# Extension of valuation

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

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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.
This makes really sense for me. For arbitary field $K$ with valuation $v$ and ${\alpha}$ is an algebraic element over $K$. Can we extend the valuation $v$ over $K[{\alpha}]$. –  Sharma Sharma Jan 6 '13 at 22:41
@SharmaSharma: There is a more general theorem: If $K$ is a field with valuation $v$ and $L|K$ is an algebraic extension, then there exists an extension of $v$ to $L$. Since your element $\alpha$ is algebraic the theorem applies to your situation. Unfortunatly the only proof I'm aware of invokes Zorn's Lemma. The construction is therefore not as explicit as in the case of Rajesh or algebraic number fields in general. –  Hans Giebenrath Jan 7 '13 at 8:00
@Hans, thank you very much. I am kind of wandering, is that extension $w$ is unique in your answer for algebraic number field $\mathbb Q$ to $\mathbb Q[2^{1/2}]$? –  Sharma Sharma Jan 7 '13 at 16:49