§ 2. General results on filtered rings and modules

1. Filtered rings and modules

Definition 1

An increasing (resp. decreasing) sequence $(G_n)_{n \in \mathbf{Z}}$ of subgroups of a group $G$ is called an increasing (resp. decreasing) filtration on $G$ . A group with a filtration is called a filtered group.

If $(G_n)_{n \in \mathbf{Z}}$ is an increasing (resp. decreasing) filtration on a group $G$ and we write $G'_n = G_{-n}$ , clearly $(G'_n)_{n \in \mathbf{Z}}$ is a decreasing (resp. increasing) filtration on $G$ . We may therefore restrict our study to decreasing filtrations and hence forth when we speak of a filtration, we shall mean a decreasing filtration, unless otherwise stated.

Given a decreasing filtration $(G_n)_{n \in \mathbf{Z}}$ on a group $G$ , clearly $\bigcap_{n \in \mathbf{Z}} G_n$ and $\bigcup_{n \in \mathbf{Z}} G_n$ are subgroups of $G$ ; the filtration is called separated if $\bigcup_{n \in \mathbf{Z}} G_n$ is reduced to the identity element and exhaustive if $\bigcup_{n \in \mathbf{Z}} G_n = G$ .

Definition 2

Given a ring $A$ , a filtration $(A_n)_{n \in \mathbf{Z}}$ over the additive group $A$ is called compatible with the ring structure on $A$ if

(1)

A_m A_n \subset A_{m+n} \quad \text{for} \quad m \in \mathbf{Z}, \quad n \in \mathbf{Z}

(2)

1 \in A_0.

The ring

A

with this filtration is then called a filtered ring.

Conditions (1) and (2) show that $A_n$ is a subring of $A$ and the $A_n$ (left and right) $A_0$ -modules. The set $B = \bigcup_{n \in \mathbf{Z}} A_n$ is a subring of $A$ and the set $n = \bigcap_{n \in \mathbf{Z}} A_n$ , a two-sided ideal of $B$ ; for if $x \in n$ and $a \in A_k$ for all $k \in \mathbf{Z}$ , $x \in A_{k-p}$ , whence $ax \in A$ , and $xa \in A$ , by (1); therefore $ax \in n$ and $xa \in n$ .

An important particular case is that in which $A_n = A$ ; then $A_n = A$ for $n \leq 0$ and all the $A_n$ are two-sided ideals of $A$ .

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

A_m E_n \subset E_{m+n} \quad \text{for} \quad m \in \mathbf{Z}, \quad n \in \mathbf{Z}.

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

A_m E_n = A_m A_n E \subset A_{m+n} E = E_{m+n},

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)

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

(A_n + b)(A_m + b) \subset A_{n+m} + b.

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

A_n(F + E_m) \subset F + A_n E_m \subset F + E_{m+n}.

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.

2. The order function

Let A be a filtered ring, E a filtered A-module and (E_n) the filtration of E. For all x ∈ E let v(x) denote the least upper bound in R of the set on integers n ∈ Z such that x ∈ E_n. Then the following equivalences hold:

\begin{cases} v(x) = -\infty \Leftrightarrow x \notin \bigcup_{n \in \mathbf{Z}} E_n \\ v(x) = p \Leftrightarrow x \in E_p \text{ and } x \notin E_{p+1} \\ v(x) = +\infty \Leftrightarrow x \in \bigcap_{n \in \mathbf{Z}} E_n \end{cases}

The mapping v : E → \overline{\mathbf{R}} is called the order function of the filtered module E. If v is known then so are the E_n, for E_n is the set of x ∈ E such that v(x) ≥ n; the fact that the E_n are additive subgroups of E implies the relation

v(x - y) \geq \inf(v(x), v(y)).

The above definition applies in particular to the filtered A-module A; let w be its order function. It follows from equation (3) of no. 1 that for a ∈ A and x ∈ E,

v(ax) \geq w(a) + v(x)

whenever the right-hand side is defined; in particular, for a ∈ A and b ∈ A,

w(ab) \geq w(a) + w(b)

whenever the right hand side is defined.

The order function is defined similarly on a filtered group G which is not necessarily commutative; the corresponding relation to (5) is then written

v(yx^{-1}) = v(xy^{-1}) \geq \inf(v(x), v(y)).

3. The graded module associated with a filtered module

Let G be a commutative group (written additively) and (G_n) a filtration on G. Let us write

\begin{align*} \operatorname{gr}_n(G) &= G_n / G_{n+1} \quad \text{for } n \in \mathbf{Z} \\ \operatorname{gr}(G) &= \bigoplus_{n \in \mathbf{Z}} \operatorname{gr}_n(G). \end{align*}

The commutative group gr(G) is then a graded group of type $\mathbf{Z}$ , called the graded group associated with the filtered group G, the homogeneous elements of degree $n$ of gr(G) being those of $\mathrm{gr}_n(G)$ .

Now let A be a filtered ring, (A_n) its filtration, E a filtered A-module and (E_n) its filtration. For all $p \in \mathbf{Z}$ , $q \in \mathbf{Z}$ , a mapping

\mathrm{gr}_p(A) \times \mathrm{gr}_q(E) \to \mathrm{gr}_{p+q}(E)

is defined as follows: given $a \in \mathrm{gr}_p(A)$ , $\xi \in \mathrm{gr}_q(E)$ , two representatives $a, a'$ of $a$ and two representatives $x, x'$ of $\xi$ , $ax \in E_{p+q}$ , $a'x' \in E_{p+q}$ and $ax \equiv a'x'$ (mod. $E_{p+q+1}$ ), for

ax - a'x' = (a - a')x + a'(x - x')

and $a - a' \in A_{p+1}$ and $x - x' \in E_{q+1}$ and hence our assertion follows from formula (3) of no. 1. We may therefore denote by $\alpha\xi$ the class in

E_{p+1}/E_{p+q+1} = \mathrm{gr}_{p+q}(E)

of the product $ax$ of any representative $a \in a$ and any representative $x \in \xi$ . It is immediate that the mapping (9) is Z-bilinear; by linearity, we derive a $\mathbf{Z}$ -bilinear mapping

\mathrm{gr}(A) \times \mathrm{gr}(E) \to \mathrm{gr}(E).

If this definition is first applied to the case $E = A_0$ , the mapping (10) is an internal law of composition on $\mathrm{gr}(A)$ , which it is immediately verified is associative and has an identity element which is the canonical image in $\mathrm{gr}_0(A)$ of the unit element of A; it therefore defines on $\mathrm{gr}(A)$ a ring structure and the graduation $(\mathrm{gr}_n(A))_{n \in \mathbf{Z}}$ is by definition compatible with this structure. The graded ring $\mathrm{gr}(A)$ (of type $\mathbf{Z}$ ) thus defined is called the graded ring associated with the filtered ring A; it is obviously commutative if A is commutative; $\mathrm{gr}_0(A)$ is a subring of $\mathrm{gr}(A)$ . The mapping (10) is on the other hand a $\mathrm{gr}(A)$ -module external law on $\mathrm{gr}(E)$ , the module axioms being trivially satisfied, and the graduation $(\mathrm{gr}_n(E))_{n \in \mathbf{Z}}$ on $\mathrm{gr}(E)$ is obviously compatible with this module structure. The graded $\mathrm{gr}(A)$ -module $\mathrm{gr}(E)$ (of type $\mathbf{Z}$ ) thus defined is called the graded module associated with the filtered A-module E.

Examples

Example 1

Let A be a commutative ring and $t$ an element of A which is not a divisor of 0. Let us give A the (t)-*adic filtration* (no. 1, Example 3). Then the associated graded ring $\mathrm{gr}(A)$ is canonically isomorphic to the polynomial ring $(A/(t))[X]$ . For $\mathrm{gr}_n(A) = 0$ for $n < 0$ and by definition the ring $\mathrm{gr}_0(A)$ is the ring $A/(t)$ . We now note that by virtue of the hypothesis on $t$ the relation $at^n \equiv 0$ (mod. $t^{n+1}$ ) is equivalent to $a \equiv 0$ (mod. $t$ ); if $\tau$ is the canonical image of $t$ in $\mathrm{gr}_1(A)$ , every element of $\mathrm{gr}_n(A)$ may then be written uniquely in the form $\alpha \tau^n$ where $\alpha \in \mathrm{gr}_0(A)$ ; whence our assertion.

Example 2

Let $K$ be a commutative ring, $A$ the ring of formal power series

K[[X_1,\ldots,X_r]]

(Algebra, Chapter IV, § 5) and $m$ the ideal of $A$ whose elements are the formal power series with no constant term. Let us give $A$ the $m$ -adic filtration (no. 1, Example 3); if $M_1,\ldots,M_s$ are the distinct monomials in $X_1,\ldots,X_r$ of total degree $n-1$ , clearly every formal power series $u$ of total order $\omega(u)\geq n$

(loc. cit., no. 2)* may be written as $\displaystyle\sum_{k=1}^{s}u_kM_k$ , where the $u_k$ belong to $m$ ; it is seen that $m^n$ is the set of formal power series $u$ such that $\omega(u)\geq n$ , which shows that $\omega$ is the orderfunction for the $m$ -adic filtration. Then clearly, for every formal power series $u\in m^n$ , there exists a unique homogeneous polynomial of degree $n$ in the $X_i$ which is congruent to $u$ mod. $m^{n+1}$ , namely the sum of terms of degree $n$ of $u$ ; we conclude that $\operatorname{gr}(A)$ is canonically isomorphic to the polynomial ring $K[X_1,\ldots,X_r]$ .

Example 3

More generally, let $A$ be a commutative ring, $b$ an ideal of $A$ and $A$ be given the $b$ -adic filtration. If we write $B=\operatorname{gr}_0(A)$ , $F=\operatorname{gr}_1(A)=b/b^2$ , we know (Algebra, Chapter III), that the identity mapping of the $B$ -module $F$ onto itself can be extended uniquely to a homomorphism $u$ from the symmetric algebra $S(F)$ of $F$ to the $B$ -algebra $\operatorname{gr}(A)$ ; it follows from the definition of $\operatorname{gr}(A)$ that $u$ is a surjective homomorphism of graded algebras; for $n\geq1$ , every element of $\operatorname{gr}_n(A)$ is a sum of classes mod $b^{n+1}$ of elements of the form $y=x_1x_2\cdots x_n$ , where $x_i\in b$ $(1\leq i\leq n)$ ; if $\xi_i$ is the class of $x_i$ mod. $b^2$ , clearly the class of $y$ mod. $b^{n+1}$ is the element $u(\xi_1)\cdots u(\xi_n)$ , whence our assertion. In particular, every system of generators of the $B$ -module $F$ is a system of generators of the $B$ -algebra $\operatorname{gr}(A)$ .

If now $E$ is an $A$ -module and $E$ is given the $b$ -adic filtration, it is seen similarly that the graded $\operatorname{gr}(A)$ -module $\operatorname{gr}(E)$ is generated by $\operatorname{gr}_0(E)=E/bE$ . To be precise, the restriction $\phi$ to $\operatorname{gr}(A)\times\operatorname{gr}_0(E)$ of the external law on the $\operatorname{gr}(A)$ -module $\operatorname{gr}(E)$ is a $\mathbf{Z}$ -bilinear mapping of $\operatorname{gr}(A)\times\operatorname{gr}_0(E)$ to $\operatorname{gr}(E)$ ; moreover $\operatorname{gr}(A)$ is a $(\operatorname{gr}_0(A),\operatorname{gr}_0(A))$ -bimodule and $\operatorname{gr}_0(E)$ a $\operatorname{gr}_0(A)$ -module; it is immediately verified that, for $a\in\operatorname{gr}(A)$ , $\alpha_0\in\operatorname{gr}_0(A)$ , $\xi\in\operatorname{gr}_0(E)$ ,

\phi(a\alpha_0,\xi)=\phi(a,\alpha_0\xi)

and hence $\phi$ defines a surjective $\operatorname{gr}_1(A)$ -linear mapping

(11)

\gamma_E:\operatorname{gr}(A)\otimes_{\operatorname{gr}_0(A)}\operatorname{gr}_0(E)\longrightarrow\operatorname{gr}(E)

which is called canonical.

Example 4

\* Let $K$ be a commutative ring, $\mathfrak g$ a Lie algebra over $K$ and $U$ the enveloping algebra of $\mathfrak g$ . An increasing filtration $(U_n)_{n\in\mathbf{Z}}$ is defined on $U$ by taking $U_n=\{0\}$ for $n<0$ and denoting by $U_n$ , for $n\geq0$ the set of elements of $U$ which can be expressed as a sum of products of at most $n$ elements of $\mathfrak g$ , then

U_0 = K and gr(U) is a commutative K-algebra (Lie Groups and Lie Algebras, Chapter I, § 2, no. 6). The canonical mapping of g to gr_1(U) = U_1/U_0 can be extended uniquely to a homomorphism h of the symmetric algebra $S(g)$ of the K-module g to the K-algebra gr(U); the homomorphism h is surjective and, if the K-module g is free, h is bijective (loc. cit., no. 7, Theorem 1).

Example 5

Let A be a graded ring of type $\mathbf{Z}$ and E a graded A-module of type $\mathbf{Z}$ ; let $A_{(i)}$ (resp. $E_{(i)}$ ) be the subgroup of homogeneous elements of degree i of A (resp. E). Let A and E be given the filtrations associated with their graduations (no. 1, Example 1) and let A' and E' denote the filtered ring and filtered A-module thus obtained. Then it is immediate that the Z-linear mapping $A \to \mathrm{gr}(A')$ which maps an element of $A_{(n)}$ to its canonical image in

\mathrm{gr}_n(A) = \left( \bigoplus_{i \geq n} A_{(i)} \right) / \left( \bigoplus_{i \geq n+1} A_{(i)} \right)

is a graded ring isomorphism. A canonical graded A-module isomorphism

E \to \mathrm{gr}(E')

is defined similarly.

