By Robin Chapman

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An ideal I of OK is maximal if I is nontrivial but the only ideals J of OK with I ⊆ J are J = I and J = OK . 4 Let K be a number field. An ideal I of OK is prime if and only if it is maximal. Proof First suppose that I is maximal. Let β, γ ∈ OK with βγ ∈ I and β∈ / I. To show that I is prime it suffices to show that γ ∈ I. Let J = I + β . Then J is an ideal and I ⊆ J, but I = J since β ∈ J. By maximality of I, J = OK . Hence 1 ∈ J so 1 = η + δβ where η ∈ I and δ ∈ OK . Then 1 ≡ δβ (mod I). Consequently, γ = 1γ ≡ δβγ ≡ 0 (mod I), as βγ ∈ I.

10 only a finite number of different ideals can occur. Hence there exist j < k with βj = βk . Thus βk = ξβj where ξ ∈ U (OK ), and ξ > 1 as βk > βj . In fact the structure of the unit group of OK is easy to determine. 3 Let K be a real quadratic field. There exists η ∈ OK such that η > 1 and such that every unit in OK has the form ±η j where j ∈ Z. 1 there exists ξ ∈ U (OK ) with ξ > 1. I claim that there are only finitely many an ε, √ units ε in OK with 1 < ε ≤ ξ. For1 such √ 1 1/ε ∈ OK . Let ε = 2 (a + b m) where a, b, m ∈ Z, then 1/ε = ± 2 (a − b m).

Similarly I2 is the m-th power of an ideal. Example As an application of these ideas, we investigate the solutions of y 3 = x2 + 6 45 (∗) √ √ with x, y ∈ Z. let us work in K = Q( −6) so that OK = Z[ −6]. From (∗) we get √ √ y 3 = (x + −6)(x − −6) and so in ideal terms y 3 = x+ √ −6 x− √ −6 . 2, we can conclude that if the ideals x + −6 and x − −6 are coprime then each is a cube √ √ √ of an ideal. √ Consider then I = x + −6 + x − −6 = x + −6, x − −6 . Then √ √ √ (x + −6) − (x − −6) = 2 −6 ∈ I and √ −6)(x + −6) = x2 + 6 ∈ I.

### Algebraic Number Theory: summary of notes [Lecture notes] by Robin Chapman

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