Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 8 (Non-commutative Algebras)

The mod-$p$ Iwasawa algebra is the completed group algebrawhere $U$ runs through open normal subgroups and the transition maps come from quotient maps. This is the mod-$p$ algebra, distinguished from the integral Iwasawa algebra ${\mathbb{Z}}_p[[K]]$.

Use the stated complete separated negative filtered ring structure, writing $F_{-n}={\mathfrak{m}}^n$ and $\nu (x)=\max \{n:x\in F_{-n}\}$ for a nonzero element. In the usual odd-prime congruence-group situation,The dimension-three assertion records the three independent entries of a trace-zero matrix; the argument below only needs the given graded polynomial-algebra property. Completeness and separatedness are with respect to this filtration and follow from the inverse-limit topology for the standard filtration. If $x,y\ne 0$, their leading symbols are nonzero, so their product in the graded domain is nonzero. Therefore $\nu (xy)=\nu (x)+\nu (y)$ and $xy\ne 0$. This proves the domain assertion by filtered-graded transfer for complete rings.

For a left ideal $I$, its leading symbols form a homogeneous left ideal $\mathrm{gr}I$ of the polynomial ring. The Hilbert basis theorem gives finitely many homogeneous generators; choose lifts $a_1,\dots ,a_s\in I$. Given $x\in I$, express its leading symbol using these generators, lift the homogeneous coefficients, and subtract the resulting $\sum r_j{a}_j$. The remainder is still in $I$ and has strictly larger valuation. Repeat. At a stage where the remainder has valuation $n$, the coefficient contribution for $a_j$ has valuation at least $n-\nu (a_j)$. Thus each accumulated coefficient is a Cauchy series, converging to an element of ${\Omega }_K$. The remainder tends to zero, and separatedness gives $x={\sum }_j{r}_j{a}_j$. Hence $I$ is algebraically generated by these finitely many lifts; no prior assumption that $I$ is closed was used. This proves $\boxed{{\Omega }_K\text{ is left Noetherian and a domain}}$ under the stated filtration hypothesis. The same proof for right ideals also gives right Noetherianity.

For the classical fractions, a left Noetherian domain satisfies the left Ore condition. Here is the common-multiple argument. If two nonzero principal left ideals $Ra,Rb$ had zero intersection, the ideals $Ra{b}^i$, $i\ge 0$, would form a direct sum: from a finite relation, its first term lies in $Ra\cap Rb$, and then cancellation of the nonzero $b$ proves the assertion inductively. Their partial sums give an infinite ascending chain, impossible in a left Noetherian ring. Thus $Ra\cap Rb\ne 0$, giving common left multiples for every two nonzero denominators. The right-handed argument also applies. Ore localization at all nonzero elements therefore produces the classical division ring of fractions, with expressions $s^{-1}a$ or $b{s}^{-1}$; every nonzero fraction is invertible.

There is a genuine prime restriction behind the auxiliary hypothesis. For $p=2$, the matrix $z=-I$ is a nonidentity element of $K$ of order two. In ${\mathbb{F}}_2[[K]]$, the nonzero element $z-1$ satisfies $(z-1{)}^2=z^2-2z+1=0$. It is nonzero already in a finite quotient separating $z$ from one. Thus the actual $p=2$ algebra is not a domain, and cannot have the supplied separated graded-domain filtration. The intended conclusion applies for odd $p$; at $p=2$ it instead applies to a suitable deeper torsion-free congruence subgroup, such as the kernel modulo four.

A uniform module is a nonzero module in which any two nonzero submodules intersect nontrivially. An essential extension $V\subseteq E$ means that $V$ is an essential submodule of $E$: every nonzero submodule of $E$ meets $V$. Since $R$ is a left Noetherian ring, the finitely generated $M$ is a Noetherian module.

Every nonzero Noetherian module contains a uniform submodule. Otherwise split a nonuniform submodule into two disjoint nonzero submodules, retain one for further splitting and set the other aside. If no uniform submodule is reached, the pieces set aside form an infinite direct sum, whose partial sums contradict the ascending chain condition. Now successively choose uniform submodules disjoint from the current direct sum. The process again must stop by Noetherianity, giving $V={⨁}_{i=1}^r{V}_i\subseteq M$. At termination no nonzero submodule is disjoint from $V$, soThis proves the finite uniform decomposition of a Noetherian module.

