TITLE: How to prove $\sum_{i=1}^{p-1}\sqrt[p]{\frac{i}{p}} $ is an algebraic integer?
QUESTION [6 upvotes]: When I read an introductory algebraic number theory textbook, there is a problem like this:
If $p$ is an odd prime, prove  $\displaystyle\sum_{i=1}^{p-1}\sqrt[p]{\frac{i}{p}} $ is an algebriac integer.
Note that it is in the first chapter, so I think that it does not require much knowledge, but I have no idea how to solve it.

REPLY [7 votes]: It is enough to show that
$$
\alpha = \sqrt[p]{\frac{i}{p}} + \sqrt[p]{\frac{p-i}{p}}
$$
is an algebraic integer for $1\leq i\leq (p-1)/2$, since the original sum is sum of such numbers. We have
$$
\alpha^{p} = \frac{(\sqrt[p]{i} + \sqrt[p]{p-i})^{p}}{p} = \frac{1}{p} \left(i + (p-i) +  \sum_{k=1}^{p-1} \binom{p}{k}i^{k/p}(p-i)^{(p-k)/p}\right) \\= 1 + \sum_{k=1}^{p} \frac{1}{p}\binom{p}{k}i^{k/p}(p-i)^{(p-k)/p}.
$$
Since $\binom{p}{k}$ is a multiple of $p$ for $1\leq k\leq p-1$ and $i^{k/p}, (p-i)^{(p-k)/p}$ are obviously algebraic integers, so is $\alpha^{p}$. So $f(\alpha^{p}) = 0$ for some monic $f(x) \in \mathbb{Z}[x]$, which implies that $g(\alpha) = 0$ for $g(x) = f(x^{p})$, which is clearly monic. Thus $\alpha$ is an algebraic integer.