Show that $\mathbb{Q}(\sqrt{-14},\sqrt{-2\sqrt{2}-1})$ has degree 8 over $\mathbb{Q}$

Question

We know that the extension $\mathbb{Q}(\sqrt{2\sqrt{2}-1}$) over $\mathbb{Q}$ has degree 4 by considering the minimal polynomial mod 3. Now I want to show that $-14$ isn't a square in this field. How do I do this in the easiest way? Please also let me know if you have a method not in this way (e.g. by considering intermediate field extensions in different orders).

Somehow feels like it ought to be true because $-14$ has a factor $7$ in it whereas $\mathbb{Q}(\sqrt{2\sqrt{2}-1}$) involves only $2$'s.

If it helps, I found that $\mathbb{Q}(\sqrt{-14},\sqrt{-2\sqrt{2}-1})$ is the splitting field of $X^4+2X^2-7$ over $\mathbb{Q}$.

Answer 1

There are many ways to see this. The easiest is to realise that $K= \mathbb Q(\sqrt{2\sqrt2-1})\subset \mathbb R$, so it cannot possibly contain a root of $-14$.

Alternatively, you could show that $K$ is not a Galois extension, so it cannot be the splitting field of an (irreducible) polynomial.

Answer 2

The splitting field of $X^4+2x^2-7$ over $\mathbb{Q}$ is what you have found. This polynomial is separable, hence the extension is Galois.Furthermore,this polynomial is irreducible. I used corollary 4.5 of this wonderful lecture by Keith Conrad : http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/cubicquartic.pdf

We can see, by what it says, that $-7 \neq 0$ and $(2^2+4*7)(-7) = -224 \neq 0$, hence the Galois group is $D_4$, wherein the extension is $8$.

For proofs of the facts, you can look at the same paper.