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Algebraic Geometry

By Stein W.A.

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For the rest of this section OK is the ring of integers of a number field K. 9 (Divides for Ideals). Suppose that I, J are ideals of OK . Then we say that I divides J if I ⊃ J. To see that this notion of divides is sensible, suppose K = Q, so OK = Z. , that there exists an integer c such that m = cn, which exactly means that n divides m, as expected. 10. Suppose I is a nonzero ideal of OK . Then there exist prime ideals p1 , . . , I divides a product of prime ideals. Proof. Let S be the set of nonzero ideals of OK that do not satisfy the conclusion of the lemma.

We finish by explaining how to use LLL to find a polynomial f (x) ∈ Z[x] such that f (α) is small, hence has a shot at being the minimal polynomial of β. Given a real number decimal approximation α, an integer d (the degree), and an integer K (a function of the precision to which α is known), the following steps produce a polynomial f (x) ∈ Z[x] of degree at most d such that f (α) is small. 1. Form the lattice in Rd+2 with basis the rows of the matrix A whose first (d + 1) × (d + 1) part is the identity matrix, and whose last column has entries K, Kα , Kα2 , .

Proof. Let S be the set of nonzero ideals of OK that do not satisfy the conclusion of the lemma. The key idea is to use that OK is noetherian to show that S is the empty set. 1. DEDEKIND DOMAINS 45 that is maximal as an element of S. If I were prime, then I would trivially contain a product of primes, so we may assume that I is not prime. Thus there exists a, b ∈ OK such that ab ∈ I but a ∈ I and b ∈ I. Let J1 = I + (a) and J2 = I + (b). Then neither J1 nor J2 is in S, since I is maximal, so both J1 and J2 contain a product of prime ideals, say p1 · · · pr ⊂ J1 and q1 · · · qs ⊂ J2 .

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