# An expression involving p-th root of unity

Let $\zeta_p$ be a $p$-th root of unity, where $p$ is an odd prime number.

I just came across the following expression:

$$\frac{(\zeta_p^2-\zeta_p+1)^3}{\zeta_p^2(\zeta_p-1)^2}.$$ Can we simplify this expression somehow? For $p=3$, I can rewrite this to $$\frac{8}{(\zeta_3-1)^2}.$$

Question. Can we simplify $$\frac{(\zeta_p^2-\zeta_p+1)^3}{\zeta_p^2(\zeta_p-1)^2} ?$$

Note that this expression does not change if we replace $\zeta_p$ by one of the elements in $\{\zeta_p, \zeta_p^{-1},1-\zeta_p, (1-\zeta_p)^{-1}, \zeta_p/(1-\zeta_p), (1-\zeta_p)/\zeta_p\}$.

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Honestly, the expression looks simple to me, particularly for large $p$. –  Tapu Oct 14 '11 at 12:18
The reason that you are successful in simplifying when $p=3$ is that the top factors as $$(\zeta_p-\zeta_6)(\zeta_p-\zeta_6^5)$$ where $\zeta_6$ is a primitive sixth root of unity; since $\zeta_6$ is $\zeta_3 + 1$ (for an appropriate choice of $\zeta_3$), you can simplify. I agree with Swapan that there doesn't seem to be any general reason why the expression would simplify for larger $p$. –  Barry Smith Oct 14 '11 at 12:55

There are at least a couple of profitable ways of re-writing this. A short triggy answer is that your expression simplifies to $$\frac{\left(2\cos\left(\frac{2\pi}{p}\right)-1\right)^3}{2\cos\left(\frac{2\pi}{p}\right)-2},$$ which for $p=3$ gives the value of $\boxed{\frac{8}{3}}$ you allude to above, and for $p=5$ gives the neatly random-looking value of

$$\frac{\left(\sqrt{5}-3\right)^3}{4(\sqrt{5} - 5)}=\frac{8\sqrt{5}-18}{\sqrt{5} - 5}=\boxed{\frac{2}{11\sqrt{5}+25}.}$$

For discussing the work that leads to the simplification, it's slightly more convenient from the view of algebraic number theory to deal with the reciprocal $$\xi_p:=\frac{\zeta_p^2(\zeta_p-1)^2}{(\zeta_p^2-\zeta_p+1)^3}.$$ First, it's not too hard to check that $\zeta_p^2-\zeta_p+1$ is a unit of $\mathbb{Z}[\zeta_p]$ for $p>3$, so writing it as $$\frac{\zeta_p^2}{(\zeta_p^2-\zeta_p+1)^3}\cdot (\zeta_p-1)^2$$ makes it clear that $\xi_p$ an algebraic integer (this is why I wanted the reciprocal), of asbolute norm $p^2$ (since $\zeta_p-1$ is a degree 1 prime above $p$ in $\mathbb{Q}(\zeta_p)$.) Second, let's take advantage of the fact that we know that $\xi_p$ is totally real, and so an element of $\mathbb{Z}[\zeta_p^+]$, where $\zeta_p^+:=\zeta_p+\zeta_p^{-1}.$ From the above re-writing, it's unreasonable to expect (actually, probably impossible) for $\xi_p$ to live in any proper subfield of $\mathbb{Q}(\zeta_p^+)$. So a reasonable interpretation for the problem of an ultimate simplication for $\xi_p$ is to write it completely in terms of $\zeta_p^+$. Let's do this now: $$\xi_p=\frac{\zeta_p^2(\zeta_p-1)^2}{(\zeta_p^2-\zeta_p+1)^3}=\frac{\zeta_p^3}{(\zeta_p^2-\zeta_p+1)^3}\cdot \frac{(\zeta_p-1)^2}{\zeta_p}=\frac{\zeta_p+\zeta_p^{-1}-2}{(\zeta_p+\zeta_p^{-1}-1)^3}=\boxed{\frac{\zeta_p^+-2}{(\zeta_p^+-1)^3}}$$ Now writing $\zeta_p^+=2\cos(2\pi/p)$ and reciprocating gives the formula at the top of this answer.

Finally, let me mention that from an algebraic number theory point of view, it might me most useful to write $\xi_p$ not in terms of $\zeta_p^+$, but in terms of a prime of $\mathbb{Z}[\zeta_p]$ above $p$, i.e., in terms of $$\mathfrak{p}:=(1-\zeta_p)(1-\zeta_p^{-1})=2-\zeta_p^+.$$ Re-writing the previous boxed expression, we finally conclude with the reasonably concise (and algebraically transparent) formulation $$\xi_p=\boxed{\frac{\mathfrak{p}}{\left(1-\mathfrak{p}\right)^3}.}$$

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