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.

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