## 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 $4-4\sqrt{-5} = 3(-5) + (1+2\sqrt{-5})(-1-2\sqrt{-5})$.