abstract algebra - How to determine the minimal polynomial of $sqrt{3 + 2sqrt{2}}$ over $mathbb{Q}$?

I first let $\alpha = \sqrt{3 + 2\sqrt{2}}$ and $\alpha^2 - 3 = 2\sqrt{2}$. This gives us $(\alpha^3 - 2)^2 = 8$. Expand the polynomial we obtain that
$x^4 - 6x^2 +1$ has $\sqrt{3 + 2\sqrt{2}}$ as a root. But apparently this polynomial is not an irreducible polynomial over $\mathbb{Q}$. How should we determine the minimal polynomial in the first place? Many thanks!

Answer

Hint A minimal polynomial must be irreducible, but as you say, the polynomial $p$ you produced is not:
$$p(x) = (x^2 + 2 x - 1) (x^2 - 2 x - 1).$$

Since $p(\alpha) = 0$, however, $\alpha$ must divide one of these factors (and since $\alpha \not\in \Bbb Q$, that factor must itself be the irreducible polynomial of $\alpha$).

To see what's going on here, expand $(1 + \sqrt{2})^2$.