By Sergey Shpectorov

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**Example text**

By the multiplicativity where 0 ≤ ai < m for all i. Note that k ≤ t log m of the valuation and by the triangle inequality (property 3 in the definition) we have that ||n||t = ||nt || ≤ ki=1 ||ai mi || = ki=1 ||ai || · ||m||i ≤ (k + 1)A log n )A. Taking the tth roots (since ||m|| ≤ 1). So we get that ||n||t ≤ (1 + t log m of both sides and letting t → ∞, we see that ||n|| ≤ 1. Since this is true for all positive integers n (and hence also for all integers), we conclude that || · || is not Archimedean, a contradiction.

The above lemma means that, say, an ideal I of ok of norm 2 would have to contain pZ for a prime p ∈ Z. Consequently, p ∈ I and so (p) ⊆ I. 4, we compute Normk/Q ((p)) = |normk/Q (p)| = p2 . By multiplicativity, we must have that 2 divides p2 . Hence p = 2. So I contains (2). Similarly, if I is an ideal of I of norm 3 then it contains (3). Thus we simply need to determine all ideals of ok above (that is, properly containing) (2) and all ideals above (3). We start with (2). By the correspondence theorem for rings, we have that all ideals above (2) bijectively correspond to the nonzero ideals in ok /(2).

Let us now turn to the√case where m is congruent to 1 modulo 4. In this case√ ok is larger than Z[ m], namely, ok consists of all complex numbers a+b m , where the integers a and b are either both even or both odd. The 2 2 2 norm is now a −mb so the units correspond to the solutions of a2 − mb2 = 4, 4 where a and b are as above. If m < −3 then we must have b = 0 and then a = ±2 , giving only the obvious units ±1. If m = −3 then in addition to the solutions (a, b) = (2, 0) and (−2, 0) we also find the solutions (1, 1), (1, −1), 31 √ (−1, 1), and (−1, −1), giving the units ±1±2 −3 .

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

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