Let $d$ be a non-square discriminant. If $I \subset \mathcal{O}_d$ is a nonzero ideal, then $I \cong \mathbb{Z}^2$ as an abelian group.

Question

For every integer $d \equiv 0,1 \mod 4,$ we define a quadratic ring $\mathcal{O}_d$ by: If $d \equiv 0 \mod 4,$ let $\mathcal{O}_d = \mathbb{Z}\left[\sqrt{\dfrac{d}{4}} \right].$ If $d$ is a non-square congruent to $1,$ then $\mathcal{O}_d = \left\{ \dfrac{a+b\sqrt{d}}{2} a \equiv b \mod 2 \right\} = \mathbb{Z}\left[\dfrac{1 + \sqrt{d}}{2} \right].$

I know that $I \cong \{0\},\mathbb{Z},$ or $\mathbb{Z}^2$ but not really sure where to go? This problem is tough.

I assume the problem is to prove all of these statements. – anon May 17 '15 at 08:14 — anon, May 17 '15 at 08:14

score 1 · Accepted Answer · answered May 17 '15 at 08:19

Assume $d$ is not a perfect square. Since $\mathcal{O}_d \cong \mathbb{Z}^2$ is an abelian group under addition, it follows that $I \subset \mathcal{O}_d$ is isomorphic to $\{0\},\mathbb{Z},$ or $\mathbb{Z}^2.$ Assume $\mathcal{O}_d \supset I \ne \{0\}.$ If $I \cong \mathbb{Z} \Longrightarrow$ there exists $\alpha \in I$ such that $I = \{k \alpha: \alpha \in \mathbb{Z} \}.$ Consider $\sqrt{d} \in \mathcal{O}_d.$ Then by definition of an ideal, $\alpha \sqrt{d} \in I.$ Hence $\alpha \sqrt{d} = n \alpha$ for some $n \in \mathbb{Z}.$ Since $d$ is not a perfect square $ \Longrightarrow \mathcal{O}_d$ is an integral domain $\Longrightarrow \sqrt{d} = k.$ This is a contradiction since we assumed $d$ is non-square. $\text{ } \Box$

Let $d$ be a non-square discriminant. If $I \subset \mathcal{O}_d$ is a nonzero ideal, then $I \cong \mathbb{Z}^2$ as an abelian group.

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