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How do we define the valuation over the algebraically closed field of rational numbers say $\bar{\mathbb Q}$ as an extension of the valuation of $\mathbb Q$ ?

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By "the" valuation, presumably you mean with respect to a fixed rational prime? –  John Martin Jan 6 '13 at 5:00
    
@John, that's true. –  Rajesh Jan 6 '13 at 5:22
    
mathoverflow.net/questions/30581/extension-of-valuation –  user26857 Jan 6 '13 at 12:17
    
See also math.stackexchange.com/questions/246627 (restrict a $p$-adic valuation on $\mathbb C$ to $\overline{\mathbb Q}$). –  user18119 Jan 6 '13 at 20:49

1 Answer 1

up vote 2 down vote accepted

For any finite Galois extension $K/\mathbb{Q}_p$, there is a unique extension of the norm that respects the $p$-adic norm on $\mathbb{Q}_p$, and this is Galois-invariant. Therefore, it must be given by $|x|_K = |Norm(x)|_p^{1/[K:\mathbb{Q}_p]}$. By uniqueness, if we have a tower of field extensions $L/K/\mathbb{Q}_p$, then restricting the norm on $|\cdot |_L$ to $K$ gives $|\cdot |_K$. Since any element of $\overline{Q}$ lives in a finite Galois extension of $\mathbb{Q}_p$, this gives a way to extend the norm to all of $\overline{Q}$.

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