By Robin Chapman
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Extra info for Algebraic Number Theory: summary of notes [Lecture notes]
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 . We claim that those uj which are nonzero form an integral basis of H. Given this claim, it is immediate that H is free abelian of rank m, where m is the number of nonzero uj , and so m ≤ n. 59 We first show that each element of H has the form nj=1 aj uj with uj ∈ Z. To do this note that if v = (0, . . , 0, cj , . . , cn ) ∈ Hj then cj = aj bj with aj ∈ Z and so v − aj uj ∈ Hj+1 if j < n and v − aj uj = 0 if j = n.
But f1 (α) = pf2 (α) ∈ p so that P1a1 P2a2 · · · Prar ⊆ p . But we have seen these ideals have the same norm, so they are equal. 37 In fact this result is true even when OK = Z[α] as long as p |OK : Z[α]|. We shall not prove this generalization. √ Example Let K = Q( m) be a quadratic field where √ m is a squarefree integer with m ≡ 1 (mod 4). Then OK = Z[α] where α = m has minimum polynomial X 2 − m. To determine the prime ideal factorization of p , where p is a prime number, in OK , we must factorize X 2 − m in Fp [X].
Then there is no prime ideal P with P ⊇ I and P ⊇ J (for otherwise P ⊇ I + J) and so no prime ideal occurs in both the factorizations of I and J. 2 Let K be a number field. Let I1 and I2 be coprime ideals of OK and suppose that I1 I2 = J m for some ideal J and integer m. Then I1 = J1m and I2 = J2m for some ideals J1 and J2 . Proof Let P be a prime ideal occurring in the factorization of I1 . Suppose that P occurs a times in this factorization. Then P does not occur in the prime ideal factorization of I2 , as I1 and I2 are coprime.
Algebraic Number Theory: summary of notes [Lecture notes] by Robin Chapman