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The group ring of an infinite cyclic group is the Laurent polynomial ring $R=\mathbb{Z}[t,t^{-1}]$, with augmentation $\epsilon (t)=1$. Multiply any element in the kernel by a power of $t$ to obtain a polynomial vanishing at one. The factor theorem makes it divisible by $t-1$. Thus the augmentation ideal is $(t-1)R$. Since $R$ is an integral domain, multiplication by $t-1$ is an injective module map, so this ideal is a free $R$-module of rank one, even though its additive group has infinite rank.

The group ring $\mathbb{Z}G$ becomes a $\mathbb{Z}G$-module under the conjugation action $g\cdot x=gx{g}^{-1}$. Its basis splits into conjugacy classes, and the span of the class of $x$ is the permutation module on $G/C_G(x)$.

Embed a module $M$ over the group ring $\mathbb{Z}[G]$ in a module $I$ with vanishing positive group cohomology, and write $M^{\prime }=I/M$. The connecting map in the long exact group cohomology sequence of $0\to M\to I\to M^{\prime }\to 0$ is an isomorphism $H^r(G,M^{\prime })\to H^{r+1}(G,M)$ for $r\ge 1$, since the groups involving $I$ on both sides vanish. For a finite group and a module $I$ with vanishing Tate cohomology of a finite group in every degree, the same argument shifts every Tate degree. Repeating this construction transports computations to a convenient resolution.

With trivial action on both end modules, an extension has underlying abelian group ${\mathbb{Z}}^r\oplus {\mathbb{Z}}^s$ and generator action $\left(\begin{array}{cc}I& T\\ 0& I\end{array}\right)$. The integer matrix $T$ is invariant under equivalences fixing the ends. The one-element projective resolution with differential $z-1$ computes the same classification over the group ring.

Suppose $A\subseteq B$ are unital rings and $B$ is a finitely generated module over $A$ on the right. If $A$ is a right Noetherian ring, then $B_A$ is a Noetherian module. Every right ideal of $B$ is an $A$-submodule, hence has finitely many generators over $A$; these also generate it over $B$. Thus $B$ is a right Noetherian ring. The same argument works on the left under the corresponding hypotheses. Finite-index group ring extensions give an important example, using coset representatives as a finite module basis.

The group ring $R[G]$ consists of finite formal sums ${\sum }_{g\in G}{r}_{g}g$ with coefficients in a ring $R$. Addition is coefficientwise and multiplication extends the group operation and multiplication in $R$ distributively. A group algebra is the case in which $R$ is a field.

The group ring of a weight lattice $P$ consists of finite sums ${\sum }_{\lambda }{a}_{\lambda }e(\lambda )$ with integer coefficients, where $e(\lambda )e(\mu )=e(\lambda +\mu )$. A finite-dimensional Lie algebra representation has formal character ${\sum }_{\lambda }(\dim V_{\lambda })e(\lambda )$ in this ring. The Weyl group acts by $e(\lambda )\mapsto e(w\lambda )$. The ambient ring need not consist of genuine characters; the representation ring of a semisimple Lie algebra of finite-dimensional modules for a complex semisimple Lie algebra identifies with the invariant subring $\mathbb{Z}[P{]}^W$.

// Target: semisimple-lie-algebra.bigb

Moving the basepoint around a loop acts on a higher homotopy group. Since ${\pi }_n(X)$ is an abelian group for $n\ge 2$, this action extends linearly to its integral group ring, making ${\pi }_n(X)$ a module. For a connected CW complex, its universal cover gives an equivalent description by deck transformations, together with change of basepoint upstairs. Left and right conventions are converted by inversion of the loop. The Hurewicz theorem respects this action whenever it identifies a higher homotopy group with the corresponding homology of the cover.

The integral cohomological dimension is the least length of a projective resolution of the trivial group ring module $\mathbb{Z}$, equivalently the largest degree in which group cohomology with some module can be nonzero. A group containing torsion has infinite integral cohomological dimension. For ${\mathbb{Z}}^r$ the dimension is $r$, realized by its torus classifying space and detected by its top cohomology.

Work over $\mathbb{C}$ with a finite-dimensional semisimple Lie algebra $\mathfrak{g}$. Choose a Cartan subalgebra $\mathfrak{h}$, its root system $R\subset {\mathfrak{h}}^{\ast }$, a positive system $R^{+}$ and its simple roots. Let $W$ be the Weyl group, generated by the corresponding root reflections, and put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the Weyl vector. The finite-dimensional irreducible highest-weight representation $L(\lambda )$ is indexed by a dominant integral weight $\lambda$.

