Suppose that $m E_{n-1} = E_n$ for $n > n_0 \geq 0$. For $i \leq n_0$, let $(e_i)_{i \leq n_0}$ be a finite system of generators of the $A$-module $E_i$. As the $A$-module $E_n \otimes_A A X^n$ is generated by the elements $e_{n,j} \otimes X^n$ for $0 \leq n \leq n_0$ and is equal to

for $n > n_0$, the $A'$-module $E'$ is generated by the elements $e_{n,j} \otimes X^n$ for $0 \leq n \leq n_0$ and $1 \leq j \leq r_n$; then it is certainly finitely generated.

Conversely, if $E'$ is a finitely generated $A'$-module, it is generated by a finite family of elements of the form $e_k \otimes X^{n(k)}$, where $e_k \in E_{n(k)}$. Let $n_0$ be the greatest of the integers $n(k)$. Then for $n \geq n_0$ and $f \in E_n$,

where $t_k \in A'$; replacing if need be $t$, by its homogeneous component of degree $n - n(k)$, we may assume that $t_k = a_k X^{n-n(k)}$, where $a_k \in m^{n-n(k)}$.

As the unique element $X^n$ forms a basis of the A-module AX", the equation $f \otimes X^n = \left( \sum_k a_k e_k \right) \otimes X^n$ implies $f = \sum_k a_k e_k$. Then $E_n \subset m^{n-n_0} E_{n_0}$; since the opposite inclusion is obvious,

whence $E_n = m E_{n-1}$ for $n > n_0$.

$A'$ is an A-algebra generated by $mX$; as A is Noetherian, $mX$ is a finitely generated A-module and the conclusion then follows from § 2, no. 10, Corollary 3 to Theorem 2.

As F is isomorphic to a submodule of E, it is finitely generated since A is Noetherian and E is finitely generated. Let $(E_n)$, $(F_n)$ be the respective filtrations on E and F, which consist of finitely generated submodules; preserving the notation of Lemma 1, we set $E' = \sum_{n \geq 0} E_n \otimes_A A X^n, F' = \sum F_n \otimes_A A X^n$; as by hypothesis $F_n$ is isomorphic to a submodule of $E_n$, we see that F' is isomorphic to a submodule of $E'$. By Theorem 1, $E'$ is a finitely generated $A'$-module and hence so is F' since A' is Noetherian (Lemma 1). Hence the conclusion by virtue of Theorem 1.

In other words, there exists an integer $n_0$ such that

for all $n \geq n_0$.

There exists n such that $a^{n+1} \cap b = a(a^n \cap b) \subset ab$ by Corollary 1 applied to $E = A, F = b$.

Corollary 1 applied to $E = A, F = A x$ shows that there exists $k$ such that, for all $n \geq k$, $m^n \cap Ax = m^{n-k}(m^k \cap Ax)$. Then, if $xy \in m^n$,

and, as $x$ is not a divisor of $0$, we deduce that $y \in m^{n-k}$.

In the notation of transporters (Chapter I, § 2, no. 10), the conclusion of Corollary 3 reads

(2)

There exists by hypothesis a surjective $A$-homomorphism $v : A^n \to E$; the mapping $u \mapsto u \circ v$ of $\mathrm{Hom}_A(E, \phi_*(F))$ to $\mathrm{Hom}_A(A^n, \phi_*(F))$ is therefore injective and, as $B$ is Noetherian, it is sufficient to prove that $\mathrm{Hom}_A(A^n, \phi_*(F))$ is a finitely generated $B$-module; which is immediate since it is isomorphic to $F^n$.

the family $(\mathrm{Hom}_A(E, F_n))_{n \in \mathbf{Z}}$ is then a filtration on $\mathrm{Hom}_A(E, F)$ compatible with its module structure over the ring $A$ filtered by the $m$-adic filtration. Since $E$ is finitely generated, there exists an integer $r > 0$ and a surjective $A$-homomorphism $u : A^r \to E$ which defines an injective $A$-homomorphism

clearly $v$ is compatible with the filtrations $(\mathrm{Hom}_A(E, F_n))$ and $(\mathrm{Hom}_A(A^r, F_n))$. As $\mathrm{Hom}_A(E, F)$ and $\mathrm{Hom}_A(A^r, F)$ are finitely generated (Lemma 2), it is sufficient by virtue of Proposition 1 to show that the filtration $(\mathrm{Hom}_A(A^r, F))$ is m-good; but this is immediate by virtue of the existence of the canonical isomorphism $\mathrm{Hom}_A(A^r, F_r) \to F_n^r$ and the fact that the relation $mF_n = F_n +$, implies $m(F_n^r) = (mF_n)^r = F_{n+1}^r$ (Algebra, Chapter II, § 3, no. 7, Remark).

