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Ian Montague
Ian Montague
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For the following, fix $m\geq 2$, and let $a_{0},\dots,a_{m-1}\in\mathbb{Z}$ be such that $\sum_{k=0}^{m-1}a_{k}=0$. I would like to find the exact conditions on the $a_{k}$ so that the following equation holds:

$$\prod_{k=0}^{m-1}(1+\zeta_{m}^{k})^{a_{k}}=1,\qquad (**)$$

where $\zeta_{m}=e^{2\pi i/m}$. In other words, I would like to find a set of relations, say $R$, among $\{a_{k}\}$ so that $(**)$ holds if and only if $R$ holds.

Here's what I was able to deduce so far:

  • For $m$ even, we must have that $a_{m/2}=0$, since otherwise the above product must be zero.
  • For $m$ an odd prime, I was able to deduce that we must have $a_{0}=0$ by noting that the norms of both sides of $(**)$ with respect to the field extension $\mathbb{Q}(\zeta_{m})/\mathbb{Q}$ must equal $1$.
  • For general $m$, one can rewrite $(**)$ as follows:

$$1=\prod_{k=0}^{m-1}(1+\zeta_{m}^{k})^{a_{k}}=2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(1+\zeta_{m}^{k})^{a_{k}}(1+\zeta_{m}^{-k})^{a_{m-k}}$$

$$=2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}((\zeta_{2m}^{k})^{a_{k}}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}})((\zeta_{2m}^{-k})^{a_{m-k}}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{m-k}})$$ $$=\Big(\zeta_{2m}^{\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})}\Big)\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}+a_{m-k}}\Big)=:A\cdot B.$$

Since $1$ and $B$ are invariant under complex conjugation, $A$ must be as well, and hence

$$\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})\equiv 0\pmod{m}.$$

Furthermore,

$$B=\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}+a_{m-k}}\Big)=\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(2\cos(\frac{k\pi}{m}))^{a_{k}+a_{m-k}}\Big)>0$$

since $0<\cos(\frac{k\pi}{m})<1$ for $1\le k\le \lfloor\frac{m-1}{2}\rfloor$, and so $A>0$, as well. Therefore

$$\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})\equiv 0\pmod{2m}.$$

I've been able to find the exact relations $R$ for $m=2,3,4,5,6$ via ad-hoc means, but I haven't been able to find them for general $m$. Is there an alternative approach I can use here, perhaps by utilizing the invariance of the product in (**) under $\text{Gal}(\mathbb{Q}(\zeta_{m})/\mathbb{Q})$? Any help would be appreciated, even for $m$ an odd prime. Thanks!

Edit: For transparency, here's a related question I posted to MSE: https://math.stackexchange.com/questions/4433781/identities-in-maximal-real-subfields-of-cyclotomic-fields

For the following, fix $m\geq 2$, and let $a_{0},\dots,a_{m-1}\in\mathbb{Z}$ be such that $\sum_{k=0}^{m-1}a_{k}=0$. I would like to find the exact conditions on the $a_{k}$ so that the following equation holds:

$$\prod_{k=0}^{m-1}(1+\zeta_{m}^{k})^{a_{k}}=1,\qquad (**)$$

where $\zeta_{m}=e^{2\pi i/m}$. In other words, I would like to find a set of relations, say $R$, among $\{a_{k}\}$ so that $(**)$ holds if and only if $R$ holds.

Here's what I was able to deduce so far:

  • For $m$ even, we must have that $a_{m/2}=0$, since otherwise the above product must be zero.
  • For $m$ an odd prime, I was able to deduce that we must have $a_{0}=0$ by noting that the norms of both sides of $(**)$ with respect to the field extension $\mathbb{Q}(\zeta_{m})/\mathbb{Q}$ must equal $1$.
  • For general $m$, one can rewrite $(**)$ as follows:

$$1=\prod_{k=0}^{m-1}(1+\zeta_{m}^{k})^{a_{k}}=2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(1+\zeta_{m}^{k})^{a_{k}}(1+\zeta_{m}^{-k})^{a_{m-k}}$$

$$=2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}((\zeta_{2m}^{k})^{a_{k}}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}})((\zeta_{2m}^{-k})^{a_{m-k}}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{m-k}})$$ $$=\Big(\zeta_{2m}^{\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})}\Big)\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}+a_{m-k}}\Big)=:A\cdot B.$$

Since $1$ and $B$ are invariant under complex conjugation, $A$ must be as well, and hence

$$\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})\equiv 0\pmod{m}.$$

Furthermore,

$$B=\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}+a_{m-k}}\Big)=\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(2\cos(\frac{k\pi}{m}))^{a_{k}+a_{m-k}}\Big)>0$$

since $0<\cos(\frac{k\pi}{m})<1$ for $1\le k\le \lfloor\frac{m-1}{2}\rfloor$, and so $A>0$, as well. Therefore

$$\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})\equiv 0\pmod{2m}.$$

I've been able to find the exact relations $R$ for $m=2,3,4,5,6$ via ad-hoc means, but I haven't been able to find them for general $m$. Is there an alternative approach I can use here, perhaps by utilizing the invariance of the product in (**) under $\text{Gal}(\mathbb{Q}(\zeta_{m})/\mathbb{Q})$? Any help would be appreciated, even for $m$ an odd prime. Thanks!

