Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 2 (Noetherian Algebras)

Put $x=I+E_{12}$, $y=I+E_{23}$ and $z=I+E_{13}$. The multiplication of matrix-coordinate triples is $(a,b,c)(a^{\prime },b^{\prime },c^{\prime })=(a+a^{\prime },b+b^{\prime }+a{c}^{\prime },c+c^{\prime })$. In particular $z$ is central, $xy=zyx$, and every element has the unique normal form $y^c{z}^b{x}^a$. ConsequentlyThe distinct normal-form monomials are the group basis, so this is an isomorphism, not merely a surjection from a presented algebra. This is the integral Heisenberg group algebra as a skew Laurent ring.

Here is the needed skew Hilbert basis theorem, including its proof. If $A$ is a left Noetherian ring and $\sigma$ an automorphism, let $J$ be a left ideal of $A[t;\sigma ]$, where $ta=\sigma (a)t$. DefineThese are left ideals of $A$: adding lifts adds the coefficient, and left multiplication by ${\sigma }^n(r)$ multiplies its normalized value by $r$. Multiplication of a lift by $t$ proves $I_n\subseteq I_{n+1}$. The chain stabilizes at some $N$. Choose finite generators $u_{nj}$ for each $I_n$, $0\le n\le N$, and lifts $f_{nj}\in J$ with $t^n$ coefficient ${\sigma }^n(u_{nj})$.

For $f\in J$ of degree $m\ge N$, write its normalized leading coefficient as ${\sigma }^{-m}(\mathrm{lc}f)={\sum }_j{r}_j{u}_{Nj}$. Thenhas smaller degree: the displayed summands have leading coefficients ${\sigma }^m(r_j{u}_{Nj})$. For $m<N$ use the generators for $I_m$ without the shift. Induction on degree expresses every element of $J$ in terms of the finitely many chosen lifts. Thus $A[t;\sigma ]$ is a left Noetherian ring. Passing to the opposite ring replaces $\sigma$ by ${\sigma }^{-1}$ and proves the corresponding right-sided result.

This proof includes the ordinary Hilbert basis theorem when $\sigma$ is the identity. Starting with the field $k$, it makes $k[y,z]$ Noetherian; localization at powers of $y,z$ makes $B$ Noetherian as well. Finally, for a left ideal $L$ in $B[x,x^{-1};\sigma ]$, its intersection $J$ with $B[x;\sigma ]$ has finitely many generators. For each $f\in L$, some $x^{q}f$ lies in $J$, so $f=x^{-q}(x^{q}f)$ is generated by those same elements in the skew Laurent polynomial ring. Right multiplication clears powers for a right ideal. Therefore $kG$ is a left Noetherian ring and a right Noetherian ring.

The commutative Laurent polynomial ring $B$ is an integral domain and $\sigma$ is invertible. Write nonzero elements asThe largest exponent in their product is $m+n$, with coefficientThere is only one pair of exponents contributing at this degree, and the coefficient cannot vanish in the integral domain $B$. Hence $rs\ne 0$. Equivalently, $rs=0$ forces $r=0$ or $s=0$. This is skew Laurent extensions of domains are domains, and the argument works over the original field in any characteristic.

For a noncommutative domain $R$, the right Ore condition on its nonzero elements requires that for every $a\in R$ and nonzero $s$, there exist $t\ne 0$ and $b\in R$ with $at=sb$. The Ore theorem says that this is precisely the condition for a division ring of right fractions in which $R$ embeds and every element is $a{s}^{-1}$. The left-handed condition similarly gives left fractions.

For completeness, a right Noetherian domain satisfies the condition. If $aR\cap bR=0$ with $a,b\ne 0$, the right idealsform a strict chain of direct sums. Indeed, in a relation ${\sum }_{i=0}^n{a}^{i}b{r}_i=0$, the first term lies in $bR$ and the rest in $aR$. Their zero intersection makes $b{r}_0=0$, hence $r_0=0$; cancel $a$ and repeat. Every new summand is nonzero in a noncommutative domain, contradicting the ascending chain condition. Thus $au=bv\ne 0$ exists, with $u,v\ne 0$, giving right common multiples. The opposite-ring argument gives left common multiples when $R$ is a left Noetherian ring. This proves right Noetherian domains satisfy the Ore condition.

