Let $K=\mathbb Q(\sqrt{N})$ be a number field, $\mathcal O$ be an order of $K$ (i.e. $\mathcal O$ is a subring of $K$ and $\mathcal O$ is a free $\mathbb Z$-module of rank 2).
In the begining of this thesis (p.7 just after the definition 1.1.3 : "De plus, si $\alpha \in \mathcal O$ [...] alors $\alpha$ est un entier algébrique de K" ), it is said that every element $\alpha\in \mathcal O$ is an algebraic integer of $K$. But I do not see why...
If I'm doing that correctly: $\mathcal O=\mathbb Z\left[\frac pq \sqrt{N}\right]$, for some integers $p,q$ such that gcd($p,q$)=1.
Then, let $\alpha\in\mathbb O$, so $\alpha=a + b \frac pq \sqrt{N}$ (with $a,b\in\mathbb Z$) is root of the polynmonial $F=q^2(X-a)^2-b^2p^2N\in\mathbb Z[X]$, but $F$ is not a unit polynomial.
Where am I wrong or what is the point I'm missing?