$\DeclareMathOperator{\char}{char}$ $\DeclareMathOperator{\aut}{Aut}$ I've been reading over the following theorem (5.7.7) from Hungerford's algebra, and two things are confusing me about the proof of the following statment:
Proposition 7.7 Let $\mathbf{F}$ be a cyclic extension field of $\mathbf{K}$ of degree $n$ and suppose $n = mp^t$ where $0\neq p = \char \mathbf{K}$ and $(m, p) = 1$. Then there is a chain of intermediate fields $F \supset E_0 \supset E_1 \supset\cdots\supset E_{t-1}\supset E_t = \mathbf{K}$ such that $\mathbf{F}$ is a cyclic extension of $E_0$ of degree $m$ and for each $0 \leq i \leq t$, $E_{i-1}$ is a cyclic extension of $E_i$ of degree $p$.
SKETCH OF PROOF. By hypothesis $F$ is Galois over $K$ and $\aut_K\!F$ is cyclic (abelian) so that every subgroup is normal. Recall that every subgroup and quotient group of a cyclic group is cyclic (Theorem I.3.5). Consequently, the Fundamental Theorem 2.5(ii) implies that for any intermediate field $E$, $F$ is cyclic over $E$ and $E$ is cyclic over $K$. It follows that for any pair $L$, $M$ is a cyclic extension of $L$; in particular, $M$ is algebraic Galois over $L$.
Let $H$ be the unique (cyclic) subgroup of order $m$ of $\aut_K\!F$ (Exercise I.3.6) and let $E_0$ be its fixed field (so that $H = H'' = E_0' = \aut_{E_0}\!F$). Then $F$ is cyclic over $E_0$ of degree $m$ and $E_0$ is cyclic over $K$ of degree $p^t$. Since $\aut_{K}\!E_0$ is cyclic of order $p^t$ it has a chain of subgroups $$1 = G_0 < G_1 < G_2 < \cdots < G_{t-1} < G_t = \aut_K\!E_0$$ with $|G_i| = p^i$, $[G_i:G_{i-1}] = p$ and $G_i/G_{i-1}$ cyclic of order $p$ (see Theorem I.3.4(vii)). For each $i$ let $E_i$ be the fixed field of $G_i$ (relative to $E_0$ and $\aut_K\!E_0$). The Fundamental Theorem 2.5 implies that: (i) $E_0 \supset E_1 \supset E_2 \supset \cdots \supset E_{t-1} \supset E_t = K$; (ii) $[E_{i -1}:E_i] = [G_i:G_{i-1}] = p$; and (iii) $\aut_{E_i}\!E_{i-1} \cong G_i/G_{i-1}$. Therefore, $E_{i-1}$ is a cyclic extension of $E_i$ of degree $p$ $(0 \leq i \leq t -1). \quad\Large\blacksquare$
Let $F$ be a cyclic extension field of $K$ of degree $n$. In view of Proposition 7.7 we may, at least in principle, restrict our attention to just two cases: (i) $n = \char K = p \neq 0$; (ii) $\char K = 0$ or $\char K = p \neq 0$ and $(p,n) = 1$ (that is, $\char K\,\nmid\,n$). The first of these is treated in
- Where does he use $\text{Char}(K)=p$ in this proof? None of the result he uses need characteristic $p$ that I can tell. The word characteristic doesn't even appear in the proof. It only appears in the remark after the proof.
- What is his comment following the proof mean exactly? It reads as if prop 7.7 eliminates some degenerate case, but it looks like he's trying to say that we only have to work about two other cases now. One where $\text{Char}(K)=p=n$, and the other where $\text{Char}(K)=0$ or $\text{Char}(K) = p$ and $\text{Char}(K) \not\mid n$, as prop 7.7 is the case where $\text{Char}(K)=p$ and $\text{Char}(K) \mid n$.
Edit: It almost seems to me like he's purposely assuming $\text{Char}(K)=p$ in the hypothesis of the theorem, even though it may hold without that assumption, simply so he can state the next results to take care of the only other two possibilities once prop 7.7 is proven.