TITLE: Extension of valuation
QUESTION [10 upvotes]: Fix a prime number $p$. Suppose that I have a valuation $v_p: \mathbb{Q} \to \mathbb{Q}$ on the rationals $\mathbb{Q}$. That is, $v_p( p^n(\frac{a}{b})) = p^{-n}$ where each of $a,b$ is not divisible by $p$. 
How can I extend $v_p$ to $v$ on the reals $\mathbb{R}$ such that $v|_\mathbb{Q} = v_p$? I am looking for an explicit description of $v$, if that is possible. I know for a fact that one can extend valuation on any field extension.
Thank you,

REPLY [7 votes]: Just in case you don't know this, the extension of $v$ to $\overline{\mathbb{Q}_p}$ is done as follows: If $x$ is algebraic over $\mathbb{Q}_p$, let $x^n + a_{n-1} x^{n-1} + \cdots a_0=0$ be the minimal polynomial of $x$. Then $v(x) = (1/n) v(a_0)$. Turning this into a valuation on $\mathbb{C}$ requires noncanonical choices of two kinds. 
For the first kind of choice, consider let $p=7$. The equation $x^2-6x+7=0$ has two roots in $\mathbb{R}$: one is roughly $4.414\ldots $ and the other is roughly $1.585\ldots $. It also has two roots in $\mathbb{Q}_7$: one is roughly $(\ldots 60)_7$ and the other is roughly $(\ldots16)_7$. You have to decide which one will be identified with which. For the second kind, there are transcendental elements of $\mathbb{C}$, such as $\pi$. You can choose their valuation freely. (More precisely, you can choose the valuations of a transcendence base freely.)