## Prove that a given ideal is maximal

Prove that $A = (1+\sqrt{-5}, 1-\sqrt{-5})$ is maximal in $\mathcal{O}_K$, where $K = \mathbb{Q}(\sqrt{-5})$.

We will prove this in slightly more generality, beginning with a lemma.

Lemma: Let $D$ be a squarefree integer with $D \equiv 3$ mod 4. Then $\{1+\sqrt{D}, 1-\sqrt{D}\}$ is a basis for $A = (1+\sqrt{D}, 1-\sqrt{D})$ over $\mathbb{Z}$. Proof: Let $\zeta \in A$. Then $\zeta = (a+b\sqrt{D})(1+\sqrt{D}) + (h+k\sqrt{D})(1-\sqrt{D})$ for some $a,b,h,k \in \mathbb{Z}$. Evidently, $\zeta = u(1+\sqrt{D}) + v(1-\sqrt{D})$, where $u = a+b\frac{D+1}{2} - k\frac{D-1}{2}$ and $v = h + b\frac{D-1}{2} - k\frac{D+1}{2}$. So $A = (1+\sqrt{D},1-\sqrt{D})_\mathbb{Z}$. Suppose now that $h(1+\sqrt{D}) + k(1-\sqrt{D}) = 0$; then $h+k = 0$ and $h-k = 0$, so that $h = k = 0$, as desired. $\square$

Let $D$ be a squarefree integer with $D \equiv 3$ mod 4. Now the ring of integers in $K = \mathbb{Q}(\sqrt{D})$ is $\mathcal{O} = \{a+b\sqrt{D} \ |\ a,b \in \mathbb{Z}\}$. We claim that $A = (1+\sqrt{D}, 1-\sqrt{D})$ is maximal in $\mathcal{O}$.

To see this, note that $2 = (1+\sqrt{D}) + (1-\sqrt{D}) \in A$. Now let $a+b\sqrt{D} \in \mathcal{O}$, and note that $a+b\sqrt{D} = a-b+b+b\sqrt{D} = a-b + b(1+\sqrt{D}) \equiv a-b$ mod $A$. In particular, if $a-b \equiv 0$ mod 2, then $a+b\sqrt{D} \equiv 0$ mod $A$, and if $a-b \equiv 1$ mod 2, then $a+b\sqrt{D} \equiv 1$ mod $A$. We claim that these cosets are distinct. Indeed, if $1 \equiv 0$ mod $A$, then $1 \in A$. In this case, by the lemma we have $h,k \in \mathbb{Z}$ such that $h(1+\sqrt{D}) + k(1-\sqrt{D}) = (h+k) + (h-k)\sqrt{D} = 1$. But then $h+k = 1$ and $h-k = 0$, so that $2h = 1$, a contradiction. So $\mathcal{O}/A = \{0+A, 1+A\}$. Now $\mathcal{O}/A$ is a unital ring with $1 \neq 0$ having only two elements, so that $\mathcal{O}/A \cong \mathbb{Z}/(2)$; that is, $\mathcal{O}/A$ is a field, and thus $A$ is a maximal ideal.

The original problem follows since $-5 \equiv 3$ mod 4.