irreducibility of a certain polynomial $\alpha =\sqrt{-1+\sqrt{-2}}$ and $\beta =\sqrt{-1-\sqrt{-2}}$ are roots of irreducible polynomial: $$f(x)=x^{4}+2x^2+3$$ over $\mathbb{Q}$. Since $\alpha ^{2}+\beta ^{2}=-2$, I want to prove that $$irr(\beta, \mathbb{Q}(\alpha ))=x^{2}+\alpha ^{2}+2$$ But I can't prove irreducibility of $x^{2}+\alpha ^{2}+2$ over $\mathbb{Q}(\alpha )$.
Help me please. Thank you in advance!
 A: First , there is an ambiguity in your notation, since $-1+\pm\sqrt{-2}$ has no canonical square root. Nevertheless, we choose such squareroots in such a way that $\alpha^*=\beta$ (where $*$ is complex conjugation).
In particular, $\alpha\beta>0$. Since $(\alpha\beta)^2=3$, we get $\alpha\beta=\sqrt{3}$.
Let $\gamma=\alpha+\beta=\alpha+\alpha^*$. If $\beta\in\mathbb{Q}(\alpha)$, then $\gamma\in\mathbb{Q}(\alpha)$. Notice that $\gamma\in\mathbb{R}$.
Now $\gamma^2=\alpha^2+\beta^2+2\alpha\beta=-2+2\sqrt{3}$.
We deduce that $\gamma^4+4\gamma^2-8=0$. Now the polynomial $f= X^4+4X^2-8$ is irreducible over $\mathbb{Q}$ because it is irreducible modulo $5$ , hence $[\mathbb{Q}(\gamma):\mathbb{Q}]=4.$ Since $[\mathbb{Q}(\alpha):\mathbb{Q}]=4,$ and $\gamma\in\mathbb{Q}(\alpha)$, we get $\mathbb{Q}(\alpha)=\mathbb{Q}(\gamma)$. In particular, $\alpha\in\mathbb{Q}(\gamma)\subset\mathbb{R}$, a contradiction.
Thus $\beta\notin\mathbb{Q}(\alpha)$, and $\beta$ has at least degree $2$ over $\mathbb{Q}(\alpha)$, which easily implies the desired irreducibility.
A: Let $\mathcal O_K$ be the ring of integers of $K$. One can see that $[K:\mathbb Q] = 4$ because $f(x+1)$ is eisenstein at $2$.
To show that $x^2 + \alpha^2 + 2$ is irreducible is equivalent to showing that $-\alpha^2 - 2$ is a non-square. Note that $-\alpha^2-2$ differs from $3$ by a square in $K$, because
$$\alpha^2(-\alpha^2-2) = - (\alpha^4 + 2\alpha) = -(-3) = 3$$
so we would also be done if we showed $3$ is a nonsquare. To prove that, it is certainly sufficient to show that the ideal $(3)$ is not a square in $\mathcal O_K$.
But this follows readily from the Dedekind-Kummer theorem: since $\alpha$ generates $K$, we want to factor its minimal polynomial mod $3$:
$$f(x) = x^2(x+1)(x+2) \mod 3$$
This has a simple factor, and so $(3)$ also has a simple factor, hence is not a square.
p.s. some versions of Dedekind-Kummer include stipulations about the index of the order generated by $\alpha$ in $\mathcal O_K$, but those are only relevant to primes associated to factors with multiplicity at least $2$. If that still worries you, you can also check that the discriminant of $f(x)$ is divisble by 3 exactly once, and so $3$ does not divide the index of the aforementioned order in $\mathcal O_K$ anyways. Another simple way to factorize (3) is to split the tower to include $\mathbb Q(\alpha^2)$ and computing the factorization of $(3)$ in steps, first factor it in the intermediate field (easy D-K again) and then the primes lying over it (being careful with the discriminant).
