I have a elementary question:If e^itπ is algebraic , is $t$ a rational number. I do not know whether it is right

closed as off topic by plusepsilon.de, quid, Alain Valette, Felipe Voloch, Benjamin Steinberg Mar 9 '13 at 13:19
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The question as stated is very easy to answer. If the question was really intended to be asked about algebraic integers, then there is still a relatively simple direct example to show that the answer is "no". The question about algebraic integers is wellstudied, with a substantial literature and the example that follows is one easy instance: If we take any real algebraic integer $s$ with $0 < s <1,$ then, $t = s + i \sqrt{1s^{2}}$ is an algebraic integer which lies on the unit circle. Now apply this with $s = \sqrt{2}1.$ Note that $t$ generates a degree $4$ extension of the rationals. If $t$ were a primitive $m$th root of unity, we would have $\phi(m) = 4$ so that $m = 8.$ Hence $t$ would be a primitive $8$th root of unity, but it is not, as each primitive $8$th root of unity has real part $\pm \frac{1}{\sqrt{2}}.$ 

