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

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

• Re: your 1st question. Doesn't the author explain exactly the reason why this claim is true in the second paragraph of page 3? IOW it is reduced to standard bits of information about algebraic integers of quadratic fields? – Jyrki Lahtonen Oct 1 '14 at 7:19
• Thanks for your answer. IOW ? Well I was a bit too quick to read the document. I found the information on the first page when he gives the form of the integers of a quadratic field. I feel a bit stupid to not have remembered this information. – Nerra Oct 1 '14 at 13:33
• Possibly related: math.stackexchange.com/questions/299710 – Watson Jul 23 '16 at 15:22

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 $\mathbb{Q}(\sqrt{m}, \sqrt{n})$.
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.
• Well done :-) ${}$ – Jyrki Lahtonen Oct 1 '14 at 14:00