If $P$ is the weight lattice, its group ring has formal symbols $e^{\mu }$ with multiplication $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. The formal character iswhere $L(\lambda {)}_{\mu }=\{v:Hv=\mu (H)v\text{ for all }H\in \mathfrak{h}\}$ is a weight space. For a weight $\nu$, define the Weyl alternant $A_{\nu }={\sum }_{w\in W}\det (w)e^{w\nu }$; the sign $\det (w)=(-1{)}^{\ell (w)}$ is the determinant of $w$ on the real root span, equivalently the parity of its expression in simple reflections.

The Weyl character formula isThe second denominator is the Weyl denominator formula. The quotient can initially be taken in the fraction field of the group ring; the formula says it is the finite character on the left, so its apparent singularities are removable. The shifts by $\rho$ make $\lambda +\rho$ strictly dominant.

Use finitely generated right projective modules throughout. The group K0 of a ring is the Grothendieck group of their direct-sum monoid, with $[P\oplus Q]=[P]+[Q]$. The group K1 of a ring is ${\mathrm{GL}}_{\infty }(A)/E_{\infty }(A)$, equivalently the abelianization of the stable general linear group. Stabilization is $g\mapsto \mathrm{diag}(g,1)$; $E_{\infty }$ is generated by elementary transvections. The stable elementary subgroup is the commutator subgroup: elementary transvections are commutators using a third index, and stable block interchange and elementary row identities make the quotient abelian.

Write $A_0$ for the lower-right ring, printed as $A^{\prime }$ in the source. A Milnor square has unital commuting maps, is a pullbackand has at least one of $j_1,j_2$ surjective. Interchange the two sides if necessary so that $j_1$ is surjective.

We first establish the patching fact used in the proof. For projective $P_i$ and an isomorphism $\varphi :P_1\otimes A_0\to P_2\otimes A_0$, setFor free modules denote the result by $M_g$. The block matrix identityexpresses $\mathrm{diag}(g,g^{-1})$ as elementary matrices. Their entries lift through $j_1$, and their elementary lifts remain invertible. A change of basis on the first side then trivializes this gluing. Thus $M_g\oplus M_{g^{-1}}\cong A^{2r}$, proving that $M_g$ is finitely generated projective. Notice that we lifted elementary matrices, not arbitrary invertible matrices.

For general $P_1,P_2$, choose $P_1\oplus Q_1\cong A_1^r$ and $P_2\oplus Q_2\cong A_2^s$. The reductions of $Q_1\oplus A_1^s$ and $Q_2\oplus A_2^r$ are isomorphic: insert the common reduced $P_i$ and reorder the two complements. Patch them by such an isomorphism. The sum with $M(P_1,P_2,\varphi )$ now patches two free modules of equal size, so is projective by the free case, making each summand projective. The same free change of basis followed by these direct-summand projections shows that extension to $A_i$ recovers $P_i$. Conversely, write an $A$-projective as the image of an idempotent over a finite free module. Since matrices over the pullback are compatible pairs of matrices, applying the compatible idempotents to the free pullback reconstructs this projective from its two reductions. Compatible module isomorphisms patch as well. This proves Milnor patching of projective modules in the form needed below.

The diagonal maps in the sequence are extension of scalars; the two maps to $A_0$ are differences $j_{1\ast }-j_{2\ast }$. Define the connecting map byIt is unchanged by stabilization or elementary multiplication: an elementary factor lifts through $j_1$ and changes a free basis in the gluing. The identity $\mathrm{diag}(g,h)=\mathrm{diag}(gh,I)\mathrm{diag}(h^{-1},h)$ and the preceding elementary factorization show $M_g\oplus M_h\cong M_{gh}\oplus A^r$ after matching sizes. Thus $\partial$ is an additive homomorphism on $K_1(A_0)$.

Exactness at $K_1(A_1)\oplus K_1(A_2)$. If the two reduced classes agree, choose stabilized representatives $g_1,g_2$ of the same size. Their reductions differ by an elementary matrix $e$ over $A_0$. Lift $e$ to an elementary matrix over $A_1$ and adjust $g_1$ without changing its $K_1$ class. Its reduction now equals that of $g_2$. The compatible pair and its compatible inverse give an invertible matrix over $A$, so the original pair of classes is in the diagonal image. The reverse containment follows from commutativity of the square.

