Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 2 / 4

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.