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consider $\mathbb{Q}\subset K$ a finite algebraic extension. Take $x\in K$ integral, why $\mid Norm_{K/\mathbb{Q}}(x)\mid \geq 1$?

Another question is: is it true that $\bar{\mathbb{Q}}_p \cong \mathbb{C}$? if it is so why?

Thank you.

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Using Zorn's lemma and transcendence degrees, it can be shown that an algebraically closed field that is uncountable is determined up to isomorphism as a field by its cardinality and characteristic. In particular, the alg. closure of Q_p has the same cardinality as the reals, as does C, and both fields are alg. closed with characteristic 0, so they are isomorphic. (The alg. closures of Q and Q(x) are both countable with characteristic 0 but are not isomorphic fields, so the uncountability condition is somewhat necessary.) –  KCd Apr 18 '11 at 10:53
What is meant by $\bar{\mathbb{Q}}_p$? Algebraic closure? –  quanta Apr 18 '11 at 10:56
Yes, the overline above the notation for a field is a standard notation for algebraic closure of that field. –  KCd Apr 18 '11 at 11:24
yes, it is algebraic closure –  user9730 Apr 18 '11 at 12:31
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1 Answer

The norm of an integer is a rational integer.

$\mathbb Q_p$ cannot be extended to $\mathbb C$ because it has a different metric.

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Your second sentence is wrong: ukn is asking about the algebraic closure of Q_p and that field is isomorphic to the complex numbers. –  KCd Apr 18 '11 at 10:51
@KCd, isomorphic as a field but not as a metric space? –  quanta Apr 18 '11 at 10:54
The p-adic abs. value on the rational numbers can be extended to the complex numbers using Zorn's lemma (but it is completely nonconstructive) and Q_p can be embedded into C by Zorn's lemma. –  KCd Apr 18 '11 at 10:56
Of course they are not isomorphic as metric spaces (one is connected in the induced metric topology and the other is not) but I suspect ukn is asking about an isomorphism as fields, not as metrized fields. In any case, it is definitely worthwhile coming to grips with the fact that the alg. closure of Q_p is isomorphic to C as abstract fields. –  KCd Apr 18 '11 at 10:58
To expand a little on quanta's answer. $x\in K$ integral means $x$ is a root of a monic polynomial with integer coefficients. Its norm is the product of its conjugates, which is (up to sign) the constant term of said monic polynomial. –  Gerry Myerson Apr 18 '11 at 13:57
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