We show first the equivalence of (a) and (b). Corollary 2 to Theorem 3, § 1, no. 6, shows immediately that (a) implies (b). Conversely, (b) implies (a), since, for every prime ideal $p$ of $A$, there exists a valuation ring of $K$ which dominates $A_p$ (Chapter VI, § 1, no. 2, Corollary to Theorem 2).

The remainder of the proof is carried out by proving the following implications:

If $A$ is a Dedekind domain and $b$ is a non-zero fractional ideal, then $bA_p = bA_p$ for every maximal ideal $p$ ($§ 1$, no. 4, Proposition 7) and hence $b = 6$ (Chapter II, $§ 3$, no. 3, Corollary 3 to Theorem 1); thus (a) implies (c).

We now show that (c) implies (d). If (c) holds, the mapping $a \mapsto \operatorname{div} a$ is a bijection of $I(A)$ onto $D(A)$ (cf. $§ 1$, no. 1); as it is a homomorphism ($§ 1$, no. 2) and $D(A)$ is a group, every element of $I(A)$ is invertible.

We show that (d) implies (e). If (d) holds, every integral ideal $\neq (0)$ of $A$ is finitely generated (Chapter 11, $§ 5$, no. 6, Theorem 4) and hence $A$ is Noetherian; as $I(A)$ is a group, $D(A)$ is a group and $A$ is therefore completely integrally closed ($§ 1$, no. 2, Theorem 1). Finally, if $p$ is a non-zero prime ideal of $A$ and $m$ is a maximal ideal of $A$ containing $p$, the ring $A_{m}$ is a principal ideal domain (Chapter II, $§ 5$, no. 6, Theorem 4); as $pA_m$ is prime and non-zero, necessarily $pA_m = mA_m$ (a principal ideal domain being a Dedekind domain) whence $p = m$ (Chapter 11, $§ 2$, no. 5, Proposition 11) and $p$ is maximal.

We now show that (e) implies (f). If $m$ is a maximal ideal of $A$ and (e) holds, $A_{m}$ is an integrally closed Noetherian domain and its maximal ideal $mA_m$ is, either (0), or the only non-zero prime ideal of $A_{m}$; hence $A_{m}$ is a field or a discrete valuation ring by Proposition 11 of $§ 1$, no. 7.

The fact that (f) implies (g) is obvious.

We show finally that (g) implies (a). As $A$ is the intersection of the $A_{m}$, where $m$ runs through the set of maximal ideals (Chapter II, $§ 3$, no. 3, Corollary 4 to Theorem 1), (g) implies that $A$ is integrally closed and Noetherian and hence that $A$ is a Krull domain ($§ 1$, no. 3, Corollary to Theorem 2). On the other hand, it can be shown that every non-zero prime ideal of $A$ is maximal as in the proof that (d) $\Rightarrow$ (e).

Let $A$ be a semi-local Dedekind domain, $K$ its field of fractions, $p_1, \ldots, p_n$ its maximal ideals and $v_1, \ldots, v$, the corresponding essential valuations; these are the only essential valuations of $A$. Let $a$ be a non-zero integral ideal of $A$. Since it is divisorial, there exists ($§ 1$, no. 4, Proposition 5) integers $q_1, \ldots, q_n$ such that $a$ is the set of $x \in K$ such that $v_i(x) \geq q_i$ for $1 \leq i \leq n$. Let $x_0$ be an element of $K$ such that $v_i(x_0) = q_i$ for $1 \leq i \leq n$ (Chapter VI, $§ 7$, no. 2, Corollary 1 to Theorem 1). Then $a$ is the set of $x \in K$ such that $v_i(xx_0^{-1}) \geq 0$ for $1 \leq i \leq n$. Thus $a = Ax_0$.

If $A$ is a Dedekind domain, it has been seen, in the proof of Theorem 1, that the group D(A) of divisors of A is identified with the group I(A) of fractional ideals $a \neq (0)$ (as A is Noetherian, every non-zero fractional ideal is finitely generated). The divisor class group C(A) of A ($§ 1$, no. 2) is then identified with the group of classes of ideals $\neq 0$ of A (defined in Chapter II, $§ 5$, no. 7).

Replacing if need be the $n_i$ by greater integers, they may be assumed all to be positive. We examine first the case where the $x_i$ are in $A$; it obviously amounts to finding an $x \in A$ satisfying the congruences

and the existence of $x$ then follows from Chapter II, § 1, no. 2, Proposition 5.

We pass now to the general case. We may write $x_i = s^{-1} y_i$ where $s, y_i$ are in $A$; writing $x = s^{-1} y$, it amounts to finding a $y \in A$ such that, on the one hand, $v_{p_i}(y - y_i) \geq n_i + v_{p_i}(s)$ and, on the other, $v_p(y) \geq v_p(s)$ for all $p \in P$ distinct from the $p_i$; as $v_p(s) = 0$ except for a finite number of indices $p$, it is thus reduced to the above case; whence the proposition.

