a,b\}$, but moves either possible chosen value. This contradicts invariance of the function. Hence choice fails, while a Quine atom witnesses failure of foundation. We have a model of $\mathrm{ZF}-\mathrm{Foundation}+\neg \mathrm{Choice}$. Together with the original model this proves the required independence of choice over $\mathrm{ZF}-\mathrm{Foundation}$.

Let $L/K$ be a finite Galois extension of number fields, with Galois group $G$, and let $\mathfrak{p}$ be a nonzero prime ideal of ${\mathcal{O}}_K$. Write its prime ideal factorization asThe Galois group acts transitively on the primes above $\mathfrak{p}$. One proof uses the Chinese remainder theorem. If $\mathfrak{Q}$ were outside the orbit of $\mathfrak{P}$, choose an integral element $a$ which is zero modulo $\mathfrak{Q}$ and one modulo every conjugate of $\mathfrak{P}$. Its field norm ${\prod }_{\sigma \in G}\sigma (a)$ belongs to $\mathfrak{Q}\cap {\mathcal{O}}_K=\mathfrak{p}$, hence to $\mathfrak{P}$. But each factor is one modulo $\mathfrak{P}$, a contradiction. Transitivity also makes all $e_j$ equal to a common ramification index $e$ and all $f_j$ equal to a common residue degree $f$.

Fix $\mathfrak{P}$ above $\mathfrak{p}$. Its decomposition group is the stabilizerThe orbit-stabilizer theorem gives $|D_{\mathfrak{P}}|=|G|/g$. Elements of $D_{\mathfrak{P}}$ act on the residue field $\kappa (\mathfrak{P})$ over $\kappa (\mathfrak{p})$; the group kernel is the inertia groupThus inertia measures the automorphisms invisible after reduction, whereas decomposition measures all automorphisms preserving the chosen prime.

The completion $L_{\mathfrak{P}}/K_{\mathfrak{p}}$ is a Galois extension with group $D_{\mathfrak{P}}$: preservation of $\mathfrak{P}$ lets an automorphism extend continuously to the completion, and the local embeddings of this normal extension arise in this way. The degree of a field extension formula is $[L_{\mathfrak{P}}:K_{\mathfrak{p}}]=ef$. Reduction gives the exact sequenceFor surjectivity, lift a primitive residue-field generator by the Hensel lemma to obtain the maximal unramified subextension of a local field extension of $L_{\mathfrak{P}}/K_{\mathfrak{p}}$. Its automorphisms are the residue-field automorphisms. They extend to $L_{\mathfrak{P}}$ because the latter is normal. Since a finite residue field extension has a cyclic Galois group of order $f$, the exact sequence yieldsThe fixed field of local inertia is the maximal unramified subextension of a local field extension, and the extension above it is totally ramified. Primes in the same global orbit have conjugate decomposition groups and inertia groups, so the orders do not depend on the chosen prime.

If $e=1$, inertia is trivial and $D_{\mathfrak{P}}$ is generated by the unique Frobenius automorphism acting on residues by $z\mapsto z^{N\mathfrak{p}}$. Its order is $f$. In a ramified extension this residue action determines only a coset modulo inertia. A prime splits completely exactly when $e=f=1$, equivalently its decomposition group is trivial. A totally ramified prime has $g=f=1$ and $I_{\mathfrak{P}}=D_{\mathfrak{P}}=G$; an unramified prime has $I_{\mathfrak{P}}=1$. In an abelian extension, conjugation does nothing, so the associated groups and unramified Artin symbol are independent of the prime above $\mathfrak{p}$.

Put $t=\sqrt{-30}$ and $K=\mathbb{Q}(t)$. The ring of integers is $\mathbb{Z}[t]$, and the fundamental discriminant is $D_K=-120$. For a reduced positive definite binary quadratic form $(a,b,c)$ of discriminant $-120$, reduction givesThe coefficient $b$ is even; checking $1\le a\le 6$ and $|b|\le a$, with the usual nonnegative sign choice at boundary equalities, leaves preciselyThe correspondence between proper quadratic forms and ideal classes therefore gives $h_K=4$. All four classes equal their inverses, since inversion replaces $b$ by $-b$. HenceThe three nontrivial classes may be represented by the ramified prime ideals ${\mathfrak{p}}_2=(2,t)$, ${\mathfrak{p}}_3=(3,t)$ and ${\mathfrak{p}}_5=(5,t)$. They square to principal ideals and satisfy ${\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_5=(t)$. None is principal: an integral generator of norm $2$, $3$ or $5$ would require $a^2+30b^2$ to equal that number, which is impossible. The relation then makes the three classes distinct, consistently with the four-form calculation.

ConsiderThe three rational square classes are independent, so $[H:\mathbb{Q}]=8$. Their radical product is $\sqrt{-30}$, so $K\subset H$ and $[H:K]=4$; the extension $H/K$ is abelian. To verify that it is unramified, rather than assuming that radical adjunctions are harmless at $2,3,5$, use the conductor-discriminant formula. The seven quadratic subfields have fundamental discriminantsThe formula givesThe discriminant tower identity now givesThus the relative discriminant is the unit ideal and no finite prime ramifies. The base has no real places, so there is no infinite ramification to check. This unramified abelian degree-four extension has degree $h_K$, and part (i) provesThis is the Hilbert class field of Q of square root minus thirty.

