Let $\omega=\exp (2 \pi i / 3)$. Show that $\alpha=\omega+2\omega^{-1}$ is algebraic over $\mathbb{Q}$ and compute its minimal polynomial. Hello I have the above exercise. I know that $\omega$ is a root of the polynomial $x^2+x+1$ and based on that and on powers of $\alpha$ I tried to compute the polynomial, but unsuccessively. Any help would be much appreciated. 
Thank you in advance.
 A: Actually, $\omega$ and $\omega^{-1}$ are the two distinct roots of $f(x)=x^2+x+1.$ 
In particular, $\omega+\omega^{-1}=-1.$
Thus $\alpha=\omega+2\omega^{-1}=-1+\omega^{-1}.$ Then $\alpha+1=\omega^{-1}$ so $\alpha$ is a root of: $$f(x+1)=(x+1)^2+(x+1)+1=x^2+3x+3.$$

Is $x^2+3x+3$ minimal? It necessarily is, since a quadratic which is not prime has rational roots, and $\alpha$ is not rational because $\omega$ is not rational.
More generally, Given a rational number $q$ and a complex number $\alpha,$ $p(x)$ is a minimal polynomial for $\alpha$ if and only if $p(x-q)$ is a minimal polynomial for $\alpha+q.$
A: Here is a systematic approach.
First, $\alpha=\omega+2\omega^{-1} \in \mathbb Q(\omega)$ and so is algebraic, because $\mathbb Q(\alpha)\subseteq\mathbb Q(\omega)$ implies that $\mathbb Q(\alpha)$ has finite dimension over $\mathbb Q$.
Next, let's write $\alpha=\omega+2\omega^{-1}$ as a polynomial in $\omega$.
Since $1+\omega+\omega^2=0$, we have $1=-(1+\omega)\omega$, and so $\omega^{-1}=-1-\omega$. Therefore, $\alpha=-2-\omega$.
Finally, let's find the minimal polynomial of the map $\mu: x \mapsto \alpha x$. The basis $\{1,\omega\}$ makes it simple. Indeed, $1 \mapsto \alpha=-2-\omega$ and $\omega \mapsto \omega\alpha=-2\omega-\omega^2=1-\omega$. The matrix of $\mu$ in that basis is thus
$$
\pmatrix{-2 & 1 \\ -1 &-1}
$$
Its characteristic polynomial is $x^2 + 3 x + 3$, which is irreducible over $\mathbb Q$ since it has no rational roots. Therefore, it is the minimal polynomial of $\alpha$.
Make sure you understand that a polynomial kills $\alpha$ iff it kills $\mu$.
A: $$\alpha=-\frac{1}{2}+\frac{\sqrt3}{2}i+2\left(-\frac{1}{2}-\frac{\sqrt3}{2}i\right)=\sqrt3\left(-\frac{\sqrt3}{2}-\frac{1}{2}i\right),$$
which says that $\alpha^6=-27.$
Now, use the following:
$$\alpha^6+27=(\alpha^2+3)(\alpha^4-3\alpha^2+9)=(\alpha^2+3)(\alpha^4+6\alpha^2+9-9\alpha^2)=$$
$$=(\alpha^2+3)(\alpha^2+3\alpha+3)(\alpha^2-3\alpha+3).$$
A: $\alpha=\omega+2\omega^{-1}=\omega+2\omega^2=\omega^2-1$ ($\because \omega^3=1 \text{ and } 1+\omega+\omega^2=0$).
\begin{align*}
\alpha & =\omega^2-1\\
\alpha+1&=\omega^2\\
(\alpha+1)^3&=1\\
\alpha^3+3\alpha^2+3\alpha&=0.
\end{align*}
Since $\alpha \neq 0$. Therefore $\color{red}{\alpha^2+3\alpha+3}$ will be the minimal polynomial.
A: we know that: $\alpha = \omega + \frac{2}{\omega}$ and $\omega^3 = 1$, so:
$$\alpha = \omega^3 \times \left(\omega + \frac{2}{\omega} \right) = \omega ^4 + 2\omega^2 = \omega + 2\omega^2 = -\omega - 2$$
Now $\omega = -\alpha-2$ and finally:
$$\omega ^2 + \omega + 1 = (\alpha +2)^2 - \alpha - 2 + 1 = \alpha ^2 + 3\alpha + 3 = 0$$
This means that $\alpha$ is algebric and the desired polynomial is $\alpha ^2 + 3\alpha + 3 $.
