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I didn't understand why this $\alpha$ is an integer. I know that $(x+y\sqrt D)(x-y\sqrt D)=4z^n$, but I don't know what to do with this information.

Note: I don't know if it's relevant but $K$ is a quadratic number field.

user42912
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    Here by an integer, we mean an algebraic integer rather than $\mathbb{Z}$. This is true because both the norm and trace are integers so it is a root of a monic polynomial with coefficients in $\mathcal{O}_{K}$, – daruma Mar 10 '17 at 05:17
  • @daruma I realized that. Thank you! – user42912 Mar 10 '17 at 05:36

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When they say $\alpha$ is an integer over $K$, they don't mean it's an element of $\Bbb Z$. They mean it's in the ring of integers of $K$. Equivalent statements: $\alpha$ is integral over $\Bbb Z \to K$; $\alpha$ is an algebraic integer.

The definition of "algebraic integer" is pretty close to the definition of being an algebraic number. An element $\alpha$ of some extension is algebraic if it's the root of some polynomial in $\Bbb Q[X]$. It is an algebraic integer if we can pick this polynomial to be in $\Bbb Z[X]$ and monic. (We can always do each of these conditions, it's making them true at the same time that's hard.)

In this case, $K = \Bbb Q (\sqrt D)$, and $p(X) = X^2 - xX + \frac{x^2 - y^2 D}{4}$.

Henry Swanson
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