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 $\mathbb Q$ ?
Asked
Active
Viewed 434 times
2
-
1By "the" valuation, presumably you mean with respect to a fixed rational prime? – John Martin Jan 06 '13 at 05:00
-
@John, that's true. – Rajesh Jan 06 '13 at 05:22
-
http://mathoverflow.net/questions/30581/extension-of-valuation – Jan 06 '13 at 12:17
-
See also http://math.stackexchange.com/questions/246627/ (restrict a $p$-adic valuation on $\mathbb C$ to $\overline{\mathbb Q}$). – Jan 06 '13 at 20:49
1 Answers
2
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 $\overline{Q}$.
Tony
- 6,718