Calculate the minimal polynomial of $\alpha=\sqrt 2e^\frac{2\pi i}{3}$ over $\mathbb Q$.
My try:
Let $x=\sqrt 2e^\frac{2\pi i}{3}\implies x-\sqrt 2(-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$.
On solving completely I got the polynomial to be $x^4+2x^2+4$ which is satisfied by $\alpha$ which is a monic polynomial .Also it is an irreducible polynomial over $\mathbb Q$.
However I am finding it difficult to check that is the minimal polynomial i.e it is the lowest degree polynomial satisfied by $\alpha $ over $\mathbb Q$.
How should I do it?
One way to check this, is to check that $x^4+2x^2+4$ is irreducible over $\Bbb{Q}$: If $\alpha$ were a root of a lower degree polynomial, then this polynomial divides $x^4+2x^2+4$. If it is irreducible, this is impossible.
Fact: Let $k$ a field. If $\alpha$ is algebraic over $k$ and $f\in k[x]$ is monic and irreducible with $f(\alpha)=0$, then $f$ is the minimal polynomial of $\alpha$ over $k$.
Proof: Let $g\in k[x]$ the minimal polynomial of $\alpha$ over $k$. Then $\deg g\leq\deg f$ by definition of the minimal polynomial, so by polynomial long division there exist $q,r\in k[x]$ such that $$f=q\cdot g+r,$$ where $\deg r<\deg g$ or $r=0$. Because $f(\alpha)=g(\alpha)=0$ it follows that $$r(\alpha)=q(\alpha)\cdot g(\alpha)+r(\alpha)=f(\alpha)=0,$$ so by minimality of $g$ it follows that $r=0$, and hence that $f=q\cdot g$. As $f$ is irreducible $q$ is a unit. Moreover, both $f$ and $g$ are monic so $q=1$ and $f=g$, so $f$ is the minimal polynomial of $\alpha$ over $k$.
Note that the same argument shows that if $\alpha$ is a root of some $h\in k[x]$, then the minimal polynomial $g$ of $\alpha$ over $k$ divides $h$, from which this fact also follows easily.