Proposition 1

Let A be a filtered ring, $(A_n)_{n \in \mathbf{Z}}$ its filtration and v its orderfunction. Suppose that $\mathrm{gr}(A)$ is a ring with no divisor of zero. Then, for every ordered pair of elements a, b of the ring $B = \bigcup_{n \in \mathbf{Z}} A_n$ , $v(ab) = v(a) + v(b)$ .

As $n = \bigcap_{n \in \mathbf{Z}} A_n$ is a two-sided ideal of the ring B, the formula holds if $v(a)$ or $v(b)$ is equal to $+\infty$ . If not, $v(a) = r$ and $v(b) = s$ are integers; the classes $\alpha$ of a mod. A, +, and $\beta$ of b mod. A, +_1 are $\neq 0$ by definition, whence by hypothesis $\alpha \beta \neq 0$ in $\mathrm{gr}(A)$ and therefore $ab \notin A_{r+s+1}$ , as $ab \in A_{r+s}$ ,

v(ab) = v(a) + v(b).

Corollary

Let A be a filtered ring and $(A_n)_{n \in \mathbf{Z}}$ its filtration; let us set $B = \bigcup_{n \in \mathbf{Z}} A_n$ , $n = \bigcap_{n \in \mathbf{Z}} A_n$ . If the ring $\mathrm{gr}(A)$ has no divisors of zero, neither has the ring $B/n$ . If a and b are elements of B not belonging to n, then $v(a) \neq +\infty$ and $v(b) \neq +\infty$ , whence $v(ab) \neq +\infty$ and therefore $ab \notin n$ .

Note that the ring A can be an integral domain and the filtration (A_n) exhaustive and separated without $\mathrm{gr}(A)$ being an integral domain (Exercise 2).

Remark

Let G be a group which is not necessarily commutative with a filtration $(G_n)_{n \in \mathbf{Z}}$ such that $G_{n+1}$ is normal in G, for all $n \in \mathbf{Z}$ ; again let $\mathrm{gr}_n(G) = G_n/G_{n+1}$ . The restricted product of the family $(\mathrm{gr}_n(G))_{n \in \mathbf{Z}}$ , that is the subgroup of the product $\prod_{n \in \mathbf{Z}} \mathrm{gr}_n(G)$ consisting of the elements $(\xi_n)$ all of whose components, except at most a finite number, are equal to the identity element, is also called the graded group associated with G and denoted by $\mathrm{gr}(G)$ .

4. Homomorphisms compatible with filtrations

Let G, G' be two commutative groups (written additively), (G_n) a filtration on G and (G'_n) a filtration on G'; a homomorphism h : G → G' is called compatible with the filtrations on G and G' if h(G_n) ⊂ G'_n for all n ∈ Z. The composite homomorphism G_n $\xrightarrow{h} G'_n \longrightarrow G'_n/G'_{n+1}$ is zero on G_{n+1} and hence defines by taking quotients a homomorphism h_n : G_n/G_{n+1} → G'_n/G'_{n+1}; there is therefore a unique additive group homomorphism gr(h) : gr(G) → gr(G') such that, for all n ∈ Z, gr(h) coincides with h_n on gr_n(G) = G_n/G_{n+1}. gr(h) is called the graded group homomorphism associated with h. If G'' is a third filtered group and h' : G' → G'' a homomorphism which is compatible with the filtrations, h' ∘ h is a homomorphism which is compatible with the filtrations and

(12)

\operatorname{gr}(h' \circ h) = \operatorname{gr}(h') \circ \operatorname{gr}(h)

Proposition 2

Let G be a filtered commutative group and H a subgroup of G; let H be given the induced filtration and G/H the quotient filtration. If j : H → G is the canonical injection and p : G → G/H the canonical surjection, j and p are compatible with the filtrations and the sequence

(13)

0 \longrightarrow \operatorname{gr}(H) \xrightarrow{\operatorname{gr}(j)} \operatorname{gr}(G) \xrightarrow{\operatorname{gr}(p)} \operatorname{gr}(G/H) \longrightarrow 0

is exact.

The first assertion is obvious; if (G_n) is the filtration on G, then

(H \cap G_n) \cap G_{n+1} = H \cap G_{n+1}

and hence gr(j) is injective; moreover the canonical mapping

G_n \to (H + G_n)/H

is surjective, hence so is gr(p) and gr(p) ∘ gr(j) = 0 by (12). Finally, let

\xi \in G_n/G_{n+1}

belong to the kernel of gr(p); then there exists x ∈ ξ such that x ∈ H + G_{n+1}; but as G_{n+1} ⊂ G,

G_n \cap (H + G_{n+1}) = (H \cap G_n) + G_{n+1}

and hence x = y + z where y ∈ H ∩ G, and z ∈ G, +_1; this proves that ξ is the class mod. G, +_1 of j(y), in other words it belongs to the image of gr(H) under gr(j).

Note that, given an exact sequence 0 → G' \xrightarrow{u} G \xrightarrow{v} G'' → 0 of filtered commutative groups, where u and v are compatible with the filtrations, the sequence 0 \longrightarrow \operatorname{gr}(G') \xrightarrow{\operatorname{gr}(u)} \operatorname{gr}(G) \xrightarrow{\operatorname{gr}(v)} \operatorname{gr}(G'') \longrightarrow 0 is not necessarily exact (Exercise 4).

If now A and B are two filtered rings and h: A → B a ring homomorphism which is compatible with the filtrations, it is immediately verified that the graded group homomorphism gr(h): gr(A) → gr(B) is also a ring homomorphism. In particular, if A' is a subring of A with the induced filtration, gr(A') is canonically identified with a graded subring of gr(A) (Proposition 2); if b is a two-sided ideal of A and A/b is given the quotient filtration, gr(A/b) is canonically identified with the quotient graded ring gr(A)/gr(b) (Proposition 2).

Finally, let A be a filtered ring, E, F two filtered A-modules and u : E → F an homomorphism compatible with the filtrations. Then it is immediate that gr(u) : gr(E) → gr(F) is a gr(A)-linear mapping and hence a homogeneous homomorphism of degree 0 of graded gr(A)-modules. Moreover, if u' : E → F is another A-homomorphism compatible with the filtrations, so is u + u' and

\operatorname{gr}(u + u') = \operatorname{gr}(u) + \operatorname{gr}(u').

Remarks

(1) Clearly filtered ring homomorphisms (resp. homomorphisms of filtered modules over a given filtered ring A) compatible with the filtrations can be taken as morphisms for the filtered ring structure (resp. filtered A-module structure) (Set Theory, Chapter IV, § 2, no. 1).

(2) Let E and F be two modules over a filtered ring A and let them have the filtrations derived from the filtration (A_n) on A (no. 1, Example 2). Then every A-linear mapping u : E → F is compatible with the filtrations, since

u(A_nE) = A_nu(E) \subset A_nF.

(3) Note that a filtered A-module homomorphism u : E → F which is compatible with the filtrations may satisfy gr(u) = 0 without being zero; this is so for example of the endomorphism x ↦ nx of the additive group $\mathbf{Z}$ with the (n)-adic filtration (for any integer n > 1). The relation gr(u) = gr(v) for two homomorphisms u, v of E to F, compatible with the filtrations, does not therefore imply necessarily u = u.

(4) The definitions at the beginning of this no. extend immediately to two groups G, G', which are not necessarily commutative and are filtered by subgroups $G_n, G'_n$ such that $G_{n+1}$ (resp. $G'_{n+1}$ ) is normal in G, (resp. $G'_n$ ). Proposition 2 is also valid with the same hypotheses on the G, and assuming that H is invariant in G, the proof remaining unchanged except for notation.

5. The topology defined by a filtration

Let G be a group filtered by a family $(G_n)_{n \in \mathbf{Z}}$ of normal subgroups of G. There exists a unique topology on G which is compatible with the group structure and for which the G_n constitute a fundamental system of neighbourhoods of the identity element e of G (General Topology, Chapter III, § 1, no. 2, Example); it is called the topology on G defined by the filtration (G_n). When we use topological notions concerning a filtered group, we shall mean, unless otherwise stated, with the topology defined by the filtration. Note that the G_n, being subgroups of G, are both open and closed (General Topology, Chapter III, § 2, no. 1, Corollary to Proposition 4).

As each G, is normal in G, the entourages of the left and right uniformities on G coincide; we deduce that G admits a Hausdorff completion group $\hat{G}$ (General Topology, Chapter III, § 3, no. 4, Theorem 1 and no. 1, Proposition 2).

For every subset M of G, the closure of M in G is equal to

\bigcap_{n \in \mathbf{Z}} (M . G_n) = \bigcap_{n \in \mathbf{Z}} (G_n . M)

(General Topology, Chapter III, § 3, no. 1, formula (1)); in particular $\bigcap_{n \in \mathbf{Z}} G$ , is the closure of $\{e\}$ ; thus it is seen that for the topology on G to be Hausdorff it is necessary and sufficient that the filtration (G_n) be separated. For the topology on G to be discrete, it is necessary and sufficient that there exist $n \in \mathbf{Z}$ such that $G_+ = \{e\}$ (in which case $G_m = \{e\}$ for $m \geq n$ ); then the filtration (G_n) is called discrete.

Since the Hausdorff group associated with G is $H = G / \left( \bigcap_{n \in \mathbf{Z}} G_n \right)$ , the associated graded groups gr(G) and gr(H) (if H is given the quotient filtration) are canonically identified.

Not let G' be another filtered group and $u : G \to G'$ a homomorphism compatible with the filtrations; the definition of the topologies on G and G' shows immediately that $u$ is continuous (*). If H is a subgroup (resp. normal subgroup) of G, the topology induced on H by that on G (resp. the quotient topology with respect to H of that on G) is the topology on H (resp. G/H) defined by the filtration induced by that on G (resp. quotient topology of that on G). The product topology of those on G and G' is the topology defined by the product of filtrations on G and G'.

Let v be the order function (no. 2) on G. The hypothesis on the G, implies that $v(xy x^{-1}) = v(y)$ and hence $v(xy^{-1}) = v(yx^{-1}) = v(x^{-1}y) = v(y^{-1}x)$ for all x, y in G. Let $\rho$ be a real number such that $0 < \rho < 1$ (for example take

(*) Throughout this chapter we shall use the words "continuous homomorphism" in the sense of what is called "continuous representation" in General Topology, Chapter III, § 2, no. 8; the word "homomorphism" will never be used in the sense of General Topology, Chapter III, § 2, no. 8, Definition 1; for this notion we shall always use the term "strict morphism" in order to avoid any confusion.

$\rho = 1/e$ and let $d(x, y) = \rho^{v(xy^{-1})}$ for all $x, y$ in G. Then $d(x, x) = 0, d(x, y) = d(y, x)$ and inequality (5') of no. 2 gives

(15) \quad d(x, y) \leq \sup(d(x, z), d(y, z))

for all $x, y, z$ in G, which implies the triangle inequality

d(x, y) \leq d(x, z) + d(y, z).

Thus $d$ is a pseudometric on G which is invariant under left and right translations and G_n is the set of $x \in G$ such that $d(e, x) < \rho^n$ ; the uniform structure defined by $d$ is then the uniform structure on the topological group G. If G is Hausdorff, G is a zero-dimensional metrizable topological space (General Topology, Chapter IX, § 6, no. 4); $d$ is a distance on G if also the filtration (G_n) is exhaustive.

Given a topological ring A, recall that a left topological A-module is an A-module E with a topology compatible with its additive group structure and such that the mapping $(a, x) \mapsto ax$ from $A \times E$ to E is continuous (General Topology, Chapter III, § 6, no. 6).

Proposition 3

Let A be a filtered ring, (A_n) its filtration, B the subring $\bigcup_{n \in \mathbf{Z}} A_n$ of A, E a filtered B-module, (E_n) its filtration and F the sub-B-module $\bigcup_{n \in \mathbf{Z}} E_n$ of E. Then the mapping $(a, x) \mapsto ax$ from $B \times F$ to F is continuous.

Let $a, \in B, x_0 \in F$ ; there exists by hypothesis integers $r, s$ such that $a_0 \in A_r$ and $x_0 \in E_s$ . The relation

ax - a_0x_0 = (a - a_0)x_0 + a_0(x - x_0) + (a - a_0)(x - x_0)

shows that if $a - a_0 \in A_i$ and $x - x_0 \in E_j$ , $ax - a_0x_0$ belongs to

E_{i+s} + E_{j+r} + E_{i+j}.

Then, given an integer $n$ , $ax - a_0x_0 \in E_n$ provided $i \geq n - s, j \geq n - r$ and $i + j \geq n$ , that is so long as $i$ and $j$ are sufficiently large.

Corollary

The ring B is a topological ring and the B-module F a topological B-module.

The first assertion is obtained by applying Proposition 3 to $F = B$ .

It is seen in particular that a filtered ring A whose filtration is exhaustive is a topological ring; if this is so every filtered A-module whose filtration is exhaustive is a topological A-module.

Proposition 4

Let A be a commutative ring filtered by an exhaustive filtration (A_n) and p an ideal of A. Suppose that the ideal $\operatorname{gr}(p) = \bigoplus_{n \in \mathbf{Z}} (p \cap A_n)/(p \cap A_{n+1})$ of the ring $\operatorname{gr}(A)$ is prime. Then the closure of p in A is a prime ideal.

