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 Marc Palm, quid, Alain Valette, Felipe Voloch, Benjamin Steinberg Mar 9 '13 at 13:19Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question. 


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 $\pm \frac{1}{\sqrt{2}}.$ 

