I am trying to show that field of all algebraic reals over $\mathbb{Q}$ has infinite degree. I guess that $$1,\sqrt{2},\sqrt[3]{2}, \sqrt[4]{2}, ...$$

are lineary independent but can't prove it.


closed as off-topic by user26857, John B, Stefan Mesken, Silvia Ghinassi, zz20s Mar 10 '16 at 2:28

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  • 1
    $\begingroup$ The correct term in English is infinite degree, not infinite order. $\endgroup$ – KCd Feb 2 '14 at 2:00

Perhaps simpler:

$$\forall\,n\in\Bbb N\;,\;\;[\Bbb Q(\sqrt[n]2):\Bbb Q]=n$$

  • $\begingroup$ Why $\sqrt[n]{2}$ cant be the root of polynomial with degree less then $n$? $\endgroup$ – Ashot Feb 1 '14 at 14:40
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    $\begingroup$ @Ashot, because $\;x^n-2\in\Bbb Q[x]\;$ is irreducible $\;\forall\,2\le n\in\Bbb N\;$ , say by Eisenstein's Criterion. $\endgroup$ – DonAntonio Feb 1 '14 at 14:42

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