How many absolute values are there? My question is the following: Are there algebraic norms on the fields $\mathbb{R}, \mathbf{Q_p}$ ''other'' than the absolute value, respectively $|\cdot|_p$?
Now phrasing more precisely: If generally $F$ is a field, an algebraic norm is a map $|\cdot| : K \to [0, \infty)$ such that
1) $|x| = 0 \iff x = 0$
2) $|xy| = |x||y| \forall x,y \in F$
3) $|x + y| \leq |x| + |y| \forall x,y \in F$
Two such norms $|\cdot|, ||\cdot||$ are called equivalent if and only if they generate the same topology. One can show that this is the case iff. there is an $s > 0$ such that $|x| = ||x||^s$ for all $x$. The question is: On the fields $\mathbb{R}, \mathbf{Q_p}$ where the latter one means the p-adic numbers, are there algebraic norms that are not equivalent to the usual absolute value, respectively the p-adic norm?
Of course, every such norm induces a norm on $\mathbb{Q}$, so restricted to $\mathbb{Q}$ it must be either trivial or equivalent to one of the norms mentioned above by the Thm. of Ostrowski. The problem is that i was unable to extend that to the whole field.
Thanks in advance
Fabian Werner
 A: It is known that $\mathbb C$ is isomorphic to $\mathbb C_p$ (completion of $({\mathbb Q}_p)^{\rm{alg}}$) as fields. See e.g. Theorem 5 in this paper. So $\mathbb R$ inherts an ultrametric absolute value from that of $\mathbb C_p$. Conversely, $\mathbb Q_p$ inherts from $\mathbb C$ an archimedian absolute value. 
A: The $p$-adic norm on $\mathbb Q$ can be extended to  $\mathbb R$.
We can consider pairs $(F,|\cdot|_F)$ where $F$ is a field with $\mathbb Q\subseteq F\subseteq \mathbb R$ and $|\cdot|_F$ is a norm on $F$ that extends $|\cdot |_p$.
The set $\mathcal F$ of such pairs is inductively ordered: 
We can define $(F,|\cdot|_F)\le (E,|\cdot|_E)$ if $F\subseteq E$ and $|\cdot|_F$ is the restriction of $|\cdot|_E$ to $F$.
If a subset $\mathcal T\subseteq \mathcal F$ is totally ordered, then we can let $L=\bigcup_{(F,|\cdot|_F)\in\mathcal T}F$ and define $|\cdot|_L\colon L\to[0,\infty)$ by letting $|x|_L:=|x|_F$ where $(F,|\cdot|_F)\in \mathcal T$ with $x\in F$. 
Then $(F,|\cdot|_F)\le (L,|\cdot|_L)$ for all $(F,|\cdot|_F)\in\mathcal T$, that is every totally ordered subset of $\mathcal F$ has an upper bound in $\mathcal F$. By Zorn's lemma, $\mathcal F$ contains a maximal element $(M,|\cdot|_M)$.
Assume there exists $\alpha\in\mathbb R\setminus M$.
Then we can define $|\cdot|$ on $M(\alpha)$:
Case (i): $\alpha$ is transcendental.
Every nonzero element $\beta\in M(\alpha)$ can be written as $\beta=\alpha^k\frac{f(\alpha)}{g(\alpha)}$ with $k\in\mathbb Z$ and $f,g\in M[X]$ with $f(0)\ne 0, g(0)=1$. Set $|\beta|:= |f(0)|_M$. One quickly verifies that this is a norm on $M(\alpha)$. 
Case (ii): $\alpha$ is algebraic with minimal polynomial $p\in M[X]$. If $p(X)=a_0+a_1X+\cdots + a_{n-1} X^{n-1}+X^n$, select $v>0$ such that $v^n=\max_{k<n}\{v^k|a_k|_M\}$.
Then define $|\sum_{k=0}^{n-1} c_k\alpha^k| = \max_{k<n}\{v^k|c_k|_M\}$.
If I'm not mistaken, this is a norm on $M(\alpha)$.
In both cases we can extend $(M,|\cdot|_M)$, contradicting the maximality of $(M,|\cdot|_M)$.
We conclude that $M=\mathbb R$.
