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

(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 $n \in \mathbf{N}$,

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 $\mathbf{Z}$.