## Compute the discriminant of an ideal in an algebraic integer ring

Let $K = \mathbb{Q}(\sqrt{-5})$ and let $\mathcal{O} = \mathbb{Z}[\sqrt{-5}]$ be the ring of integers in $K$. Let $A = (3,1+\sqrt{-5})$ and $B = (2)$. Exhibit bases for $A$ and $B$, compute $\mathcal{O}/A$ and $\mathcal{O}/B$, and compute the discriminant of $A$ and $B$.

We claim that $\{3, 1+\sqrt{-5}\}$ is a basis for $A$ over $\mathbb{Z}$. It is clear that $(3, 1+\sqrt{-5})_\mathbb{Z} \subseteq A$, and if $\zeta = 3(a+b\sqrt{-5}) + (1+\sqrt{-5})(h+k\sqrt{-5})$, then evidently $\zeta = 3(a-b-2k) + (1 + \sqrt{-5})(3b+k+h)$. So $(3, 1+\sqrt{-5})_\mathbb{Z} = A$. Now if $a,b \in \mathbb{Z}$ and $3a + (1+\sqrt{-5})b = 0$, then $b = 0$, so that $a = 0$. Hence $\{3, 1+\sqrt{-5}\}$ is a basis for $A$ over $\mathbb{Z}$. The discriminant of $A$ is then $\mathsf{det}^2 \left[ \begin{array}{cc} 3 & 1+\sqrt{-5} \\ 3 & 1-\sqrt{-5} \end{array} \right] = -180$.

We now claim that $A$ is proper. To see this, suppose to the contrary that $1 \in A$; then $1 = 3a + (1+\sqrt{-5})b$ for some integers $a$ and $b$; comparing coefficients, we have $b = 0$ and $3a+b = 1$. But then $3a = 1$, which has no integer solutions; thus $A$ is proper.

Now let $a+b\sqrt{-5}$ be arbitrary in $\mathcal{O}$ and let $a-b \equiv k$ mod 3, where $k \in \{0,1,2\}$. Now $a+b\sqrt{-5} = a-b+b(1+\sqrt{-5}) \equiv a-b \equiv k$ mod $A$, so that $\mathcal{O}/A = \{\overline{0}, \overline{1}, \overline{2}\}$. Since $A$ is proper, $0 \not\equiv 1$ and $2 \not\equiv 1$ mod $A$. If $2 \equiv 0$, then we have $3-2 = 1 \in A$, a contradiction; so $2 \not\equiv 0$ mod $A$. Thus $\mathcal{O}/A \cong \mathbb{Z}/(3)$.

We claim now that $\{2,2\sqrt{-5}\}$ is a basis for $B$ over $\mathbb{Z}$. Certainly $(2,2\sqrt{-5})_\mathbb{Z} \subseteq B$, and if $\zeta = 2(a+b\sqrt{-5}) \in (2)$, then $\zeta = 2a + 2\sqrt{-5}b$. So $(2,2\sqrt{-5})_\mathbb{Z} = B$. Moreover, if $2a + 2b\sqrt{-5} = 0$, then $a = b = 0$. So $\{2,2\sqrt{-5}\}$ is a basis for $B$ over $\mathbb{Z}$. The discriminant of $B$ is then $\mathsf{Det}^2 \left[ \begin{array}{cc} 2 & 2\sqrt{-5} \\ 2 & -2\sqrt{-5} \end{array} \right] = -320$.

We claim that $B$ is proper. Indeed, if $1 \in B$, then $1 = 2a$ for some integer $a$, a contradiction.

Now let $a+b\sqrt{-5}$ be arbitrary in $\mathcal{O}$. Say $a \equiv a_0$ and $b \equiv b_0$ mod 2, where $a_0, b_0 \in \{0,1\}$. Then $a+b\sqrt{-5} \equiv a_0 + b_0 \sqrt{-5}$ mod $B$. Thus $\mathcal{O}/B = \{\overline{0}, \overline{1}, \overline{\sqrt{-5}}, \overline{1+\sqrt{-5}}\}$. Since $B$ is proper, we have $0 \not\equiv 1$ and $\sqrt{-5} \not\equiv 1+\sqrt{-5}$ mod $B$. If $\sqrt{-5} \in B$, then we have $1 = 2b$ for some integer $b$, a contradiction. Thus $\sqrt{-5} \not\equiv 0$ and $1+\sqrt{-5} \not\equiv 1$ mod $B$. Likewise, if $1+\sqrt{-5} \in B$, we have a contradiction, so that $1+\sqrt{-5} \not\equiv 0$ and $\sqrt{-5} \not\equiv 1$ mod $B$.

We claim that $\mathcal{O}/B \cong \mathbb{F}_2[x]/(x^2)$. To see this, define $\varphi : \mathbb{F}_2[x] \rightarrow \mathcal{O}/B$ by $x \mapsto 1+\sqrt{-5}$. Certainly $\varphi$ is surjective. Now $\varphi(x^2) = 0$, so that $(x^2) \subseteq \mathsf{ker}\ \varphi$. Now if $p(x) = a_0 + a_1x + p^\prime(x)x^2 \in \mathsf{ker}\ \varphi$, then $\varphi(p(x)) = (a_0+a_1) + a_1\sqrt{-5} \equiv 0$. Thus $a_1 \equiv a_0 \equiv 0$ mod 2. By the first isomorphism theorem for rings, $\mathcal{O}/B \cong \mathbb{F}_2(x)/(x^2)$.