We know that gr(A/p) is isomorphic to gr(A)/gr(p) (no. 4, Proposition 2) and hence an integral domain; we conclude that $A / \bigcap_{n \in \mathbf{Z}} (p + A_n)$ is an integral domain (no. 3, Corollary to Proposition 1). Then the closure $\bigcap_{n \in \mathbf{Z}} (p + A_n)$ of $p$ is a prime ideal.

Let $A$ be a ring and $m$ a two-sided ideal of $A$ ; the topology defined on $A$ by the $m$ -adic filtration (no. 1, Example 3) is called the m-adic topology; as the m-adic filtration is exhaustive, $A$ is a topological ring with this topology (Corollary to Proposition 3). Similarly, for every $A$ -module $E$ , the topology defined by the $m$ -adic filtration is called the m-adic topology on $E$ ; $E$ is a topological $A$ -module under this topology.

Let $m'$ be another two-sided ideal of $A$ ; for the $m'$ -adic topology on $A$ to be finer than the $m$ -adic topology, it is necessary and sufficient that there exist an integer $n > 0$ such that ${m'}^n \subset m$ ; the condition is necessary and, if it is fulfilled, ${m'}^{hn} \subset m^h$ for all $h > 0$ and hence the condition is sufficient. If $A$ is a commutative Noetherian ring, it amounts to the same to say that $V(m) \subset V(m')$ in the prime spectrum of $A$ (Chapter II, § 4, no. 3, Corollary 2 to Proposition 11 and § 2, no. 6, Proposition 15).

6. Complete filtered modules

Proposition 5

Let $G$ be a filtered group whose filtration $(G_n)$ consists of invariant subgroups of $G$ . The following conditions are equivalent:

(a) $G$ is a complete topological group.

(b) The associated Hausdorff group $G' = G / (\bigcap_{n \in \mathbf{Z}} G_n)$ is complete.

If $G$ is commutative and written additively, these conditions are also equivalent to the following:

(d) Every family $(x_\lambda)_{\lambda \in L}$ of elements of $G'$ which converges to 0 with respect to the filter $\mathscr{F}$ of complements of finite subsets of $L$ is summable in $G'$ .

For a filter on $G$ to be a Cauchy filter (resp. a convergent filter), it is necessary and sufficient that its image under the canonical mapping $G \to G'$ be a Cauchy (resp. convergent) filter (General Topology, Chapter II, § 3, no. 1, Proposition 4); whence first of all the equivalence of (a) and (b); on the other hand, as $G'$ is metrizable, the equivalence of (b) and (c) follows from Proposition 9 of General Topology, Chapter IX, § 2, no. 6.

Suppose now that $G$ is commutative. Suppose that $G'$ is complete and let $(x_\lambda)_{\lambda \in L}$ be a family of elements of $G'$ which converge to 0 with respect to $\mathscr{F}$ . For every neighbourhood $V'$ of 0 in $G$ which is a subgroup of $G'$ , there exists a finite subset $J$ of $L$ such that the condition $\lambda \in L - J$ implies $x_\lambda \in V'$ ; then $\sum_{\lambda \in H} x_\lambda \in V'$ for every finite subset $H$ of $L$ not meeting $J$ , which shows that the family $(x_\lambda)_{\lambda \in L}$ is summable (General Topology, Chapter III, § 5, no. 2, Theorem 1).

Conversely, suppose that condition (d) holds and let $(x_n)$ be a Cauchy sequence on $G'$ ; the family $(x_{n+1} - x_n)$ is then summable and in particular the series with general term $x_{n+1} - x_n$ is convergent and hence the sequence $(x_n)$ is convergent.

Let $G$ be a filtered group whose filtration $(G_r)$ consists of normal subgroups of $G$ ; the quotient groups $G/G_n$ are discrete and hence complete, since the $G_r$ are open in $G$ . Let $f_n$ be the canonical mapping $G \to G/G_n$ and for $m \leq n$ let $f_{mn}$ be the canonical mapping $G/G_n \to G/G_m$ ; $(G/G_n, f_{mn})$ is an inverse system of discrete groups with $\mathbf{Z}$ as indexing set (General Topology, Chapter III, § 7, no. 3). Let $\tilde{G}$ be the topological group the inverse limit of this inverse system and for all $n$ let $g_n : \tilde{G} \to G/G_n$ be the canonical mapping; let $f : G \to \tilde{G}$ be the inverse limit of the inverse system of mappings $(f_n)$ such that $f_n = g_n \circ f$ for all $n$ ; finally, let $j$ be the canonical mapping of $G$ to its Hausdorff completion $\hat{G}$ ; as the $G/G_n$ are complete, there exists a unique topological group isomorphism $i : \hat{G} \to \tilde{G}$ such that $f = i \circ j$ (loc. cit., Corollary 1 to Proposition 2); we shall call it the canonical isomorphism of $\hat{G}$ onto $\tilde{G}$ .

Let $H$ be another filtered group whose filtration $(H_r)$ consists of normal subgroups of $H$ and let $u : G \to H$ be a homomorphism compatible with the filtrations (no. 4). Set $\tilde{H} = \lim H/H_n$ ; for all $n$ , $u$ defines by taking quotients a homomorphism $u_n : G/G_n \to H/H_n$ and the $u_n$ obviously form an inverse system of mappings; set $\tilde{u} = \lim u_n$ . Moreover let $\hat{H}$ be the Hausdorff completion of $H$ and $ti : \hat{G} \to H$ the homomorphism derived from $u$ by passing to the Hausdorff completions (General Topology, Chapter II, § 3, no. 7, Proposition 15). It follows immediately from the definitions that if $\hat{G}$ is identified with $\tilde{G}$ and $\hat{H}$ with $\tilde{H}$ by means of the canonical isomorphisms, $ti$ is identified with $\tilde{u}$ . We conclude in particular that, if, for all $n$ , $u_n$ is an isomorphism, then $ti$ is an isomorphism of topological groups.

Examples of complete filtered groups and rings

Example 1

Let $G$ be a complete filtered group. Every closed subgroup of $G$ with the induced filtration is complete (General Topology, Chapter II, § 3, no. 4, Proposition 8). Every quotient group of $G$ with the quotient filtration is complete (General Topology, Chapter IX, § 3, no. 1, Remark 1).

Example 2

Let $A$ be a filtered commutative ring whose filtration we denote by $(a_n)_{n \in \mathbf{Z}}$ ; let $A'$ be the ring of formal power series $A[[X_1, \ldots, X_s]]$ . For all $e = (e_1, \ldots, e_s) \in \mathbf{N}^s$ , we write $|e| = \sum_{i=1}^s e_i$ , $X^e = \prod_{i=1}^s X_i^{e_i}$ so that every element $P \in A'$ can be written uniquely $P = \sum_{e \in \mathbf{N}^s} \alpha_{e,P} X^e$ where $\alpha_{e,P} \in A$ . For all $n \in \mathbf{Z}$ , let $a'_n$ denote the set of $P \in A'$ such that $\alpha_{e,P} \in a_{n-|e|}$ for all $e \in \mathbf{N}^s$ ; we show that $a'_n$ is an ideal of $A'$ . Clearly $a'_n$ is an additive subgroup of $A'$ ; on the other hand, if $P \in a'_n$ and $Q \in A'$ , then, for all $e \in \mathbf{N}^s$ , $\alpha_{e,PQ} = \sum_{e'+e''=e} \alpha_{e',Q}\alpha_{e'',P}$ ; as the relation $e' + e'' = e$ implies $|e''| \leq |e|$ , $PQ \in a'_n$ . Moreover, if $Q \in a'_m$ , then, for $e' + e'' = e$ , $\alpha_{e',Q}\alpha_{e'',P} \in a_{m-|e'|}a_{n-|e''|} \subset a_{m+n-|e|}$ , which proves that $(a'_n)_{n \in \mathbf{Z}}$ is a filtration compatible with the ring structure on $A'$ (for obviously $1 \in a'_0$ ). When in future we speak of $A'$ as a filtered ring, we shall mean, unless otherwise stated, with the filtration (ah). Clearly $\bigcap_{n \in \mathbf{Z}} a'_n$ is the set of formal power series all of whose coefficients belong to $\bigcap_{n \in \mathbf{Z}} a_n$ ; then, if $A$ is Hausdorff, so is $A'$ . If $a_0 = A$ , then $a'_0 = A'$ .

Proposition 6

With the above notation, suppose that $a_0 = A$ and let $h$ denote the mapping $P \mapsto (\alpha_{e,P})_{e \in \mathbf{N}^s}$ . Then $h$ is an isomorphism of the additive topological group $A$ onto the additive topological group $A^{\mathbf{N}^s}$ . The polynomial ring $A[X_1, \ldots, X_s]$ is dense in $A'$ ; if $A$ is complete, so is $A'$ .

Clearly $h$ is bijective; $V_n = h(a'_n)$ is the set of $(a_e) \in A^{\mathbf{N}^s}$ such that $a_e \in a_{n-|e|}$ for all $e \in \mathbf{N}^s$ such that $|e| \leq n$ ; as these elements $e$ are finite in number, $V_n$ is a neighbourhood of 0 in $A^{\mathbf{N}^s}$ . Conversely, if $V$ is a neighbourhood of 0 in $A^{\mathbf{N}^s}$ , there is a finite subset $E$ of $\mathbf{N}^s$ and an integer $v$ such that the conditions $a_e \in a_v$ for all $e \in E$ imply $(a_e) \in V$ ; if then $n$ is the greatest of the integers $v + |e|$ for $e \in E$ , then $h(a'_n) \subset V$ , which proves the first assertion of Proposition 6.

Moreover, with $n$ and $E$ defined as above, $h(P - \sum_{e \in E} \alpha_{e,P}X^e) \in V$ for all $P \in A'$ , which shows that $A[X_1, \ldots, X_s]$ is dense in $A'$ . The last assertion follows from the first and the fact that a product of complete spaces is complete.

Let $m$ be an ideal of $A$ and suppose that $(a_i)$ is the $m$ -adic filtration; then, if $n$ is the ideal of $A'$ generated by $m$ and the $X_i$ ( $1 \leq i \leq s$ ), the filtration $(a_i)$ is the $n$ -adic filtration. For clearly, for all $k \geq 0$ , $n^k$ is generated by the elements $aX^e$ such that $a \in m^{k-|e|}$ for all $e \in \mathbf{N}^s$ such that $|e| \leq k$ , whence $n^k \subset a \&$ . Let us prove conversely that $a'_k \subset n^k$ . For all $P \in a'_k$ , $P = P' + P''$ , where $P' = \sum_{|e| < k} \alpha_{e,P}X^e$ , $P'' = \sum_{|e| \geq k} \alpha_{e,P}X^e$ . Clearly it is possible to write $P'' = \sum_{|e|=k} X^e Q_e$ , where the $Q_e$ are elements of $A'$ , whence $P'' \in n^k$ ; on the other hand, clearly $\alpha_{e,P}X^e \in n^k$ for all $e \in \mathbf{N}^s$ , whence $P' \in n^k$ . Then $n^k = a'_k$ .

Corollary

Let $A$ be a commutative ring,

A' = A[[X_1, \ldots, X_s]]

the ring of formal power series in $s$ indeterminates over $A$ and $n$ the ideal of $A'$ consisting of the formal power series with no constant term. The ring $A'$ is Hausdorff and complete with the n-adic topology and the polynomial ring $A[X_1, \ldots, X_s]$ is everywhere dense in $A'$ .

It is sufficient to apply what has just been said to the case $m = \{0\}$ .

7. Linear compactness properties of complete filtered modules

Recall that, if E is an A-module, an affine subset (or an affine linear variety) of E is any subset F which is empty or of the form $a + M$ , where $a \in E$ and M is a submodule of E called the direction of F (Algebra, Chapter II, § 9, nos. 1 and 3).

Proposition 7

Let A be a filtered ring, E a filtered A-module and $(E_n)$ filtration of E; suppose that $E_0 = E$ , that the $E_n$ are submodules of E, that the A-modules $E/E_n$ are Artinian and finally that the topological group E is Hausdorff and complete. Then the intersection of a decreasing sequence of non-empty closed affine subsets of E is non-empty.

We have seen in no. 6 that, since E is Hausdorff and complete, it is identified with $\tilde{E} = \lim_{\leftarrow} E/E_n$ . Let $(W_p)$ be a decreasing sequence of non-empty closed affine subsets of E and, for all $n \geq 0$ , let $W_{p,n}$ be the canonical image of $W_p$ in $E/E_n$ ; we are going to construct a sequence $x = (x_n) \in \tilde{E}$ such that $x_n \in W_{p,n}$ for all $p$ and all $n$ ; hence $x \in W_p + E_n$ for all $p$ and all $n$ and, as the $W_p$ are closed, $x \in W_p$ for all $p$ (no. 5), which will prove the proposition.

As $E/E_0 = 0$ , we shall take $x_0 = 0$ . Suppose that the $x_i$ are defined for $0 \leq i \leq n - 1$ and let $W'_{p,n}$ be the set of elements of $W_{p,n}$ whose canonical image in $E/E_{n-1}$ is $x_{n-1}$ ; as $x_{n-1} \in W_{p,n-1}$ and $W_{p,n-1}$ is the canonical image of $W_{p,n}$ , $W'_{p,n}$ is non-empty and is obviously an affine subset of $E/E_n$ ; moreover the sequence $(W'_{p,n})_{p \in \mathbf{N}}$ is decreasing. As $E/E_n$ is Artinian, this sequence is stationary (otherwise the sequence of submodules of $E/E_n$ which are the directions of the $W'_{p,n}$ would be strictly decreasing, which is absurd). It is sufficient then to take $x_n$ to be an element of $\bigcap_{p \in \mathbf{N}} W'_{p,n}$ and the construction of $(x_n)$ can then be performed by induction.

