Is every absolute value defined on $\Bbb R$ or $\Bbb C$ not differentiable at zero? An absolute value $|\cdot|$ of field $F$ is a function $F \to \Bbb R$ such that 

  
*
  
*$|x|≥0$ for any $x∈V$ and $|x|=0$ iff $x=0$
  
*$|xy|=|x||y|$
  
*$|x+y|≤|x|+|y|$ for any $x,y∈F$
  

An absolute value can be viewed as a norm on $F$. The absolute value of $\Bbb R$ and the modulus of $\Bbb C$ are both not differentiable at 0. I am wondering if all absolute values defined on subfields of $\Bbb C$ are not differentiable at zero.
 A: The answer to the question in the title is yes. More generally --

I am wondering if all absolute values defined on subfields of $\mathbb{C}$ are not differentiable at zero.

Yes, this is the case. Any subfield of $\mathbb{C}$ contains $\mathbb{Q}$. We show that as long as $| \cdot |$ is defined on $\mathbb{Q}$, it is not differentiable at $0$ in the following sense:
$$
\lim_{\substack{h \to 0 \\ h \in \mathbb{Q}}} \frac{|h| - |0|}{h - 0}
= \lim_{\substack{h \to 0 \\ h \in \mathbb{Q}}} \frac{|h|}{h} \tag{*}
$$
does not exist as a complex number. (That is to say, the limit above does not exist with respect to the usual topology on complex numbers.)
First, note that property (2) implies $|1|^2 = |1|$, so $|1| \in \{0,1\}$, and by property (1) $|1| = 1$. Then by property (2) $|-1|^2 = |1| = 1$ so $|-1| = 1$. By property (2) we conclude that $|-x| = |x|$ for all $x$.
Consider two sequences converging to $0$: $a_n = \frac{1}{2^n}$, and $b_n = \frac{-1}{2^n}$.
On the one hand,
$$
\frac{|a_n|}{a_n} = \frac{|1/2|^n}{(1/2)^n} = 2^n \left|\frac12\right|^n
= \left( \left|\frac12\right| + \left|\frac12\right| \right)^n 
\ge (|1|)^n = 1,
$$
by property (3). On the other hand,
$$
\frac{|b_n|}{b_n} = \frac{|1/2|^n}{-(1/2)^n} = -2^n \left|\frac12\right|^n
= -\left( \left|\frac12\right| + \left|\frac12\right| \right)^n 
\le -(|1|)^n = -1.
$$
We have two sequences which cannot possibly converge to the same number, so the limit in (*) does not exist.
