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I know that every square root $\sqrt{n}$ of an integer is contained in some cyclotomic extension $\Bbb Q[\omega_m]$. If we know $\sqrt{n}\in\Bbb Q[\omega_m]$ and that $n$ is square-free, can we conclude that for its prime divisor $p|n$ we have $\sqrt{p}\in\Bbb Q[\omega_m]$ as well?

For example, could anyone give me some $\Bbb Q[\omega_m]$ such that $\sqrt{6}$ lies inside but $\sqrt{2}$ or $\sqrt{3}$ does not?

user108580
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    if $n=m=pq$ where $p\neq q$ are primes congruent to $3$ mod $4$ (e.g. $m=21$) then $\sqrt{n}$ is in your field but $\sqrt{p}$ and $\sqrt{q}$ are not; however, $\sqrt{-p}$ and $\sqrt{-q}$ are in the field – user8268 Mar 28 '24 at 15:54
  • For a simple proof of the claim in the first sentence, see https://math.stackexchange.com/questions/3380845/every-quadratic-number-field-is-contained-in-a-cyclotomic-field – lhf Mar 28 '24 at 23:43

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