Proposition 8

Suppose that A and E satisfy the hypotheses of Proposition 7. Let $(F_p)$ be a decreasing sequence of closed submodules of E such that $\bigcap F_p = 0$ . Then, for every neighbourhood V of 0 in E, there exists $p$ such that $F_p \subset V$ (in other words, the base of the filter $(F_p)$ converges to 0).

We may assume that V is one of the $E_n$ , in which case $E/V$ is Artinian. Let us write $F'_p = (F_p + V)/V$ ; as the $F'_p$ form a decreasing sequence of submodules of $E/V$ , there exists an integer $j$ such that $F'_p = F'_j$ for all $p > j$ . We shall see that $F'_j = \{0\}$ , which will complete the proof. Let $x \in F'_j$ and let W, be the set of elements of F, whose image in $E/V$ is $x$ ( $p \geq j$ ); by definition of j, the W, are non-empty closed affine subsets of E and obviously $W_{p+1} \subset W$ ;

then it follows from Proposition 7 that there exists an element y belonging to all the W,. As W, $\subset F,$ and $\bigcap_{p \in \mathbf{N}} F_p = \{0\}, y = 0$ ; since x is the canonical image of y in E/V, $x = 0$ (cf. Exercises 15 to 21).

8. The lift of homomorphisms of associated graded modules

Theorem 1

Let X, Y be two filtered groups whose filtrations $(X_n), (Y_n)$ consist of normal subgroups; let $u : X \to Y$ be a homomorphism compatible with the filtrations.

(i) Suppose that the filtration $(X_n)$ is exhaustive. For $\mathrm{gr}(u)$ to be injective, it is necessary and sufficient that $\bar{u}^{-1}(Y_n) = X_n$ for all $n \in \mathbf{Z}$ .

(ii) Suppose that one of the following hypotheses holds: (α) X is complete and Y Hausdorff; (β) Y is discrete. Then, for $\mathrm{gr}(u)$ to be surjective, it is necessary and sufficient that $Y_n = u(X_n)$ for all $n \in \mathbf{Z}$ .

(i) To say that the mapping $\mathrm{gr}_n(u)$ is injective means that

X_n \cap \bar{u}^{-1}(Y_{n+1}) \subset X_{n+1}.

This is obviously the case if

(Y_{n+1}) = X_{n+1}

. Conversely, if

X_n \cap \bar{u}^{-1}(Y_{n+1}) \subset X_{n+1}

for all

n

, we deduce by induction on

k

that

X_{n-k} \cap \bar{u}^{-1}(Y_{n+1}) \subset X_{n+1}

for all

n \in \mathbf{Z}

and all

k \geq 0

. As the filtration

(X_n)

is exhaustive, we see that, for all

n

X_n

is the union of the

X_{n-k}

(

k \geq 0

), hence

\bar{u}^{-1}(Y_{n+1}) \subset X_{n+1}

for all

n

and therefore

X_{n+1} \subset \bar{u}^{-1}(Y_{n+1})

, which completes the proof.

(ii) To say that the mapping $\mathrm{gr}_n(u)$ is surjective means that

Y_n = u(X_n)Y_{n+1}.

This is obviously the case if

Y_n = u(X_n)Y_{n+1}

for all

n \in \mathbf{Z}

. Let

n

be an integer and

y

an element of

Y_n

; we shall define a sequence

(x_k)_{k \geq 0}

of elements of

X_n

such that

x_k \in X_n

x_{k+1} \equiv x_k \pmod{X_{n+k}}

and

u(x_k) \equiv y \pmod{Y_{n+k}}

for all

k \geq 0

. We shall take

x_0

equal to the identity element of X, which certainly gives

u(x_0) \equiv y \pmod{Y_n}

. Suppose that an

x_k \in X_n

has been constructed such that

u(x_k) \equiv y \pmod{Y_{n+k}}

; then

(u(x_k))^{-1}y \in Y_{n+k}

; the hypothesis implies that there exists

t \in X_{n+k}

such that

u(t) \equiv (u(x_k))^{-1}y \pmod{Y_{n+k+1}}

and hence

u(x_k t) \equiv y \pmod{Y_{n+k+1}}

; it is sufficient to take

x_{k+1} = x_k t

to carry out the induction. This being so, if Y is discrete, there exists

k \geq 0

such that

Y_{n+k} = \{e'\}

(the identity element of Y), whence

u(x_k) = y

and hence in this case it has been proved that

u(X_n) = Y

, for all

n

. Suppose now that X is complete and Y Hausdorff. As

x_h^{-1} x_k \in X_{n+k}

for

h \geq k \geq 0

(x_k)

is a Cauchy sequence in

X_n

; as

X_n is closed in X and hence complete, this sequence has at least one limit x in X_n. By virtue of the continuity of u, u(x) is the unique limit of the sequence (u(x_k)) in Y, Y being Hausdorff. But the relations u(x_k) \equiv y (\text{mod. } Y_{n+k+1}) show that y is also a limit of this sequence, whence $u(x) = y$ and it has also been proved that $u(X_n) = Y_n$ .

Corollary 1

Suppose that X is Hausdorff and its filtration exhaustive. Then, if gr(u) is injective, u is injective.

Let e, e' be the identity elements of X and Y respectively. Then

\bar{u}^{-1}(e') \subset \bigcap_n \bar{u}^{-1}(Y_n) = \bigcap_n X_n = \{e\}

by hypothesis, whence the corollary.

Corollary 2

Suppose that one of the following hypotheses holds :* $(\alpha)$ X is complete, Y is Hausdorff and its filtration is exhaustive; $(\beta)$ Y is discrete and its filtration is exhaustive. *Then, if gr(u) is surjective, u is surjective.

In this case $Y = \bigcup_n Y_n = \bigcup_n u(X_n) \subset u(X)$ .

Corollary 3

Suppose that X and Y are Hausdorff, the filtrations of X and Y exhaustive and X complete. Then, if gr(u) is bijective, u is bijective.

Let A be a local ring, m its maximal ideal and M an A-module; let A and M be given the m-adic filtrations and let gr(A) and gr(M) be the graded ring and the graded gr(A)-module associated with A and M. We have seen (no. 3, Example 3) that the canonical mapping (11) is always surjective; we are going to consider the following property of M: (GR) The canonical mapping

\gamma_M : \operatorname{gr}(A) \otimes_{\operatorname{gr}_0(A)} \operatorname{gr}_0(M) \to \operatorname{gr}(M)

is bijective.

Proposition 9

Let A be a local ring, m its maximal ideal, M, N two A-modules and u : N \to M an A-homomorphism. M and N are given the m-adic filtrations and suppose that: (1) M satisfies property (GR); (2) \operatorname{gr}_0(u) : \operatorname{gr}_0(N) \to \operatorname{gr}_0(M) is injective. Then \operatorname{gr}(u) : \operatorname{gr}(N) \to \operatorname{gr}(M) is injective, N and P = \operatorname{Coker}(u) satisfy property (GR) and $m^n N = \bar{u}^{-1}(m^n M)$ for every integer $n > 0$ .

It is immediately verified that the diagram

\begin{array}{ccc} \operatorname{gr}(A) \otimes_{\operatorname{gr}_0(A)} \operatorname{gr}_0(N) & \xrightarrow{1 \otimes \operatorname{gr}_0(u)} & \operatorname{gr}(A) \otimes_{\operatorname{gr}_0(A)} \operatorname{gr}_0(M) \\ \downarrow \gamma_N & & \downarrow \gamma_M \\ \operatorname{gr}(N) & \xrightarrow{\operatorname{gr}(u)} & \operatorname{gr}(M) \end{array}

is commutative. As $\mathrm{gr}_0(A) = A/m$ is a field, the hypothesis implies that $1 \otimes \mathrm{gr}_0(u)$ is injective; as by hypothesis $\gamma_M$ is injective, so is $\gamma_M \circ (1 \otimes \mathrm{gr}_0(u))$ . This implies first that $\gamma_N$ is injective and hence bijective and therefore that $\mathrm{gr}(u)$ is injective. The formula $\tilde{u}^1(m^nM) = m^nN$ is then a consequence of Theorem 1 (i).

Also, let us write $N' = u(N)$ and let $j : N' \to M$ be the canonical injection. If $p : M \to P = M/N'$ is the canonical homomorphism, then in the commutative diagram

\begin{array}{ccccccccc} \mathrm{gr}(A) \otimes \mathrm{gr}_0(N') & \xrightarrow{1 \otimes \mathrm{gr}_0(j)} & \mathrm{gr}(A) \otimes \mathrm{gr}_0(M) & \xrightarrow{1 \otimes \mathrm{gr}_0(p)} & \mathrm{gr}(A) \otimes \mathrm{gr}_0(P) & \to & 0 \\ \downarrow \gamma_{N'} & & \downarrow \gamma_M & & \downarrow \gamma_P & & \\ \mathrm{gr}(N') & \xrightarrow{\mathrm{gr}(j)} & \mathrm{gr}(M) & \xrightarrow{\mathrm{gr}(p)} & \mathrm{gr}(P) & \to & 0 \end{array}

the lower row is exact (no. 4, Proposition 2) and so is the upper row by virtue of Proposition 2 of no. 4 and the fact that $\mathrm{gr}_0(A)$ is a field. Moreover, $\mathrm{gr}(j)$ is injective (no. 4, Proposition 2) and hence $\mathrm{gr}_0(j)$ is injective, The first part of the argument applied to $j$ shows that $\gamma_{N'}$ is bijective; as $\gamma_M$ is also bijective by hypothesis, we conclude that $\gamma_P$ is bijective (Chapter I, § 1, no. 4, Corollary 2 to Proposition 2).

Corollary

Under the hypotheses of Proposition 9, if we assume also that $N$ is Hausdorf with the $m$ -adic filtration, then $u$ is injective.

This follows from the fact that $\mathrm{gr}(u)$ is injective (Corollary 1 to Theorem 1).

\* Remark. Suppose that the hypotheses of Proposition 9 hold and also one of the following conditions: (1) $m$ is nilpotent; (2) $A$ is Noetherian and $P$ is ideally Hausdorff (cf. § 5, no. 1); then $P$ is a flat $A$ -module. This follows from the fact that $\gamma_P$ is bijective and § 5, no. 2, Theorem 1 (iv), since $A/m$ is a field. \*

9. The lift of families of elements of an associated graded module

Let $A$ be a filtered commutative ring, $(A_n)_{n \in \mathbf{Z}}$ its filtration and $C$ a subring of $A$ , such that $C \cap A_1 = \{0\}$ . The restriction to $C$ of the canonical mapping $A_0 \to A_0/A_1 = \mathrm{gr}_0(A)$ is then injective, which allows us to identify $C$ with a submodule of $\mathrm{gr}_0(A)$ ; this is what we shall usually do in similar cases. If $A_1 \neq A$ , and $K$ is any subfield of $A$ , then $K \cap A_1 = (0)$ since $K \cap A_1$ is an ideal of $K$ not containing 1 ; then $K$ may be identified with a subfield of $\mathrm{gr}_0(A)$ .

Proposition 10

Let $A$ be a filtered commutative ring and $(A_n)$ its filtration; suppose that there exists a subring $C$ of $A_0$ such that $C \cap A_n = \{0\}$ and $C$ is identified with a subring of $\mathrm{gr}_0(A)$ . Let $(x_i)_{1 \leq i \leq q}$ be a finite family of elements of $A$ ; suppose that $x_i \in A_n$ , for $1 \leq i \leq q$ and let $\xi_i$ be the class of $x_i$ in $\mathrm{gr}_{n_i}(A)$ for $1 \leq i \leq q$ .

(i) If the family $(\xi_i)$ of elements of $\mathrm{gr}(A)$ is algebraically free over $C$ , the family $(x_i)$ is algebraically free over $C$ .

(ii) If the filtration on $A$ is exhaustive and discrete and $(\xi_i)$ is a system of generators of the $C$ -algebra $\mathrm{gr}(A)$ , then $(x_i)$ is a system of generators of the $C$ -algebra $A$ .

Let $A'$ be the polynomial algebra $\mathbf{C}[X_1, \ldots, X_q]$ over $\mathbf{C}$ ; let $A'$ be given the graduation $(A'_n)$ of type $\mathbf{Z}$ where $A_n$ is the set of $\mathbf{C}$ -linear combinations of the monomials $X_1^{s(1)} \ldots X_q^{s(q)}$ such that $\sum_{i=1}^q n_i s(i) = n$ . Let $u$ be the homomorphism $f \mapsto f(x_1, \ldots, x_q)$ from the $\mathbf{C}$ -algebra $A'$ to the $\mathbf{C}$ -algebra $A$ ; by definition, $u(A'_n) \subset A_n$ for all $n \in \mathbf{Z}$ and hence $u$ is compatible with the filtrations $(A')$ being given its filtered ring structure associated with its graded ring structure, cf. no. 1, Example 1). This being so, the hypothesis of (i) means that

\mathrm{gr}(u): A' = \mathrm{gr}(A') \to \mathrm{gr}(A)

is injective; as the filtration on $A'$ is exhaustive and separated, Corollary 1 to Theorem 1 of no. 8 may be applied and $u$ is injective, which proves the conclusion of (i). Similarly, the hypothesis (ii) on the (\&) means that $\mathrm{gr}(u)$ is surjective; as $A$ is discrete and its filtration is exhaustive, Corollary 2 to Theorem 1 of no. 8 may be applied and $u$ is surjective, which proves the conclusion of (ii).

