Show that $K(\sqrt{2\sqrt{2}-1})$ is a degree $4$ Abelian extension of $K=\Bbb Q (i\sqrt{14})$

Question

I have to show that $K(\sqrt{2\sqrt{2}-1})$ is a a degree $4$ Abelian extension of $K=\Bbb Q (i\sqrt{14})$.

Now, $a=\sqrt{2\sqrt{2}-1}$ has minimal polynomial $f(x)=x^4+2x^2-7$ over $\Bbb Q$, which can be checked. I have to show that that $f$ is irreducible over $K$ and that the extension is normal. Then it'll follow that it's Galois and hence Abelian since it has degree 4. WolframAlpha tells me that the roots are $\pm \sqrt{2\sqrt2-1},\pm i\sqrt{2\sqrt{2}+1}$. So, the field is also normal as $\alpha \in L \implies \sqrt{2} \in L$ and then $i\sqrt{2\sqrt{2}+1} = {i\sqrt{14} \over \sqrt{2} \alpha} \in L$.

Then the only thing left to prove is that $f$ is irreducible over $K$. Please help me out.

EDIT : Writing $K_1=K(\sqrt{2})$, we get that $[K_1:K]=2$ as otherwise $K_1=K$ and $[\Bbb Q{\sqrt{2}}:\Bbb Q]=2=[K:\Bbb Q]$ mean $K=\Bbb Q{\sqrt{2}}$ but that's absurd as $\Bbb Q{\sqrt{2}}$ is real while $K$ is not. So, we'll be done if we can prove that $[L:K_1]=2$. The obvious choice for minimal polynomial is $x^2+1-2\sqrt{2}$ and it'll be enough if we can show that it is indeed irreducible over $K_1$.

Answer 1

You have found that $[\Bbb{Q}(a):\Bbb{Q}]=4$. Because $K$ is Galois over $\Bbb{Q}$, we have${}^1$ $$[K(a):K]<4\qquad\Leftrightarrow\qquad[\Bbb{Q}(a)\cap K:\Bbb{Q}]>1,$$ which is of course equivalent to $K\subset\Bbb{Q}(a)$. Now it suffices to check that $\sqrt{-14}\notin\Bbb{Q}(a)$.

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1: It is very much worth while trying to prove a general version of this equivalence; draw a diagram and try to prove when exactly the degree of a compositum of fields is precisely the product of their degrees. See also corollary 3.19 on page 41 of Milne's notes, and the diagram there.

Answer 2

Let $a=\sqrt{2\sqrt{2}-1}, b=i\sqrt{14}$ and then we know that $[\mathbb{Q} (a) :\mathbb{Q}] = 4$ and $[\mathbb{Q} (b) :\mathbb {Q}] = 2$.

If we assume that these fields are subfields of $\mathbb {C} $ then we can observe that $b\notin \mathbb{Q} (a) $ and hence $[\mathbb{Q} (a, b) :\mathbb {Q} (a)] =2$ so that $[\mathbb{Q} (a, b) :\mathbb{Q}] =8$ and thus $[\mathbb{Q} (a, b) :\mathbb{Q} (b)] =4$ and we are done.