A sketch of the fraction construction also explains the theorem. Pairs $(a,s)$ represent $a{s}^{-1}$. They represent the same fraction when a common right denominator $su=tv\ne 0$ also satisfies $au=bv$. Addition uses $(au+bv)(su{)}^{-1}$. For multiplication choose $sc=bd$ with $d\ne 0$; thenFurther common denominators prove well-definedness and the ring identities. The original map is injective because all nonzero denominators are regular; a zero fraction numerator would satisfy $au=0$ for $u\ne 0$. Every nonzero fraction has inverse $s{a}^{-1}$, so the ring is a division ring. This construction has the universal property of inverting all nonzero elements.

Apply this to the two-sided Noetherian noncommutative domain $kG$. Its left Ore property expresses any right fraction as a left one: choose $u\ne 0$ with $ua=bs$, yielding $a{s}^{-1}=u^{-1}b$. Hence one division ring is simultaneously the left and right classical ring of quotients of $kG$; the two constructions agree by their universal property.

Work over $\mathbb{C}$, or another field of characteristic zero. The Weyl algebra $A_n$ is generated by $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ with commuting $x$'s, commuting $\partial$'s and $[{\partial }_i,x_j]={\delta }_{ij}$. Its ordered monomial basis of a Weyl algebra consists of the monomials $x^{\alpha }{\partial }^{\beta }$. One way to verify independence is to apply a finite operator relation to the formal exponential $e^{\mathbf{t}\cdot \mathbf{x}}$: after canceling that exponential the relation becomes a polynomial identity $\sum c_{\alpha \beta }{\mathbf{x}}^{\alpha }{\mathbf{t}}^{\beta }=0$, so every coefficient vanishes. The relations already reorder every word into that basis.

Give both generators degree one. The Bernstein filtration isThus $\dim F_m{A}_n=\left(\genfrac{}{}{0ex}{}{m+2n}{2n}\right)$. Commutator terms have lower degree, accounting for the commutative symbol algebra. Since that algebra is Noetherian, leading-symbol generation followed by finite downward degree cancellation proves that $A_n$ is Noetherian on both sides.

A left finitely generated module $M$ has a good filtration, for example $M_m={\sum }_j{F}_m{A}_n{m}_j$ for a finite generating set of the module. More generally one allows fixed degree shifts of the generators. The associated graded module is finite over the symbol algebra. DefineThis is the characteristic variety of a Weyl algebra module for the Bernstein algebra filtration. It records where the module's leading symbols remain nonzero and makes noncommutative module growth accessible to commutative algebra.

For a fixed algebra filtration it is independent of the good filtration. Any two such filtrations differ by bounded shifts $M_i\subseteq {\tilde{M}}_{i+c}$ and conversely, since each finite set of generators occurs in bounded degree in the other filtration. If a degree-$d$ operator has symbol annihilating $\mathrm{gr}M$, its $r$th power maps $M_i$ into $M_{i+rd-r}$. For $r>2c$, bounded-shift comparison forces the power of that symbol to annihilate $\mathrm{gr}\tilde{M}$. Reversing the argument gives equal radicals of ideals of annihilators. This is independence of characteristic support from a good module filtration. Changing the algebra filtration itself, for example to the order filtration of a Weyl algebra with the $x_i$ in degree zero, need not preserve the same variety; the convention matters.

The Hilbert-Serre theorem gives an eventual cumulative polynomialHere $d$ is also the Gelfand–Kirillov dimension of a module, and $e(M)>0$ is an integer multiplicity of a filtered module for nonzero $M$. Finite generation bounds $d\le 2n$.

Here is a proof sketch of the essential lower bound, rather than an appeal to its name. For a nonzero $f\in F_m{A}_n$, choose a nonzero monomial of greatest total degree in its ordered expansion, say $x^{\alpha }{\partial }^{\beta }$. The operations $\mathrm{ad}{\partial }_i$ and $\mathrm{ad}(-x_i)$ differentiate this expansion with respect to $x_i$ and ${\partial }_i$. Apply them ${\alpha }_i$ and ${\beta }_i$ times. Lower-degree terms die, and every other monomial of the same degree also dies unless it has exactly those exponents. The result is a nonzero scalar multiple of the chosen coefficient. Characteristic zero is essential for the factorials. Expanding the repeated commutators therefore givesChoose $0\ne v\in M_0$. If $f$ acted as zero on all $M_m$, every $f{v}_{j}v$ would vanish, contrary to $v={\sum }_j{u}_{j}f{v}_{j}v$. Hence the linear action mapis injective. It follows that $\dim F_m{A}_n\le (\dim M_m)(\dim M_{2m})$. Comparing polynomial degrees gives $2n\le 2d$. ThusThis proves Bernstein inequality for Weyl algebra modules through the faithful finite-step action of a Weyl algebra.

