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.

  • $\begingroup$ Localize and use the fact that DVRs are maximal amongst domination. $\endgroup$ Aug 19, 2015 at 11:55
  • $\begingroup$ @AlexYoucis I am not sure what to localise and how that solve my question. $\endgroup$
    – CYC
    Aug 19, 2015 at 12:50

1 Answer 1


In non-archimedean case, absolute value corresponds to prime ideals.


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