Proposition 2 may be interpreted as a density theorem. To be precise, for all $p \in P$, let $\hat{K}_p$ (resp. $\hat{A}_p$) be the completion of $K$ (resp. $A$) with respect to the discrete valuation $v_p$ and consider the product $\prod_{p \in P} \hat{K}_p$; an element $x = (x_p)$ of this product is called a restricted adèle of $A$ if $x_p \in \hat{A}_p$ for all $p \in P$ with the exception of a finite number of them. Clearly the set $A$ of restricted adtles is a subring of $\prod_{p \in P} \hat{K}_p$, which contains the product ring $A, = \prod_{p \in P} \hat{A}_p$. Consider on $A$, the product topology, with respect to which $A,$ is complete; there is on $A$ a unique topology $\mathscr{T}$ which is compatible with its additive group structure and for which the neighbourhoods of 0 in $A$, form a fundamental system $\mathscr{G}$ of neighbourhoods of 0. The topology $\mathscr{T}$ is compatible with the ring structure on $A$; for clearly axiom (AVII) of General Topology, Chapter 111, § 6, no. 3, holds, the topology induced by $\mathscr{T}$ on $A,$ being compatible with the ring structure on $A,$. On the other hand, for all $x \in A$ there exists a finite subset $J$ of $P$ such that, if we write $J' = P - J, \ K_J \leq \prod_{p \in J} \hat{K}_p, \ A_{J'} = \prod_{p \in J'} \hat{A}_p,$ then x \in K_J \times A_{J'} and, as $\hat{A}_p$ is open in $\hat{K}_p$ for all p, $\mathscr{G}$ is a fundamental system of neighbourhoods of 0 for the product topology on $K_J \times A_{J'}$; since the latter is compatible with the ring structure on this product, axiom (AV_1) of General Topology, Chapter III, loc. cit. is also seen to hold, which proves our assertion. Clearly A, is an open subring of A and hence A is also a complete ring (General Topology, Chapter III, § 3, no. 3, Proposition 4).

For all $x \in K$, let $\Delta(x)$ be the element $(x_p) \in \prod_{p \in P} \hat{K}_p$ such that $x_p = x$ for all $p \in P$; as $x_p \in \hat{A}_p$ except for a finite number of values of p, $\Delta(x) \in A$; hence we have thus defined a homomorphism $A : K \to A$ which is injective if $P \neq \varnothing$ (that is if A is not a field); the elements of $\Delta(K)$ are called principal restricted adèles and clearly $\Delta(A) \subset A$.

It is immediate that the topology considered on A (or K) is Hausdorff. Taking account of no. 3, the assertion concerning A, follows from Chapter 111, § 2, no. 13, Proposition 17. This shows therefore that $\Delta(A)$ is dense in A,; to see similarly that $\Delta(K)$ is dense in A, note that for all $x = (x_i) \in A$ there is only a finite number of $p \in P$ such that $v_p(x_p) < 0$; by § 1, no. 5, Proposition 9 there is therefore an $s \in K$ such that $sx_p \in \hat{A}_p$ for all $p \in P$, in other words $\Delta(s)x \in A$, and, as multiplication by $\Delta(s)$ is a homomorphism from A to itself, it suffices to apply the fact that $\Delta(A)$ is dense in A , to deduce that $\Delta(K)$ is dense in A.

We could of course also prove that $\Delta(K)$ is dense in A by using Proposition 2.

Consider now the multiplicative group $\mathbf{SL}(n, A)$ consisting of the matrices $U \in \mathbf{M}_n(A)$ such that $\det(U) = 1$; if $\mathbf{M}_n(A) = A^{n^2}$ is given the product topology, it induces on $\mathbf{SL}(n, A)$ a topology compatible with the group structure on $\mathbf{SL}(n, A)$. It suffices to verify that the mapping $U \mapsto U^{-1}$ is continuous on $\mathbf{SL}(n, A)$; but as U is unimodular, it is known (Algebra, Chapter III, § 6, no. 5, formula (17)) that the elements of $U^{-1}$ are minors of U and hence polynomials in the elements of U, which proves our assertion. If K is identified with a subring of A by means of A, the group $\mathbf{SL}(n, K)$ is a subgroup of $\mathbf{SL}(n, A)$.

Let G be the closure of $\mathbf{SL}(n, K)$ in $\mathbf{SL}(n, A)$; as K is dense in A (Proposition 3), G contains all the matrices of the form $I + a . E_{ij}$ for $i \neq j$ and $a \in A$. For all $p \in P$ and all $\lambda \in \hat{K}_p$, let $\lambda(p)$ be the restricted adèle $x = (x_q)_{q \in P}$ such that $x_p = \lambda$ and $x_q = 0$ for $q \neq p$; the above shows that G contains the matrices $I + \lambda(p)E_{ij}$ for $i \neq j$. But we know that the matrices of the form $I + \lambda E_{ij}$ for $\lambda \in \hat{K}_p$ generate the group $\mathbf{SL}(n, \hat{K}_p)$ (Algebra, Chapter 111). For every matrix $U \in \mathbf{SL}(n, A)$ let $U_p$ denote the canonical image of $U$ in $\mathbf{SL}(n, \hat{\mathbf{K}}_p)$; it is therefore seen that, for all $p \in P$, $G$ contains the matrices $U \in \mathbf{SL}(n, A)$ such that $U_q = I$ for all $q \# p$. Since $G$ is a group, it also contains all the matrices $U \in \mathbf{SL}(n, A)$ such that $U_p = I$ except for a finite number of $p \in P$; now, the definition of the topology on $A$ shows immediately that the set of these matrices is dense in $\mathbf{SL}(n, A)$.