An injective hull $E(M)$ is an injective module containing $M$ essentially. It is equivalently a minimal injective extension; we use the general existence theorem for injective hulls. To prove uniqueness, let $E_1,E_2$ be two hulls of $M$. Injectivity of $E_2$ extends the identity on $M$ to $f:E_1\to E_2$. Its kernel is disjoint from $M$ and therefore zero by essentiality. Its image is injective, so its inclusion splits in $E_2$. The complementary summand is disjoint from $M$, hence zero. Thus $f$ is an isomorphism fixing $M$.

A finite direct sum of injective modules is injective, and a finite direct sum of essential inclusions is essential. For the latter assertion, start with a nonzero vector and process its coordinates in turn: if a coordinate lies outside its specified essential submodule, multiply by a ring element taking that coordinate nontrivially into that submodule. Already processed coordinates stay inside their submodules; the chosen coordinate keeps the vector nonzero. Essentiality is also transitive. Consequently both $E(M)$ and ${⨁}_{i}E(V_i)$ are injective hulls of $V$, so uniqueness givesThe essential extension of a uniform module is uniform, since intersecting any two nonzero submodules with the original uniform module already gives a nonzero common intersection.

Write $E=E(V_i)$ and $S={\mathrm{End}}_R(E)$. If $f\in S$ is injective, its image is isomorphic to the injective module $E$, hence is a direct summand. Uniformity leaves no nonzero complementary summand, so $f$ is an automorphism. Therefore the nonunits are exactly the noninjective endomorphisms. Let $J$ be their set. Two such maps have nonzero kernels, whose intersection is nonzero by uniformity; their sum kills this intersection. For a product, composition on the left preserves the original kernel. Composition on the right either has a noninjective right factor, or that factor is an automorphism and transports the original nonzero kernel. Thus $J$ is a proper two-sided ideal. Every element outside $J$ is a unit, so every nonzero residue class is invertible and every proper left or right ideal is contained in $J$. This proves the local endomorphism ring of a uniform injective module assertion:For the whole hull, let $E_i=E(V_i)$. Its endomorphism ring is the ring of finite block matriceswith multiplication given by summing compositions. The diagonal entries have the local rings just proved. Off-diagonal entries are homomorphism groups, not necessarily zero; in particular this ring is not generally the product of its diagonal rings. Repeated isomorphic summands give matrix blocks over their common local endomorphism ring.

For the left regular module of $A_1(\mathbb{C})$, the Bernstein filtration has graded ring $\mathbb{C}[x,\xi ]$, so finite leading-term cancellation proves that $A_1$ is left and right Noetherian and a domain. Let $Q$ be its classical division ring of fractions. It is an essential extension of $A_1$: a nonzero fraction $q=s^{-1}a$ satisfies $sq=a\ne 0$ in $A_1$.

To prove that $Q$ is injective, use the Baer criterion. For a homomorphism $f:I\to Q$ on a nonzero left ideal, fix $a\in I\setminus \{0\}$. For any nonzero $x\in I$, take common left multiples $sx=ta$ with $s,t\ne 0$. Then $sf(x)=tf(a)$, so in the division ringThus right multiplication by $a^{-1}f(a)$ extends $f$ from $I$ to $A_1$. The zero ideal is harmless. Hence $Q$ is injective and $E({}_{A_1}{A}_1)=Q$. For any $A_1$-linear endomorphism $F$ of $Q$, clearing the left denominator in $q$ gives $F(q)=qF(1)$. These are all right multiplications, and their composition reverses the multiplying elements. ThereforeThe opposite ring in the injective hull of an Ore domain computation is essential for left-module conventions.

