I'm working on the proof of Theorem 4.26 (pg. 215) in Rotman's "Advanced Modern Algebra". The theorem is stated as follows:
Let $f(x)\in k[x]$, where $k$ is a field, and let $E$ be a splitting field of $f(x)$ over $k$. If $f(x)$ is solvable by radicals, then its Galois group $\operatorname{Gal}(E/k)$ is a solvable group.
Here $E$ is the splitting field of $f(x)$ over $k$. The proof relied on Theorem 4.20 (pg.213)
Let $k$ be a field and let $f(x)\in k[x]$ be solvable by radicals, so there is a radical extension $k=K_0\subseteq K_1\subseteq \dots \subseteq K_t$ with $K_t$ containing a splitting field $E$ of $f(x)$. If each $K_i/K_{i-1}$ is a pure extension of prime type $p_i$, where $p_i\neq \operatorname{char}(k)$, and if $k$ contains all $p_ith$ roots of unity, then the Galois group $\operatorname{Gal}(E/k)$ is a quotient of a solvable group.
(Hence it is a solvable group.)
In the proof, he constructed an extension $E^*$ of $E$ and an extension $k^*$ of $k$ so that $k^*$ contains appropriate roots of unity, thus $\operatorname{Gal}(E^*/k^*)$ is solvable by Theorem 4.20. The problem is, I don't know why he can get rid of the extra hypothesis $p_i\neq \operatorname{char}(k)$ for all $i$. I read the chapter twice, but still no clue.
Best Answer
I have the second edition of Rotman's book. The material is now slightly different.
The theorem in question is now Theorem 3.27, pp. 189:
The proof relies on Lemma 3.21, which is the counterpart of what you quote. This one now reads:
So, the assumption on the characteristic of $k$ has been dropped. This Lemma in turn relies on Theorems 3.17, Lemma 3.18, and Lemma 3.19. They are:
So the assumption on the characteristic of $k$ has been dropped entirely in the Second edition.