Question

Let $q$ and $r$ be distinct prime numbers. I noticed (computing a few cases) that $\zeta_{2q} + \zeta_{2q}^{-1} + \zeta_{2r} + \zeta_{2r}^{-1}$ is a unit (in $\mathbb{Z}[\zeta_{2qr}]$, say). Is this always true? Why is that?

Answer 1

I assume you want $q$ and $r$ to be odd primes. Also, note that I will be using the notation that $\zeta_m$ means an arbitrary primitive $m$-th root of unity (but the same one every time it appears in an equation), and will be proving the statement in that generality.

Lemma: For any odd $m>1$ and any $\zeta_m$, the number $\zeta_m+1$ is a unit.

Proof: Let $r$ be such that $m | 2^r-1$. We'll abbreviate $\zeta_m$ to $\zeta$.

Then $\zeta^{2^r} = \zeta$ so $$1 = \left( \frac{\zeta^{2}-1}{\zeta -1} \right) \left( \frac{\zeta^{4}-1}{\zeta^{2} -1} \right) \cdots \left( \frac{\zeta^{2^r}-1}{\zeta^{2^{r-1}} -1} \right)=$$ $$\left( \zeta+1 \right) \left( \zeta^{2} + 1 \right) \cdots \left( \zeta^{2^{r-1}} +1 \right),$$ exhibiting an explicit inverse for $\zeta+1$.

Let $\eta$ be a primitive $2qr$ root of unity. Then your proposed unit is $\eta^{r}+\eta^{-r} + \eta^q + \eta^{-q}$ and factors as $$\eta^r (1+\eta^{q-r})(1+\eta^{-q-r}).$$ Since $q$ and $r$ are odd and relatively prime, $\eta^{q-r}$ and $\eta^{q+r}$ are primitive $qr$-th roots of unity and we are done by the lemma.

Answer 2

The answer is also yes if one of the primes, say $r$, is $2$, because then $\zeta_{2r}+\zeta_{2r}^{-1}=0$ and $\zeta_{2q}+\zeta_{2q}^{-1}=\zeta_{2q}(1+\zeta_{2q}^{-2})$ is a unit (as $\zeta_{2q}^{-2}$ is a primitive $q$th root of $1$ and $q$ is an odd prime).

Edit:

(1) Note that if both primes are odd then $\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1}$ is also a unit. Indeed, $-\zeta_q$ is a primitive $2q$th root of $1$ (this relies on $q$ being odd), so let's call it $\zeta_{2q}$, and likewise for $r$. Then $-(\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1})=\zeta_{2q}+\zeta_{2q}^{-1}+\zeta_{2r}+\zeta_{2r}^{-1}$, and we know that the latter is a unit.

(2) Note also that if one of the primes, say $r$, is $2$ then $\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1}$ is not a unit: it is equal to $\zeta_q+\zeta_q^{-1}-2$, so it is a unit times the square of $\zeta_q-1$, but the latter (unlike $\zeta_q +1$) is not a unit because it goes to $0$ under the unique ring homomorphism $\mathbb Z [\zeta_q]\to \mathbb Z/q$, which takes $\zeta_q$ to $1$.