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

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.