I was trying to find the minimal polynomial of $\sqrt{i + \sqrt{2}}$ over $\mathbb{Q}$. I went as follows:
$a = \sqrt{i + \sqrt{2}} $
$a^{2} = i + \sqrt{2} $
$a^{4} = 1 + 2\sqrt{2}i $
$(a^{4} - 1)^{2} = -8 $
$a^{8} - 2a^{4} + 9 = 0 $
I'm guessing that $p(x) = x^{8} - 2x^{4} + 9$ is the minimal polynomial. I was then thinking about how to show this (or discover that it is reducible). I could try the rational roots theorem to show that there are no rational roots of $p(x)$. Then any possible polynomial factors would be of degree 2 and 6, 3 and 5, or 4 and 4 with coefficients in $\mathbb{Z}$. I could set up a homomorphism with $\mathbb{Z_{n}}$ and use Gauss's lemma. Either way it seems I'd have to check multiple systems of equations to investigate whether it is irreducible or not. Are there any ways of doing this quicker? It seems like a bit of a slog to have to work through all the equations I'd have to check - e.g. to check for degree 3 and 5 I'd have to check if there exists $a(x) = x^{5} + ax^{4} + bx^{3} + cx^{2} + dx + e$
and
$b(x) = x^{3} + fx^{2} + gx + h$
such that $p(x) = a(x)b(x)$.