By Stein W.A.

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The other inclusion is obvious by definition of ideal. 2. 3. Suppose I1 , . . , Is are pairwise coprime ideals. Then I1 is coprime to the product I2 · · · Is . Proof. 11); instead, we give a direct general argument. It suffices to prove the lemma in the case s = 3, since the general case then follows from induction. By assumption, there are x1 ∈ I1 , y2 ∈ I2 and a1 ∈ I1 , b3 ∈ I3 such x1 + y2 = 1 and a1 + b3 = 1. Multiplying these two relations yields x1 a1 + x1 b3 + y2 a1 + y2 b3 = 1 · 1 = 1.

Since p is prime, OK /p is an integral domain. Every finite integral domain is a field (see Exercise 10), so p is maximal, which completes the proof. If I and J are ideals in a ring R, the product IJ is the ideal generated by all products of elements in I with elements in J: IJ = (ab : a ∈ I, b ∈ J) ⊂ R. Note that the set of all products ab, with a ∈ I and b ∈ J, need not be an ideal, so it is important to take the ideal generated by that set (see Exercise 11). 6 (Fractional Ideal). A fractional ideal is a nonzero OK -submodule I of K that is finitely generated as an OK -module.

2 (LLL-reduced basis). The basis b1 , . . , bn for a lattice L ⊂ Rn is LLL reduced if for all i, j, 1 |µi,j | ≤ 2 and for each i ≥ 2, |b∗i |2 ≥ 3 − µ2i,i−1 |b∗i−1 |2 4 For example, the basis b1 = (1, 2), b2 = (3, 4) for a lattice L is not LLL reduced because b∗1 = b1 and b2 · b∗ 11 1 µ2,1 = ∗ 1∗ = > . b1 · b1 5 2 However, the basis b1 = (1, 0), b2 = (0, 2) for L is LLL reduced, since µ2,1 = b2 · b∗1 = 0, b∗1 · b∗1 and 22 ≥ (3/4) · 12 . LLL() [1 0] [0 2] 38 CHAPTER 2. 2 What LLL really means The following theorem is not too difficult to prove.