Let $(K,u)$ be a complete valued field, $u$ be its *discrete* absolute value (corresponds to a discrete valuation on $K$), then:

($\ast)$

Let $E/K$ is a finite separable field extension, then the absolute value $w$ of $E$ which extends $u$ is unique.

In the book *Algebraic Number Theory* by *Fröhlich, A., and Taylor, M.J.*, it first proves the above property, then gives an alternate proof as follows:

Let $\mathfrak o$ denotes the valuation ring in $K$ associates with $u$, $\mathfrak o_E$ the integral closure of $\mathfrak o$ in $E$,

then in order to prove ($\ast$), it suffices to show that $\mathfrak o_E$ has the unique prime ideal $\mathfrak B$ above $\mathfrak o$.($\mathfrak B$ is above $\mathfrak o$ means $\mathfrak B \cap \mathfrak o=\mathfrak p$ is an unique prime ideal in $\mathfrak o$.)

My question is, how the uniqueness of $\mathfrak B$ implies the uniqueness of extension $w$, why it suffices?

I know there exists a absolute value $w'$ of $E$ extends $u$ which gives $\mathfrak {B} =\{ x; w'(x)<1 \}$, but this only works for such specific $w'$, for an arbitrary extension $w$, not even being discrete, I am not sure how it gives a prime ideal in $\mathfrak o_E$ to obtain the uniquemess.