# 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.

• @DietrichBurde Sorry, I am new to this. I saw that an algebraic integer is a root of some monic polynomial with $\mathbb{Z}$ coefficients, but $i/p$ is not an integer, why is it an algebraic integer? I also don't get what you mean by "$\overline{Q}$". Can you explain it more? Thank you. – Eric Sun May 27 at 9:24
• Which textbook, please? – Gerry Myerson May 27 at 9:39
• Eric, I thought you meant algebraic numbers $\overline{\Bbb {Q}}$. For algebraic integers see for example this question. – Dietrich Burde May 27 at 9:41
• @GerryMyerson A Conversational Introduction to Algebraic Number Theory written by Paul Pollack – Eric Sun May 27 at 9:48
• @DietrichBurde The method in that link does work I think. It is not just a finite sum. $p$ can be arbitrarily large. – Eric Sun May 27 at 9:50

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.

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:

1. 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 $$Q(\alpha) = 0$$.
2. Any quantity of the form $$\sqrt[a]{b}$$ for $$a$$ and $$b$$ integers is an algebraic integer: it is the root of $$X^a - b$$.
3. 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.