In a quadratic field, every element of norm 1 is a quotient of an algebraic integer by its conjugate

Let K = \mathbb{Q}(\sqrt{D}) be a quadratic field, and let \epsilon \in K be an element with N(\epsilon) = 1. Prove that there exists an algebraic integer \gamma such that \epsilon = \gamma/\gamma^\prime, where \prime denotes conjugate.

Let \epsilon = a+b\sqrt{D}. If b = 0, then a = \pm 1, and either \epsilon = 1/1 or \epsilon = -\sqrt{D}/\sqrt{D}; similarly if a = -1. So we assume that b \neq 0 and a \neq -1. Now 1 = a^2 - Db^2. Rearranging, we have \frac{a-1}{bD} = \frac{b}{a+1}. Choose some integers k and h such that \frac{k}{h} is equal to this common ratio. Let \gamma = h+k\sqrt{D}; certainly \gamma is an algebraic integer in K. We claim that \epsilon = \gamma/\gamma^\prime.

To that end, note that \gamma^\prime \epsilon = (ah-bkD) + (bh - ak)\sqrt{D}. From \frac{k}{h} = \frac{b}{a+1}, we see that bh - ka = k, and likewise since \frac{k}{h} = \frac{a-1}{bD} we have ha - bkD = h. So \gamma^\prime\epsilon = \gamma, and thus \epsilon = \gamma/\gamma^\prime.

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