Extension of valuation

Question

We define a valuation on the field of rational number $\mathbb Q$ as follows. For example if we choose a prime number $2$ then for $x \neq 0\in \mathbb Q$, $v(x) = v(2^{n}a/b)= n$ where $n$ is an integer and $a$ and $b$ are relatively prime integer. I have a question, how do we extend this valuation explicitly over the field $\mathbb Q(2^{1/2})$. Thanks

Answer 1

The ring of integers of $\mathbb{Q}(\sqrt 2)$ is given by $\mathbb{Z}[\sqrt 2]$. The valuations of $\mathbb{Q}(\sqrt 2)$ extending $v$ can be deduced from the prime ideal factorization of $(2)$ in the ring $\mathbb{Z}[\sqrt 2]$. More precisely we have $$ (2) = \mathfrak p^2 $$ with $\mathfrak p = (\sqrt 2)$ a prime ideal of $\mathbb{Z}[\sqrt 2]$. Given $x \in \mathbb{Z}[\sqrt 2]$ denote by $v_\mathfrak p(x)$ the integer $n$ such that $x \in \mathfrak p^n$ and $x \notin \mathfrak p^{n+1}$. For arbitrary $\frac{x}{y} \in \mathbb Q(\sqrt 2)$ with $x,y \in \mathbb Z[\sqrt 2]$ set $v_\mathfrak p(\frac x y) = v_\mathfrak p(x) - v_\mathfrak p(y)$. A valuation $w$ of $\mathbb Q(\sqrt 2)$ extending $v$ is then given by $w(x) = \frac{v_\mathfrak p(x)}{2}$ (The $2$ in the denominator being the inertia degree of $\mathfrak p$ over $2$). I hope this is explicitly enough.