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If 3 | x then we have y 3 ≡ 3 (mod 9) which is impossible. 2. 1, J is a principal ideal. √ Write J = β . Then x + −6 = β 3 . This means that x + −6 = ηβ 3 √ where η is a unit of OK . But the only units of OK are ±1 so that x + −6 = √ ±β 3 = (±β)3 = γ 3 where γ = ±β ∈ OK . Write γ = a + b −6 where a, b ∈ Z. Then (x − √ x = a3 − 18ab2 and 1 = 3a2 b − 6b3 . The latter equation is absurd as it implies that 3 | 1. Hence there are no integers x and y satisfying (∗). 1, by regarding a number field K of degree n as a subset of Rn .

Cn ∈ Z} ∩ H. That is Hj is the set of all vectors in H where the first j − 1 entries vanish. We also let Kj = {cj : (0, 0, . . , 0, cj , cj+1 , . . , cn ) ∈ Hj } for each j. That is, Kj is the set of j-th entries of vectors in Hj . It is easy to see that Hj is a subgroup of H (and so of Zn ) and consequently that Kj is a subgroup of Z. Then Jj = Zbj = {abj : a ∈ Z} for some integer bj . Define vectors u1 , . . , un ∈ H as follows. If bj = 0 let uj be the zero vector, otherwise let uj be any vector in Hj whose j-th entry is bj .

Then I ⊆ J1 and I ⊆ J2 , but I = J1 and I = J2 . Hence N (J1 ) < N (I) and N (J2 ) < N (I). But J1 J2 = βγ + βI + γI + I 2 ⊆ I as βγ ∈ I. By the inductive hypothesis, J1 ⊇ P1 · · · Pr and J2 ⊇ Q1 · · · Qs where the Pj and Qk are prime. Hence I ⊇ J1 J2 ⊇ P1 · · · Pr Q1 · · · Qs as required. As a technical convenience we extend the notion of ideals in OK to that of fractional ideal. It will turn out that the set of fractional ideals forms a group under multiplication, which it is clear that the set of ideals do not.

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