# field extension with degree $q^r$

Let $K\subset \mathbb{C}$ be a subfield, and $q$ a prime number. Suppose, for all field extension $L/K$ with finite degree, $q|[L:K]$. Then, for all finite extension $L$, there exists $r\in \mathbb{N} \cup \{0\}$ s.t., $[L:K]=q^r$.

If there exists $L$ with $[L:K]=nq^r,\;(q\not \mid n)$, its Galois closure is finite extension. So, I can prove by contradiction when Galois group is abelian, since structure theorem of finite abelian group. But I can't prove general case.

• What is the difference between $\Bbb N$ and $\Bbb N\cup\{0\}$ ? – Bernard May 26 '18 at 13:22
• $0 \not \in \mathbb{N}$ Of course,if $[L:K]=1$, $K=L$. – J.smith May 26 '18 at 13:28
• But $\Bbb N$ has a zero element! See Peano's axioms. – Bernard May 26 '18 at 13:58
• It depends on definition. And, $0\not \in \mathbb{N}$ according to original Peano's paper.archive.org/details/arithmeticespri00peangoog – J.smith May 26 '18 at 14:10

If there exists $L$ with $[L:K] = nq^r$, then its Galois closure $L'$ is a finite extension and $[L':K] = n'q^{r'}$ for some $n'$ and $r'$ with $q \nmid n'$ (and of course, $n| n'$ and $r' \geq 0$). The Galois group $Gal(L' : K)$ is of order $n'q^{r'}$.
By Sylow's theorems, $Gal(L':K)$ has a subgroup of order $q^{r'}$. This subgroup corresponds to an intermediate field $E$ (with $L' \supset E \supset K$) such that $[L' : E] = q^{r'}$. But then, $[E : K] = n'$, contradicting the assumption that $q | [E : K]$.