is there any bound on the absolute number of algebraic integer in terms of its degree?

If Z is a sum of t distinct roots of unity and |Z| is a rational integer, can someone find a bound on |Z| in terms of k=deg(Q(Z):Q))?

Clearly we need to have distinct roots of unity otherwise this won't work!

Correction: Let assume that Z is not rational itself otherwise obviously it's wrong. Here I hope to extend the proof of Kronecker thm! I have "Z is a sum of t distinct roots of unity and |Z| is a rational integer" I conjecture that either Z is rational or a root of unity!

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Not sure what you are asking but you might want to check: Loxton, J. H. On two problems of R. W. Robinson about sums of roots of unity. Acta Arith. 26 (1974/75), 159–174. –  Felipe Voloch Oct 9 '12 at 2:04
Even with the correction, all you have to do is multiply Will Sawin's sum by, say, a nonreal cube root of 1, $\omega$. Then $Z=-n\omega$ is not rational, $|Z|=n$ is a rational integer, and $k=2$. –  Gerry Myerson Oct 10 '12 at 4:06
@Gerry: sorry I edited my answer to fix it before seeing your identical comment. –  Will Sawin Oct 14 '12 at 17:20

If you multiply all the primitive roots of unity for the first $n$ primes by $-i$, then add them, you get $in$, whose degree is $2$ and whose absolute value is $n$, rational and unbounded.

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Dear Will, Doesn't the question ask that $|Z|$ (where $Z$ is the sum of roots of unity in question) be an integer? Regards, –  Emerton Oct 9 '12 at 2:25
I edited it to a more elegant solution that deals with that issue. –  Will Sawin Oct 9 '12 at 3:17
In fact, this construction allows you to write any sum of roots of unity as a sum of distinct roots of unity. –  Will Sawin Oct 14 '12 at 22:46