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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.)2011-04-18
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    What is meant by $\bar{\mathbb{Q}}_p$? Algebraic closure?2011-04-18
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    Yes, the overline above the notation for a field is a standard notation for algebraic closure of that field.2011-04-18
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    yes, it is algebraic closure2011-04-18

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