I am now reading Lang's "Algebraic number theory" (http://gen.lib.rus.ec/book/index.php?md5=9D32C9B248831979ADE79FACDC40129B) and I am having problems with understanding the statement of the theorem 2 of the $\S$1 of Chapter 2 on page 38. It states:

Let $K$ be a number field, $v$ one of its canonical absolute values, $E$ a finite extension of $K$. Two embeddings $\sigma$, $\tau$: $E \to \overline{K}_v$ over $K$ give rise to the same absolute value on E if and only if they are conjugate over $K_v$. (By conjugate over $K_v$ we mean that there exists an isomorphism $\lambda$ of $\sigma E \cdot K_v$ onto $\tau E \cdot K_v$ which is the indentity on $K_v$.)

(Notations: $K_v$ is the comletion of $K$ with respect to the norm $v$, $\overline K$ is an algebraic closure of $K$)

What is unclear to me is how two embeddings can give rise to different absolute values on E if the continuation of the absolute value from $K$ to $E$ is unique. Perhaps I do not understand the meaning of the word "absolute value" in a right way. Could you please help me?

  • 2
    $\begingroup$ The continuation of the absolute value from $K_v$ (a complete field) to any of its finite extensions is unique, but definitely not from $K$ to any of its finite extensions, in general. For example, take $K = \mathbf Q$, $E = \mathbf Q(i)$ and $v$ the 5-adic absolute value on $\mathbf Q$. The absolute value $v$ on $K$ has two different extensions to $E$, corresponding to the two different embeddings of $E$ into $K_v = \mathbf Q_5$. $\endgroup$ – KCd Apr 3 '15 at 12:27
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    $\begingroup$ The two different extensions of the 5-adic absolute value from $\mathbf Q$ to $\mathbf Q(i)$ are the $(1+2i)$-adic and $(1-2i)$-adic absolute values, which correspond to the two different primes in the factorization $5 = (1+2i)(1-2i)$. More generally, the extensions of a $\mathfrak p$-adic absolute value on a number field $K$ to a finite extension $E$ correspond to each of the prime ideals in $\mathcal O_E$ appearing in the prime factorization of $\mathfrak p\mathcal O_E$. $\endgroup$ – KCd Apr 3 '15 at 12:35

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