Would anyone be able to give me an outline or a hint towards the proof that the field of algebraic numbers is an infinite extension of the field of rationals? Many Thanks
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3If you consider the field as an vector space over $\mathbb Q$, if the vector space was finite-dimensional of dimension $n$, that would mean that every algebraic number satisfies a rational polynomial of degree $n$. So you just need to show that for any $n$ there is an irreducible rational polynomial of degree $>n$. – Thomas Andrews Mar 21 '13 at 16:59
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See also http://math.stackexchange.com/questions/151586/infinite-degree-algebraic-field-extensions. – lhf Nov 14 '13 at 11:43
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If you know Eisenstein's criterion, you might consider the algebraic numbers $\sqrt[n]{2}$, for which one has $$ \lvert \Bbb{Q} [\sqrt[n]{2}] : \Bbb{Q} \rvert = n, $$ since the minimal polynomial of $\sqrt[n]{2}$ over $\Bbb{Q}$ is $x^n - 2$, as the latter is irreducible over $\Bbb{Q}$ by Eisenstein.
Since $n \ge 1$ is arbitrary, this shows that the degree of the field of the algebraic numbers over the rationals cannot be finite.
Andreas Caranti
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+1 About accepting it I'd say the OP can still wait a little, since perhaps some other answer(s) may come that (s)he will love better, but it is always nice to upvote any answer that helps even a little... – DonAntonio Mar 21 '13 at 17:09
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Anyway @AndreasCaranti, I'd remind the OP later...many times either they don't know or they forget. – DonAntonio Mar 21 '13 at 17:13