# 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.

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.

• Thanks in advance! His new treatment indeed solve the problem. Now I see it, he uses the assumption on the characteristic of $k$ to conclude that $K_i/K_{i-1}$ is a splitting field of a separable polynomial, then $G_{i-1}/G_i=\operatorname{Gal}(K_i/K_{i-1})$ is a cyclic group of prime order $p_i$. In the case where $\text{char}(K_0)=p_i$, the Galois group is $\{1\}$, as we see in the statement of Lemma 3.19. It does not effect the solvability of Galois group, so the assumption is not required. – withoutfeather Apr 30 at 6:12