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If $a$ is a transcendental number (i.e., a number s.t. there does not exist a polynomial $P(x)$ s.t. $P(a) = 0$), is $a^n$ also transcendental?

It would seem to me that it should be, but I can't figure out why. How would I prove this?

N. F. Taussig
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user2738036
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    Is $n$ supposed to be an integer? – Akiva Weinberger Nov 04 '15 at 12:24
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    You should specify if $n$ is supposed to be integer, rational, or something else. For example this if false if we can take $n$ real, e.g. if $a = e$ and $n = \log(2)$, then $a^n = 2$ even though $a$ is transcendental. – A.P. Nov 04 '15 at 15:08

2 Answers2

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If $p(a^n)=0$ for some polynomial $p(x)$, let $q(x)=p(x^n)$ and observe that $$q(a)=p(a^n)=0.$$ Hence, if $a^n$ is algebraic, so is $a$.

David Hill
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If $a^n$ isn't transcendental then $a=(b)^{1/n}$ for some nontranscendental number $b\in R$ thus $f(b)=\sum\limits_{i=0}^{k}c_i b^i=0$ for some polynomial equation where $c_i\in \mathbb{Q}$. So $g(a)=f(a^n)=\sum\limits_{i=0}^{k}c_i a^{ni}=0$. So thus $a$ isn't trancendental.

user160110
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