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Math Help - Show that alpha is a perfect square in the quadratic integers.

  1. #1
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    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})).
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  2. #2
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    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<br />
    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.

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