Proposition 11

Let $A$ be a complete Hausdorff filtered commutative ring, $C$ a subring of $A_0$ such that $C \cap A_n = 0$ and $(x_i)_{1 \leq i \leq q}$ a finite family of elements of $A$ such that $x_i \in A_{n_i}$ where $n_i > 0$ for $1 \leq i \leq q$ ; let $\xi_i$ be the class of $x_i$ in $\mathrm{gr}_{n_i}(A)$ for $1 \leq i \leq q$ .

(i) There exists a unique $C$ -homomorphism $v$ from the algebra of formal power series $A'' = \mathbf{C}[[X_1, \ldots, X_q]]$ to $A$ such that $v(X_i) = x_i$ for $1 \leq i \leq q$ .

(ii) If the family $(\xi_i)$ is algebraically free over $C$ , the homomorphism $v$ is injective.

(iii) If the filtration on $A$ is exhaustive and the family $(\xi_i)$ is a system of generators of the $C$ -algebra $\mathrm{gr}(A)$ , the homomorphism $v$ is surjective.

As $n_i \geq 1$ for all $i$ , $\sum_{i=1}^q n_i s(i) \geq \sum_{i=1}^q s(i)$ for every monomial $X_1^{s(1)} \ldots X_q^{s(q)}$ and on the other hand $\sum_{i=1}^q n_i s(i) \leq r \cdot \sum_{i=1}^q s(i)$ if $r$ is the greatest of the $n_i$ . If $A''_n$ denotes the set of formal power series whose non-zero terms $a_s X_1^{s(1)} \ldots X_q^{s(q)}$ satisfy $\sum_{i=1}^q n_i s(i) \geq n$ , it follows from no. 6, Corollary to Proposition 6 that $A''$ is Hausdorff and complete with the exhaustive filtration $(A''_n)$ and that

A' = \mathbf{C}[X_1, \ldots, X_q]

is dense in $A''$ ; moreover the homomorphism $u$ defined in the proof of Proposition 10 is continuous on $A'$ and can be extended uniquely to a continuous homomorphism $v : A \to A$ , since $A$ is Hausdorff and complete (General Topology, Chapter III, § 3, no. 3, Proposition 5), which proves (i); also, $\mathrm{gr}(A'') = \mathrm{gr}(A')$ and $\mathrm{gr}(v) = \mathrm{gr}(u)$ ; (ii) and (iii) then follow respectively from Corollaries 1 and 2 to Theorem 1 of no. 8 in view of the hypotheses on $A$ .

The conclusion of (ii) (resp. (iii)) of the proposition is sometimes expressed by saying that the family $(x_i)$ is formally free over $C$ (resp. a formal system of generators of $A$ ).

Proposition 12

Let $A$ be a filtered ring, $E$ a filtered $A$ -module and $(A_n)$ and $(E_n)$ the respective filtrations on $A$ and $E$ . Suppose that $A$ is complete and the filtration $(E_n)$ is exhaustive and separated. Let $(x_i)_{i \in I}$ be a finite family of elements of $E$ and for $i \in I$ let $n(i)$ be an integer such that $x_i \in E_{n(i)}$ ; finally let $\xi_i$ be the class of $x_i$ in $\mathrm{gr}_{n(i)}(E)$ . Then, if $(\xi_i)$ is a system of generators of the $\mathrm{gr}(A)$ -module $\mathrm{gr}(E)$ , $(x_i)$ is a system of generators of the $A$ -module $E$ .

In the $A$ -module $L = A'$ , let $L_0$ denote the set $(a_i)$ such that $a_i \in A_{n - n(i)}$ for all $i \in I$ ; if $p$ and $q$ are the least and greatest of the $n(i)$ , then $A^I_{n-q} \supset L_0 \supset A^I_{n-p}$ and the topology defined on $L$ by the definition $(L_n)$ is the same as the product topology; hence $L$ is a complete filtered $A$ -module. As $L$ is free, there exists an $A$ -linear mapping $u : L \to E$ such that $u((a_i)) = \sum_{i \in I} a_i x_i$ and it is obviously compatible with the filtrations; we must prove that $u$ is surjective and for this it is sufficient, by virtue of Corollary 2 to Theorem 1, no. 8, to show that

\mathrm{gr}(u) : \mathrm{gr}(L) \to \mathrm{gr}(E)

is surjective or also that, for all $x \in E_n$ , there exist a family $(a_i)$ such that $a_i \in A_{n-n(i)}$ for all $i \in I$ and $x \equiv \sum_{i \in I} a_i x_i \pmod{E_{n+1}}$ . Let $\xi$ be the class of $x$ in $\mathrm{gr}_n(E)$ ; since the $\xi_i$ generate the $\mathrm{gr}(A)$ -module $\mathrm{gr}(E)$ , there exist $\alpha_i \in \mathrm{gr}(A)$ such that $\xi = \sum_{i \in I} \alpha_i \xi_i$ and we may assume that $\alpha_i \in \mathrm{gr}_{n-n(i)}(A)$ by replacing if need be $\alpha_i$ by its homogeneous component of degree $n - n(i)$ . Then $\alpha_i$ is the image of an element $a_i \in A_{n-n(i)}$ and the family $(a_i)$ has the required property.

Corollary 1

Let $A$ be a complete filtered ring and $E$ a filtered $A$ -module whose filtration is exhaustive and separated. If $\mathrm{gr}(E)$ is a finitely generated (resp. Noetherian) $\mathrm{gr}(A)$ -module, then $E$ is a finitely generated (resp. Noetherian) $A$ -module.

If $\mathrm{gr}(E)$ is finitely generated, there is a finite system of homogeneous generators and Proposition 12 shows that $E$ is finitely generated. Suppose now that $\mathrm{gr}(E)$ is Noetherian and let $F$ be a submodule of $E$ ; the filtration induced on $F$ by that on $E$ is exhaustive and separated and $\mathrm{gr}(F)$ is identified with a sub-gr(A)-module of gr(E) (no. 4, Proposition 2) and hence is finitely generated by hypothesis; we conclude that F is a finitely generated A-module and hence E is Noetherian.

Corollary 2

Let A be a complete Hausdorff filtered ring whose filtration is exhaustive. If gr(A) is a left Noetherian ring, so is A.

It is sufficient to apply Corollary 1 with E = A.

Corollary 3

Let A be a complete filtered ring, (A_n) its filtration, E a Hausdorff filtered A-module, (E_n) its filtration and F a finitely generated submodule of E; suppose that A_∞ = A and E_0 = E.

(i) If, for all k ≥ 0, E_k = E_{k+1} + A_k F, then F = E. (ii) If it is further supposed that the filtration on E is derived from that on A (no. 1, Example 2), the relation E = E_1 + F implies F = E.

Let ξ_i (1 ≤ i ≤ n) be the classes mod. E_1 of a finite system of generators of F. It follows from the given hypothesis that for all k ≥ 0 every element of gr_k(E) can be expressed in the form $\sum_{i=1}^n \alpha_i \xi_i$ where $\alpha_i \in \mathrm{gr}(A)$ ; the $\xi_i$ therefore generate the gr(A)-module gr(E), which proves (i) by virtue of Proposition 12. If the filtration on E is derived from that on A, the relation E = E_1 + F implies

E_k = A_k E = A_k E_1 + A_k F = A_k A_1 E + A_k F \subset A_{k+1} E + A_k F \\ = E_{k+1} + A_k F \subset E_k,

whence (ii).

Proposition 13

Let A be a ring, m a two-sided ideal of A contained in the Jacobson radical of A and E an A-module. Let A and E be given the m-adic filtrations (no. 1, Example 3). Suppose that one of the following conditions holds: (a) E is a finitely generated A-module and A is Hausdorff; (b) m is nilpotent.

For E to be a free A-module, it is necessary and sufficient that E/mE be a free (A/m)-module and that E satisfy property (GR) (no. 8).

If E is a free A-module and $(e_\lambda)$ a basis of E, $m^k E$ is the direct sum of the submodules $m^k e_\lambda$ of E for all $k > 0$ (Algebra, Chapter II, § 3, no. 7, Remark); then $m^k E / m^{k+1} E$ is identified with the direct sum of the $m^k e_\lambda / m^{k+1} e_\lambda$ (Algebra, Chapter 11, § 1, no. 6, Proposition 7). We deduce first (for $k = 0$ ) that the classes $1 \otimes e_\lambda$ of the $e_\lambda$ in $E / mE = (A / m) \otimes_A E$ form a basis of the (A/m)-module $E / mE$ , since the canonical mapping

(m^k / m^{k+1}) \otimes_A (E / mE) \to m^k E / m^{k+1} E

is bijective for all $k \geq 0$ ; hence $\gamma_E$ is bijective. Note that this part of the proof uses neither condition (a) nor condition (b).

Suppose conversely that the conditions of the statement hold and let $(x_i)_{i \in I}$ be a family of elements of E whose classes mod. $mE$ form a basis of the $(A/m)$ -module $E/mE$ ; let L be the free A-module $A_s^{(I)}$ , $(f_i)_{i \in I}$ its canonical basis and $u : L \to E$ the A-linear mapping such that $u(f_i) = x_i$ for all $i \in I$ . The hypotheses already imply that $u$ is surjective (Chapter II, § 3, no. 2, Corollary 1 to Proposition 4) and it remains to prove that $u$ is injective. Now, each of the hypotheses (a) and (b) implies that A is Hausdorff and hence so is L with the m-adic filtration, since $m^kL = (m^k)^{(I)}$ (Algebra, Chapter II, § 3, no. 7, Remark) and $\operatorname{gr}(L)$ is identified with $\operatorname{gr}(A) \otimes_{A/m} (L/mL)$ from the first part of the proof; the homomorphism $u$ is compatible with the filtrations and it is possible to write $\operatorname{gr}(u) = \gamma_E \circ v$ where $v$ is the bijection of $\operatorname{gr}(L)$ onto

\operatorname{gr}(A) \otimes_{A/m} (E/mE)

mapping the class off, mod. mM onto

1 \otimes \bar{x}_i

, where

\bar{x}_i

is the class of

x_i

mod.

mE

. The hypothesis then implies that

\operatorname{gr}(u)

is injective and the conclusion follows with the aid of Corollary 1 to Theorem 1, no. 8.

10. Application: examples of noetherian rings

Lemma 1

Let A be a graded ring of type $\mathbf{Z}$ , whose graduation $(A_n)$ is such that $A_n = 0$ for all $n < 0$ or $A_n = 0$ for all $n > 0$ . Let M be a graded A-module of type $\mathbf{Z}$ . For M to be a Noetherian A-module, it is necessary and sufficient that every graded submodule of M be finitely generated.

As $n \mapsto -n$ is an automorphism of the group $\mathbf{Z}$ , we may restrict our attention to the case $A_n = 0$ for all $n > 0$ . Let A' and M' denote the ring A and the module M with the filtrations associated with their respective graduations (no. 1, Example 1), which are exhaustive and separated; the hypothesis on A implies that A' is discrete and hence complete. If E is a sub-A-module of M, the filtered A'-module E' obtained by giving E the induced filtration is Hausdorff and its filtration is exhaustive; moreover $\operatorname{gr}(E')$ is identified with a graded sub-A-module of $M = \operatorname{gr}(M')$ and hence is finitely generated by hypothesis. The conclusion then follows from Corollary 1 to Proposition 12 of no. 9.

Theorem 2

Let A be a graded ring of type $\mathbf{N}$ , M a graded A-module of type $\mathbf{N}$ and $(A_n)$ and $(M_n)$ their respective graduations. Suppose that there exists an element $a \in A_0$ such that $A_n = A_0 a^n$ and $M_n = a^n M_0$ for all $n > 0$ . Then, if M, is a Noetherian $A_0$ -module, M is a Noetherian A-module.

By virtue of Lemma 1 it is sufficient to prove that every graded submodule N of M is finitely generated. For all $r \geq 0$ , let $N_r = N \cap M_r$ and let L_r be the set of $m \in M_0$ such that $a^r m \in N_r$ . As

a^r A_0 \subset A_r = A_0 a^r, \quad a^r A_0 L_r \subset A_0 a^r L_r \subset A_0 N_r \subset N_r

and hence the L, are sub-A,-modules of M,; moreover,

aN_r \subset N \cap aM_r = N \cap M_{r+1} = N_{r+1}

and hence the sequence $(L_r)_{r \geq 0}$ is increasing. The hypothesis implies that there exists an integer $n \geq 0$ such that $L_r = L_n$ for $r \geq n$ . For each $r \leq n$ , let $(m_{r,s})_{1 \leq s \leq k_r}$ be a system of generators of the A,-module $L_r$ . We shall show that the elements $a^r m_{r,s}$ for $1 \leq s \leq k_r$ , $0 \leq r \leq n$ form a system of generators of the A-module N. As $M_r = a^r M_0$ for all $r$ , $N_r = a^r L_r$ for all $r$ by definition of $L_r$ . Then, for $r \leq n$ ,

N_r = a^r L_r = \sum_{s=1}^{k_r} a^r A_0 m_{r,s} \subset \sum_{s=1}^{k_r} A_0 a^r m_{rs},

and, for $r > n$ ,

N_r = a^r L_n = \sum_{s=1}^{k_n} a^r A_0 m_{n,s} \subset \sum_{s=1}^{k_n} A_0 a^r m_{n,s} \subset \sum_{s=1}^{k_n} A_0 a^{r-n} \cdot (a^n m_{n,s})

which completes the proof (cf. Exercise 10).

Corollary 1 (Hilbert’s Theorem)

For every commutative Noetherian ring C, the polynomial ring C[X] is Noetherian* (cf. Exercise 10).

Corollary 2

For every commutative Noetherian ring C and every integer $n > 0$ , the polynomial ring C[X_1, ..., X_n] is Noetherian.

