# $2$-adic valuation on $\mathbb Q (\sqrt{-15})$

I was reading the book "The structure of groups of prime power order". In the book (page 226, Example 10.1.18) it is proved that

$$-15$$ has a square root is $$\mathbb Z_2$$ where $$\mathbb Z_2$$ denotes the ring of $$2$$-adic integers.

[I have attached a screen-shot of Example 10.1.18 at the end of my post]

So suppose $$a=\sqrt{-15}$$, then for any integers $$x,y$$ we have $$x+ya\in \mathbb Z_2$$ since $$\mathbb Z \subset \mathbb Z_2$$. So my question is

Given $$x,y\in \mathbb Z$$, is it possible to determine explicitly $$\bar x, \bar y \in \mathbb Z$$ and a maximum possible $$l\ge 0$$ such that

$$x+ya = 2^l(\bar x + \bar y a) \quad \text{ and } \quad \bar x + \bar y a \text{ is a unit in } \mathbb Z_2$$

Though not exactly related to the question but I think I should also say why I have this question; I am thinking of finding some kind of normal form of a matrix whose entries are from $$\mathbb Z_2$$ and has the form $$x+ya$$. So if I get a method for my above mentioned question then it will possibly help to write an algorithm for the normal form.

But I apologise for not showing much effort from my side for the question, I am actually not sure how to handle this. Any help will be greatly appreciated.

Thanks.

Screen-shot of the example:

A way to do this is to write $$\sqrt{-15}=\sqrt{1-16}$$ from which you get the $$2$$-adic expansion of $$a$$, and thus of $$x+ya$$. To get an a priori bound on $$l$$, you can use the fact that $$v_2(x+ya) \leq v_2((x+ya)(x-ya)) \leq v_2(x^2+15y^2)$$ which you can compute since $$x^2+15y^2 \in \mathbf{Z}$$.
• Thanks. It will be really helpful if you give a hint how to get $2$-adic expansion from $\sqrt{1-16}$. And then do I need to use computational method to get exact $l$? Thanks again. – usermath Nov 29 '18 at 9:59
• @usermath Use $\sqrt{1+x} = 1+\frac{x}{2}-\frac{x^2}{8} + \cdots$ with $x=-16=-2^4$. My answer gives an upper bound for $l$, so you can get the exact value by computing finitely many terms of the $2$-adic expansion of $x+ya$ en.wikipedia.org/wiki/P-adic_number#p-adic_expansions Your question would have been more suitable on math.stackexchange.com – François Brunault Nov 29 '18 at 11:55