$\xi = e^{\frac{\pi}4i}$
$K=\mathbb Q (\sqrt[4]{5} \xi , \sqrt[4]{5} \xi^3 , \sqrt[4]{5} \xi^5 , \sqrt[4]{5} \xi^7 )= \mathbb Q (\sqrt[4]{5} \xi , \xi^2)=\mathbb Q (\sqrt[4]{-5}, \xi^2 )$
I am pretty sure these above equalities hold.
$L=\mathbb Q (\sqrt[4]{-5})$, $J=\mathbb Q (\xi^2)$
Also let $Q = \mathbb Q$
$[K:Q]=[K: L] \cdot [L:Q]$ and $[K:Q]=[K: J] \cdot [J:Q]$
$[L:Q]=4$ and $[J:Q]=2$, which I am pretty confident with. So we have:
$[K:Q]=4[K: L] $ and $[K:Q]=2[K: J] $
Can someone tell me how to work out either $[K:J]$ or $[K:L]$ (in a lot of detail please).
Consider the tower of fields $\mathbb Q \subseteq \mathbb Q(i) \subseteq \mathbb Q(i,\sqrt5) \subseteq \mathbb Q(i,\sqrt[4]{-5})$ and show that these are strict inclusions. It then follows that $[\mathbb Q(i,\sqrt[4]{-5}): \mathbb Q] = 8$. It is obvious that $i\notin \mathbb Q$ and $\sqrt5\notin \mathbb Q(i)$, so I restrict to show that $\sqrt[4]{-5}\notin \mathbb Q(i,\sqrt5)$.
Assume for a contradiction that $\sqrt[4]{-5}\in \mathbb Q(i,\sqrt5)$. Then there exist $a,b,c,d\in \mathbb Q$ such that $\sqrt[4]{-5} = a + ib + \sqrt 5 c + \sqrt{-5}d$. Hence we calculate $$ \begin{align*} \sqrt{-5} &= (\sqrt[4]{-5})^2 = (a+ib +\sqrt5c+\sqrt{-5}d)^2\\ &= a^2-b^2+5c^2-5d^2 + i(ab+5cd) + \sqrt{5}(ac-bd) + \sqrt{-5}(ad+bc). \end{align*} $$ From this we get the following four equations:
$$ \begin{gather} a^2-b^2 +5c^2-5d^2 = 0 \tag{1}\\ ab+5cd=0\tag{2}\\ ac-bd=0\tag{3}\\ ad+bc=1\tag{4}\\ \end{gather}$$ Now, we show that $a,b,c,d\neq 0$: If $a=0$, then (4) implies $b,c\neq 0$ and hence $d=0$ by (3). But then (1) says $-b^2+5c^2=0$ or equivalently $5 = (b/c)^2$ which is a contradiction to $\sqrt{5}\notin \mathbb Q$. Hence $a\neq 0$. Quite the same argument shows that $b,c,d\neq 0$.
Multiplying (2) by $d$ and substituting $ac = bd$ (by (3)), we obtain $$ \begin{align*} 0 = abd+5cd^2 = a^2c + 5cd^2 = c(a^2+5d^2). \end{align*} $$ Since $c\neq 0$ we have $a^2+5d^2=0$ in $\mathbb Q$. But this is only possible if $a=d=0$, which contradicts $a,d\neq 0$.
Thus, we have shown $\sqrt[4]{-5}\notin \mathbb Q(i,\sqrt5)$.