This follows from Corollary 1 by induction on $n$ .

Corollary 3

If C is a commutative Noetherian ring, every finitely generated commutative C-algebra is a Noetherian ring.

Such an algebra is isomorphic to a quotient of a polynomial algebra C[X_1, ..., X_n] (§ 1, no. 1).

Corollary 4

Let A be a graded commutative ring of type $\mathbf{N}$ and let $(A_n)$ be its graduation. For A to be Noetherian, it is necessary and sufficient that $A_n$ be Noetherian and that A be ajinitely generated A,-algebra.

The condition is sufficient by Corollary 3. Conversely, suppose A is Noetherian; $m = \sum_{n \geq 1} A_n$ which is an ideal of A, is then finitely generated; then A is a finitely generated A,-algebra (§ 1, no. 2, Corollary to Proposition 1); on the other hand $A_n$ , which is isomorphic to $A/m$ , is a Noetherian ring.

Corollary 5

Let A be a commutative ring and m an ideal of A such that $A/m$ is Noetherian, $m/m^2$ is ajinitely generated $(A/m)$ -module and A is Hausdorff and complete with the m-adic topology. Then gr(A) and A are Noetherian.

gr(A) is an $(A/m)$ -module generated by $m/m^2$ (no. 3, Example 3) and hence the ring gr(A) is Noetherian by Corollary 3. From this we deduce that A itself is Noetherian (no. 9, Corollary 2 to Proposition 12).

Corollary 6

For every commutative Noetherian ring C and every integer n > 0, the ring of formal power series $\mathbf{C}[[X_1, \ldots, X_n]]$ is Noetherian.

This follows from Corollary 5 and no. 6, Corollary to Proposition 6, for if m is the ideal of $A = \mathbf{C}[[X_1, \ldots, X_n]]$ consisting of the formal power series with no constant term, $A/m$ is isomorphic to C and $m/m^2$ to the C-module $\mathbf{C}"$ .

Remarks

(1) Corollaries 2, 3 and 6 apply in particular if C is a commutative field.

\* (2) Let g be a Lie algebra over a commutative Noetherian ring C and suppose that g is a finitely generated C-module. Let the enveloping algebra U of g be given the increasing filtration $(U_n)$ defined in no. 3, Example 4. With the corresponding topology, U is discrete and hence Hausdorff and complete; the associated graded ring gr(U) is a finitely generated C-algebra, being a quotient of the symmetric algebra $\mathscr{S}(g)$ , hence gr(U) is a Noetherian ring (Corollary 3) and we deduce that U is a left and right Noetherian ring (no. 9, Corollary 2 to Proposition 12). \*

11. Complete m-adic rings and inverse limits

We have seen in no. 6 that, if A is a commutative ring and m an ideal of A such that A is Hausdorf and complete with the m-adic topology, then the topological ring $A$ is canonically identified with the inverse limit of the discrete rings $A_i = A/m^{i+1} \ (i \in \mathbf{N})$ with respect to the canonical mappings

h_{ij}: A/m^{j+1} \to A/m^{i+1} \quad (i \leq j);

note that $h_{ij}$ is surjective and that, if $n_{ij}$ is its kernel, then

n_{ij} = m^{i+1}/m^{j+1} = (m/m^{j+1})^{i+1} =

in particular $(n_{0j})^{j+1} = 0$ . Conversely:

Proposition 14

Let $(A_i, h_{ij})$ be an inverse system of discrete commutative rings, whose indexing set is $\mathbf{N}$ and let $(M_i, u_{ij})$ be an inverse system of modules over the inverse system of rings $(A_i, h_{ij})$ . Let $n_j$ denote the kernel of $h_{0j}: A \to A$ , and set $A = \lim A_i$ , $M = \lim M_i$ . Suppose that

(a) for all $i \in \mathbf{N}$ , $h_{ii}$ is the identity mapping on A, and, for $i \leq j$ , $h_{ij}$ and $u_{ij}$ are surjective; (b) for $i \leq j$ , the kernels of $h_{ij}$ and $u_{ij}$ are $n_j^{i+1}$ and $n_j^{i+1} M_j$ respectively.

Then: (i) $A$ is a complete Hausdorff topological ring, $M$ is a complete Hausdorff topological A-module and the canonical mappings $h_i : A \to A_i$ , $u_i : M \to M_i$ are surjective. (ii) If $M_0$ is a finitely generated A,-module, $M$ is a finitely generated A-module;

to be precise, every finite subset S of M such that $u_0(S)$ generates M, is a system of generators of M.

The assertions in (i) follow from General Topology, Chapter II, § 3, no. 5, Corollary to Proposition 10 and Corollary 1 to Theorem 1.

For all $i \in \mathbf{N}$ , let us write $m_{i+1} = \mathrm{Ker}(h_i), N_{i+1} = \mathrm{Ker}(u_i)$ ; then

m_{i+1} = \lim_{k \geq 0} h_{i,t+k}^{-1}(0) = \lim_{\leftarrow} n_{i+1}^{t+k}

and $N_{i+1} = \lim_{\leftarrow} n_{i+k}^{t+1} M_{i+k}$ ; as $h_{i+k}$ and $u_{i+k}$ are surjective,

h_{i+k}(m_{i+1}) = n_{i+k}^{t+1}, \quad u_{i+k}(N_{i+1}) = n_{i+k}^{t+1} M_{i+k}.

Let us show that $m_i N_j \subset N_{i+j}$ for $i \geq 1$ and $j \geq 1$ , which amounts to proving that $u_{i+j-1}(m_i N_j) = 0$ ; now

u_{i+j-1}(m_i N_j) = h_{i+j-1}(m_i) u_{i+j-1}(N_j)

is equal to $n_{i+j-1}^{t'}(n_{i+j-1}^{t'} M_{i+j-1}) = 0$ , as, for all $k \geq 0$ , $n_k^{k+1}$ , which is the kernel of $h_{kk}$ , is equal to 0. We see similarly that $m_i m_j \subset m_{i+j}$ . If for $i \leq 0$ we set $m_i = A$ and $N_i = M$ , $(m_i)_{i \in \mathbf{Z}}$ is a filtration of A and $(N_i)_{i \in \mathbf{Z}}$ a filtration of M compatible with the filtration on A; the topologies on A and M are obviously those defined by these filtrations. This being so, let $a$ be an ideal of A such that $h_1(a) = n_1$ and $M'$ the submodule of M generated by S; we are going to prove that

N_i = a^i M' + N_{i+1} \quad \text{for } i \geq 0.

Let us write $a_i = h_i(a), M'_i = u_i(M')$ ; it is sufficient to show that

u_i(N_i) = a_i^i M'_i.

This is true if $i = 0$ , for $N_0 = M$ and $M'_0 = M$ , by hypothesis. If $i \geq 1$ , then $u_i(N_i) = n_i^i M_i$ by (16). As $h_{1i}$ is surjective and $h_{0i} = h_{01} \circ h_{1i}$ , $h_{1i}$ maps the kernel $n_i$ of $h_i$ , onto the kernel $n_1$ of $h_1$ , and $n_i = h_{1i}(n_1)$ ; then

h_{1i}(a_i) = h_1(a) = n_1 = h_{1i}(n_i)

and, as the kernel of $h_{1i}$ is $n_i^2$ , $n_i \subset a_i + n_i^2$ and $a_i \subset n_i$ , whence $n_i = a_i + n_i^2$ . Moreover $u_{0i}(M'_i) = u_0(M') = M, = u_{0i}(M_i)$ and, as $\mathrm{Ker}(u_{0i}) = n_i M_i$ , $M_i = M'_i + n_i M_i$ ; whence

n_i^i M_i = (a_i + n_i^2)^i (M'_i + n_i M_i).

Now, $a_i^k n_i^{i+1-k} \subset n_i^{i+1} = 0$ for $0 \leq k \leq i$ ; then it certainly follows that $u_i(N_i) = n_i M_i = a_i^i M'_i$ , which proves (17).

Moreover $m_1 = h_1(n_1)$ , whence $a \subset m_1$ and therefore $a^i \subset m_1^i \subset m_i$ , whence

$N_i \subset m_i M' + N_{i+1}$ ; on the other hand obviously $m_i M \subset N_i$ and hence $N_i = m_i M' + N_{i+1}$ for all $i \geq 0$ ; then it follows from Corollary 3 to Proposition 12 of no. 9 that $M' = M$ , which completes the proof.

Corollary 1

With the notation and hypotheses of Proposition 14 suppose further that $M_0$ is a finitely generated $A$ -module and that the ideal $n_1$ of $A$ , is finitely generated. Let $m_1$ be the kernel of $h_1$ ; the topologies on $A$ and $M$ are then the $m_1$ -adic topologies on this ring and this module respectively; to be precise, for all $i \geq 0$ , the kernels of $h_i$ and $u_i$ are $m_1^{i+1}$ and $m_1^{i+1} M$ respectively; further $m_1 / m_1^2$ is a finitely generated $A$ -module.

We preserve the notation of the proof of Proposition 14; the hypotheses here allow us to assume that the ideal $a$ is finitely generated. Let $i \geq 0$ be any integer; for all $j \geq 0$ , by (17), $N_{i+j} = a^j(a^i M) + N_{i+j+1} \subset m_j(a^i M) + N_{i+j+1}$ ; conversely, $m_j(a^i M) \subset m_j m_i M \subset m_{i+j} M \subset N_{i+j}$ , whence

N_{i+j} = m_j(a^i M) + N_{i+j+1}.

As $a$ and $M$ are finitely generated $A$ -modules, so is $a^i M$ . Applying Corollary 3 to Proposition 12 of no. 9 to the module $N_i$ with the filtration $(N_{ij})_{j \in \mathbf{Z}}$ defined by $N_{ij} = N_i$ , if $j < 0$ , $N_{ij} = N_{i+j}$ if $j \geq 0$ , we obtain $N_i = a^i M$ , whence $N_i \subset m_1^i M$ . But also $m_1^i M \subset m_i M \subset N_i$ , whence $N_i = m_1^i M$ . Applying this to the case where $M = A$ , $u_{ij} = h_{ij}$ , we obtain $m_i = m_1^i$ . Moreover, $m_1 = a + m_1^2$ by (17), which proves the last assertion of the corollary.

Corollary 2

Under the hypotheses of Corollary 1, for $A$ to be Noetherian, it is necessary and sufficient that $A$ , be so.

The condition is necessary since $A$ , is isomorphic to a quotient of $A$ ; it is sufficient by virtue of no. 10, Corollary 5 to Theorem 2.

12. The Hausdorff completion of a filtered module

Let $G$ be a filtered group whose filtration $(G_n)$ consists of normal subgroups of $G$ ; we have already recalled (no. 6) that the Hausdorff completion $\hat{G}$ of the topological group $G$ is canonically identified with the inverse limit $\lim \leftarrow G/G_n$ of the discrete groups $G/G_n$ , the canonical homomorphism $i : G \to \hat{G}$ having image the Hausdorff group associated with $G$ (everywhere dense in $\hat{G}$ ) and kernel the closure $\bigcap G_n$ of $\{0\}$ in $G$ . The Hausdorff completion $\hat{G}_n$ of the subgroup $G_n$ of $G$ is identified with the closure of $i(G_n)$ in $\hat{G}$ (General Topology, Chapter II, § 3, no. 9, Corollary 1 to Proposition 18) and, since $G$ , is closed in $\hat{G}$ ,

G, = i^{-1}(\hat{G}_n) = i^{-1}(\hat{G}_n \cap i(G)).

Moreover, the $\hat{G}_n$ form a fundamental system of neighbourhoods of 0 in $\hat{G}$ (General Topology, Chapter 111, § 3, no. 4, Proposition 7) and are therefore normal open subgroups of $\hat{G}$ (General Topology, Chapter 111, § 2, no. 3, Proposition 8); the topology on $\hat{G}$ is defined by the filtration $(\hat{G}_n)$ , which is always separated by definition. As $i(G)$ is dense in $\hat{G}$ and $\hat{G}_n$ is open,

\hat{G} = i(G) \cdot \hat{G}_n

and similarly

= i(G_{n-1}) \cdot \hat{G}_n.

We deduce from (18) and (19) that the filtration $(\hat{G}_n)$ is exhaustive if and only if $(G_n)$ is.

The second isomorphism theorem (Algebra, Chapter I, § 6, no. 13, Theorem 6 (d)) and equations (18), (19) and (20) show that the canonical homomorphisms

G_{n-1}/G_n \to \hat{G}_{n-1}/\hat{G}_n, \quad G/G_n \to \hat{G}/\hat{G}_n,

are bijective and hence so is the canonical homomorphism

\operatorname{gr}(G) \to \operatorname{gr}(\hat{G}).

Now let $A$ be a filtered ring, $E$ a filtered $A$ -module and $(A_n)$ and $(E_n)$ the respective filtrations of $A$ and $E$ ; we shall assume that these filtrations are exhaustive so that for the corresponding topologies $A$ is a topological ring and $E$ a topological $A$ -module (no. 5, Proposition 3). Then we have defined (General Topology, Chapter 111, §6, nos. 5 and 6) $\hat{A}$ as a topological ring and $\hat{E}$ as a topological $\hat{A}$ -module. If $i : A \to \hat{A}$ is the canonical homomorphism, then $i(A_m)i(A_n) \subset i(A_{m+n})$ , whence by the continuity of multiplication in $A$ ,

\hat{A}_m \hat{A}_n \subset \hat{A}_{m+n}

since $A_n$ is the closure of $i(A_n)$ in $\hat{A}$ . It can be similarly shown that

\hat{A}_m \hat{E}_n \subset \hat{E}_{m+n};

in other words:

Proposition 15

Let $A$ be a filtered ring and $E$ a filtered $A$ -module, the respective filtrations $(A_n), (E_n)$ of $A$ and $E$ being exhaustive. Then $(\hat{A}_n)$ is a filtration compatible with the ring structure on $\hat{A}$ and $(\hat{E}_n)$ a filtration compatible with the module structure on $\hat{E}$ over the filtered ring $\hat{A}$ ; moreover these filtrations are exhaustive and define respectively the topologies on $\hat{A}$ and $\hat{E}$ . Finally, the canonical mappings $\operatorname{gr}(A) \to \operatorname{gr}(\hat{A})$ and $\operatorname{gr}(E) \to \operatorname{gr}(\hat{E})$ of graded $\mathbf{Z}$ -modules are respectively a graded ring isomorphism and a graded $\operatorname{gr}(A)$ -module isomorphism.