The bound is sharp. The polynomial representation $\mathbb{C}[x_1,\dots ,x_n]\cong A_n/{\sum }_i{A}_n{\partial }_i$ has the zero section as its characteristic variety of a Weyl algebra module, namely $\xi =0$ and dimension $n$. It is a simple module: differentiate a nonzero polynomial sufficiently to obtain a nonzero constant, then multiply by the $x_i$ to generate the whole module. The module $A_n/{\sum }_i{A}_n{x}_i$ similarly has characteristic variety of a Weyl algebra module $x=0$, a fiber of the cotangent bundle. A nonzero free module has the whole cotangent bundle and dimension $2n$. No nonzero finite-dimensional module exists for $n\ge 1$, also seen by taking the trace of $[{\partial }_1,x_1]=I$.

Modules attaining dimension $n$ are holonomic Weyl algebra modules (with the zero module included conventionally). Every nonzero subquotient remains of dimension $n$ by the lower bound and monotonicity of growth. Compatible filtrations in a short exact sequence make the leading multiplicity of a filtered module additive. Since each nonzero factor has positive integer multiplicity of a filtered module, the number of factors in any strict submodule chain is bounded by $e(M)$. Together with Noetherianity, this proves finite composition length. It illustrates why characteristic variety of a Weyl algebra module and multiplicity of a filtered module provide useful structure beyond a list of presentations. In positive characteristic there are finite-dimensional Weyl modules, so the characteristic-zero inequality and this discussion must not be asserted without its field hypothesis.

Whenever the graded pieces $N_i$ are finitely generated over $S_0$, define the Poincare series of a graded module byAdditivity means $\lambda (B)=\lambda (A)+\lambda (C)$ for every short exact sequence $0\to A\to B\to C\to 0$ in the indicated category. Dimension over a field is a basic example; length is another when the modules have finite length.

The Hilbert-Serre theorem needs finite-generation hypotheses: take $S_0$ commutative Noetherian and $S$ generated over it by homogeneous $x_1,\dots ,x_r$ of positive degrees $d_1,\dots ,d_r$. For finitely generated graded $N$, every $N_i$ is then finite over $S_0$, andIf negative grading indices are allowed for the module, the numerator is a Laurent polynomial instead. In standard degree one, the coefficients are eventually polynomial functions of the index.

To prove this, regard $N$ as finite over the surjective polynomial presentation $S_0[X_1,\dots ,X_r]\to S$. Induct on $r$. At $r=0$ only finitely many homogeneous generator degrees occur, so the series is a polynomial. For the last variable $x$ of degree $d$, put $K=(0{:}_{N}x)$ and $Q=N/xN$. The polynomial ring is Noetherian by the Hilbert basis theorem, hence $K$ is finite; both $K$ and $Q$ are finite modules over the ring with that variable removed. The degree-$i$ exact sequence isAdditivity, followed by summing over $i$, givesThe induction hypothesis supplies the remaining denominator factors. This proves Hilbert-Serre theorem for an additive coefficient function. If all $d_i=1$, expanding $(1-t{)}^{-r}$ proves eventual polynomial coefficients directly.

An arbitrary positively graded ring in the introductory wording is not enough by itself. For example $S=k[X_1,X_2,\dots ]$ with every variable of degree one and $N=S$ is a cyclic graded module, but $N_1$ is not finite-dimensional over $k$, so the dimension-valued series is not even defined. The theorem above states the standard hypotheses that make the requested definition and rationality valid.

Choose a basis $x_1,x_2,x_3$ of the Lie algebra and filter its universal enveloping algebra by word length: $F_{0}R=\mathbb{C}$ and $F_{d}R$ is spanned by products of at most $d$ basis elements. The relations $x_j{x}_i-x_i{x}_j=[x_j,x_i]$ replace a degree-two commutator by degree one, so the degree-one symbols commute.

