By Sergey Shpectorov

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Additional info for Algebraic Number Theory [Lecture notes]

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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. ) We say that two valuations, || · ||1 and || · ||2 , are equivalent if and only if ||α||2 = ||α||c1 for a fixed c ∈ R+ and all α ∈ K. 1 Suppose || · || is a valuation on k. Then the set {α ∈ k | ||α|| < 1} coincides with the set of all α ∈ k for which the sequence αn converges (with respect to the topology induced by || · ||) to 0k .

3. Clearly, we only need to consider integers m with |m| > 1. Without loss of generality we can restrict ourselves to just the positive integers, so let us take m, n ∈ Z with m, m > 1. 3, pick a positive power t and write nt = ki=0 ai mi , for a suitable ai with 0 ≤ ai < m log n . Using multiplicativity and the triangle inequality we get: and k ≤ t log m k t ||ai || · ||m||i ≤ (1 + t ||n|| ≤ i=0 log n log n log n )A · ||m||k ≤ (1 + t )A · ||m||t log m , log m log m where A is the maximum of ||a|| for 0 ≤ a < m.

3 that dk = 2(−10) = −20. 1 with C = √12 we get that every ideal class √ √ contains an integral ideal of norm at most √12 20 = 10 < 4. Hence we need to find all ideals of norm 1, 2, and 3. Clearly, there is only one ideal of norm 1—ok itself. 5) every ideal whose norm is a prime number is prime. So we are looking at prime ideals only. We will need the following fact about prime ideals. 3 If I is a nonzero prime ideal of ok then I ∩ Z is a nonzero prime ideal of Z. Proof: Let J = I ∩ Z. If a, b ∈ Z \ J then a, b ∈ ok \ I and hence, since I is prime, we have that ab ∈ I.

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Algebraic Number Theory [Lecture notes] by Sergey Shpectorov


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