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For which totally complex number fields $K$ with embeddings $\{ \sigma_1, \dots, \sigma_m\}$ do we have the equality:

$$ |\sigma_1(x)| = |\sigma_2(x)| = \dots = |\sigma_m(x)|, $$

for all $x \in K$ where $|\cdot|$ corresponds to the complex absolute value $|x| = (x\bar{x})^{1/2}$? In other words, which number fields have embeddings that do not affect the distance of a coordinate on an argand diagram?

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  • $\begingroup$ What do the units have to be? They would also have to have absolute value 1. $\endgroup$
    – TCiur
    Apr 4 '18 at 21:51
  • $\begingroup$ @TCiur be that as it may, how does it help answer my question? $\endgroup$
    – Chris
    Apr 4 '18 at 22:03
  • $\begingroup$ @TCiur I believe the quadratic extension $\mathbb{Q}(\sqrt{d})$ where $d$ is a squarefree negative integer satisfies the condition too. $\endgroup$
    – Chris
    Apr 4 '18 at 22:18
  • $\begingroup$ Your condition requires that all archimedean places of K coincide, but this does not occur. The only possibility is when there is only 1 archimedean place, and this happens when we have a pair of complex embeddings, I.E an imaginary quadratic field $\endgroup$
    – TCiur
    Apr 4 '18 at 23:53
  • $\begingroup$ I might be showing my ignorance here, but don't they all satisfy that condition? $\endgroup$ Apr 5 '18 at 3:29
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Assuming your condition, we must have that every galois action acts via multiplication by some complex unit:

$|x| = |\sigma(x)| \implies \sigma(x) = u \cdot x$ for some complex unit $u$.

Assume $\sigma(x) = u \cdot x$, then $\sigma(x+1) = u\cdot x + 1 = w\cdot x + w$ for some other unit $w$. This would imply that $|x+1| = |u\cdot x + 1|$. This is absurd unless $u\cdot x = \bar{x}$ or $x$. Hence the galois group must consist only of the identity and the complex conjugate (only imaginary quadratic fields fit the condition).

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  • $\begingroup$ The embeddings on cyclotomic fields don't affect the absolute value of a cyclotomic number either, unless I am mistaken, as they correspond to shifts in the argument of the number in the complex plane. $\endgroup$
    – Chris
    Apr 5 '18 at 10:04
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    $\begingroup$ In any cyclotomic field, non-complex-conjugate galois actions on $\zeta + \zeta^{-1}$ will change the absolute value (think geometrically). $\endgroup$
    – TCiur
    Apr 5 '18 at 16:36
  • $\begingroup$ Sorry for the late reply - you're absolutely correct. Thanks for the response! $\endgroup$
    – Chris
    Apr 8 '18 at 14:50

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