Work over $\mathbb{C}$ in characteristic zero. The second Weyl algebra has generators $x_1,x_2,{\partial }_1,{\partial }_2$ and relations $[{\partial }_i,x_j]={\delta }_{ij}$, $[x_i,x_j]=[{\partial }_i,{\partial }_j]=0$. Its Bernstein filtration gives every generator degree one. The ordered monomials $x_1^{a_1}{x}_2^{a_2}{\partial }_1^{b_1}{\partial }_2^{b_2}$ form a basis, soFor a nonzero finitely generated module, choose a finite-dimensional generating subspace $M_0$ and put $M_r=F_r{A}_2{M}_0$. Its graded module is finite over this polynomial ring. The Hilbert polynomial theorem gives an eventual polynomial $\dim M_r$, whose degree $d(M)$ is the Bernstein growth dimension of a Weyl algebra module, equal to its Gelfand–Kirillov dimension of a module. Two choices of finite generators bound each other after fixed filtration shifts, proving independence of the choice. The characteristic variety of a Weyl algebra module is the support of this graded module; its algebraic dimension is $d(M)$. In particular $d(M)$ is an integer at most four.

Here is a growth proof of the lower bound, rather than just quoting the Bernstein inequality for Weyl algebra modules. A nonzero $u\in F_r{A}_2$ can be turned into a nonzero scalar by at most $r$ successive commutators with the four generators: choose a monomial of maximal total degree, differentiate it to a constant using $[{\partial }_i,\cdot ]$ and $[x_i,\cdot ]$, and all different monomials are killed by that derivative multi-index. This uses the ordered-monomial expressions and characteristic zero. Expanding the iterated commutator expresses that scalar as ${\sum }_j{v}_{j}u{w}_j$ with $v_j,w_j\in F_r{A}_2$.

If $u$ acted as zero on $M_r$, every $u{w}_j{M}_0$ would be zero, so the scalar would kill the nonzero $M_0$, a contradiction. Therefore the action mapis injective. This is the faithful finite-step action of a Weyl algebra. Comparing growth ingives $4\le 2d(M)$. Thus the only possible dimensions for nonzero modules are $\boxed{d(M)\in \{2,3,4\}}$.

All three values occur in cyclic modules. On $\mathbb{C}[x_1,x_2]$, let $x_i$ multiply and ${\partial }_i$ differentiate. The module is generated by $1$, its filtered pieces are precisely polynomials of degree at most $r$, and $\dim M_r=\left(\genfrac{}{}{0ex}{}{r+2}{2}\right)$. Hence $\boxed{d(\mathbb{C}[x_1,x_2])=2}$; equivalently it is $A_2/(A_2{\partial }_1+A_2{\partial }_2)$.

For $M=A_2/A_2{\partial }_2$, the ordered basis shows that the left ideal $A_2{\partial }_2$ is spanned by monomials with positive ${\partial }_2$ exponent. The remaining basis is $x_1^{a_1}{x}_2^{a_2}{\partial }_1^{b_1}$. Its degree-$\le r$ piece has dimension $\left(\genfrac{}{}{0ex}{}{r+3}{3}\right)$, so $\boxed{d(A_2/A_2{\partial }_2)=3}$. Finally the regular module has the four-variable ordered basis above, giving $\boxed{d(A_2)=4}$.

The dimension-two modules are the holonomic Weyl algebra modules. Exact sequences satisfy $d(M)=\max \{d(M^{\prime }),d(M^{\prime \prime })\}$, by using compatible good filtrations and the resulting Hilbert polynomials. Therefore nonzero subquotients of a holonomic module remain holonomic. Their positive integer multiplicities, namely $2!$ times the leading Hilbert-polynomial coefficient, add in exact sequences. Every nonzero successive factor consumes at least one unit of multiplicity, bounding the length of a submodule chain; hence holonomic modules have finite composition length. The polynomial module above is even simple: differentiating a nonzero polynomial to a constant and then multiplying by arbitrary polynomials generates the whole module. In contrast the regular and dimension-three examples exhibit the larger growth allowed for nonholonomic modules. These examples and the growth bound explain both the range and its representation-theoretic significance.

For a finitely generated algebra over $k$, choose a finite-dimensional generating subspace $V$ containing one. Its Gelfand–Kirillov dimension isChanging generators bounds the two filtrations by constant rescalings of the index, so the value is independent of $V$.

