I'm trying to prove the following.
Suppose that $\alpha$ is an $n$th root of unity whose real part is an algebraic integer. Then $\alpha^4 = 1$.
I've thought about this for a while, and have the following ideas/questions.
If I can show that the algebraic norm $\frac{1}{2}(\alpha + \alpha^{-1})$ is not an integer for $n> 4$, then I will be done. Since $x^2-(\alpha +\alpha^{-1})x +1$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}(\alpha+\alpha^{-1})$, we have that
$$
N\left(\frac{1}{2}(\alpha + \alpha^{-1})\right) = \frac{1}{2^{\phi(n)/2}}N(\alpha+\alpha^{-1})
$$
where $\phi(n)$ is the Euler $\phi$ function. So we just need to consider $N(\alpha+\alpha^{-1})$. Is there a way that we can bound $N(\alpha + \alpha^{-1})$? If we can show that this norm is less than $2^{\phi(n)/2}$ then we're done.
Another thought is that $\frac{1}{2}(\alpha+\alpha^{-1})$ is a root of $T_n(x)-1$ where $T_n$ is the $n$th Chebyshev polynomial. This however is not generally irreducible, so the fact that $T_n$ is not monic does not tell us that $\frac{1}{2}(\alpha+\alpha^{-1})$ is not an algebraic integer.
Thanks for any thoughts or hints.
Best Answer
Here is a more direct argument that shows that the norm of $(\alpha + \alpha^{-1})/2$ is less than one unless $\alpha^2 = 1$. The norm is the product of the conjugates. Note that the conjugates of this are all of the form $(\beta + \beta)/2$ for some other root of unity $\beta = \alpha^k$. But note by the triangle inequality that
$$\left| \frac{\beta + \beta^{-1}}{2} \right| \le \frac{1}{2} + \frac{1}{2} = 1.$$
In particular, either the norm has absolute value strictly less than one, or equality holds for all conjugates and particular for $\beta = \alpha$. But the triangle inequality $|x + y| \le |x| + |y|$ for complex numbers is strict unless $\arg(x) = \arg(y)$. Since $\alpha$ and $\alpha^{-1}$ both have absolute value one, that means the triangle inequality is strict unless $\alpha = \alpha^{-1}$, which implies that $\alpha^2 = 1$.
To finish the original problem, if $(\alpha + \alpha^{-1})/2$ is an algebraic integer then its norm is an algebraic integer. Since the norm is less than $1$ unless $\alpha^2 = 1$, the only remaining possibility is that the norm is $0$, in which case $\alpha + \alpha^{-1} = 0$ and $\alpha^2 = -1$, or $\alpha^4 = 1$.