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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?

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Somewhat related:… – S.C. Aug 2 '11 at 14:41
up vote 17 down vote accepted

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.

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As a rule of thumb, the only easy way to find get units in cyclotomic fields is to play with ratios of the form $(\eta^a-1)/(\eta-1)$ where $a$ is relatively prime to the order of $\eta$, so it would be very surprising if an argument like this didn't work. – David Speyer Aug 2 '11 at 15:31
Thanks! I guess I should have tried to play with the expression a little more... – expmat Aug 2 '11 at 16:09
@David: how precise can that rule of thumb be made? Is it known for what polynomials $P$ it's true that $P(\zeta_n)$ is always a unit for $n$ not contained in a union of arithmetic progressions, or something like that? – Qiaochu Yuan Aug 3 '11 at 1:38
@Qiaochu: Great question! I don't know. – David Speyer Aug 3 '11 at 2:42

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).


(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 $1$.

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Sorry, but I don't understand your second paragraph... – darij grinberg Aug 2 '11 at 17:10
@darij: I have re-edited for clarity and details. – Tom Goodwillie Aug 2 '11 at 18:30
I still don't get it. Your first paragraph proves it for $r=2$, not for $r$ being even... but I assume that I am misunderstanding you on a more general scale. – darij grinberg Aug 3 '11 at 11:16
In the question $q$ and $r$ are primes. David Speyer answered it when neither of them is $2$. I answer it when one of them is $2$. I also point out that a related statement is true when neither of them is $2$ but false when one of them is $2$. – Tom Goodwillie Aug 3 '11 at 12:33
Ah! I thought your post was supposed to be independent of David's. Everything is clear now. – darij grinberg Aug 3 '11 at 14:17

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