Compositum of all Galois extensions of prime power degree Let $\mathbb{K}$ a field of characteristic $\not =p$ prime, and $\mathbb{K}_p$ the compositum of all the Galois extensions of $\mathbb{K}$ whose degree is a power of $p$.
I have to prove that for every finite extension $\mathbb{K}\subseteq\mathbb{L}\subseteq\mathbb{K}_p$, the degree of $\mathbb{L}$ over $\mathbb{K}$ is a power of $p$.
My conjecture is that every Galois finite extension of $\mathbb{K}$ laying under $\mathbb{K}_p$ has this property, and then the thesis would follow from the multiplicativity of degrees. But I don't know how to prove this (or the thesis itself) because I have no clue at all about whar $\mathbb{K}_p$ looks like.
Do you have any hint or idea? I've been thinking about this problem for days but I'm totally stuck.
 A: Take a finite set of generators for $L/K$. (In fact, a single generator will suffice: why?) Each of these lives in $K_p$, which means it lives in a compositum of finitely many Galois extensions of $p$-power degree.
Then $L$ lives in a sub-extension of a finite compositum of $p$-power extensions.
We want to show that $L$ has $p$-power degree. It will suffice to show that the Galois closure of $L$ is of $p$-power degree, which we will do by induction:
Suppose $L=K_1K_2$, and that $K_1$ and $K_2$ are each Galois of $p$-power order over $K$. Then $L$ is also Galois (why?).
Now consider the lattice of fields containing $L$, $K_1$, $K_2$, and $K_1\cap K_2$. This corresponds to a lattice of Galois groups over $K_1\cap K_2$.
Now $\operatorname{Gal}(L/K_1)$ and $\operatorname{Gal}(L/K_2)$ generate $\operatorname{Gal}(L/K_1\cap K_2)$, and they are both normal and disjoint in $\operatorname{Gal}(L/K_1\cap K_2)$ (why?). So you have a direct product, and $[L:K_1\cap K_2]$ is of $p$-power order. It follows that $[L:K]$ is too.
