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Let $\zeta$ be a complex number on the unit circle $\{z\in \mathbb{C}: |z|=1\}$.Suppose that $[\mathbb{Q}(\zeta):\mathbb{Q}] < \infty$.Is it true that $\zeta ^n=1$ for some positive integer $n$?

Ben
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    No, take almost any algebraic number an normalize it. – Alex Youcis Apr 12 '14 at 09:53
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    In fact the interesting fact is that there are many algebraic integers on the unit circle which are not roots of unity, see http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity – Ferra Apr 12 '14 at 10:18

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In the spirit of Alex Youcis' comment let $$ z=\frac{2+i}{2-i}=\frac35+\frac45i. $$ Then $[\Bbb{Q}[z]:\Bbb{Q}]=2<\infty.$

The numbers $2+i$ and $2-i$ generate distinct prime ideals in the ring of Gaussian integers $\Bbb{Z}[i]$. But $\Bbb{Z}][i]$ is a unique factorization domain and it only has the four obvious units. If $z^n=1$ then $(2+i)^n=(2-i)^n$ violating unique factorization.

Jyrki Lahtonen
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An example (from which you can deduce many others):

Take the non-real algebraic number $\;z:=1+\sqrt2\,i\;$ and look at

$$\omega:=\frac z{|z|}=\frac1{\sqrt5}\left(1+\sqrt2\,i\right)=e^{i\arctan\sqrt2}$$

Then clearly $\;\omega\in S^1\;$ and algebraic , yet

$$\forall\;n\in\Bbb N\;\;:\;\;\omega^n=e^{ni\arctan\sqrt2}=1\iff n\arctan\sqrt2=2k\pi\;,\;\;k\in\Bbb Z$$

Now read the answers here...and the comment below the accepted one: ArcTan(2) a rational multiple of $\pi$?

Community
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DonAntonio
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Roots of unit are algebraic integers. Take an algebraic number that don't be algebraic integer.

Martín-Blas Pérez Pinilla
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  • And do what with it? We want $|z|=1$. – anon Apr 13 '14 at 05:47
  • Obviously, algebraic number of modulus 1. – Martín-Blas Pérez Pinilla Apr 13 '14 at 17:39
  • So just pick an algebraic number of modulus $1$ that's not an algebraic integer? If one knew that was possible one would have already known the answer to this question. Your post seems rather useless. – anon Apr 13 '14 at 18:08
  • (a) Algebraic integers are not mentioned in the original post. (b) Finding algebraic numbers of modulus 1 that not be algebraic integers is very easy, as proves the answer of Jyrki Lahtonen. – Martín-Blas Pérez Pinilla Apr 13 '14 at 18:20
  • Yes, but an answer to a question that, to be understood, requires a reader to already know the answer to the question (and has no other redeeming value like insightful commentary, history, tips and tricks, intuition, humour, illustration of the power of higher methods, etc.) is useless. That is why I have downvoted this answer: the reader cannot get anything from it. – anon Apr 13 '14 at 18:24
  • Then, why the question of the OP wasn't "there are algebraic numbers of modulus 1 that not are algebraic integers?" – Martín-Blas Pérez Pinilla Apr 13 '14 at 18:39
  • I also downvote for feigning dumb and being obtuse, but I've already used my vote on this answer and I don't stalk profiles for these purposes. Perhaps this concludes our discussion. – anon Apr 13 '14 at 18:45