Valuation associated to a non-zero prime ideal of the ring of integers I have a question from Frohlich & Taylor's book 'Algebraic Number Theory', p.64. I will keep the notation used there. 
Let $K$ be a number field, $\mathcal o$ its ring of integers. Let $\mathfrak p$ be a non-zero prime ideal of $\mathcal o$ and $v=v_\mathfrak p$ the valuation of $K$ associated to $\mathfrak p$. Suppose that $\rho$ is a field automorphism of $K$ such that $v(x^\rho)=v(x)$ for all $x\in K$.

Why is this equivalent to the condition $\mathfrak p^\rho=\mathfrak p$?

Would it help to show that $(x\mathfrak o)^\rho=x^\rho\mathfrak o$ for all $x\in K$? - although I am not sure if this is even true. 
Thanking you in advance. 

$v(x)$ is the number $n$ such that $x \in \mathfrak{p}^n-\mathfrak{p}^{n+1}$ now $v(x)\geq 1$ iff $x \in \mathfrak{p}$.
So we see that if $v(x^{\rho})=v(x)$ then 
$$x^{\rho} \in \mathfrak{p} \Leftrightarrow x \in \mathfrak{p}$$
which means that $\mathfrak{p}^{\rho}=\mathfrak{p}$.
Conversely if $\mathfrak{p}^{\rho}=\mathfrak{p}$ then it is easy to see
$$x^{\rho} \in \mathfrak{p}^n-\mathfrak{p}^{n+1}\Leftrightarrow x \in \mathfrak{p}^n-\mathfrak{p}^{n+1}$$
and so $v(x^{\rho})=v(x)$.