Exactness at $K_1(A_0)$. A gluing obtained from a change of basis over either upper ring patches a free module, so the difference-map image is killed by $\partial$. Conversely, $\partial [g]=0$ means that $M_g$ becomes free after adding a common projective summand. Add a complement of that summand to obtain an actual isomorphism $M_g\oplus A^s\cong A^{r+s}$. Its two local bases express $g\oplus I_s=j_2(v_2)j_1(v_1{)}^{-1}$. Therefore $[g]=j_{2\ast }[v_2]-j_{1\ast }[v_1]$, which lies in the image of the difference map. This uses stable equality of projective classes, not cancellation of arbitrary projectives.

Exactness at $K_0(A)$. Connecting classes reduce to zero on both sides. Conversely, write a class in the kernel as $[P]-[A^r]$ by adding a complement to its negative projective term. Its reductions have the same classes as $A_i^r$. Equality in $K_0(A_i)$ becomes an isomorphism after a finite common free summand, by taking a complement to the auxiliary projective. Use one sufficiently large free summand on both sides. The stabilized $P$ is then locally free with gluing $g$, so the original class is $[M_g]-[A^{r+s}]=\partial [g]$.

Exactness at $K_0(A_1)\oplus K_0(A_2)$. Write a kernel pair as $([P_1]-[A_1^r],[P_2]-[A_2^r])$, using complements and padding to make the free sizes equal. Their reduced positive classes agree. Adding a common finite free summand makes those reductions isomorphic. Patch the resulting projectives, and subtract the matching free $A$-module. Its restrictions are the desired pair. The reverse containment is immediate because both reductions to $A_0$ agree.

These arguments prove the Mayer-Vietoris sequence in algebraic K-theoryNo zero is inserted at either end; this fragment does not assert injectivity at the first term or surjectivity at the last.

For the group ring of the cyclic group, take the cyclotomic Milnor squareThe maps send the cyclic generator to $1$ and to $\zeta$; the lower map sets $\zeta =1$, and the right map reduces modulo $p$. Since ${\Phi }_p(1)=p$, $\mathbb{Z}[\zeta ]/(\zeta -1)={\mathbb{F}}_p$. To verify the pullback, represent an element by a polynomial of degree at most $p-1$. If both evaluations vanish, it is a constant multiple of the minimal polynomial of an algebraic element ${\Phi }_p$, and evaluation at one forces that constant to be zero. Conversely, represent $b\in \mathbb{Z}[\zeta ]$ by $f$ of degree below $p-1$. For a compatible integer $a$, $a-f(1)$ is divisible by $p$; replacing $f$ by $f+[(a-f(1))/p]{\Phi }_p$ yields the pair $(a,b)$. The right map is surjective, so this is a Milnor square.

The hinted cyclotomic units are $u_k=1+\zeta +\cdots +{\zeta }^{k-1}$. If $kl\equiv 1(\mathrm{mod}p)$ with $1\le l<p$, the geometric sum ${\sum }_{j=0}^{l-1}{\zeta }^{kj}$ is their inverse, because ${\zeta }^{kl}=\zeta$. Their residues are $k\in {\mathbb{F}}_p^{\times }$, covering every unit of the common field. Gaussian elimination identifies $K_1({\mathbb{F}}_p)$ with ${\mathbb{F}}_p^{\times }$ by determinant. Thus $K_1(\mathbb{Z}[\zeta ])\to K_1({\mathbb{F}}_p)$ is surjective, and the connecting map vanishes.

The remaining exact sequence injects $K_0(\mathbb{Z}[C_p])$ into $\mathbb{Z}\oplus K_0(\mathbb{Z}[\zeta ])$, with image $\{(r,x):r=\mathrm{rank}x\}$. Here $K_0(\mathbb{Z})=K_0({\mathbb{F}}_p)=\mathbb{Z}$, and reduction of a projective over the integral domain $\mathbb{Z}[\zeta ]$ has dimension its rank: localize at $(\zeta -1)$, where the projective is free, and compare its residue-field dimension with its fraction-field dimension. Projection onto the second factor is therefore a bijection on this kernel. Hence extension along the cyclotomic evaluation givesThe essential step is reduction of cyclotomic units at the ramified prime; surjectivity of a ring map alone would not imply surjectivity on $K_1$.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.

