By Robin Chapman
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From the PREFACE. THE ebook of this tract has been behind schedule by means of numerous factors, and i'm now pressured to factor it with no Dr Riesz's assist in the ultimate correction of the proofs. This has at any expense, one virtue, that it supplies me the potential for announcing how unsleeping i'm that no matter what price it possesses is due more often than not to his contributions to it, and particularly to the actual fact, that it comprises the 1st systematic, account of his attractive concept of the summation of sequence through 'typical means'.
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Extra info for Algebraic Number Theory: summary of notes [Lecture notes]
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