Understand a proof in Galois theory 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.
 A: 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:

Theorem 3.27 (Galois). 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 $\mathrm{Gal}(E/k)$ is a solvable group.

The proof relies on Lemma 3.21, which is the counterpart of what you quote. This one now reads:

Lemma 3.21. Let $k$ be a field,  let $f(x)\in k[x]$ be solvable by radicals, and let $k=K_0\subseteq K_1\subseteq\cdots\subseteq K_t$ be a tower with $K_i/K_{i-1}$ a pure extension of prime type $p_i$ for all $i$. If $K_t$ contains a splitting field $E$ of $f(x)$ and $k$ contains all the $p_i$th roots of unity, then the Galois group $\mathrm{Gal}(E/k)$ is a quotient of a solvable group.

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:

Theorem 3.17. Let $k\subseteq B\subseteq E$ be a tower of fields. If $B/k$ and $E/k$ are normal extensions, then $\sigma(B)=B$ for all $\sigma\in \mathrm{Gal}(E/k)$, $\mathrm{Gal}(E/B)\triangleleft \mathrm{Gal}(E/k)$, and $\mathrm{Gal}(E/k)/\mathrm{Gal}(E/B)\cong \mathrm{Gal}(B/k)$.
Lemma 3.18. (i) If $B=k(u_1,\ldots,u_t)/k$ is a finite extension field, then there is a normal extension $E/k$ containing $B$; that is, $E$ is a splitting field for some $f(x)\in k[x]$. If each $u_i$ is seprable over $k$, then $f(x)$ is a separable polynomial and, if $G=\mathrm{Gal}(E/k)$, then $E=k(\sigma(u_1),\ldots,\sigma(u_t)\colon \sigma\in G)$.
(ii) If $B/k$ is a radical extension, then the normal extension $E/k$ is a radical extension.
Lemma 3.19. Let $k=K_0\subseteq K_1\subseteq K_2\subseteq \cdots\subseteq K_t$ be a tower with each $K_i/K_{i-1}$ a pure extension of prime type $p_i$. If $K_t/k$ is a normal extension and $k$ contains all the $p_i$th roots of unity, for $i=1,\ldots,t$, then there is a sequence of groups $\mathrm{Gal}(K_t/k)=G_0\supseteq G_1\supseteq G_2\supseteq\cdots\supseteq G_t=\{1\}$, with each $G_{i+1}\triangleleft G_i$ and $G_i/G_{i+1}$ cyclic of prime order $p_{i+1}$ or $\{1\}$.

So the assumption on the characteristic of $k$ has been dropped entirely in the Second edition.