For the trefoil relator, with $\alpha (x)=\alpha (y)=t$, direct differentiation givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor a connected sum of knots, join Seifert surfaces in separated balls. Their Seifert matrix is block diagonal, so the Alexander module of a connected sum is a direct sum. Thus ${\mathcal{A}}_{T\#T}\cong (\Lambda /(f){)}^2$.

This module is not cyclic. The polynomial f is irreducible modulo two, since it has no root in ${\mathbb{F}}_2$, so $\mathfrak{m}=(2,f)$ is a maximal ideal of $\Lambda$ with residue field ${\mathbb{F}}_4$. The quotient ${\mathcal{A}}_{T\#T}/\mathfrak{m}{\mathcal{A}}_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a cyclic module has dimension at most one.

Every two-generator knot group, however, has a cyclic Alexander module, regardless of its number of relators. Indeed its primitive abelianization vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one module on the second edge, and any quotient by lifted relators is cyclic. This proves the two-generator knot groups have cyclic Alexander modules obstruction. Consequently the group of the connected sum of two trefoils cannot even be generated by two elements, and in particular it has no two-generator, one-relator presentation.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.

For the trefoil relator, with $\alpha (x)=\alpha (y)=t$, direct differentiation givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor a connected sum of knots, join Seifert surfaces in separated balls. Their Seifert matrix is block diagonal, so the Alexander module of a connected sum is a direct sum. Thus ${\mathcal{A}}_{T\#T}\cong (\Lambda /(f){)}^2$.

This module is not cyclic. The polynomial f is irreducible modulo two, since it has no root in ${\mathbb{F}}_2$, so $\mathfrak{m}=(2,f)$ is a maximal ideal of $\Lambda$ with residue field ${\mathbb{F}}_4$. The quotient ${\mathcal{A}}_{T\#T}/\mathfrak{m}{\mathcal{A}}_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a cyclic module has dimension at most one.

Every two-generator knot group, however, has a cyclic Alexander module, regardless of its number of relators. Indeed its primitive abelianization vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one module on the second edge, and any quotient by lifted relators is cyclic. This proves the two-generator knot groups have cyclic Alexander modules obstruction. Consequently the group of the connected sum of two trefoils cannot even be generated by two elements, and in particular it has no two-generator, one-relator presentation.

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.

Both relators have zero exponent sums, so abelianization gives $H_1(Y)={\mathbb{Z}}^3$, with meridian variables $x,y,z$ corresponding to $a,b,c$. The universal abelian cover has deck transformation group ${\mathbb{Z}}^3$ and coefficient group ringThe generalized Heegaard diagram gives a two-dimensional spine with one vertex, three edges and two faces. Its lifted cellular chain complex iswhere chosen lifts of the cells giveThe two columns are the abelianized Fox derivatives of $[a,bc]$ and $[ab,c]$. The Fox calculus identity gives $d_1{d}_2=0$; this can also be checked by multiplying the displayed matrices. There is no three-cell in this spine. One may use the lifted spine because its deformation retraction from $Y$ lifts to the universal abelian cover.

The maximal minors of the Alexander matrix, in row-pair order $(1,2),(1,3),(2,3)$, areTheir greatest common divisor in $R$ is $xyz-1$, since $x-1,y-1,z-1$ have no common nonunit divisor. Accordingly the multivariable Alexander polynomial isThe allowed units are $\pm x^i{y}^j{z}^k$. The single-variable specialization convention can introduce extra $(t-1)$ factors; the answer here is the genuinely multivariable Alexander polynomial.

Choose a projective presentation $0\to K\stackrel{j}{\to }P\stackrel{\pi }{\to }C\to 0$, with $P$ a projective module. Continuing it to a projective resolution showsIndeed a degree-one cocycle in the Hom functor applied to the projective resolution descends to $K$, while the degree-one coboundaries are exactly restrictions of maps from $P$. This establishes the description of the Ext functor without first assuming the extension classification.

For a module extension with inclusion $i:A\to B$ and quotient map $r:B\to C$, projectivity lifts $\pi$ to $s:P\to B$. Since $rsj=0$, there is a unique map $h:K\to A$ with $ih=sj$. A different lift changes $h$ by $\ell j$ for some $\ell :P\to A$. An equivalence of module extensions, which is the identity on both end modules, also preserves this class. We have therefore defined a map from extension classes to ${\mathrm{Ext}}_R^1(C,A)$.

