with sequences. This reduces the problem to whether is continuous on However, this is clear, because of the reverse triangle inequality. Definition. We define the set of -idele as: is a subgroup of Since it is a closed subset of Finally the -topology on equals the subset-topology of on Artin's Product Formula. for all Proof. We prove the formula for number fields, the case of global function fields can be proved similarly. Let be a number field and We have to show: For a finite place for which the corresponding prime ideal does not divide we have and therefore This is valid for almost all We have: In going from line 1 to line 2, we used the identity where is a place of and is a place of lying above Going from line 2 to line 3, we use a property of the norm. We note the norm is in so without loss of generality we can assume Then possesses a unique integer factorisation: where is for almost all By Ostrowski's theorem all absolute values on are equivalent to the real absolute value or a -adic absolute value. Therefore: Lemma. There exists a constant depending only on such that for every satisfying there exists such that for all Corollary. Let be a place of and let be given for all with the property for almost all Then there exists so that for all Proof. Let be the constant from the lemma. Let be a uniformizing element of Define the adele via with minimal, so that for all Then for almost all Define with so that This works, because for almost all By the Lemma there exists so that for all Theorem. is discrete and cocompact in Proof. Since is discrete in it is also discrete in To prove the compactness of let is the constant of the Lemma and suppose satisfying is given. Define: Clearly is compact. We claim the natural projection is surjective. Let be arbitrary, then: and therefore It follows that By the Lemma there exists such that for all and therefore proving the surjectivity of the natural projection. Since it is also continuous the compactness follows. Theorem. There is a canonical isomorphism Furthermore, is a set of representatives for and is a set of representatives for Proof. Consider the map This map is well-defined, since for all and therefore Obviously is a continuous group homomorphism. Now