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

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

2. $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 $r_i$ should appear twice in the left side of the equation. it cannot appear once in
$\alpha$ and once in $\bar{\alpha}$ because then they will have a common factor, so each factor $r_i$ in $\alpha$ is squared, meaning $\alpha = \beta (r_{n_1} r_{n_2} \cdots r_{n_k})^2 = \beta \gamma^2$ 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 $\mathbb{Q}(\sqrt{d})$ that I'm forgetting.