Find the minimal polynomial $f$ of $√5+i$ over $\mathbb Q$ The candidate is $x^4-8x^2+36=0$ but we cant use Eisenstein here to prove irreducibility. What do we do?
 A: Note that $ \mathbb{Q}(\sqrt{5}, i)/\mathbb{Q} $ is Galois of degree 4, and the stabilizer of $ \sqrt{5} + i $ in the Galois group is the trivial group, which means that $ \mathbb{Q}(\sqrt{5} + i) = \mathbb{Q}(\sqrt{5}, i) $ so that $ \sqrt{5} + i $ has degree $ 4 $ over $ \mathbb{Q} $. This means that the polynomial you've found is irreducible, and it is indeed the minimal polynomial.
A more elementary argument can be given by noting that
$$ \frac{6}{\sqrt{5} + i} = \frac{6(\sqrt{5} - i)}{6} = \sqrt{5} - i $$
so that we have $ \sqrt{5} - i \in \mathbb{Q}(\sqrt{5} + i) $, and by extension $ \sqrt{5}, i \in \mathbb{Q}(\sqrt{5} + i) $. The rest of the argument proceeds similarly.
To find the minimal polynomial, the easiest way is to note that we have
$$ m_{\alpha}(x) = \prod_{\sigma} (x - \sigma(\alpha)) $$
for any element $ \alpha $ of the extension, where the product runs over distinct $ \mathbb{Q} $-conjugates. Using this formula, we have for $ \alpha = \sqrt{5} + i $:
$$ m_{\alpha}(x) = (x - (\sqrt{5} + i))(x + (\sqrt{5} + i))(x - (\sqrt{5} - i))(x + (\sqrt{5} - i)) = (x^2 - (\sqrt{5} + i)^2)(x^2 + (\sqrt{5} - i)^2) = x^4 - 8x^2 + 36 $$
A: My “easiest” way of finding the $\Bbb Q$-irreducible polynomial for $\alpha=\sqrt5+i$ is this:
Certainly the $\Bbb Q(\sqrt5\,)$-irreducible polynomial for $\alpha$ is $(X-\sqrt5\,)^2+1=X^2-2\sqrt5X+6$. And so $(X^2-2\sqrt5X+6)(X^2+2\sqrt5X+6)=X^4-8X^2+36$ has the following three properties. It is the only factorization of the polynomial you found  into monic $\Bbb Q(\sqrt5\,)$-irreducibles; it is a $\Bbb Q$-polynomial; it is therefore $\Bbb Q$-irreducible.
