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,
As you point out, it follows on general grounds that there is an extension of $v_p$ to a valuation on ${\mathbb R}$ (in fact, there are uncountably many such extensions), but it is impossible to give an "explicit" description. Indeed, not only will any such extension by discontinuous with respect to the usual Euclidean topology on the reals, but it will not be a measurable function.
You can extend $v_p$ to $\overline{\mathbb{Q}}_p$ in a unique way, and then the general theory of fields tells you that $\overline{\mathbb{Q}}_p$ is isomorphic to $\mathbb{C}$ which gives you the extension you want. The fields $\overline{\mathbb{Q}}_p$ and $\mathbb{C}$ are isomorphic because they are both algebraically closed and have the same transcendence degree over $\mathbb{Q}$. However as Thomas said, this construction is not explicit, indeed it uses the axiom of choice.
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.)
I guess everybody knows this, I am just mentioning it: in Atiyah-MacDonald's Commutative Algebra, the proof for the extension of valuation is done by using maps $f$ from the original ring A (in our case $\mathbb{Z}_p$, corrected to ℤ(p)) to an algebraic closure (say $\mathbb{C}$ or as Laurent Berger mentioned $\overline{\mathbb{Q}_p}$, or $\overline{\mathbb{F}_p}$ in our case), and then for any $a\not\in A$, there is a way to decide whether we can extend the valution to $A[a]$ or $A[a^{-1}]$, using the maps previously defined. To me, this method seems a little bit more explicit, which is actually not.
