# Thread: Show that alpha is a perfect square in the quadratic integers.

1. ## Show that alpha is a perfect square in the quadratic integers.

Suppose $\displaystyle \mathbb{Q}(\sqrt{d})$ is a unique factorization domain, and $\displaystyle \alpha$ is an integer in $\displaystyle \mathbb{Q}(\sqrt{d})$ so that $\displaystyle \alpha$ and $\displaystyle \bar{\alpha}$ have no common factor, but $\displaystyle N(\alpha)$ (norm of alpha) is a perfect square in $\displaystyle \mathbb{Z}$. Show that $\displaystyle \alpha$ is a perfect square in the quadratic integers in $\displaystyle \mathbb{Q}(\sqrt{d})$.

I'm not sure if I'm approaching this correctly but if $\displaystyle \alpha$ and $\displaystyle \bar{\alpha}$ have no common factor, then $\displaystyle \alpha=\pi_{1}\pi_{2}\cdots\pi_{k}$ and $\displaystyle \bar{\alpha}=\pi'_{1}\pi'_{2}\cdots\pi'_{j}$ where $\displaystyle \pi_{i}$ and $\displaystyle \pi'_{i}$ are prime in $\displaystyle \mathbb{Q}(\sqrt{d})$. But $\displaystyle N(\alpha) = \alpha\bar{\alpha}=\pi_{1}\pi_{2}\cdots\pi_{k}\pi' _{1}\pi'_{2}\cdots\pi'_{j}=n^{2}$ where $\displaystyle n \in \mathbb{Z}$. Somehow I need to show that $\displaystyle \alpha=\beta^{2}$ where $\displaystyle \beta$ is a quadratic integer in $\displaystyle \mathbb{Q}(\sqrt{d})$ (i.e., $\displaystyle \alpha$ is a perfect square in the quadratic integers in $\displaystyle \mathbb{Q}(\sqrt{d})$).

2. $\displaystyle N(\alpha) = \alpha\bar{\alpha}=\pi_{1}\pi_{2}\cdots\pi_{k}\pi' _{1}\pi'_{2}\cdots\pi'_{j}=n^{2}= (r_1 r_2 \cdots r_m )^2$
so up to associates, each $\displaystyle r_i$ should appear twice in the left side of the equation. it cannot appear once in
$\displaystyle \alpha$ and once in $\displaystyle \bar{\alpha}$ because then they will have a common factor, so each factor $\displaystyle r_i$ in $\displaystyle \alpha$ is squared, meaning $\displaystyle \alpha = \beta (r_{n_1} r_{n_2} \cdots r_{n_k})^2 = \beta \gamma^2$ where beta is a unit. I'm not sure how to show that beta is also a square. maybe there is a special property of $\displaystyle \mathbb{Q}(\sqrt{d})$ that I'm forgetting.