Algebraic Number Theory [Lecture notes] by Sergey Shpectorov

By Sergey Shpectorov

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If B = ∅ then also B −1 = {α ∈ k | ||α||i > 1} is empty. This means that both valuations are trivial (hence equivalent). So we can assume that B = ∅. Select α0 ∈ B and define c ∈ R+ via: ||α0 ||2 = ||α0 || 1c . Note that c > 0, since ||α0 ||2 < 1. Now consider an arbitrary α ∈ B. Let λi , i = 1, 2, be defined by: ||α||i = ||α0 ||λi i . Clearly, both λ1 and λ2 are positive. Suppose λ1 < m ∈ Q, where n m αm ||α || ||α0 ||m 0 1 both m and n are positive integers. Then || α0n ||1 = ||α||n1 = nλ1 = 1 1 ||α0 ||m−nλ .

We will refer to the property 2 above as to multiplicativity of the valuation. Note that this property means that the valuation induces a homomorphism from the multiplicative group k # into R+ = {a ∈ R | r > 0}. In particular, ||1k || = 1 and if α = 0k then ||α−1 || = ||α||−1 . In general, ||αn || = ||α||n for all n ∈ Z. Each valuation defines a metric on k, via d(α, β) = ||α − β||, and the metric defines a topology. For example, the trivial valuation defines the discrete metric: d(α, β) = 1 if and only if α = β and the discrete topology on k, whereby every one-point subset of k is open.

This ideal in non-principal; √ we skip the details as they are quite similar to the case of k = Q( −5). The situation is quite different for p = 3. Namely, ok /(3) ∼ = Z3 [x]/(x2 + 1) 2 (since 13 is the same as 1 modulo 3). Since x + 1 is irreducible in Z3 [x], the factor ring Z3 [x]/(x2 + 1) is a field (the unique field of size 32 = 9). In particular, it has no proper nonzero ideals, which means that there no ideals of norm 3 in ok . Hence (3) is prime. For p = 5 we have a similar situation: ok /(5) ∼ = Z5 [x]/(x2 + 3), which is a field.

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