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In a recent theorem we have naturally come across this condition, that seems to be important, but rarely satisfied: $\sqrt{\frac 1 4 + a^m} \in \mathbb Q(\zeta_m, a)$ where $a\in\mathbb C^*$ and $\zeta_m$ is a primitive $m$-th root of unity.

We are interested in an equivalent condition for any $a$, but we learned here that the condition for some $a$ seems to be just a technical obstacle of our method of proof and not intrinsic to the problem.

We are especially interested in the case where $a$ is a root of unity. We have checked that the containment condition is fulfilled, whenever $a^m=1$ and $5\mid m$ as this reduces to $\sqrt 5\in\mathbb Q(\zeta_m,a)$ or $a^m=-1$ and $3\mid m$ as this reduces to $\sqrt{-3}\in\mathbb Q(\zeta_m, a)$. Numerically these seem to be the only examples at $a$ root of unity and they all lead to different results in our theorem. So we believe that the condition is natural.

Is there a way to prove that these conditions are equivalent to the containment condition for $a$ root of unity - maybe using algebraic number theory?

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    $\begingroup$ You want this to be impossible for $a \in \mathbf{C}^*$ and $|a| < 1/4$, with no other conditions on $a$? Then you're out of luck. Let $\rho$ be a rational number a tiny little bit bigger than 1/2 (but smaller than $\sqrt{1/4 + 1/4^m}$). Then there's a unique real algebraic number $a \in (0, 1/4)$ such that $\sqrt{1/4 + a^m} = \rho$, and for this $a$ the condition that $\sqrt{1/4 + a^m} \in \mathbf{Q}(\zeta_m, a)$ is vacuously satisfied because $\rho \in \mathbf{Q}$. $\endgroup$
    – David Loeffler
    Commented Mar 30, 2015 at 15:30
  • $\begingroup$ (PS: I suspect this isn't an answer to the question you meant to ask, so can you refine the question to rule out such silliness?) $\endgroup$
    – David Loeffler
    Commented Mar 30, 2015 at 15:32

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Your characterization is correct. When $a$ is a root of unity, the question reduces to understanding when $\mathbb{Q}(\sqrt{1 + 4 \zeta})$ is an abelian extension (and $\zeta$ is a root of unity). This occurs only for $\zeta = +1$ or $-1$. As you suspect, one can prove this using [elementary] algebraic number theory. Writing the details on an iPhone is a bit annoying, so I will leave that to someone else.

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    $\begingroup$ Please flesh out this answer once you are off your iPhone and at a computer, when you get a chance. $\endgroup$
    – Todd Trimble
    Commented Apr 1, 2015 at 20:02
  • $\begingroup$ What's a computer? $\endgroup$
    – Sent from my iphone
    Commented Apr 3, 2015 at 19:21
  • $\begingroup$ Whatever instrument you would use to post a proper MO answer. (I'm not playing games here.) $\endgroup$
    – Todd Trimble
    Commented Apr 3, 2015 at 22:00

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