By Stein W.

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This result generalizes to rings of integers of number fields. 1. If I and J are coprime ideals in OK , then I ∩ J = IJ. Proof. The ideal I ∩ J is the largest ideal of OK that is divisible by (contained in) both I and J. , I ∩ J ⊂ IJ. By definition of ideal IJ ⊂ I ∩ J, which completes the proof. 2. This lemma is true for any ring R and ideals I, J ⊂ R such that I + J = R. For the general proof, choose x ∈ I and y ∈ J such that x + y = 1. If c ∈ I ∩ J then c = c · 1 = c · (x + y) = cx + cy ∈ IJ + IJ = IJ, so I ∩ J ⊂ IJ, and the other inclusion is obvious by definition.

For any prime p ∈ Z, the following (sketch of an) algorithm computes the set of maximal ideals of O that contain p. 1. FACTORING PRIMES Sketch of algorithm. Let K = Q(a) be a number field given by an algebraic integer a as a root of its minimal monic polynomial f of degree n. We assume that an order O has been given by a basis w1 , . . , wn and that O that contains Z[a]. Each of the following steps can be carried out efficiently using little more than linear algebra over Fp . 5]. 1. 3, we easily factor pO.

Discriminant(x^5 + 7*x^4 + 3*x^2 - x + 1); 2945785 We have x5 + 7x4 + 3x2 − x + 1 ≡ (x + 2) · (x + 3)2 · (x2 + 4x + 2) (mod 5), which yields the factorization of 5OK given before the theorem. If we replace a by b = 7a, then the index of Z[b] in OK will be a power of 7, which is coprime to 5, so the above method will still work. f); [ , , ] 54 CHAPTER 8. FACTORING PRIMES Thus 5 factors in OK as 5OK = (5, 7a + 1)2 · (5, 7a + 4) · (5, (7a)2 + 3(7a) + 3). If we replace a by b = 5a and try the above algorithm with Z[b], then the method fails because the index of Z[b] in OK is divisible by 5.

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