Definition 3

Let $A$ be a filtered ring, $(A_n)_{n \in \mathbf{Z}}$ its filtration and $E$ an $A$-module. A filtration $(E_n)_{n \in \mathbf{Z}}$ on $E$ is called compatible with its module structure over the filtered ring $A$ if

The $A$-module $E$ with this filtration is called a filtered module.

The $E_n$ are all $A_n$-modules; if $B = \bigcup_{n \in \mathbf{Z}} A_n$, clearly $\bigcup_{n \in \mathbf{Z}} E_n$ is a $B$-module and so is $\bigcap_{n \in \mathbf{Z}} E_n$ by the same argument as above for $\bigcap_{n \in \mathbf{Z}} A_n$. If $A_n = A$, all the $E_n$ are submodules of $E$.

Examples

(1) Let $A$ be a graded ring of type $\mathbf{Z}$; for all $i \in \mathbf{Z}$, let $A_{(i)}$ be the subgroup of homogeneous elements of degree $i$ in $A$. Let us write $A_n = \sum_{i \geq n} A_{(i)}$; then it is immediate that $(A_n)$ is an exhaustive and separated decreasing filtration which is compatible with the ring structure on $A$; this filtration is said to be associated with the graduation $(A_{(i)})_{i \in \mathbf{Z}}$ and the filtered ring $A$ is said to be associated with the given graded ring $A$.

Now let $E$ be a graded module of type $\mathbf{Z}$ over the graded ring $A$ and for all $i \in \mathbf{Z}$ let $E_{(i)}$ be the subgroup of homogeneous elements of degree $i$ of $E$. Let us write $E_n = \sum_{i \geq n} E_{(i)}$; then $(E_n)$ is an exhaustive and separated decreasing filtration which is compatible with the module structure on $E$ over the filtered ring $A$; this filtration is said to be associated with the graduation $(E_{(i)})_{i \in \mathbf{Z}}$ and the filtered module $E$ is said to be associated with the given graded module $E$.

(2) Let $A$ be a filtered ring, $(A_n)_{n \in \mathbf{Z}}$ its filtration and $E$ an $A$-module. Let us write $E_n = A_n E$; it follows from (1) that

and from (2) that $E_0 = E$; then $(E_n)$ is an exhaustive filtration which is compatible with the $A$-module structure on $E$. This filtration is said to be derived from the given filtration $(A_n)$ on $A$; note that it is not necessarily separated, even if $(A_n)$ is separated and $E$ and the $A_n$ are finitely generated $A$-modules (cf. , Exercise 2; see however § 3, no. 3, Proposition 5 and no. 2, Corollary to Proposition 4)

(3) Let $A$ be a ring and $m$ a two-sided ideal of $A$. Let us write $A_n = m^n$ for $n \geq 0$, $A_n = A$ for $n < 0$; it is immediate that $(A_n)$ is an exhaustive filtration on $A$, called the $m$-adic filtration. Let $E$ be an $A$-module; the filtration $(E_n)$ derived from the $m$-adic filtration on $A$ is called the $m$-adic filtration on $E$; in other words, $E_n = m^n E$ for $n \geq 0$ and $E_n = E$ for $n < 0$.

If $A$ is commutative and $B$ is an $A$-algebra, $n = mB$ is a two-sided ideal of $B$ and, for every $B$-module $F$, $n^k F = m^k F$ and hence the $n$-adic filtration on $F$ coincides with the $m$-adic filtration (if $F$ is considered as an $A$-module).

(4) If $A$ is a filtered ring and $(A_n)$ its filtration, the left $A$-module $A_n$ is a filtered $A$-module with the filtration $(A_n)$. On the other hand, clearly $(A_n)$ is a filtration which is compatible with the ring structure on the opposite ring $A^0$ and $A_n$ is a filtered (left) $A^0$-module with the filtration $(A_n)$.

(5) On a ring $A$ the sets $A_n$ such that $A_n = 0$ for $n > 0$, $A_n = A$ for $n \leq 0$ form what is called a trivial filtration associated (Example 1) with the trivial graduation on $A$; on an $A$-module $E$, every filtration $(E_n)$ consisting of sub-$A$-modules is then compatible with the module structure on $E$ over the filtered ring $A$. Then it is possible to say that every filtered commutative group $G$ is a filtered $\mathbf{Z}$-module, if $\mathbf{Z}$ is given the trivial filtration.

Let $G$ be a filtered group and $(G_n)_{n \in \mathbf{Z}}$ its filtration; clearly, for every subgroup $H$ of $G$, $(H \cap G_n)_{n \in \mathbf{Z}}$ is a filtration said to be induced by that on $G$; it is exhaustive (resp. separated) if that on $G$ is. Similarly, if $H$ is a normal subgroup of $G$, the family $((H.G_n)/H)_{n \in \mathbf{Z}}$ is a filtration on the group $G/H$, called the quotient under $H$ of the filtration on $G$; it is exhaustive if $(G_n)$ is. If $G'$ is another filtered group and $(G'_n)_{n \in \mathbf{Z}}$ its filtration, $(G_n \times G'_n)$ is a filtration on $G \times G'$ called the product of the filtrations on $G$ and $G'$, which is exhaustive (resp. separated) if $(G_n)$ and $(G'_n)$ are.

Now let $A$ be a filtered ring and $(A_n)$ its filtration; on every subring $B$ of $A$, clearly the filtration induced by that on $A$ is compatible with the ring structure on $B$. If $b$ is a two-sided ideal of $A$, the quotient filtration on $A/b$ of that on $A$ is compatible with the structure of this ring, for

If $A'$ is another filtered ring, the product filtration on $A \times A'$ is compatible with the structure of this ring.

Finally let $E$ be a filtered $A$-module and $(E_n)$ its filtration; on every submodule $F$ of $E$, the filtration induced by that on $E$ is compatible with the $A$-module structure on $F$ and, on the quotient module $E/F$, the quotient filtration of that on $E$ is compatible with the $A$-module structure, as

Note that if the filtration on E is derived from that on A (Example 2), so is the quotient filtration on E/F, but not in general the filtration induced on F (Exercise 1; see however , no. 2, Theorem 2)

If E' is another filtered A-module, the product filtration on E × E' is compatible with its A-module structure. If the filtrations on E and E' are derived from that on A (Example 2), so is their product filtration.