How to prove $\sum_{i=1}^{p-1}\sqrt[p]{\frac{i}{p}} $ is an algebraic integer? 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.
 A: 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. 
A: To show this sum (call it $S$) is an algebraic integer, you want to find a monic polynomial that makes this sum vanish. You can use a few facts about algebraic integers:


*

*If for some integer $k$, a number $\alpha^k$ is an algebraic integer, then $\alpha$ is an algebraic integer: take $P$ monic with integer coefficient such that $P(\alpha^k) = 0$, then $Q = P \circ X^k$ is also monic, with integer coefficients, such that $Q(\alpha) = 0$.

*Any quantity of the form $\sqrt[a]{b}$ for $a$ and $b$ integers is an algebraic integer: it is the root of $X^a - b$.

*The sum of two algebraic integers is an algebraic integer. You will probably find a proof of this fact in your textbook, the main ingredient is that a number $\alpha$ is an algebraic integer if and only if $\mathbb{Z}[\alpha]$ is a finitely generated $\mathbb{Z}$-module, so that $\mathbb{Z}[\alpha + \beta]$ is a submodule of $\mathbb{Z}[\alpha][\beta]$, which is finitely generated, so that $\mathbb{Z}[\alpha + \beta]$ is finitely generated, hence $\alpha + \beta$ is an algebraic integer.


Now, to show $S$ is an algebraic integer, it is enough to show that S so a suitable power is a sum of known algebraic integers. This suitable power is obviously $p$: \begin{align}
S^p &= \left(\sum\limits_{k=1}^{p-1}\sqrt[p]{\frac{k}{p}}\right)^p\\
&= \sum\limits_{i_1 + \cdots + i_{p-1} = p}\dbinom{p}{i_1,\cdots,i_{p-1}}\prod\limits_{j=1}^{p-1}\left(\sqrt[p]{\frac{j}{p}}\right)^{i_j}\\
&= \sum\limits_{i_1 + \cdots + i_{p-1} = p}\dbinom{p}{i_1,\cdots,i_{p-1}}\left(\sqrt[p]{\prod\limits_{j=1}^{p-1}\frac{j^{i_j}}{p^{i_j}}}\right)\\
&= \sum\limits_{i_1 + \cdots + i_{p-1} = p}\dbinom{p}{i_1,\cdots,i_{p-1}}\left(\sqrt[p]{\frac{\prod\limits_{j=1}^{p-1}j^{i_j}}{p^{\sum\limits_{j=1}^{p-1}i_j}}}\right)\\
&= \sum\limits_{i_1 + \cdots + i_{p-1} = p}\dbinom{p}{i_1,\cdots,i_{p-1}}\sqrt[p]{\frac{\prod\limits_{j=1}^{p-1}j^{i_j}}{p^{p}}}\\
&= \sum\limits_{i_1 + \cdots + i_{p-1} = p}\frac{1}{p}\dbinom{p}{i_1,\cdots,i_{p-1}}\sqrt[p]{\prod\limits_{j=1}^{p-1}j^{i_j}}\\
\end{align}
But $p$ divides $\dbinom{p}{i_1,\cdots,i_{p-1}}$ if there is no $i_j$ equal to $p$, indeed $\dbinom{p}{i_1,\cdots,i_{p-1}} = \frac{p!}{i_1!i_2!\cdots i_{p-1}!}$, looking at $p$-valuations, if no $i_j$ is $p$, then $p$ divides this coefficient.
For the remaining coefficients, one has the sum \begin{align}
\sum\limits_{j=1}^{p-1}\frac{j}{p} &= \frac{p(p-1)}{2p}\\
&= \frac{p-1}{2}
\end{align} which is an integer because $p$ is odd.
Thus the $p$-th power of S is the sum of agebraic integers. Hence an algebraic integer, by the first fact listed above, $S$ is an algebraic integer. 