In what follows, for every uniform space $X$ , $j_X$ will denote the canonical mapping from $X$ to its Hausdorff completion $\hat{X}$ and $X_0 = j_X(X)$ the uniform subspace of $\hat{X}$ , which is the Hausdorff space associated with $X$ . Recall that the topology on $X$ is the inverse image under $j_X$ of that on $X_0$ (General Topology,

Chapter II, § 3, no. 7, Proposition 12). Recall also that, for every uniformly continuous mapping $f : X \to Y$ , $\hat{f}$ denotes the uniformly continuous mapping from $\hat{X}$ to $\hat{Y}$ such that $\hat{f} \circ j_X = j_Y \circ f$ (loc. cit., Proposition 15); if $X$ is a uniform subspace of $Y$ and $\mathscr{E}$ the canonical injection, $\hat{X}$ is identified with a uniform subspace of $\hat{Y}$ and $\hat{f}$ is the canonical injection of $X$ into $\hat{Y}$ (loc. cit., no. 9. Corollary 1 to Proposition 18).

Lemma 2

Let $X \xrightarrow{f} Y \xrightarrow{g} Z$ be an exact sequence of strict morphisms of topological groups (Algebra, Chapter II, § 1, no. 4, Remark). Suppose that $X, Y, Z$ admit Hausdorff completion groups and that the identity elements $\mathscr{E} X, Y, Z$ admit countable fundamental systems of neighbourhoods. Then $\hat{X} \xrightarrow{\hat{f}} \hat{Y} \xrightarrow{\hat{g}} \hat{Z}$ is an exact sequence of strict morphisms.

Let $N_1, N_g$ be the respective kernels off and $g$ ; let us write

f = f_3 \circ f_2 \circ f_1

where $f_1$ is the canonical mapping $X \to X/N_f$ , $f_2$ is an isomorphism of $X/N_f$ onto $N$ , and $f_3$ is the canonical injection $N \to Y$ . We already know that $f_2$ is an isomorphism of $(X/N_f)^*$ onto $\hat{N}_g$ and we have just recalled that $f_1$ is an injective strict morphism of $N$ , to $\hat{Y}$ ; if we show that $\hat{f}_1$ is a surjective strict morphism, it will follow that $\hat{f}$ is a strict morphism (General Topology, Chapter III, § 2, no. 8, Remark 2). Let $g_1$ be the canonical mapping $Y \to Y/N_g$ ; if we show that $\hat{g}_1$ is a surjective strict morphism of the kernel $\hat{N}_g$ , we shall see as above that $\hat{g}$ is a strict morphism and the sequence $\hat{X} \xrightarrow{\hat{f}} \hat{Y} \xrightarrow{\hat{g}} \hat{Z}$ will be exact. Thus we have reduced the problem to proving that, if $Y = X/N$ (where $N$ is a normal subgroup of $X$ ) and $f : X \to Y$ is the canonical mapping, $\hat{f}$ is a surjective strict morphism with kernel $\hat{N}$ .

Let $f_0 : X_0 \to Y_0$ be the mapping which coincides with $f$ on $X_0$ ; as $j_X$ (resp. $j_Y$ ) is a surjective strict morphism of $X$ onto $X$ , (resp. $Y$ onto $Y_0$ ), $f_0$ is a surjective strict morphism (General Topology, Chapter III, § 2, no. 8, Remark 3). Now $X_0$ and $Y_0$ are metrizable (General Topology, Chapter IX, § 3, no. 1, Proposition 1); then it follows from General Topology, Chapter IX, § 3, no. 1, Corollary 1 to Proposition 4 and Lemma 1, that $\hat{f}_0 = \hat{f}$ is a surjective strict morphism and has kernel the closure $\hat{N}_0'$ in $X$ of the kernel $N_0'$ of $f_0$ . Then it will be sufficient for us to prove that $\hat{N}_0' = \hat{N}$ . Now $N_0'$ obviously contains $N_0 = j_X(N)$ ; it will be sufficient to show that $N_0'$ is contained in the closure $\overline{N}_0$ of $N_0$ in $X$ . Now,

u = j_X^{-1}(X_0 - \overline{N}_0) = X - j_X^{-1}(\overline{N}_0)

is an open set in $X$ which does not meet $N$ ; as $f$ is a surjective strict morphism, $V = f(U)$ is an open set in $Y$ not containing the identity element $e'$ of $Y$ and hence not meeting the closure of $e'$ ; then $j_Y(V)$ does not contain the identity element of $Y_0$ . But $j_Y(V) = f_0(X_0 - \overline{N}_0)$ and hence $N'_0 \subset \overline{N}_0$ , which complete the proof of Lemma 2.

Proposition 16

Let $A$ be a filtered ring, ( $A,$ ) its filtration, $E$ an $A$ -module and ( $E_n$ ) the filtration on $E$ derived from that on $A$ consisting of the $E_n = A_n E$ . Suppose that the filtration ( $A,$ ) is exhaustive and the module $E$ is finitely generated. If $i : E \to \hat{E}$ is the canonical mapping, then, for all $n \in \mathbf{Z}$ ,*

8_n = \hat{A}_n \hat{E} = \hat{A}_n i(E) \quad \text{and} \quad \hat{E} = \hat{A}.i(E).

*In particular

\hat{E}

is a finitely generated

\hat{A}

-module.

The equation $A_n E = E_n$ implies, by virtue of the continuity of the external law on the $A$ -module $8, \hat{A}_n \hat{E} \subset 8$ , and obviously $\hat{A}_n \hat{E} \supset \hat{A}_n i(E)$ . By hypothesis there exists a surjective homomorphism $u : L \to E$ , where $L = A_s^I$ , $I$ being a finite set; let $L$ be given the product filtration, consisting of the $L_n = A_n^I$ , which define on $L$ the product topology; then $\hat{L} = \hat{A}_s^I$ and $\hat{L}_n = \hat{A}_n^I$ (General Topology, Chapter II, § 3, no. 9, Corollary 2 to Proposition 18). Let $j : L \to \hat{L}$ be the canonical mapping and $(e_i)_{i \in I}$ the canonical basis of $L$ ; for an element $\sum_{i \in I} a_{ij}(e_i)$ (where $a_i \in \hat{A}$ for all $i \in I$ ) to belong to $\hat{L}_n$ , it is necessary and sufficient that $a_i \in \hat{A}_n$ for all $i$ ; then $\hat{L}_n = \hat{A}_n . j(L)$ . This being so, by definition $u(\hat{L}_n) = A_n E = E_n$ and hence $u$ is a strict morphism of $L$ onto $E$ (General Topology, Chapter III, § 2, no. 8, Proposition 24). Lemma 2 then shows that $\hat{u} : \hat{L} \to \hat{E}$ is a surjective strict morphism. As $\hat{L}_n$ is an open subgroup off, $\hat{u}(\hat{L}_n)$ is an open (and therefore closed) subgroup of $\hat{E}$ ; but $\hat{u}(\hat{L}_n) = \hat{A}_n \hat{u}(j(L)) = \hat{A}_n i(E)$ and, as $i(E_n) \subset A_n i(E) \subset \hat{A}_n i(E)$ , finally $\hat{E}_n \subset \hat{A}_n i(E) \subset \hat{A}_n \hat{E} \subset \hat{E}_n$ and therefore $\hat{E}_n = \hat{A}_n \hat{E} = \hat{A}_n i(E)$ ; setting $n = 0$ , we obtain the second formula of (25).

Corollary 1

Under the conditions of Proposition 16, if $A$ is complete, so is $E$ .

As the canonical mapping $A \to \hat{A}$ is then surjective (no. 6, Proposition 5), $\hat{E} = i(E)$ by (25) and the conclusion follows by Proposition 5 of no. 6.

Corollary 2

Let $A$ be a commutative ring, $m$ a finitely generated ideal of $A$ and $\hat{A}$ the Hausdorff completion of $A$ with respect to the $m$ -adic topology. Then $\hat{m}^n = (\hat{m})^n = m^n . \hat{A}$ for every integer $n > 0$ and the topology on $\hat{A}$ is the $\hat{m}$ -adic topology.

Let us write $A_n = m^n$ , which is a finitely generated ideal of $A$ . The formula $m^p A_n = m^{n+p}$ shows that the topology induced on $A$ , by the $m$ -adic topology coincides with the $m$ -adic topology on the $A$ -module $A_n$ (no. 1, Example 3).

By Proposition 16 applied to $E = A_n$ , $\hat{A}_n = \hat{A}.A_n$ in other words $\hat{m}^n = m^n . \hat{A}$ . In particular $m = m . \hat{A}$ , whence

(\hat{m})^n = m^n . \hat{A} = \hat{A}.A_n

(cf. Exercise 12).

Examples of Hausdorff completions of filtered rings

(1) Let $A$ be a graded ring of type $\mathbf{N}$ and let $(A_n)_{n \geq 0}$ be its graduation; let it be given the associated filtration which is separated and exhaustive (no. 1, Example 1). The additive group $A$ is canonically identified with a subgroup of $B = \prod_{n \in \mathbf{N}} A_n$ ; if $B$ is given the topology the product of the discrete topologies, the topology induced on $A$ is the topology defined by the filtration on $A$ ; also $B$ is a complete topological group and $A$ is dense in $B$ (General Topology, Chapter III, §2, no. 9, Proposition 25). The additive topological group $B$ is then identified with the completion $\hat{A}$ of the Hausdorff additive group $A$ and it follows from Proposition 15 that it has a unique ring structure which makes it the completion of the topological ring $A$ . To define multiplication in this ring, note that, if we write $A'_n = \sum_{i > n} A_i$ , the closure in $B$ of the two-sided ideal $A'_n$ is the set $B$ , of $x = (x_i) \in B$ such that $x_i = 0$ for $i \leq n$ . Then let $x = (x_i), y = (y_i)$ be two elements of $B$ and $z = (z_i)$ their product. Then, for all $n > 0$ , $x \equiv x'_n$ (mod. $B_n$ ), $y \equiv y'_n$ (mod. $B_n$ ), where $x'_n = (x_i)_{0 \leq i \leq n}$ , $y'_n = (y_i)_{0 \leq i \leq n}$ , whence $z \equiv x'_n y'_n$ (mod. $B_n$ ). But $x'_n$ and $y'_n$ belong to $A$ and it is therefore seen that, for all $n \in \mathbf{N}$ ,

z_n = \sum_{j=0}^n x_j y_{n-j}.

In particular, we again obtain the Corollary to Proposition 6 of no. 6 : if $C$ is a commutative ring, the completion of the polynomial ring $C[X_1, \ldots, X_r]$ , with the filtration associated with its usual graduation (by total degree) is canonically identified with the ring of formal power series $C[[X_1, \ldots, X_r]]$ (cf. Algebra, Chapter IV, §5, no. 10).

(2) Let $K$ be a complete commutative field with a valuation. The completion of the ring of convergent series in $r$ variables over $K$ is canonically identified with the ring of formal power series $K[[X_1, \ldots, X_r]]$ . \

(3) Let $a$ be a non-zero non-invertible element of a principal ideal domain; the (a)-adic topology on $A$ is also called the $a$ -adic topology; it is Hausdorff, for the intersection of the ideals $(a^n)$ reduces to 0 (Algebra, Chapter VII, §1, no. 3). Note that the completion of $A$ with respect to this topology is not necessarily an integral domain (cf. no. 13, Remark 3). The associated graded ring $\mathrm{gr}(A) = \mathrm{gr}(\hat{A})$ is canonically isomorphic to $(A/a)[X]$ (no. 3, Example 1). If $A = \mathbf{Z}$ , the completion of $\mathbf{Z}$ with respect to the n-adic topology ( $n > 1$ ) is denoted by $\mathbf{Z}_n$ and its elements are called $n$ -adic integers.

Every element of $\mathbf{Z}/n^k \mathbf{Z}$ admits a unique representative of the form $\sum_{i=0}^{k-1} a_i n^i$ where $0 \leq a_i \leq n - 1$ for all $i$ ; moreover, its canonical image in $\mathbf{Z}/n^{k-1} \mathbf{Z}$ is the class of $\sum_{i=0}^{k-2} a_i n^i$ . These remarks and the fact that $Z$ , is canonically identified with the inverse limit $\lim_{\leftarrow} \mathbf{Z}/n^k \mathbf{Z}$ show immediately that every element of $Z$ , can be written uniquely in the form $\sum_{i=0}^{\infty} a_i n^i$ where $0 \leq a_i < n$ and conversely that such a series is convergent in $\mathbf{Z}$ .

