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If $\alpha \in \mathbb{C},$ algebraic (over $\mathbb{Q}$) and $ \ \beta \in \mathbb{N} \ $then is $ \sqrt[\beta]{\alpha}$ algebraic?

This is my attempt at a proof:

Given that $\alpha$ is algebraic then there is $p \in \mathbb{Q}[x]: p(\alpha)=0$

Let $p=\sum\limits_{n=0}^\infty a_nx^n,$

Then if $q=\sum\limits_{n=0}^\infty a_nx^{n\beta}, \ q\in \mathbb{Q}[x] \ $and:

$q(\sqrt[\beta]{\alpha})=\sum\limits_{n=0}^\infty a_n(\sqrt[\beta]{\alpha})^{n\beta}=\sum\limits_{n=0}^\infty a_n(\alpha)^n=p(\alpha)=0 $

Which means $\sqrt[\beta]{\alpha}$ is algebraic.

Is this correct? Is there a easier way of demonstrating this result?

King Ghidorah
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    Looks good. The one thing I might add is to note how $p(x) \ne 0$ implies $q(x) \ne 0$. – aschepler Apr 20 '15 at 20:11
  • All elementary algebraic operation with an algebraic element gives another one because we are working in a field – Piquito Apr 21 '15 at 13:25

1 Answers1

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If $\alpha$ is algebraic over $\mathbb Q$, then $\mathbb Q(\sqrt[\beta]{\alpha})$ is an algebraic (even finite) extension of $\mathbb Q(\alpha)$ since $\sqrt[\beta]{\alpha}$ is a root of the polynomial $X^{\beta} - \alpha \in \mathbb Q(\alpha)[X]$. Therefore the extension $\mathbb Q(\sqrt[\beta]{\alpha})$ is algebraic over $\mathbb Q$, which means $\sqrt[\beta]{\alpha}$ is algebraic over $\mathbb Q$.

The "behind the scenes" of my argument is your argument. But my argument uses the general theory so you get your hands less dirty in the process ("getting your hands dirty" meaning "writing down polynomials explicitly").

Hope that helps,

Patrick Da Silva
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