It follows immediately from the definitions that (a) implies (c). The fact that (b) implies (c) is a consequence of § 1, no. 3, Lemma 1; conversely, if (c) holds, clearly $\mathrm{gr}(E)$ is generated as a $\mathrm{gr}(A)$-module by the sum of the $\mathrm{gr}_p(E)$ for $p \leq n_0$ and hence by hypothesis admits a finite system of generators. It remains to prove that (c) implies (a); as the $\mathrm{gr}_n(E)$ are finitely generated and $E_n = E$, clearly first, by induction on $n$, $E/E_n$ is a finitely generated $A$-module for all $n$; it will therefore be sufficient to prove that, for $n > n_0$, $E_n$ is a finitely generated $A$-module and that $mE_n = E_{n+1}$. Now, consider the $A$-module $E_{n+1}$ with the exhaustive and separated filtration $(E_{n+k})(k \geq 1)$; $mE_n \subset E_{n+1}$; hypothesis (c) implies that the image of $mE_n$ in $\mathrm{gr}_{n+1}(E) = E_{n+1}/E_{n+2}$ is equal to $\mathrm{gr}_{n+1}(E)$ and generates the graded $\mathrm{gr}(A)$-module $\mathrm{gr}(E_{n+1})$. As $\mathrm{gr}_{n+1}(E)$ is by hypothesis a finitely generated $A$-module, it follows from § 2, no. 9, Proposition 12 that $mE_n = E_{n+1}$ and that $E_{n+1}$ is a finitely generated $A$-module.

Let $(E_n)$ be an m-good filtration on $E$. As this filtration is exhaustive, every element of $E$ belongs to one of the $E_n$ and, as $E$ is finitely generated and the $E_n$ are $A$-modules, there exists an integer $n_1$ such that $E_{n_1} = E$. On the other hand let $n_0$ be such that $mE_n = E_{n+1}$ for $n \geq n_0$; then, for $n > n_0 - n_1$, $m^n E \subset E_{n+n_1} = m^{n+n_1-n_0} E_{n_0} \subset m^{n+n_1-n_0} E$, which proves the proposition.

Theorem 2 (Krull). *Let $A$ be a commutative Noetherian ring, $m$ an ideal of $A$, $E$ a finitely generated $A$-module and $F$ a submodule of $E$. Then the $m$-adic topology on $F$ is induced by the $m$-adic topology on $E$.

It follows from no. 1, Proposition 1 that the filtration induced on $F$ by the $m$-adic filtration on $E$ is m-good and the conclusion then follows from Proposition 4.

As $u(m^nE) = m^n u(E)$, $u$ is a strict morphism of $E$ onto $u(E)$ for the $m$-adic topologies on these two modules and the $m$-adic topology on $u(E)$ is induced by the $m$-adic topology on $F$ by Theorem 2.

If $x = mx$ where $m \in m$, $x = m^n x \in m^nE$ for every integer $n \geq 0$ and hence $x \in F = \bigcap_{n=0}^m m^nE$. Conversely, if $x \in F$, $Ax$ is contained in the intersection of the neighbourhoods of 0 in $E$; it then follows from Theorem 2 that the $m$-adic topology on $Ax$, which is induced by that on $E$, is the coarsest topology; as $mx$ is by definition a neighbourhood of 0 with this topology, $mx = Ax$ and hence there exists $m \in m$ such that $x = mx$.

We may also say that $mF = F$ since

as $A$ is Noetherian, $F$ is a finitely generated $A$-module; it is then sufficient to apply Chapter II, § 2, no. 2, Corollary 3 to Proposition 4.

It is sufficient to apply Proposition 5 to $E = A$.

Let us show that (a) implies (b). Suppose that $m$ is contained in the Jacobson radical of $A$ and let $E$ be a finitely generated $A$-module. If $x \in E$ and $m \in m$ are such that $(1 - m)x = 0$, then $x = 0$, for $1 - m$ is invertible in $A$. Then (no. 2, Proposition 5) $E$ is Hausdorff with the $m$-adic topology.

Let us prove that (b) implies (c). Suppose (b) holds. Let $E$ be a finitely generated $A$-module and $F$ a submodule of $E$. Then $E/F$ is Hausdorff with the $m$-adic topology, which is the quotient topology of the $m$-adic topology on $E$; then $F$ is closed in $E$.

Clearly (c) implies (d). Let us show finally that (d) implies (a). It follows from (d) that, for every maximal ideal $a$ of $A$, the $A$-module $A/a$ is Hausdorff with the $m$-adic topology. This implies $m(A/a) \neq A/a$, unless the $m$-adic topology on $A/a$ were the coarsest topology and $A/a$ were reduced to 0, which is absurd since $A/a$ is ajield. The canonical image of $m$ in $A/a$ is therefore an ideal of $A/a$ distinct from $A/a$ and hence reduced to 0; then $m \subset a$, which proves that $m$ is contained in the Jacobson radical of $A$.

