suppose Then there exists such that By considering the infinite place we see proving injectivity. To show surjectivity let The absolute value of this element is and therefore Hence and we have: Since we conclude is surjective. Theorem. The absolute value function induces the following isomorphisms of topological groups: Proof. The isomorphisms are given by: Relation between ideal class group and idele class group Theorem. Let be a number field with ring of integers group of fractional ideals and ideal class group We have the following isomorphisms where we have defined Proof. Let be a finite place of and let be a representative of the equivalence class Define Then is a prime ideal in The map is a bijection between finite places of and non-zero prime ideals of The inverse is given as follows: a prime ideal is mapped to the valuation given by The following map is well-defined: The map is obviously a surjective homomorphism and The first isomorphism follows from fundamental theorem on homomorphism. Now, we divide both sides by This is possible, because Please, note the abuse of notation: On the left hand side in line 1 of this chain of equations, stands for the map defined above. Later, we use the embedding of into In line 2, we use the definition of the map. Finally, we use that is a Dedekind domain and therefore each ideal can be written as a product of prime ideals. In other words, the map is a -equivariant group homomorphism. As a consequence, the map above induces a surjective homomorphism To prove the second isomorphism we have to show Consider Then because for all On the other hand, consider with which allows to write As a consequence, there exists a representative, such that: Consequently, and therefore We have proved the second isomorphism of the theorem. For the last isomorphism note that induces a surjective group homomorphism with Remark. Consider with the idele topology and equip with the discrete topology. Since is open for each is continuous. It stands, that is open, where so that Decomposition of the idele group and idele class group of K Theorem. Proof. For each place of so that for all belongs to the subgroup of generated by Therefore for each is in the subgroup of generated by Therefore the image of the homomorphism is a discrete subgroup of generated by Since this group is non-trivial, it