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For an increasing filtration of a module $M_n$ compatible with a filtered algebra $A$, set $\mathrm{gr}M={⨁}_n{M}_n/M_{n-1}$. Multiplication on the quotients makes it a graded module over $\mathrm{gr}A$. Its cumulative homogeneous dimensions equal $\dim M_n$ for a finite-dimensional exhaustive nonnegative filtration of a module.

For a nonzero right module $M$ that is a finitely generated module, choose a finite-dimensional generating vector subspace $W$ and define ${\mathrm{GKdim}}_{A}M={\mathrm{limsup}}_{n\to \infty }\log ({\dim }_{k}W{V}^n)/\log n$, where $V$ generates the algebra and contains $1$. For a left module use $V^{n}W$. It is independent of these choices and measures polynomial-order growth of a generating filtration of a module.

For a filtered algebra $A$ with finite-dimensional filtration pieces, a good filtration of a module on a finitely generated module is compatible with multiplication and has finitely generated associated graded module over the associated graded ring. Taking bounded-degree translates of a finite set of module generators supplies such a filtration of a module when the associated graded ring is Noetherian.

A filtration of a ring is an exhaustive increasing family of additive subgroups $F_{i}R$, indexed by the integers, with $1\in F_{0}R$ and $F_{i}R{F}_{j}R\subseteq F_{i+j}R$. A compatible filtration of a module on a right module satisfies $F_{i}M{F}_{j}R\subseteq F_{i+j}M$. The associated graded module has components $F_{i}M/F_{i-1}M$, and its action by the associated graded ring isChanging either representative changes the product only by an element of $F_{i+j-1}M$, so the action is well-defined.

The subspace filtration on $N$ is $F_{i}N=N\cap F_{i}M$. The quotient filtration is $F_i(M/N)=(F_{i}M+N)/N$. Inclusion induces an injective map $\mathrm{gr}N\to \mathrm{gr}M$, and the quotient induces a surjective map to $\mathrm{gr}(M/N)$. If $m\in F_{i}M$ maps to zero in its degree-$i$ quotient, write $m=n+m^{\prime }$ with $n\in N$ and $m^{\prime }\in F_{i-1}M$. Then $n\in N\cap F_{i}M$, so the kernel is exactly the image of $\mathrm{gr}N$. We have the short exact sequencewhich is exactness of associated graded modules for induced filtrations.

For the negative case, $F_{i}R=R$ for $i\ge 0$. Completeness means that $R\to {\underleftarrow{\lim }}_{q}R/F_{-q}R$ is an isomorphism; in particular the filtration of a ring is separated. If $x\in F_{-1}R$ and $r\in R$, then $(xr{)}^q\in F_{-q}R$. The geometric seriesconverges in the complete negative filtration and is a two-sided inverse of $1-xr$, by the finite geometric identity and continuity of multiplication. The unit criterion for the Jacobson radical now gives

Let $I$ be any right ideal. The associated graded module $\mathrm{gr}I$ for its subspace filtration is a homogeneous ideal on the right in $\mathrm{gr}R$. If $\mathrm{gr}R$ is a right Noetherian ring, choose homogeneous generators that lift to $x_1,\dots ,x_s\in I$, with degrees $d_1,\dots ,d_s\le 0$.

For $x\in I\cap F_{d}R$, cancel its degree-$d$ symbol by a sum ${\sum }_i{x}_i{a}_{i,0}$ with $a_{i,0}\in F_{d-d_i}R$. The remainder belongs to $I\cap F_{d-1}R$. Repeat the cancellation in each succeeding degree. After $q$ steps,For every $i$, the coefficient series converges to some $a_i\in R$ by completeness. Passing to the limit gives $x={\sum }_i{x}_i{a}_i$. Thus these finitely many $x_i$ generate the actual right ideal $I$; no assumption that $I$ is closed is needed. Since every right ideal is finitely generated, $R$ is right Noetherian, by complete negative filtered-graded transfer of Noetherianity.