(i) For the $m'$-adic topology on $A'$ to be Hausdorff, it is necessary and sufficient that the elements of $1 + h(m)$ be not divisors of $0$ in $A'$.

(ii) If $A$ with the $m$-adic topology is a Zariski ring, then $A'$ with the $m'$-adic topology is a Zariski ring.

(iii) If $h$ is injective (thus identifying $A$ with a subring of $A'$), the $m'$-adic topology on $A'$ induces on $A$ the $m$-adic topology.

Recall that the $m'$-adic filtration on $A'$ coincides with the $m$-adic filtration on the A-module $A'$ (§ 2, no. 1, Example 3). Assertion (i) is thus a special case of Proposition 5 of no. 2 and assertion (iii) a special case of Theorem 2 of no. 2. Finally let us show (ii). Suppose that $A$ is a Zariski ring with the $m$-adic topology and let $E'$ be a finitely generated $A'$-module; it is also a finitely generated $A$-module and the $m$-adic and $m'$-adic filtrations on $E'$ coincide; then $E'$ is Hausdorff with the $m'$-adic topology. Finally the $A$-module $A'$ is Noetherian and hence the ring $A'$ is Noetherian, which completes the proof that $A'$ is a Zariski ring.

(i) If $E \xrightarrow{u} F \xrightarrow{v} G$ is an exact sequence of A-linear mappings, the sequence $\hat{E} \xrightarrow{\hat{u}} \hat{F} \xrightarrow{\hat{v}} \hat{G}$ obtained by passing to the Hausdorff completions (with respect to the m-adic topologies) is exact.

(ii) The canonical A-linear mapping $\alpha_E : \hat{A} \otimes_A E \to \hat{E}$ is bijective.

(iii) The A-module $\hat{A}$ is flat.

We have seen that $u$ and $v$ are strict morphisms of topological groups (no. 2, Corollary to Theorem 2). Assertion (i) then follows from § 2, no. 12, Lemma 2. Assertion (ii) is obvious when $E = A$ and the case where E is a free finitely generated A-module can be immediately reduced to that. In the general case, E admits a finite presentation

(Chapter I, § 2, no. 8, Lemma 8). We derive a commutative diagram

The first row is exact (Chapter I, § 2, no. 1, Lemma 1) and so is the second by (i). We already know that $\alpha_E$ is surjective (§ 2, no. 12, Proposition 16); on the other hand, as $\alpha_L$ and $\alpha_{L'}$ are bijective and $1 \otimes v$ is surjective, $\alpha_E$ is injective by virtue of Chapter I, § 1, no. 4, Corollary 2 to Proposition 2; this shows (ii).

Then it follows from (i) and (ii) that, if $a$ is an ideal of A (necessarily finitely generated), the canonical mapping $\hat{A} \otimes_A a \to \hat{A}$ is injective, being the composition of $\hat{a} \to \hat{A}$ and $\alpha_E$, which proves that $\hat{A}$ is a flat $A$-module (Chapter I, § 2, no. 3, Proposition 1).

Under the conditions of Theorem 3 $\hat{A} \otimes_A E$ is often identified with $E$ by means of the canonical mapping $\alpha_E$. If $u : E \to F$ is a homomorphism of finitely generated $A$-modules, $\hat{u} : E \to \hat{F}$ is then identified with $1 @ u$ by virtue of the commutativity of diagram (4).

Moreover, if $a$ and $b$ are two ideals of $A$ and $c = ab, \hat{c} = \hat{a}\hat{b}$.

By Theorem 3, $\hat{E}, \hat{F}, \hat{G}$ are canonically identified with $\hat{A} \otimes_A E, \hat{A} \otimes_A F, \hat{A} \otimes_A G$, which establishes the first two formulae. The third and fourth follow respectively from Chapter I, § 2, no. 6, Proposition 6 and no. 10, Proposition 12. Finally, as $\hat{a} = \hat{A}i(a), \hat{b} = \hat{A}i(b), \hat{c} = \hat{A}i(c)$,

As $\hat{A}$ is a flat $A$-module, the corollary is a special case of Chapter I, § 2, no. 4, Proposition 3 (i).

It is the completion of the polynomial ring

with respect to the $m$-adic topology, where $m$ is the set of polynomials with no constant term ($§ 2,$ no. 12, Example 1); as $B$ is Noetherian ($§ 2,$ no. 10, Corollary 2 to Theorem 2), $A[[X_1, \ldots, X_n]]$ is a flat $B$-module by Theorem 3 and, as $B$ is free $A$-module, $A[[X_1, \ldots, X_n]]$ is a flat $A$-module (Chapter I, § 2, no. 7, Corollary 3 to Proposition 8).

(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 $A_n$.