Suppose w^(2n)=1 (w is a complex number). For which n (if any) \sqrt(w) \in Q(w) ?
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$\begingroup$ Perhaps the title should say something about the field Q(w)... how about "Is root of unity w a square in Q(w)?" $\endgroup$– Ilya NikokoshevCommented Oct 22, 2009 at 8:01
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$\begingroup$ This could be homework for a second course in number theory, but it also could be a lemma needed by someone in a field far from number theory. I would leave this open. $\endgroup$– David E SpeyerCommented Oct 22, 2009 at 10:44
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2 Answers
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The key point is to understand the field Q(w) for w a primitive kth root of unity. Call this field Qk. In particular, you want to know that Q4n \neq Q2n.
The key fact here is that the field extension Qk/Q has degree phi(k), where phi(k) is the Euler phi function, and phi(4k) \neq phi(2k). For a proof that Qk/Q has degree phi(k), see the early parts of any book on cyclotomic fields. This is probably also done in many Galois theory books but I don't know which ones.
answered Oct 22, 2009 at 10:52
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$\begingroup$ Ok, thanks. Yes I need this for some lemma. $\endgroup$– user966Commented Oct 22, 2009 at 12:16
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If w is a primitive 2n-th root, then the answer is "none". If w is not primitive, then Q(w) has a square root of w if and only if and odd power of w is 1.
answered Oct 22, 2009 at 6:45
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$\begingroup$ I want w to be a primitive 2n-th root. Can you give me a hint of how to prove the fact that you mentioned? $\endgroup$– user966Commented Oct 22, 2009 at 6:59