$\mathbb{Q}(\sqrt{m}, \sqrt{n})$ : ring of integers, integral basis and discriminant

Question

In the following document, http://people.math.carleton.ca/~williams/papers/pdf/033.pdf, I found three results about biquadratic fields and their ring of integers. It's the proof of the first theorem that bugs me (page 2).

1. Page 3 at the beginning : We consider the cases [...] Hence $2a_{i}$ are all integers [...]. How can we say that about the $\boldsymbol{2a_{i}}$ ? How can we write equation (5) ?
2. Page 4 middle : We now consider the case [...] are either all integers or all all halves of odd integers. Same problem. Why can't they be something else ?

This message is not pretty but writting everything here would be pointlessly long. And perhaps there is a simpler way of dealing with this theorem ?

Jérôme

Answer 1

Well, thanks to Jyrki Lahtonen, I read the document from the very beginning again. I found the answers to my questions all in one place, the first page when they give the form of the integeres in a quadratic field.

If it can help someone looking for a similar question, here's the result used (I feel a bit ashamed to not have seen that immediately).

If $k$ is a squarefree integer then the integers of $\mathbb{Q}(\sqrt{k})$ are given by $\frac{1}{2}(x_{0} + x_{1} \sqrt{k})$, where $x_{0}, x_{1}$ are integeres such that $x_{0} \equiv x_{1} \!\!\pmod{2}$, if $k \equiv 1 \!\!\pmod{4}$; and by $x_{0} + x_{1}\sqrt{k}$, where $x_{0}, x_{1}$ integers, if $k \equiv 2 \text{ or } 3 \!\!\pmod{4}$. Thus we know the integers of the subields $\mathbb{Q}(\sqrt{m}), \mathbb{Q}(\sqrt{n}), \mathbb{Q}(\sqrt{mn})$ of $\mathbb{Q}(\sqrt{m}, \sqrt{n})$.

For a proof of this, see for example the one given by Siddharth Prasad.

Because the quantities of (4) are in subfields of $\mathbb{Q}(\sqrt{m}, \sqrt{n})$, we can say things about the $2a_{i}$ using the previous result.