maps on the adele ring, and they fulfil the usual equations: Furthermore, for and are identical to the trace and norm of the field extension For a tower of fields the result is: Moreover, it can be proven that: Properties of the adele ring Theorem. For every set of places is a locally compact topological ring. Remark. The result above also holds for the adele ring of vector-spaces and algebras over Theorem. is discrete and cocompact in In particular, is closed in Proof. Prove the case To show is discrete it is sufficient to show the existence of a neighbourhood of which contains no other rational number. The general case follows via translation. Define is an open neighbourhood of It is claimed that Let then and for all and therefore Additionally, and therefore Next, to show compactness, define: Each element in has a representative in that is for each there exists such that Let be arbitrary and be a prime for which Then there exists with and Replace with and let be another prime. Then: Next, it can be claimed that: The reverse implication is trivially true. The implication is true, because the two terms of the strong triangle inequality are equal if the absolute values of both integers are different. As a consequence, the (finite) set of primes for which the components of are not in is reduced by 1. With iteration, we deduce there exists such that Now select such that Then The continuous projection is surjective, therefore as the continuous image of a compact set, is compact. Corollary. Let be a finite-dimensional vector-space over Then is discrete and cocompact in Theorem. The following are assumed: is a divisible group. is dense. Proof. The first two equations can be proved in an elementary way. By definition is divisible if for any and the equation has a solution It is sufficient to show is divisible but this is true since is a field with positive characteristic in each coordinate. For the last statement note that because the finite number of denominators in the coordinates of the elements of can be reached through an element As a consequence, it is sufficient to show is dense, that is each open subset contains an element of Without loss of generality, we can assume because is a neighbourhood system of in By Chinese Remainder Theorem there exists such that Since powers of distinct