Conversely, for $h:K\to A$, form the pushout of a module extensionThe map $A\to B_h$ sends $a$ to $[a,0]$, and $B_h\to C$ sends $[a,p]$ to $\pi (p)$. The first is injective because $j$ is injective. If $\pi (p)=0$, write $p=j(k)$; then $[a,p]=[a+h(k),0]$, proving exactness in the middle. The last map is surjective. Thus $B_h$ is a module extension. If $h^{\prime }=h+\ell j$, the map $[a,p]\mapsto [a-\ell (p),p]$ is an equivalence of module extensions from $B_h$ to $B_{h^{\prime }}$. Finally $[a,p]\mapsto i(a)+s(p)$ identifies the pushout of a module extension built from an existing extension with its middle module. These two constructions are inverse, proving the classification:The zero class corresponds to a split short exact sequence. Fixed end modules matter: equivalence does not permit arbitrary automorphisms of $A$ or $C$.

In the second calculation the acting group is the infinite cyclic group. Its modules are modules over the group ring $S=\mathbb{Z}[z,z^{-1}]$, not merely over the underlying ring $\mathbb{Z}$. The trivial representation has $z$ acting as the identity. Its projective resolution isMultiplication by $z-1$ is injective, and the augmentation quotient is $\mathbb{Z}$. Applying the Hom functor into the trivial module makes the differential zero, so ${\mathrm{Ext}}_S^1(\mathbb{Z},\mathbb{Z})\cong \mathbb{Z}$. Taking two copies gives $\boxed{\mathcal{E}({\mathbb{Z}}^2,\mathbb{Z})\cong {\mathbb{Z}}^2}$.

An explicit representative of this extension of trivial modules for an infinite cyclic group is the abelian group $B_{m,n}=\mathbb{Z}u\oplus \mathbb{Z}v\oplus \mathbb{Z}w$, with injection $1\mapsto u$, quotient $v\mapsto (1,0)$, $w\mapsto (0,1)$, and actionThis is an invertible action: the inverse subtracts the same multiples of $u$. Every underlying abelian-group extension splits because ${\mathbb{Z}}^2$ is free, so any group-module extension has this form after choosing lifts of the quotient basis. Replacing those lifts by multiples of $u$ does not alter $m,n$. An equivalence fixing the ends has exactly such changes of lifts, so two representatives are equivalent precisely when their ordered pairs agree. Only $(0,0)$ splits as a group-module extension; overlooking the group action would incorrectly give a single class.

A Wirtinger presentation from a connected knot diagram has one generator per arc and one relator per crossing, with one relator redundant. Its presentation complex is a finite two-dimensional CW complex $X$ with one zero-cell, $n$ one-cells, and $n-1$ two-cells, and the usual diagrammatic construction gives a homotopy equivalence $X\simeq E_K$.

The abelianization of the knot group is $H_1(E_K;\mathbb{Z})\cong \mathbb{Z}$, generated by a meridian of a knot. Every homomorphism to the cyclic group $\mathbb{Z}/2$ factors through this abelianization, and reduction modulo two is its unique surjection. Thus the requested map $\alpha$ is unique.

Its kernel determines a two-sheeted covering space $\hat{X}$. The nontrivial deck transformation acts on cellular chains and homology, giving them module structures over the group ringLift one copy of every cell of $X$; its two deck translates form a free $\hat{R}$-basis. Hence $C_0\cong \hat{R}$, $C_1\cong {\hat{R}}^n$, and $C_2\cong {\hat{R}}^{n-1}$. If $\tilde{X}$ is the infinite cyclic cover, its cellular chains are free over $R=\mathbb{Z}[t^{\pm 1}]$, and imposing $t^2=1$ gives

For an odd prime $p$, the two idempotents $e_{+}=(1+t)/2$ and $e_{-}=(1-t)/2$ split the group algebraThe plus summand is the cellular chain complex of $X$ with ${\mathbb{F}}_p$ coefficients, while the minus summand is $C_{-}=C_{\ast }^{\mathrm{cell}}(\tilde{X}){\otimes }_R{R}_{-}$. Therefore

On the minus summand, the boundary $t-1:C_1\to C_0$ becomes multiplication by $-2$, which is invertible in ${\mathbb{F}}_p$, so $H_0(C_{-})=0$. After the corresponding cancellation, the remaining square boundary matrix is an Alexander matrix specialized at $t=-1$. It is singular over ${\mathbb{F}}_p$ exactly whenor equivalently when $p$ divides the knot determinant $\det K$. Thus $H_{\ast }(C_{-})$ is nonzero exactly in that case.