The Poincaré-Birkhoff-Witt theorem says the ordered monomials $x_1^{a_1}{x}_2^{a_2}{x}_3^{a_3}$ are a basis. Its relevant proof can be seen by reordering an inverted neighboring pair using the displayed relation. Reordering terminates because same-length terms have fewer inversions and bracket terms have smaller word length. Disjoint swaps commute; the only overlapping three-letter reorderings differ by $[x_i,[x_j,x_k]]+[x_j,[x_k,x_i]]+[x_k,[x_i,x_j]]=0$. The Jacobi identity therefore makes the reduced ordered expression independent of the reordering path. Ordered monomials have no remaining relations and form the basis. Their degree-$d$ symbols giveFor generators $m_1,\dots ,m_q$ of the finitely generated module $M$, put $M_d={\sum }_j{F}_{d}R{m}_j$. This is a good filtration, and $\mathrm{gr}M$ is finite over the three-variable polynomial ring. Define the dimension of a filtered module as the Krull dimension of its associated graded module, equivalently the degree of the eventual cumulative polynomial ${\dim }_{\mathbb{C}}{M}_d$, or the Gelfand–Kirillov dimension of a module. Two good filtrations have bounded shifts, so the growth degree is independent of the generators. Sinceits polynomial degree is at most three. Therefore every nonzero finitely generated $R$-module has dimension at most $3$. The zero module may be assigned the usual separate empty-support convention. This bound also follows immediately from its support of a module inside affine three-space.

Let $I=(a_1,\dots ,a_s)$ and form the Rees algebra $\mathcal{R}={⨁}_{j\ge 0}{I}^j{t}^j$. It is generated over $R$ by $a_{1}t,\dots ,a_{s}t$, so the Hilbert basis theorem makes it Noetherian. The Rees module $\mathcal{M}={⨁}_{j\ge 0}{I}^{j}M{t}^j$ is finite over it, generated by finitely many generators of the module $M$ in degree zero. Hence its graded submodulehas finitely many homogeneous generators, all in degrees at most some $k$.

For $i\ge k$, each degree-$d$ generator contributes a multiple from $I^{i-d}(N\cap I^{d}M)$. This is contained in $I^{i-k}(N\cap I^{k}M)$, because multiplication by $I^{k-d}$ puts its coefficient both in $N$ and in $I^{k}M$. Thus $N\cap I^{i}M\subseteq I^{i-k}(N\cap I^{k}M)$. The reverse inclusion follows directly by multiplication: it stays in $N$ and in $I^{i}M$. ConsequentlyThis is the Artin-Rees lemma, proved through finite graded generation rather than an assumption that the intersection filtration initially has that form.

Write the negative filtration as $F_{0}R=R\supseteq F_{-1}R\supseteq F_{-2}R\supseteq \cdots$, with $F_{i}R{F}_{j}R\subseteq F_{i+j}R$. Completeness is used in the separated sense $R\cong \underleftarrow{\lim }R/F_{-n}R$; in particular the intersection of all filtration terms is zero.

For any left ideal $J$, its induced associated graded ideal is a left ideal of $\mathrm{gr}R$. Choose finite homogeneous generators and lift them to $j_1,\dots ,j_r\in J$, with orders $n_i\ge 0$, meaning $j_i\in F_{-n_i}R\setminus F_{-n_i-1}R$. Let $j\in J$. At each order $s$, the leading symbol of the current residual is a sum of the generator symbols with homogeneous coefficients. Lift these coefficients to $c_{i,s}\in F_{-(s-n_i)}R$ (zero when $n_i>s$) and subtract ${\sum }_i{c}_{i,s}{j}_i$. The new residual lies in $F_{-(s+1)}R$.

Iterating expresses $j$ as partial sums of ${\sum }_i{c}_i{j}_i$, with residuals tending to zero. For fixed $i$ the successive coefficient increments have orders tending to infinity. Separated completeness therefore supplies $c_i={\sum }_s{c}_{i,s}\in R$. Multiplication is continuous by the multiplicative filtration, soThe sum on the right is finite, despite the convergent process used to obtain its coefficients. It belongs to the actual left ideal generated by the lifts; no assumption that $J$ is closed was made. Every left ideal is consequently finitely generated, and $R$ is a left Noetherian ring. This is the left-handed complete negative filtered-graded transfer of Noetherianity; the separated-completeness hypothesis is essential to the limiting argument.

