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Is that true that an arbitrary algebraic number raised to a power is again algebraic? That's false, there are many examples to found. Now, is that true that an algebraic number raised to a rational power is necessarily algebraic?

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The first question in answered by the Gelfond–Schneider theorem:

If $a$ and $b$ are algebraic with $a \ne 0,1$ and $b$ irrational, then $a^b$ is transcendental.

In particular, $2^{\sqrt 2}$ is transcendental.

For the second question, if $a$ is algebraic, then $b=a^{m/n}$ is algebraic because it is a root of $x^n-a^m=0$, an equation with algebraic coefficients.

Indeed, $[\mathbb Q(b):\mathbb Q] = [\mathbb Q(b):\mathbb Q(a)] \ [\mathbb Q(a):\mathbb Q] < \infty$, since both factors are finite.

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  • $\begingroup$ Yes, we don´t know explicitily wich is the polinomyal that a satisfies. $\endgroup$ – Julio May 28 '14 at 11:44
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The answer is yes, a rational power of an algebraic number is again algebraic:

Let $\alpha$ be any algebraic number, meaning that there exists $f(X)\in\Bbb{Q}[X]$ such that $f(\alpha)=0$, and let $q=\tfrac{a}{b}\in\Bbb{Q}$ be any rational number. Then $\alpha^{\tfrac{1}{b}}$ is a root of $f(X^b)$, so it is an algebraic number. It follows that $\alpha^q=\alpha^{\tfrac{a}{b}}=(\alpha^{\tfrac{1}{b}})^a$ is also an algebraic number.

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