Compute the discriminant of a basis for a given ideal in a quadratic integer ring

Let \mathcal{O} be the ring of integers in \mathbb{Q}(\sqrt{D}). Find a basis for the ideal (4) over \mathbb{Z}. Then compute the discriminant of this basis.


Suppose first that D \not\equiv 1 mod 4.

Note that (4) = \{4a + 4b\sqrt{D} \ |\ a,b \in \mathbb{Z} \}. We claim that \{4,4\sqrt{D}\} is a basis for (4). Certainly (4,4\sqrt{D})_\mathbb{Z} = (4). Now if 4a+4b\sqrt{D} = 0 and b \neq 0, then \sqrt{D} = -a/b is rational, a contradiction. So 4a = 0, and a = b = 0. So \{4,4\sqrt{D}\} is a basis for (4) over \mathbb{Z}. Evidently, the discriminant of this basis is \Delta[4,4\sqrt{D}] = 1024D. Recall from Theorem 6.11 in TAN that the discriminant of \mathbb{Q}(\sqrt{D}) is \Delta_\mathcal{O} = 4D. Thus \Delta[4,4\sqrt{D}] = 16^2 \Delta_\mathcal{O}.

Now suppose D \equiv 1 mod 4.

Now (4) = \{ 4a + 2b(1+\sqrt{D}) \ |\ a,b \in \mathbb{Z}\}. We claim that \{4,2+2\sqrt{D}\} is a basis. Certainly (4,2+2\sqrt{D})_\mathbb{Z} = (4). If 4a + 2b + 2b\sqrt{D} = 0 and b \neq 0, then -(2a+b)/b = \sqrt{D}, a contradiction. So b = 0, and thus a = b = 0. Hence \{4,2+2\sqrt{D}\} is a basis for (4). Evidently the discriminant of this basis is \Delta[4,2+2\sqrt{D}] = 256D. Again recall from Theorem 6.11 in TAN that \Delta_\mathcal{O} = D; hence \Delta[4,2+2\sqrt{D}] = 16^2 \Delta_\mathcal{O}.

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