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Unformatted text preview: Math 676. The lattice of Sintegers Let K be a number field, and let S be a finite set of places of K containing the set S ∞ of archimedean places. Recall that we define the ring O K,S of Sintegers in K to be the set of a ∈ K such that a is vintegral for all (necessarily nonarchimedean!) v 6∈ S ; that is,  a  v ≤ 1 for all v 6∈ S . For a ∈ O K,S , we have a ∈ O × K,S if and only if a 6 = 0 and a, 1 /a ∈ K × each lie in O K,S . That is, a ∈ O × K,S if and only if a ∈ K × and  a  v ,  1 /a  v ≤ 1 for all v 6∈ S . This final condition says  a  v = 1 for all v 6∈ S . Hence, O × K,S = { x ∈ K ×  x  v = 1 for all v 6∈ S } . 1. Preliminaries We first wish to show that O K,S can be concretely constructed from O K and knowledge of the class number. For each nonarchimedean place of K , we let p v denote the corresponding prime ideal of O K . Since the class group is killed by the class number, for all nonarchimedean v the ideal p h ( K ) v in O K is principal. Hence, the finite product Q v ∈ S S ∞ p h ( K ) v has the form a S O K , so 1 /a S ∈ K × is nonintegral at precisely those nonarchimedean v that lie in S (if S = S ∞ then this product is empty and we may interpret the product over the empty set S S ∞ to be the unit ideal O K ; a S is an element of O × K in this case). Having constructed one such element, we now show that any such element allows us to construct O K,S as a localization of O K : Lemma 1.1. For a ∈ O K { } , we have O K,S = O K [1 /a ] if and only if the finite set of nonarchimedean v for which  1 /a  v > 1 is exactly the set S S ∞ . ( Equivalently, the condition is that the prime factors of a O K are exactly the primes p v for v ∈ S S ∞ ) . Proof. If O K,S = O K [1 /a ] then 1 /a is vintegral for all v 6∈ S , so  1 /a  v ≤ 1 for all v 6∈ S . We wish to show that  1 /a  v > 1 for the other nonarchimedean places, namely those v ∈ S S ∞ . Suppose otherwise, so  1 /a  v ≤ 1 for some v ∈ S S ∞ . That is, assume 1 /a is vintegral for some nonarchimedean v ∈ S . Since all elements of O K are also vintegral, it follows that all elements of O K,S = O K [1 /a ] are vintegral. However, this is not true: by finiteness of the class group we have p h ( K ) v = a O K for some a ∈ O K { } , and clearly 1 /a ∈ O K,S (since a is even a local unit at all places not in S ) yet 1 /a is not vintegral for the place v ∈ S (as  1 /a  v > 1 due to the prime factorization of a O K )....
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 Fall '11
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 Number Theory, Topology, Integers, Topological space, Topological group, OK,S, v∈S Kv

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