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By Stein W.A.

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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 . Then p1 · · · pr · q1 · · · qs ⊂ J1 J2 = I 2 + I(b) + (a)I + (ab) ⊂ I, so I contains a product of primes. This is a contradiction, since we assumed I ∈ S. Thus S is empty, which completes the proof. We are now ready to prove the theorem. 8. 8 in the case when R = OK is the ring of integers of a number field K.

This is of great value even without proof, since often in practice once you know a potential minimal polynomial you can verify that it is in fact right. Exactly this situation arises in the explicit construction of class fields (a more advanced topic in number theory) and in the construction of Heegner points on elliptic curves. As we will see, the LLL algorithm provides a polynomial time way to solve this problem, assuming α has been computed to sufficient precision. 1 LLL Reduced Basis Given a basis b1 , .

This is less nonsensical; however, fast factoring is not known to break all commonly used public-key cryptosystem. For example, there are cryptosystems based on the difficulty of computing discrete logarithms in F∗p and on elliptic curves over Fp , which (presumably) would not be broken even if one could factor large numbers quickly. 1. 2 51 Examples The following Sage session shows the commands needed to compute the factorization of pOK for K the number field defined by a root of x5 + 7x4 + 3x2 − x + 1 and p = 2 and 5.

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