By Sergey Shpectorov

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**Example text**

6 we know that for some nonzero a ∈ Z we have that β = aα is an algebraic integer, that is, β ∈ ok . Since ||a|| = 1, we must now have that ||β|| = ||a|| · ||α|| = ||α|| < 1. Thus, β ∈ p, and so p is nonzero. We conclude this section with a statement classifying non-Archimedean valuations, which is given without a proof. We need some notation. For a nontrivial non-Archimedean valuation || · ||, let p be the corresponding prime ideal in ok and for a nonzero fractional ideal F let vp (F ) be the exponent of p in the unique factorization of F .

Since this is true for all positive integers n (and hence also for all integers), we conclude that || · || is not Archimedean, a contradiction. Next we show that the field k = Q has up to equivalence only one Archimedean valuation. 4 Every Archimedean valuation on k = Q is equivalent to the absolute value mapping. Proof: We first show that ||m|| = |m|c for a fixed positive c ∈ R and all m ∈ Z. 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.

Proof: Consider a large integer t and write ||α + β||t = ||(α + β)t || = t t i t−i || ti=0 ti αi β t−i || ≤ . Here we used multiplicai=0 || i || · ||α|| · ||β|| tivity, then the triangular inequality and then again multiplicativity. Since the valuation is non-Archimedean, || ti || ≤ 1. Also, clearly, ||α||, ||β|| ≤ max(||α||, ||β||). Hence every summand is at most max(||α||, ||β||)t . We conclude that ||α + β||t ≤ (t + 1) max(||α||, ||β||)t . Taking the tth root of both sides and letting t go to infinity yields the desired inequality.

### Algebraic Number Theory [Lecture notes] by Sergey Shpectorov

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