Proposition 8

Let $A$ be a commutative Noetherian ring, $m$ an ideal of $A$, $\hat{A}$ the Hausdorff completion of $A$ with respect to the $m$-adic topology and $j$ the canonical mapping from $A$ to $\hat{A}$. The: (i) $\hat{A}$ is a Zariski ring and $\hat{m} = \hat{A}.j(m)$ is a defining ideal of $\hat{A}$.

(ii) The mapping $n \mapsto A = \hat{A}.j(n)$ is a bijection of the set of maximal ideals of $A$ containing $m$ onto the set of maximal ideals of $\hat{A}$ and $q \mapsto j^{-1}(q)$ is the inverse bijection.

(iii) Let $n$ be a maximal ideal of $A$ containing $m$. The homomorphism $j': A_n \to \hat{A}_{\hat{n}}$ derived from $j$ is injection; if $A$, is identified by means of $j$ with a subring of $\hat{A}$, the $(nA_n)$*-adic topology on* $A_n$ is induced by the $A$*-adic topology on* $\hat{A}_{\hat{n}}$ and $A_n$ is dense in $A$, with the $A$*-adic topology*.

Let us show (i). As $m$ is a finitely generated ideal, $(m^n)^{\wedge} = (\hat{m})^n = m^n \hat{A}$ (§ 2, no. 12, Corollary 2 to Proposition 16) and the topology on $\hat{A}$ is the $m$-adic topology. As $\hat{A}/\hat{m}$ is isomorphic to $A/m$, it is a Noetherian ring and $\hat{m} = m \hat{A}$ is a finitely generated $A$-module and therefore $\hat{A}$ is Noetherian (§ 2, no. 10, Corollary 5 to Theorem 2); finally, as $\hat{A}$ is Hausdorff and complete with respect to the $m$-adic topology, $\hat{A}$ is a Zariski ring (no. 3, Example 1).

Assertion (ii) follows immediately from the fact that the canonical homomorphism $A/m + \hat{A}/\hat{m}$ derived from $j$ is bijective and the fact that every maximal ideal of $\hat{A}$ contains $\hat{m}$, since $\hat{A}$ is a Zariski ring and the Jacobson radical of $\hat{A}$ then contains $m$ (no. 3, Proposition 6).

Finally let us prove (iii). As $n = j^{-1}(A)$, $j(A - n) \subset \hat{A} - \hat{n}$ and $j$ certainly defines a homomorphism $j': A_n \to \hat{A}_{\hat{n}}$ (Chapter II, § 2, no. 1, Proposition 2). Let us show that $j'$ is injective; let $a \in A, s \in A - n$ be such that

then there exists $s' \in \hat{A} - \hat{n}$ such that $s'j(a) = 0$ (Chapter II, § 2, no. 1, Remark 3) and the annihilator of $j(a)$ in $\hat{A}$ is therefore not contained in $\hat{n}$. Now, if $b$ is the annihilator of $a$ in $A$, the annihilator of $j(a)$ in $\hat{A}$ is $b$ (Corollary 1 to Theorem 3); hence $b \notin n$, which shows that $a/s = 0$.

Moreover, there is a commutative diagram

where $h$ and $h'$ are derived from $j$ and $j'$ respectively and the horizontal arrows are the canonical isomorphisms of Chapter II, § 3, no. 3, Proposition 9. As $n^k$ is an open ideal of $A$ (since it contains $m^k$), $h$ is bijective and hence so is $h'$. This shows first that $(nA_n)^k = j'((\hat{n}\hat{A}_{\hat{n}})^k)$ and hence the topology on $A_n$ is induced by that on $A_n$; moreover, $\hat{A}_{\hat{n}} = A_n + (\hat{n}\hat{A}_{\hat{n}})^k$ for all $k > 0$ and hence $A_n$ is everywhere dense in $A_n$.