## 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 $\epsilon = \gamma/\gamma^\prime$.