13. The Hausdorff completion of a semi-local ring

Proposition 17

Let $A$ be a commutative ring and $(m_{\lambda})_{\lambda \in L}$ a family of ideals of $A$ , distinct from $A$ , such that $m_{\lambda}$ and $m_{\mu}$ are relatively prime for $\lambda \neq \mu$ . For every family $s = (s(\lambda))_{\lambda \in L}$ of integers $\geq 0$ , of finite support, set $a_s = \bigcap_{\lambda \in L} m_{\lambda}^{s(\lambda)}$ (equal to the product of the $m_{\lambda}^{s(\lambda)}$ for the $\lambda$ such that $s(\lambda) \neq 0$ ; cf. Chapter 11, § 1, no. 2, Propositions 3 and 5); the $a_s$ form a fundamental system of neighbourhoods of 0 with respect to a topology $\mathscr{T}$ compatible with the ring structure on $A$ ; let $\hat{A}$ be the Hausdorff completion of $A$ with respect to this topology. On the other hand, for all $\lambda \in L$ , let $A_{\lambda}$ be the ring $A$ with the $m_{\lambda}$ -adic topology and let $\hat{A}_{\lambda}$ be its Hausdorff completion. If $u : A \to \prod_{\lambda \in L} A_{\lambda}$ denotes the diagonal homomorphism, $u$ is continuous and the corresponding homomorphism $\hat{u} :$

\hat{A} \to \left( \prod_{\lambda \in L} A_{\lambda} \right)^{\hat{}} = \prod_{\lambda \in L} \hat{A}_{\lambda}

(General Topology, Chapter III, § 6, no. 5 and Chapter II, § 3, no. 9, Corollary 2 to Proposition 18) is a topological ring isomorphism.

The first assertion follows from General Topology, Chapter III, § 6, no. 3, Example 3. Let us set $B = \prod_{\lambda \in L} A_{\lambda}$ ; as the topology on $A$ is finer than each of the $m_{\lambda}$ -adic topologies, the mappings $\mathrm{pr}_{\lambda} \circ u$ are continuous and hence $u$ is continuous. Also, $u(a_s)$ is the intersection of the diagonal $A$ of $B$ and the open set $\bigcap_{\lambda \in L} \mathrm{pr}_{\lambda}^{-1}(m_{\lambda}^{s(\lambda)})$ of $B$ ; it follows that $u$ is a strict morphism from the additive group $A$ to $B$ with image $A$ . Now $\Delta$ is dense in $B$ . For let $b = (a_{\lambda})_{\lambda \in L}$ be an element of $B$ ; every neighbourhood of $b$ in $B$ contains a set of the form $b + V$ , where $V = \bigcap_{\lambda \in L} \mathrm{pr}_{\lambda}^{-1}(m_{\lambda}^{s(\lambda)})$ for a family $s = (s(\lambda))_{\lambda \in L}$ with finite support of integers $\geq 0$ . As the $m_{\lambda}^{s(\lambda)}$ are relatively prime in pairs (Chapter II, § 1, no. 2, Proposition 3), there exists $x \in A$ such that $x \equiv a_{\lambda} \pmod{m_{\lambda}^{s(\lambda)}}$ for all $\lambda$ (loc. cit., Proposition 5) and hence $(b + V) \cap A \neq \varnothing$ . The Hausdorff completion of the group $B/\Delta$ is then $\{0\}$ ; applying Lemma 2 of no. 12 to the exact sequences $0 \to A \xrightarrow{u} B, A \xrightarrow{u} B \to B/\Delta$ , we see that $\hat{u}$ is an isomorphism of $\hat{A}$ onto $\hat{B}$ .

Corollary

Let $A$ be a principal ideal domain and $P$ a representative system of extremal elements of $A$ (Algebra, Chapter VII, § 1, no. 3). The topology on $A$ with respect to which the ideals $\neq 0$ of $\mathbf{A}$ form a fundamental system of neighbourhoods of 0, which is compatible with the ring structure on $\mathbf{A}$ , is Hausdorff and the completion of $\mathbf{A}$ with this topology is canonically isomorphic to the product of the completions of $\mathbf{A}$ with respect to the $x$ -adic topologies, where $x$ runs through $P$ .

The principal ideals $(x)$ where $\pi \in P$ are maximal and distinct and hence relatively prime, we have already seen (no. 12, Example 3) that the $x$ -adic topologies are Hausdorff and hence so is the topology defined in the statement of Proposition 17, which is finer than each of the $x$ -adic topologies.

If the Corollary to Proposition 17 is applied when $\mathbf{A} = \mathbf{Z}$ , we denote by $\hat{\mathbf{Z}}$ the completion of $\mathbf{Z}$ with respect to the topology for which all the ideals $\neq 0$ of $\mathbf{Z}$ form a fundamental system of neighbourhoods of 0, the ring isomorphic to the product $\prod_{p \in P} \mathbf{Z}_p$ of the rings of $p$ -adic integers (P being the set of prime numbers).

Remarks

(1) Clearly, under the conditions of Proposition 17, the topology $\mathscr{T}$ is the least upper bound of the $m_\lambda$ -adic topologies on $\mathbf{A}$ .

(2) Every closed ideal $a$ of $\prod_{\lambda \in L} \hat{A}_\lambda$ is identical with the product of its projections $a_\lambda = \mathrm{pr}_\lambda(a)$ , which are closed ideals in the $A_\lambda$ ; for $\hat{A}_\lambda$ is canonically identified with a closed ideal $A'_\lambda$ of $\prod_{\lambda} \hat{A}_\lambda$ and $a_\lambda$ with $a \cap A'_\lambda$ (Algebra, Chapter I, §8, no. 10, Proposition 6), the sum of the $a_\lambda$ is dense in the product $\prod_\lambda a_\lambda$ (General Topology, Chapter III, §2, no. 9, Proposition 25) and the latter is closed in $\prod_\lambda \hat{A}_\lambda$ , whence our assertion.

Proposition 18

Let $\mathbf{A}$ be a commutative ring, $(m_i)_{1 \leq i \leq q}$ a finite family of distinct maximal ideals of $\mathbf{A}$ , t the product ideal $m_1 m_2 \ldots m_q = m_1 \cap m_2 \cap \ldots \cap m_q$ and $S$ the multiplicative subset $\bigcap_{i=1}^q (\mathbf{A} - m_i)$ . Let $\mathbf{A}$ be given the $r$ -adic topology, the ring $B = S^{-1} \mathbf{A}$ the $rB$ -adic topology and each of the local rings $A_n$ the $(m_i A_{m_i})$ -adic topology. Let $u : \mathbf{A} \to B$ , $v_i : B \to A_{m_i}$ be the canonical homomorphism (Chapter II, §2, no.1, Corollary 2 to Proposition 2) and $v$ the homomorphism $(v_i) : B \to \prod_{i=1}^q A_n$ . The homomorphisms $u$ and $v$ are continuous and the corresponding homomorphism $\hat{u} : \hat{\mathbf{A}} \to$ and $\hat{v} : \hat{B} \to \prod_{n} (A_n)^*$ are topological ring isomorphisms.

$m_i \cap S = \varnothing$ for $1 \leq i \leq q$ , hence the ideal $m'_i = m_i B$ of $B$ is maximal (Chapter II, §2, no. 5, Proposition 11) and

rB = m'_1 \cap m'_2 \cap \cdots \cap m'_q

(Chapter II, § 2, no. 4); finally, $B_{m'_i} = A_i$ , up to a canonical isomorphism (Chapter 11, § 2, no. 5, Proposition 11). As $\bar{u}^1(rB) = r$ and $\bar{v}_i^1(m_iA_{m_i}) \supset rB$ , $u$ and $v$ are continuous. Then it is sufficient to prove that,

w = v \circ u : A \to \prod_{i=1}^4 A_{m_i},

$\hat{w}$ is an isomorphism of $\hat{A}$ onto $\prod_{i=1}^q \hat{A}_{m_i}$ , for this result applied to $B$ and the $m'_i$ will show that $\hat{v}$ is an isomorphism and therefore also $\hat{u}$ . Note that every product of powers of the $m_i$ contains a power of $c$ and hence the $r$ -adic topology is the least upper bound of the $m_i$ -adic topologies; moreover, if $A_i$ denotes the ring $A$ with the $m_i$ -adic topology and $\phi : A \to \prod_{i=1}^q A$ , the diagonal mapping, $\hat{\phi} : \hat{A} \to \prod_{i=1}^4 \hat{A}_i$ is an isomorphism (Proposition 17). Then it all amounts to proving that, if $u_i : A_i \to A_{m_i}$ is the canonical mapping, $\hat{u}_i : \hat{A}_i \to \hat{A}_{m_i}$ is an isomorphism. Now, for all $n$ , the mapping

u_{i,n} : A/m_i^n \to A_{m_i}/m_i^n A_{m_i}

derived from $u_i$ by taking quotients is an isomorphism (Chapter II, §3, no. 3, Proposition 9); our assertion follows from the fact that $\hat{A}_i$ (resp. $\hat{A}_{m_i}$ ) is the inverse limit of the discrete rings $A/m_i^n$ (resp. $A_{m_i}/m_i^n A_{m_i}$ ) (cf. no. 6).

Remark (3). We see that an integral domain $A$ can be such that its Hausdorff completion $\hat{A}$ admits non-zero divisors of zero.

Proposition 19

Let $A$ be a commutative ring and $m$ a maximal ideal of $A$ . The Hausdorff completion $\hat{A}$ of $A$ with respect to the $m$ -adic topology is a local ring whose maximal ideal is $m$ .

If $a = \bigcap_{k \geq 1} m^k$ , $\hat{A}$ is the completion of the Hausdorff ring $A/a$ associated with $A$ and, as $m/a$ is maximal in $A/a$ , we may assume that $A$ is Hausdorff with respect to the $m$ -adic topology. As $A/m$ and $\hat{A}/\hat{m}$ are isomorphic rings (no. 12, formula (21)), $m$ is maximal in $\hat{A}$ . As the topology on $\hat{A}$ is defined by the filtration $(m^n)^*$ (no. 12), the proposition will be a consequence of the following lemma:

Lemma 3

Let $A$ be a complete Hausdorf topological ring, in which there exists a fundamental system $\mathscr{G}$ of neighbourhoods of 0 consisting of additive subgroups of $A$ . (i) For all $x \in A$ such that $\lim_{n \to m} x^n = 0$ , $1 - x$ is invertible in $A$ and its inverse is equal to $\sum_{n=0}^\infty x^n$ . (ii) Let $a$ be a two-sided ideal of $A$ such that $\lim_{n \to \infty} x^n = 0$ for all $x \in a$ . For an element y of $\mathbf{A}$ to be invertible, it is necessary and sufficient that its class mod. a be invertible in $\mathbf{A}/a$ ; in particular a is contained in the Jacobson radical of $\mathbf{A}$ .

(i) As

(1 - x)(1 + x + \cdots + x^n) = (1 + x + \cdots + x^n)(1 - x) = 1 - x^{n+1},

it all amounts to proving that the series with general term $x^n$ is convergent in $\mathbf{A}$ ; now, by hypothesis, for every neighbourhood $V \in \mathfrak{S}$ of 0 in $\mathbf{A}$ , there exists an integer $p > 0$ such that $x^n \in V$ for all $n \geq p$ . We conclude that

x^p + x^{p+1} + \ldots + x^q \in V

for all $q \geq p$ and our assertion then follows from Cauchy's criterion (General Topology, Chapter III, § 5, no. 2, Theorem 1).

(ii) Suppose that there exists $y' \in \mathbf{A}$ such that $yy' \equiv 1$ (mod. a) and $y'y \equiv 1$ (mod. a). The hypothesis on a implies, by (i), that $yy'$ and $y'y$ are invertible in $\mathbf{A}$ and hence $y$ is invertible in $\mathbf{A}$ . In particular, every $x \in a$ is such that $1 - x$ is invertible in $\mathbf{A}$ and, as a is a two-sided ideal of $\mathbf{A}$ , it is contained in the Jacobson radical of $\mathbf{A}$ (Algebra, Chapter VIII, § 6, no. 3, Theorem 1).

Having established this lemma, it is sufficient to apply it to the topological ring $\hat{\mathbf{A}}$ and the ideal $\hat{\mathfrak{m}}$ , as, for all $x \in \hat{\mathfrak{m}}$ , $x^n \in (\hat{\mathfrak{m}})^n \subset (\mathfrak{m}^n)^{\wedge}$ and the sequence $(x^n)$ therefore tends to 0.

If we take $\mathbf{A} = \mathbf{Z}$ , every maximal ideal of $\mathbf{Z}$ is of the form $p\mathbf{Z}$ where $p$ is prime. The ring of $p$ -adic numbers $\mathbf{Z}_p$ is then a local ring of which $p\mathbf{Z}_p$ is the maximal ideal (Corollary 2 to Proposition 16) and whose residue field is isomorphic to $\mathbf{Z}/p\mathbf{Z} = \mathbf{F}_p$ , and $\mathbf{Z}_{(p)}$ with the $p\mathbf{Z}_{(p)}$ -adic topology is identified with a topological subring of $\mathbf{Z}$ , containing $\mathbf{Z}$ .

Corollary

Let $\mathbf{A}$ be a semi-local ring (Chapter II, § 3, no. 5), $m_i$ its distinct maximal ideals $(1 \leq i \leq q)$ and

r = m_1 \cap m_2 \cap \cdots \cap m_q

its Jacobson radical. The Hausdorff completion $\hat{\mathbf{A}}$ of $\mathbf{A}$ with respect to the r-adic topology is a semi-local ring, canonically isomorphic to the product $\prod_{i=1}^q \hat{\mathbf{A}}_{m_i}$ , where $\hat{\mathbf{A}}_{m_i}$ is the Hausdorff completion ring of the local ring $\mathbf{A}_{m_i}$ with respect to the $(m_i \mathbf{A}_{m_i})$ -adic topology.

Exercises

See the exercises for § 2.