For the following, fix $m\geq 2$, and let $a_{0},\dots,a_{m-1}\in\mathbb{Z}$ be such that $\sum_{k=0}^{m-1}a_{k}=0$. I would like to find the exact conditions on the $a_{k}$ so that the following equation holds:

$$\prod_{k=0}^{m-1}(1+\zeta_{m}^{k})^{a_{k}}=1,\qquad (**)$$

where $\zeta_{m}=e^{2\pi i/m}$. In other words, I would like to find a set of relations, say $R$, among $\{a_{k}\}$ so that $(**)$ holds if and only if $R$ holds.

Here's what I was able to deduce so far:

  • For $m$ even, we must have that $a_{m/2}=0$, since otherwise the above product must be zero.
  • For $m$ an odd prime, I was able to deduce that we must have $a_{0}=0$ by noting that the norms of both sides of $(**)$ with respect to the field extension $\mathbb{Q}(\zeta_{m})/\mathbb{Q}$ must equal $1$.
  • For general $m$, one can rewrite $(**)$ as follows:

$$1=\prod_{k=0}^{m-1}(1+\zeta_{m}^{k})^{a_{k}}=2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(1+\zeta_{m}^{k})^{a_{k}}(1+\zeta_{m}^{-k})^{a_{m-k}}$$

$$=2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}((\zeta_{2m}^{k})^{a_{k}}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}})((\zeta_{2m}^{-k})^{a_{m-k}}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{m-k}})$$ $$=\Big(\zeta_{2m}^{\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})}\Big)\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}+a_{m-k}}\Big)=:A\cdot B.$$

Since $1$ and $B$ are invariant under complex conjugation, $A$ must be as well, and hence

$$\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})\equiv 0\pmod{m}.$$

Furthermore,

$$B=\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}+a_{m-k}}\Big)=\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(2\cos(\frac{k\pi}{m}))^{a_{k}+a_{m-k}}\Big)>0$$

since $0<\cos(\frac{k\pi}{m})<1$ for $1\le k\le \lfloor\frac{m-1}{2}\rfloor$, and so $A>0$, as well. Therefore

$$\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})\equiv 0\pmod{2m}.$$

I've been able to find the exact relations $R$ for $m=2,3,4,5,6$ via ad-hoc means, but I haven't been able to find them for general $m$. Is there an alternative approach I can use here, perhaps by utilizing the invariance of the product in (**) under $\text{Gal}(\mathbb{Q}(\zeta_{m})/\mathbb{Q})$? Any help would be appreciated, even for $m$ an odd prime. Thanks!

Edit: For transparency, here's a related question I posted to MSE: https://math.stackexchange.com/questions/4433781/identities-in-maximal-real-subfields-of-cyclotomic-fields

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Ian Montague
Ian Montague
  • 299
  • 1
  • 7

Finding when a certain product in a cyclotomic field is equal to one

For the following, fix $m\geq 2$, and let $a_{0},\dots,a_{m-1}\in\mathbb{Z}$ be such that $\sum_{k=0}^{m-1}a_{k}=0$. I would like to find the exact conditions on the $a_{k}$ so that the following equation holds:

$$\prod_{k=0}^{m-1}(1+\zeta_{m}^{k})^{a_{k}}=1,\qquad (**)$$

where $\zeta_{m}=e^{2\pi i/m}$. In other words, I would like to find a set of relations, say $R$, among $\{a_{k}\}$ so that $(**)$ holds if and only if $R$ holds.

Here's what I was able to deduce so far:

  • For $m$ even, we must have that $a_{m/2}=0$, since otherwise the above product must be zero.
  • For $m$ an odd prime, I was able to deduce that we must have $a_{0}=0$ by noting that the norms of both sides of $(**)$ with respect to the field extension $\mathbb{Q}(\zeta_{m})/\mathbb{Q}$ must equal $1$.
  • For general $m$, one can rewrite $(**)$ as follows:

$$1=\prod_{k=0}^{m-1}(1+\zeta_{m}^{k})^{a_{k}}=2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(1+\zeta_{m}^{k})^{a_{k}}(1+\zeta_{m}^{-k})^{a_{m-k}}$$

$$=2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}((\zeta_{2m}^{k})^{a_{k}}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}})((\zeta_{2m}^{-k})^{a_{m-k}}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{m-k}})$$ $$=\Big(\zeta_{2m}^{\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})}\Big)\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}+a_{m-k}}\Big)=:A\cdot B.$$

Since $1$ and $B$ are invariant under complex conjugation, $A$ must be as well, and hence

$$\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})\equiv 0\pmod{m}.$$

Furthermore,

$$B=\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(\zeta_{2m}^{k}+\zeta_{2m}^{-k})^{a_{k}+a_{m-k}}\Big)=\Big(2^{a_{0}}\prod_{k=0}^{\lfloor\frac{m-1}{2}\rfloor}(2\cos(\frac{k\pi}{m}))^{a_{k}+a_{m-k}}\Big)>0$$

since $0<\cos(\frac{k\pi}{m})<1$ for $1\le k\le \lfloor\frac{m-1}{2}\rfloor$, and so $A>0$, as well. Therefore

$$\sum_{k=1}^{\lfloor\frac{m-1}{2}\rfloor}k(a_{k}-a_{m-k})\equiv 0\pmod{2m}.$$

I've been able to find the exact relations $R$ for $m=2,3,4,5,6$ via ad-hoc means, but I haven't been able to find them for general $m$. Is there an alternative approach I can use here, perhaps by utilizing the invariance of the product in (**) under $\text{Gal}(\mathbb{Q}(\zeta_{m})/\mathbb{Q})$? Any help would be appreciated, even for $m$ an odd prime. Thanks!