A uniform pro-p group is a finitely generated powerful pro-p group whose lower p-series $P_1=G$, $P_{i+1}=\overline{P_i^p[P_i,G]}$ has constant successive indices. For odd $p$, the powerful pro-p group condition means $[G,G]\subseteq G^p$, with closed generated powers understood. For $p=2$ it means $[G,G]\subseteq G^4$. For finitely generated powerful pro-p groups, uniformity is equivalent to being a torsion-free group.

It is convenient to use general matrix size $d$, then put $d=p$. Let $U_m=1+p^m{M}_d({\mathbb{Z}}_p)$. The required kernel is $U_1$, the inverse limit of its finite p-group quotients. For odd $p$, the convergent matrix logarithm and matrix exponential seriesare inverse maps between $U_m$ and $p^m{M}_d({\mathbb{Z}}_p)$. The valuation bounds for $A,B\in p{M}_d$ ensure convergence and integrality. They give $\log (u^p)=p\log u$ and hence a bijective power map $U_m\to U_{m+1}$: the unique root of $v\in U_{m+1}$ is $\exp (p^{-1}\log v)$.

Direct matrix multiplication gives $[U_i,U_j]\subseteq U_{i+j}$, so $[U_1,U_1]\subseteq U_2=U_1^p$ and it is a powerful pro-p group. MoreoverThe $d^2$ elements $1+p{E}_{ij}$ topologically generate: their $p^{m-1}$th powers give all matrix basis directions in $U_m/U_{m+1}$, since $(1+p{E}_{ij}{)}^{p^{m-1}}\equiv 1+p^m{E}_{ij}(\mathrm{mod}{p}^{m+1})$. Successive correction of each matrix entry, followed by completeness, approximates every element by words in these generators.

Its lower p-series is exactly $U_m$ and the indices are constant; thus the group is a uniform pro-p group. Alternatively, the logarithm excludes nontrivial p-power torsion. Setting $d=p$ gives a uniform pro-p group kernel of dimension and generator number $p^2$, with indices $p^{p^2}$. This is the odd-prime principal p-adic congruence subgroup is uniform result. The odd-prime condition matters: $-I$ is nontrivial order-two torsion in $1+2M_d({\mathbb{Z}}_2)$.

Use $[g]$ for the basis element of the group algebra $A={\mathbb{F}}_p[{\mathbb{Z}}_p]$, so $[g][h]=[g+h]$ and $[0]=1$. The kernel of the augmentation map is the augmentation ideal $I$. This algebra must not initially be identified with a Laurent polynomial ring: the additive group ${\mathbb{Z}}_p$ is not abstractly generated by the integer $1$.

For $r\ge 1$, map $A$ onto ${\mathbb{F}}_p[{\mathbb{Z}}_p/p^r{\mathbb{Z}}_p]$ and call its kernel $J_r$. It is generated by $[g]-1$ for $g\in p^r{\mathbb{Z}}_p$: differences of basis elements in one coset have the form $[h]([g]-1)$. Since each such $g=p^{r}h$,by the characteristic-$p$ binomial identity. Thus $J_r\subseteq I^{p^r}$.

The quotient group is a cyclic group of order $p^r$, generated by the image of $1$. Its group algebra isThe image of $I$ is $(T)$ and its $p^r$th power vanishes, proving the reverse inclusion. HenceThis is cyclic finite quotients of the p-adic augmentation filtration.

For any $n\ge 0$, choose $p^r>n+1$. The finite quotient then identifies $I^n/I^{n+1}$ with $(T^n)/(T^{n+1})$, a one-dimensional space generated by the image of $([1]-1{)}^n$. The degree-one symbol $X$ of $[1]-1$ consequently gives a surjective graded map ${\mathbb{F}}_p[X]\to {\mathrm{gr}}_{I}A$, and no power of $X$ vanishes in its degree. ThusThe same finite quotients show that the I-adic completion is ${\mathbb{F}}_p[[T]]$, and its associated graded ring is identical. The proof therefore covers the completed group algebra interpretation too, while proving the literal ordinary-algebra statement without assuming completion. Applying part ii to a ring would require its actual completeness, not just this polynomial associated graded ring.