Next translate prime representation into an ideal class. For $p\nmid 30$, we claimIf a representation exists, $\alpha =2x+yt$ lies in ${\mathfrak{p}}_2$ and has $N_{K/\mathbb{Q}}\alpha =2p$. The integral ideal $(\alpha ){\mathfrak{p}}_2^{-1}$ has norm $p$, so it is a prime ideal $\mathfrak{P}$. Its class is $[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]$. Conversely, if $[\mathfrak{P}]=[{\mathfrak{p}}_2]$, the product ${\mathfrak{p}}_2\mathfrak{P}$ is principal, say $(\alpha )$. Since this ideal lies in ${\mathfrak{p}}_2$, its generator has the form $\alpha =2x+yt$ with integers $x,y$. Taking norms gives $4x^2+30y^2=2p$, proving the reverse implication. This is prime representation via an ideal class, and it proves sufficiency as well as necessity.

Identify the required class in $\mathrm{Gal}(H/K)$ through the Artin map. We can use the explicit representation $17=2+15$: $2+t$ has norm $34$, so the prime above $17$ obtained from $(2+t){\mathfrak{p}}_2^{-1}$ has class $[{\mathfrak{p}}_2]$. Its Frobenius in the multiquadratic extension acts on the three radicals with signsTheir product is one, so this automorphism fixes $K$. Because $H/\mathbb{Q}$ is abelian, for any unramified rational prime $p$ that splits in $K$, the Frobenius at either prime of $K$ of norm $p$ is its rational Frobenius restricted to $H/K$. The Artin isomorphism therefore makes the required class condition equivalent toConversely, these signs already imply splitting in $K$, since their product is $(\frac{-30}{p})=1$. No extra splitting condition is missing.

The supplementary law for $2$, quadratic reciprocity for $-3$ and $5$, and then the Chinese remainder theorem turn these sign conditions intoTheir four simultaneous residue classes modulo $120$ are $17,23,47,113$. Finally, $2$ is represented by $(x,y)=(\pm 1,0)$, whereas $3$ and $5$ cannot be represented: $y\ne 0$ gives a value at least $15$, and $y=0$ gives twice a square. We have proved the complete criterion

For the indicated cyclic Galois resolvent, reindexing its terms and using $g^m=1$ givesThere exists $w$ with $R(w)\ne 0$ because distinct field automorphisms are linearly independent as functions. A short proof is as follows. Take a nonzero relation with the fewest nonzero terms, normalize one coefficient, evaluate it at $uv$ and subtract a scalar multiple of the relation at $v$. Choose $u$ on which two participating automorphisms differ. This produces a nonzero relation with fewer terms, a contradiction.

Set $\alpha =R(w)\ne 0$. Then $g^j\alpha ={\zeta }^j\alpha$ are $m$ distinct conjugates. Also $g({\alpha }^m)={\alpha }^m$, so $a={\alpha }^m\in K$. The orbit has size $m$, giving $[K(\alpha ):K]=m$ and hence $K(\alpha )=M$. All the roots ${\zeta }^j\alpha$ belong to $M$, soThis also treats $m=1$ with a nonzero $w\in K$.

Suppose $1\ne N⊲G$ is proper. The defining property makes $N$ cyclic of order $p$. Conjugation gives a homomorphismwhose kernel is the centralizer $C_G(N)$. The automorphism group has order $p-1$, so the kernel has finite index in $G$. It contains $N$, because $N$ is abelian. If this kernel were proper, it would be an order-$p$ subgroup, and its finite index would force $G$ itself to be finite. Therefore $C_G(N)=G$, so $N$ is central.

Choose $h\notin N$. Part (i) shows that $h$ has order $p$. Its cyclic subgroup meets $N$ trivially, since two order-$p$ subgroups with nontrivial intersection are equal. Since $N$ is central,This subgroup has order $p^2$ and is proper because $G$ is infinite, contradicting the assumed order of every proper nontrivial subgroup. Thus no such $N$ exists:Together with part (i), this proves that Tarski monster groups are two-generated and simple.

Conjugation ${\alpha }_g(x)=gx{g}^{-1}$ is an automorphism of $G$ sending $A$ to $gA{g}^{-1}$. It induces the bijectionbetween their left coset sets; conjugation by $g^{-1}$ gives the inverse. ThereforeThis holds whether or not $A$ is normal, and whether the index is finite or infinite.

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.

Take a recursively enumerable set $S\subseteq {\mathbb{N}}_{>0}$ that is not recursive, for example the diagonal halting set from Question 2 shifted by one. To obtain a finitely generated group with recursive presentation but undecidable word problem for a group, first define the countable Right-angled Artin groupTranslation of the indices gives an automorphism $a_i\mapsto a_{i+1}$. The semidirect product with its infinite cyclic shift is generated by $a=a_0,t$ and has presentationTo verify this presentation, the semidirect product satisfies its relators, while the reverse map sends $a_i$ to $t^{i}a{t}^{-i}$; all translated commutation relations follow by conjugation. These maps are inverse on their generators. Thus $B_S$ embeds in $G_S$.

For $n\in S$, the word $[a_0,a_n]$ is trivial. For $n\notin S$, map $a_0,a_n$ to the free generators $x,y$ of $F(x,y)$ and every other $a_i$ to one. Every defining commutator of $B_S$ maps to one, since the only possible relation involving both surviving generators would have separation $n$, which is absent. The resulting group homomorphism sends $[a_0,a_n]$ to the nontrivial reduced word $[x,y]$. HenceThis proves the shift-invariant graph groups encode arbitrary sets construction rather than supposing that a missing relator automatically means a nontrivial word. A word algorithm for $G_S$ would decide $S$, so none exists. Since $S$ is enumerable, the displayed two-generator presentation is a recursive presentation of a group. By the Higman embedding theorem, $G_S$ embeds in a finitely presented group $G_0$. A word algorithm for $G_0$, composed with substitution of the two fixed generator-image words, would solve the word problem for $G_S$. Therefore $\boxed{G_0\text{ is finitely presented with insoluble word problem}}$.