For the noncommutative Krull dimension, set the zero module's dimension to $-1$. A nonzero module has dimension zero exactly when it is an Artinian module. Recursively, dimension at most $d$ means that in every descending chain of submodules the successive factors eventually have dimension less than $d$; use the ordinal version of this definition if necessary. The left Krull dimension of $R$ is that of its left regular module. For commutative Noetherian rings this agrees with the usual prime-chain Krull dimension, but prime-chain counting alone is not its definition for a noncommutative ring.

Write $U=U({\mathfrak{sl}}_2(\mathbb{C}))$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. By the Poincaré-Birkhoff-Witt theorem, $f^i{h}^j{e}^k$ form a basis. In the total-degree filtration there are $\left(\genfrac{}{}{0ex}{}{n+3}{3}\right)$ such monomials of degree at most $n$, soThe commutators lower degree, giving $\mathrm{gr}U\cong \mathbb{C}[f,h,e]$, which also proves that $U$ is left and right Noetherian by finite leading-term cancellation.

To determine the module-theoretic Krull dimension, use the generalized Weyl algebra description and verify the dimension criterion, rather than substituting the three-dimensional graded ring. PutCommutation with $h$ is immediate; with $e$, the contributions are $4he-2(eh+he)-4e=0$, using $he-eh=2e$. The corresponding calculation with $f$ also vanishes. Thus $z$ is a Casimir element. The subalgebra $D=\mathbb{C}[h,z]$ is a polynomial ring in two variables: for a nonzero polynomial in $z$ with coefficients in $\mathbb{C}[h]$, its highest $z$ power contributes a nonzero PBW term with that many copies of both $f$ and $e$, and no lower power can cancel it.

Take $\sigma (h)=h-2$, $\sigma (z)=z$ and $t=(z-h^2-2h)/4$. The relations becomeThey identify $U$ with $D(\sigma ,t)$. Surjectivity follows from its generators; injectivity follows from PBW: the graded pieces $D$, $D{e}^r$ and $D{f}^r$ for $r>0$ are linearly independent, using the difference between the $e$ and $f$ exponents and then the highest paired exponent. Every PBW monomial lies in one of these pieces by replacing paired $f,e$ powers with polynomials in $h,z$.

We use the following general Krull dimension criterion for rank-one generalized Weyl algebras: for a commutative Noetherian base $D$ of dimension $d$, the generalized Weyl algebra has Krull dimension $d$ provided no height-$d$ maximal ideal is periodic under $\sigma$ and none contains infinitely many translates ${\sigma }^n(t)$. This is Theorem 5.3, specialized to the commutative base, in Krull dimension of Generalized Weyl Algebras with non-commutative coefficients.

Here every maximal ideal is ${\mathfrak{m}}_{\lambda ,\mu }=(h-\lambda ,z-\mu )$ and has height two. Its translates have $h$ coordinate $\lambda +2n$, so no nonzero power of $\sigma$ fixes it. Further,This is a nonzero quadratic polynomial in $n$, so it vanishes for at most two integers. Both exceptional mechanisms are excluded for every maximal ideal, and the base has dimension two. The criterion consequently givesThe orbit and root calculations are needed for this upper bound on the entire left-ideal lattice; merely exhibiting a length-two chain of prime ideals would only prove a lower bound.

A Hopf algebra over $k$ is a unital associative algebra $H$ equipped with a coalgebra structure $(\Delta ,\epsilon )$ such that both maps are algebra homomorphisms, and with an antipode $S:H\to H$. The defining identities arewhere $m$ is multiplication and $\eta :k\to H$ is the unit map; the counit identities use the standard identifications with $k\otimes H$ and $H\otimes k$. The first compatibility conditions define a bialgebra; the antipode provides its convolution inverse. Multiplication commutativity and coproduct cocommutativity are independent conditions, as the two examples below demonstrate.

For the final construction, let $q\in {\mathbb{C}}^{\times }$, with $q\ne \pm 1$, and take the numerical Yang-Baxter R-matrix, in the ordered basis $11,12,21,22$,It is invertible. Its four action rules are $11\mapsto q11$, $12\mapsto 12+(q-q^{-1})21$, $21\mapsto 21$, $22\mapsto q22$. Applying these rules on the eight basis vectors of $V^{\otimes 3}$ verifies $R_{12}{R}_{13}{R}_{23}=R_{23}{R}_{13}{R}_{12}$, the Yang-Baxter equation.

