absolute value of norm of integral element

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

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