Given an element a of an ideal in an algebraic integer ring, give a two element generating set containing a

Find \alpha such that (180) = (\alpha, 16200) in \mathbb{Z}. Find \alpha such that (3, 1+2\sqrt{-5}) = (9,\alpha) in \mathbb{Z}[\sqrt{-5}].


Note that (160)(90) = (16200). Using Theorem 9.3, it suffices to find an element \alpha \in (180) and an ideal C such that (180)C = (\alpha) and (C, (90)) = (1). To that end, note that (90) factors into prime ideals as (90) = (2)(3)^2(5), with distinct factors (2), (3), and (5). Let A_1 = (180)(3)(5) = (2700), A_2 = (180)(2)(5) = (1800), and A_3 = (180)(2)(3) = (1080). Now let \alpha_1 = 8100, \alpha_2 = 3600, and \alpha_3 = 2160. Evidently, A_1(3) = (8100), A_2(2) = (3600), and A_3(2) = (2160), and (3,2) = (2,5) = (3,5) = (1). Following the proof of Theorem 9.3, let \alpha = \alpha_1 + \alpha_2 + \alpha_3 = 13860. Then (13860,16200) = (180). Indeed, we have 180 = -7 \cdot 13860 + 6 \cdot 16200, 13860 = 180 \cdot 77, and 16200 = 180 \cdot 90.

We first need to find an ideal D such that (3,1+2\sqrt{-5})D = (9). In a previous exercise we found that D = (9, 3-6\sqrt{-5}) is such an ideal. Next we wish to find the prime factorization of D. To this end, we note that D = (3)(3,1-2\sqrt{-5}) = (3,1+2\sqrt{-5})(3,1-2\sqrt{-5})^2 (since (1+2\sqrt{-5})(1+2\sqrt{-5}) - 2 \cdot 3 \cdot 3 = 3).

Claim: (3,1+2\sqrt{-5}) is proper. Proof of claim: Suppose to the contrary that this ideal contains 1. THen we have 1 = 3(a+b\sqrt{-5}) + (1+2\sqrt{-5})(h+k\sqrt{-5}), so that 1 = 3a + h - 10k and 0 = 3b + 2h + k. This leads to a contradiction mod 3, so that (3,1+2\sqrt{-5}) is proper.

Claim: (3,1+2\sqrt{-5}) is maximal. Proof of claim: Let a+b\sqrt{-5} \in \mathbb{Z}[\sqrt{-5}], and say b = 2b_1 + b_0 and a-b_1 = 3a_1 + a_0 where b_0 \in \{0,1\} and a_0 \in \{0,1,2\}. Mod (3,1+2\sqrt{-5}), we have a+b\sqrt{-5} \equiv a_0 + b_0 \sqrt{-5}. Now 2+\sqrt{-5} \equiv (2+\sqrt{-5}) + (1+2\sqrt{-5}) \equiv 3(1+\sqrt{-5}) \equiv 0 mod (3,1+2\sqrt{-5}). Thus \sqrt{-5} \equiv 1 and 1+\sqrt{-5} \equiv 2. In particular, \mathcal{O}/(3,1+2\sqrt{-5}) = \{ \overline{0}, \overline{1}, \overline{2} \}. Now 1 \not\equiv 0 since this ideal is proper, and so also 2 \not\equiv 1. If 2 \equiv 0, then 3-2 = 1 \in (3,1+2\sqrt{-5}), a contradiction. So \mathcal{O}/(3,1+2\sqrt{-5}) \cong \mathbb{Z}/(3) is a field, and thus (3,1+2\sqrt{-5}) is maximal.

We can show very similarly that (3,1-2\sqrt{-5}) is proper and also maximal. Here we can show that 2-\sqrt{-5} \equiv 0.

Thus we have the prime factorization of D, with distinct factors P_1 = (3,1+2\sqrt{-5}) and P_2 = (3,1-2\sqrt{-5}). Let A_1 = P_1P_2 and A_2 = P_1^2. We see that A_1 = (3) and A_2 = (9,1-4\sqrt{-5}).

Next we find \alpha_1 \in A_1 and \alpha_2 \in A_2 such that ((\alpha_1)/A_1, (3,1+2\sqrt{-5})) = ((\alpha_2)/A_2, (3,1-2\sqrt{-5})) = (1). By our proof of Theorem 9.2, it suffices to exhibit \alpha_1 \in A_1 \setminus A_1(3,1+2\sqrt{-5}), and likewise for \alpha_2. Noting that A_2(3,1-2\sqrt{-5}) = (9,3+6\sqrt{-5}), we see that \alpha_1 = 3 and \alpha_2 = 1-4\sqrt{-5} are such elements. Let \alpha = \alpha_1 + \alpha_2 = 4-4\sqrt{-5}. By the proof of Theorem 9.3 in TAN, (3,1+2\sqrt{-5}) = (9,4-4\sqrt{-5}).

We can verify this by noting that 3 = 9(-1-4\sqrt{-5}) + (4-4\sqrt{-5})(-7+2\sqrt{-5}), 1+2\sqrt{-5} = 9(-3-2\sqrt{-5}) + (4-4\sqrt{-5})(-3+2\sqrt{-5}), 9 = 3 \cdot 3, and 4-4\sqrt{-5} = 3(-5) + (1+2\sqrt{-5})(-1-2\sqrt{-5}).

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