Exercise 8.6.7

Proof. We follow the proof of (b) $\Rightarrow$ (a) in the proof of Theorem 8.3.3. The part "A Special Case" is unchanged.

Assume that $\mathrm{Gal}(L∕F)$ is solvable and that $p\nmid [L:F]$.

A Special Case. Assume first that $F$ satisfies the following special hypothesis:

We will prove that $F\subset L$ is radical in this situation. Since $\mathrm{Gal}(L∕F)$ is solvable, we have subgroups $\{1_L\}=G_n\subset \cdots \subset G_0=\mathrm{Gal}(L∕F)$ as in Definition 8.1.1. Then consider the fixed fields

Since the Galois correspondence is inclusion-reversing, this gives the fields

Furthermore, since $G_i$ is normal in $G_{i-1}$, the Galois correspondance together with Theorem 7.2.7 implies that

Since $[G_{i-1}:G_i]$ is prime, $\mathrm{Gal}(F_i∕F_{i-1})\simeq \mathbb{Z}∕\mathit{q\mathbb{Z}}$ for a prime $q$. By Exercise 5, we know that $q=|\mathrm{Gal}(F_i∕F_{i-1})|$ divides $|\mathrm{Gal}(L∕F)|$. By (8.12), $F$ and hence $F_{i-1}$ contain a primitive $q$th root of unity.

It follows that $F_{i-1}\subset F_i$ satisfies the conditions of Lemma 8.3.2. Thus $F_i$ is obtained from $F_{i-1}$ by adjunction of a $q$th root of an element of $F_{i-1}$:

This proves that $F\subset L$ is a radical extension when $F$ satisfies (8.12).

The General Case. Finally, we consider what happens when we only assume that $F\subset L$ is a Galois extension with solvable Galois group.

Write $m=[L:F]=|\mathrm{Gal}(L∕F)|$, then $m,p$ are relatively prime, so $m$ is invertible in $F$.

If $h=x^m-1$, then $h^{\prime }=m{x}^{m-1}$, where $m$ is invertible in $F$, so the Bézout’s relation $x{m}^{-1}{h}^{\prime }+h(-1)=x(x^{m-1})-(x^m-1)=1$ proves that $h\wedge h^{\prime }=1$, so $h$ is separable. Let $K$ a splitting field of $h$ over $L$. Let

the set of the roots of $h$ in $K$. Then ${𝕌}_m$ is a subgroup of $K^{\text{∗}}$, so ${𝕌}_m$ is cyclic (Proposition A.5.3), and as $h$ is separable, $|{𝕌}_m|=m$, hence ${𝕌}_m\simeq \mathbb{Z}∕\mathit{m\mathbb{Z}}$, so ${𝕌}_m$ has a generator $\zeta$, i.e. a $m$th primitive root of unity, and

(this is Exercise 8.3.1 in characteristic $p$, where $p\nmid m$.)

Thus $K=L(\zeta )$ is the splitting field of $\zeta$ over $L$, and the proof of Exercise 8.3.2 remains unchanged in characteristic $p$, so Lemma 8.3.1 is also valid in characteristic $p$.

So $F\subset L(\zeta )$ is a Galois extension and $\mathrm{Gal}(L(\zeta )∕F(\zeta )$ is solvable since $\mathrm{Gal}(L∕F)$ is, and

This isomorphisms comes from the homomorphism

given by restricting an automorphism of $L(\zeta )$ to $L$. Since $\mathrm{Gal}(L(\zeta )∕F(\zeta ))$ is a subgroup of $\mathrm{Gal}(L(\zeta )∕F)$, we have a homomorphism

also given by restriction to $L$. But the kernel of this map is the identity, since elements of the kernel are the identity on both $L$ and $F(\zeta )$. Thus $\phi$ is injective, which by Lagrange’s Theorem implies that

Now let $q$ be a prime dividing $|\mathrm{Gal}(L(\zeta )∕F(\zeta ))|$. Then $q$ divides $m$, so $q\ne p$. Since $\zeta$ is a primitive $m$th root of unity, ${\zeta }^{m∕p}$ is a primitive $p$th root of unity (see Exercise 8.3.6).

Since ${\zeta }^{m∕p}\in F(\zeta )$, we conclude that $F(\zeta )\subset L(\zeta )$ satisfies (8.12) with $F$ and $L$ replaced by $F(\zeta )$ and $L(\zeta )$, respectively. It follows that $F(\zeta )\subset L(\zeta )$ is radical by the Special Case. But $F\subset F(\zeta )$ is obviously radical ( ${\zeta }^m=1\in F$), so that $F\subset L(\zeta )$ is radical by part (a) of Proposition 8.2.7. Hence