Filtered Algebras

We give a criterion for a filtered algebra to be a polynomial ring.

We will allow our rings to be non-commutative today. In life, one hopes to have gradings.

Definition 1. Fix an \(R\)-algebra \(A\). Then we say that \(A\) is graded if and only if it is equipped with a multiplicative grading. In other words, there is a decomposition \[A=\bigoplus_{i=0}^\infty A_i\] of \(A\) as an \(R\)-module, where \(R\subseteq A_0\) and \(A_i\cdot A_j\subseteq A_{i+j}\). A morphism of graded algebras is one preserving the grading.

Example 2. Any polynomial ring \(R[x_1,\ldots,x_n]\) is graded by degree.

However, one frequently must settle for a weaker notion.

Definition 3. Fix an \(R\)-algebra \(A\). Then we say that \(A\) is filtered if and only if it is equipped with an exhaustive, multiplicative ascending filtration. In other words, we have an ascending chain \(F^0A\subseteq F^1A\subseteq\cdots\) of \(R\)-submodules so that \(R\subseteq F^0A\) and \(\bigcup_nF^nA=A\) and \(F^nA\cdot F^mA\subseteq F^{m+n}A\). A morphism of filtered algebras is one preseving the filtration.

We take a moment to emphasize that the filtrations in this blog post are ascending, rather than descending.

Example 4. If \(A\) is a graded \(R\)-algebra, then we may filter \(A\) by \[F^nA=\bigoplus_{i=0}^nA_i.\]

Luckily, there is a way to pass from a filtered algebra to a graded one.

Definition 5. Fix a filtered \(R\)-algebra \(A\). Then the associated graded ring \(\operatorname{gr}A\) of \(A\) is defined to be \[\operatorname{gr}A=\bigoplus_{n=0}^\infty\frac{F^nA}{F^{n-1}A},\] where \(F^{-1}A=0\) by convention. This construction is functorial in \(A\).

Remark 6. The given decomposition turns out to make \(\operatorname{gr}A\) into a graded \(R\)-algebra. For example, the \(R\)-module structure is well-defined because the filtered pieces are \(R\)-submodules. The multiplication is well-defined because the filtration is multiplicative.

However, it is a feature of this construction that taking the associated graded tends to remove a lot of information. Here is what is probably the canonical example, though it requires using Lie algebras.

Example 7. Fix a Lie algebra \(\mathfrak g\). Then there is a universal enveloping algebra \(U\mathfrak g\), which is simply the free (noncommutative) algebra on \(\mathfrak g\) modulo the relations \(XY-YX=[X,Y]\) whenever \(X,Y\in\mathfrak g\). In particular, one can filter the free (noncommutative) algebra on \(\mathfrak g\) by degree, but \(U\mathfrak g\) turns out to be merely a filtered algebra (by degree) because elements of the form \(XY-YX\) can be presented with smaller degree. The Poincaré–Birkhoff–Witt theorem shows that the associated graded ring of \(U\mathfrak g\) is the commutative polynomial ring on \(\mathfrak g\).

However, the associated graded can at least remember isomorphisms.

Lemma 8. Let \(f\colon A\to B\) be a map of filtered \(R\)-algebras which induces an isomorphism \(\operatorname{gr}A\to\operatorname{gr}B\). Then \(f\) is an isomorphism of filtered algebras.

Proof. We follow sx3048120. It is enough to show that we have a bijection. Because the filtrations are exhaustive, it is enough to show that \(f\) induces a bijection \(F^iA\to F^iB\) for all \(i\). We may show this by induction, where the case \(i=-1\) follows because then both groups vanish.

For the induction, we use the Five lemma. Suppose we have an isomorphism on the \(i\)th filtered piece. Then \(f\) induces a morphism \[\begin{array}{cccccccccc} 0 &\to& F^iA &\to& F^{i+1}A &\to& \operatorname{gr}_{i+1}A &\to& 0 \\\
& & \downarrow & & \downarrow & & \downarrow \\\
0 &\to& F^iB &\to& F^{i+1}B &\to& \operatorname{gr}_{i+1}B &\to& 0 \end{array}\] of short exact sequences (of \(R\)-modules), where the left vertical arrow is an isomorphsim by the induction, and the right vertical arrow is an isomorphism by hypothesis. The result follows. \(\blacksquare\)

Proposition 9. Let \(A\) be a filtered \(R\)-algebra such that \(R\subseteq Z(A)\). If \(\operatorname{gr}A\) is isomorphic to \(R[x]\), then \(A\) is isomorphic to \(R[x]\).

Proof. By Lemma 8, the main point is to figure out how to produce a map \(R[x]\to A\). Well, \(\operatorname{gr}A\) is isomorphic to \(R[x]\), so we may find a generating class \(y\in A\) inducing the isomorphism \(R[y]\to\operatorname{gr}A\). However, the choice of \(y\in A\) actually lifts this to a map \(R[y]\to A\) which becomes an isomorphism upon passing to the associated graded algebras. (One needs \(y\) to commute with \(R\) for this map to exist.) It follows that the map \(R[y]\to A\) is actually an isomorphism of filtered algebras. \(\blacksquare\)

Remark 10. Of course, one can also increase the number of variables and obtain a similar statement.