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

(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,

$\hat{w}$ 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

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.