As $M$ is a torsion module, every prime ideal associated with $M$ is $\neq (0)$ and therefore maximal. The lemma then follows from Chapter IV, § 2, no. 5, Proposition 7.

$\operatorname{long}_A(T_i) \leq \operatorname{long}_A(T)$ for all $i$. The lemma is obvious if no integer exceeds the $\operatorname{long}_A(T_i)$, both sides then being infinite. Otherwise, let $i_0$ be an index for which $\operatorname{long}_A(T_{i_0})$ takes its greatest value; then $T_{i_0} = T$ since the family $(T_i)$ is directed; whence our assertion in this case.

M/aM onto aA/a^2M; similarly for L; whence, by induction on n, the formulae:

Taking account of (7) we deduce:

for all $n > 0$; as L is isomorphic to A', $\text{long}_A(L/aL) = r \text{long}_A(A/Aa)$; whence (6) by letting $n$ tend to infinity in (9).

We now pass to the general case. Let $(M_i)$ be the family of finitely generated submodules of M. The module $T = M/aM$ is the union of the submodules $T_i = (M_i + aM)/aM = M_i/(M_i \cap aM)$. Now, T, is isomorphic to a quotient of $M_i/aM_i$ and hence

by what we have just proved. Whence

by Lemma 2.

Let b be a non-zero ideal of B. We shall show that B/b is an A-module of finite length (hence, a fortiori, a B-module of finite length) and that b is a finitely generated B-module.

A non-zero element y of b satisfies an equation of the form:

This equation shows that $a_1 \in By \subset b$. Applying Lemma 3 to $M = B$, it is seen that $B/a_0B$ is an A-module of finite length; so is B/b which is a quotient module of it. Further the B-module b contains, as a submodule, $a_0B$ which is finitely generated; as $b/a_0B$ is of finite length (as a submodule of $B/a_0B$) and hence finitely generated, b is certainly a finitely generated B-module.

The above shows first that B is Noetherian. On the other hand, if $p$ is a non-zero prime ideal of B, the ring $B/p$ is an integral domain and of finite length and hence is a field (Algebra, Chapter VIII, § 6, no. 4, Proposition 9), so that $p$ is maximal.

Suppose first that $p = (0)$; then the only prime ideal $q$ of $B$ such that $q \cap A = (0)$ is $(0)$; otherwise, writing $S = A - \{0\}, S^{-1}q$ would be a non-zero prime ideal of $S^{-1}B$ (Chapter 11, § 2, no. 5, Proposition 11) and $S^{-1}B$ is just the field of fractions of $B$, for it is a subring of $L$ containing $K$ (\emph{Algebra}, Chapter V, § 3, no. 2, Proposition 3); whence an absurd conclusion. If now $p \neq (0)$, it follows from Proposition 5 that $B/pB$ is a finite-dimensional vector space over the field $A/p$, hence an \emph{Artinian} ring and therefore has only a finite number of prime ideals (Chapter IV, § 2, no. 5, Proposition 9), which proves that there is only a finite number of prime ideals of $B$ containing $p$.

This integral closure is an integrally closed Noetherian domain all of whose non-zero prime ideals are maximal; it suffices therefore to apply Theorem 1 of no. 2.

In particular:

the decomposition of the ideal $Bp$ as a product of prime ideals. Then: (a) the prime ideals of $B$ lying above $p$ are the $p_i$ such that $e(i) > 0$; (b) the valuations $v_i$ on $L$ corresponding to these ideals $p_i$ are, up to equivalence, the valuations on $L$ extending $v$; (c) $[B/p_i : A/p] = f(v_i/v);$ (d) $e_i = e(v_i/v)$ (cf. Chapter VI, § 8, no. 1, Definitions 1 and 2).

(a) To say that a prime ideal $q$ of $B$ lies above $p$ amounts to saying that $q \supseteq p$, hence that $q \supseteq Bp$ and that $q$ contains one of the $p_i$ such that $e(i) > 0$ (Chapter 11, § 1, no. 1, Proposition 1). (b) This follows, taking account of (a), from § 1, no. 8, Corollary to Proposition 12. (c) The residue field of $v$ is identified with $A/p$ and that of $v_i$ with $B/p_i$ (§ 1, no. 4, Corollary 1 to Proposition 6).

(d) Let $a$ (resp. $a_i$) be a uniformizer for $v$ (resp. $v_i$). Then