Units of $\overline{\mathbb{Z}}$ 
What are the units of $ \overline{\mathbb{Z}} $ (the ring of algebraic integers)? 

I know all roots of monic polynomials with constant term 1 are units, but are there any others?
 A: Let $\alpha$ be an algebraic integer with minimal polynomial $X^n + a_1 X^{n-1} + \dots + a_{n-1}X+ a_n$.
We have that  $\alpha^{-1}$ is a root of $a_nX^{n} + a_{n-1}X^{n-1}+ \dots a_1 X+ 1 $ and thus of (of cours $a_n$ is nonzero by the irreducibility of the above polynomial)  $$X^{n} + \frac{a_{n-1}}{a_n}X^{n-1}+ \dots \frac{a_1}{a_n} X+ \frac{1}{a_n}.$$
Now, just note that since $\alpha$ and $\alpha^{-1}$ have the same degree (certainly the generate the same field over the rationals), the polynomial
$$
X^{n} + \frac{a_{n-1}}{a_n}X^{n-1}+ \dots \frac{a_1}{a_n} X+ \frac{1}{a_n}
$$
is the minimal polynomial of $\alpha^{-1}$. 
Thus, $\alpha^{-1}$ is integral if and only if this polynomial has integer coefficients. 
This is equivalent to $a_n$ being invertible, that is $a_n= \pm 1$.
Thus,  $\alpha$ is a unit in the ring of algebraic integers if and only if the constant coefficient of its minimal polynomial (over the integers) is $+1$ or $-1$.
A: If $\alpha \in \overline{\mathbb{Z}}$ $($this is a pretty bad abuse of notation$)$, then $1/\alpha$ is a root of a monic polynomial if and only if the minimum polynomial of $\alpha$ has constant term $\pm 1$. This is because $1/\alpha$'s minimum polynomial must be $\alpha$'s minimum polynomial with its coefficients reversed $($obviously $1/\alpha$ is a root of this, furthermore it is not hard to check this polynomial is irreducible$)$.
Another way to do this problem is to adjoin all the roots of $\alpha$'s minimum polynomial to $\mathbb{Q}$ and then use norms to deduce the result instantly, if my knowledge of Galois theory is not lacking.