In the RTT construction, introduce $T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and impose $R{T}_1{T}_2=T_2{T}_{1}R$, where $T_1=T\otimes I$, $T_2=I\otimes T$. Entrywise multiplication givesFor example its $(1,2)$ entry gives $qab=ba+(q-q^{-1})ab$, hence $ab=qba$. DefineThese maps descend to the RTT algebra. Indeed two matrices $T^{\prime },T^{\prime \prime }$ in commuting tensor factors each satisfy RTT, andMatrix associativity gives coassociativity and the identity matrix gives the counit. Thus RTT first gives a bialgebra; the R-matrix alone does not automatically provide an antipode.

Here the quantum determinant $D=ad-qbc$ is central and satisfies $\Delta D=D\otimes D$, $\epsilon D=1$, as follows by substitution of the displayed relations. Impose $D=1$. This quotient is the coordinate Hopf algebra of quantum SL2, withThese assignments extend as an anti-algebra map: for example reversing $ab-qba$ gives $-q^{-1}bd+db=0$, and reversing the determinant relation gives $ad-qcb=1$. The remaining relations are checked in the same way. Direct multiplication gives $TS(T)=S(T)T=I$: the first diagonal entry of $TS(T)$ is $ad-qbc=1$, its second is $da-q^{-1}cb=1$, and the off-diagonal entries vanish by the $q$-commutation relations. These are precisely the antipode identities on generators, proving that the quotient is a Hopf algebra.

To verify both required failures of symmetry, map this algebra onto the quantum Laurent plane with generators $u,u^{-1},v$ and relation $uv=qvu$, by $a\mapsto u$, $b\mapsto v$, $c\mapsto 0$, $d\mapsto u^{-1}$. Construct this algebra directly on the vector space with basis $u^i{v}^j$, $i\in \mathbb{Z}$, $j\ge 0$, and product $(u^i{v}^j)(u^k{v}^{\ell })=q^{-jk}{u}^{i+k}{v}^{j+\ell }$. The exponent identity $-jk-(j+\ell )m=-\ell m-j(k+m)$ makes multiplication associative, and the basis products give $uv=qvu$ with $u$ invertible. Since $uv\ne vu$ for $q\ne 1$, the Hopf algebra is noncommutative. Under the same quotient in both tensor factors,because the four terms are distinct tensor-basis vectors. Hence it is also noncocommutative. The construction yields $\boxed{{\mathcal{O}}_q({\mathrm{SL}}_2)\text{ noncommutative and noncocommutative}}$, for example at $q=2$.

Take $H={\mathbb{C}}^{S_3}$, the function Hopf algebra of a finite group. Multiplication is pointwise, andThe group associativity, identity and inverse laws respectively verify coassociativity, the counit and the antipode; pointwise products show that $\Delta ,\epsilon$ are algebra homomorphisms. Thus $H$ is a commutative Hopf algebra.

For $g=(12)$ and $h=(23)$, using right-to-left permutation composition, $gh=(123)$ whereas $hg=(132)$. The delta function $f={\delta }_{(123)}$ has $(\Delta f)(g,h)=1$ but $({\Delta }^{\mathrm{op}}f)(g,h)=0$. Therefore $\boxed{{\mathbb{C}}^{S_3}\text{ is commutative but noncocommutative}}$.

Take the group algebra $H=\mathbb{C}[S_3]$ withon its group basis, extended linearly. Multiplication of group basis elements makes $\Delta ,\epsilon$ algebra homomorphisms, and $g^{-1}g=g{g}^{-1}=1$ verifies the antipode identities. The coproduct is fixed by the tensor flip on every basis element, so this is a cocommutative Hopf algebra. But $(12)(23)$ and $(23)(12)$ are distinct basis elements, proving noncommutativity. Thus $\boxed{\mathbb{C}[S_3]\text{ is noncommutative but cocommutative}}$.