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For $n$ a positive integer, let us write $\zeta_n = e^\frac{2 \pi i}{n}$, a primitive $n$th root of unity. It is clear that, if $m$ divides $n$, then we have an inclusion of cyclotomic fields $$ \mathbb{Q}(\zeta_m) \subseteq \mathbb{Q}(\zeta_n).$$ On the other hand, these inclusions are not always strict. For example, since $\zeta_3 = \frac{1+i \sqrt{3}}{2}$ and $\zeta_6 = \frac{-1 + i \sqrt{3}}{2}$, we have $$\mathbb{Q}(\zeta_3) = \mathbb{Q}(\zeta_6) = \mathbb{Q}(i\sqrt{3}).$$ Does this sort of thing happen infinitely often, or are there just some coincidences among small numbers? If this continues, is there some way to know when a particular inclusion of cyclotomic fields is strict?

Mike F
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  • Related: https://math.stackexchange.com/questions/4749611/prove-that-mathbb-q-cos-tfrac-pi7-neq-mathbb-q-cos-tfrac-pi9 – mr_e_man Aug 01 '24 at 02:39

1 Answers1

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The equality $\mathbb{Q}(\zeta_m)=\mathbb{Q}(\zeta_n)$ holds if and only if (WLOG) $m$ is odd and $n=2m$.

The following is from Marcus's Number Fields, p.19:

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Zev Chonoles
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  • Hmm OK I have an argument for one direction, I think. Say $m$ is odd and $n=2m$. Consider the intersection of the multiplicative group of $n$th roots of unity with $\mathbb{Q}(\zeta_m)$. This includes all the $m$th roots of unity but, in addition, the element $-1$. Now we have $n+1$ elements of a group of order $2n$, so we have generated the whole group and get $\mathbb{Q}(\zeta_n) = \mathbb{Q}(\zeta_m)$, as desired. I'll have to think more about the converse.... – Mike F Apr 22 '16 at 03:46
  • For the other way, you have $L/K/Q$ and neither extension is generated by a degree 1 polynomial but nevertheless one of them is degenerate. Try working with the degrees of the extensions. – Stella Biderman Apr 22 '16 at 04:09