Proposition 3

The ring A, (resp. A ) is identified with the completion of A (resp. K) with respect to the ring topology for which a fundamental system of neighbourhoods & 0 consists & all the integral ideals $\neq (0)$ of 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 $\mathbf{SL}(n, A)$.