Exercise 8.3.7

Proof. $F\subset L$ a solvable Galois extension, with $F$ of characteristic 0. $p_1,\cdots ,p_r$ are the distinct prime numbers which divide $[L:F]$.

(a)

Lemma 1. Suppose that two elements $a,b$of an Abelian group $G$are of respective order $p,q$, where $p,q$are relatively prime. Then the order of $\mathit{ab}$is $\mathit{pq}$.

Proof of Lemma 1:

$a^p=b^q=e$, therefore ${(\mathit{ab})}^{\mathit{pq}}={(a^p)}^q{(b^q)}^p=e$.

For all $k\in \mathbb{Z}$, since $p\wedge q=1$,

$${(\mathit{ab})}^k=e\Rightarrow {(\mathit{ab})}^{\mathit{qk}}=e\Rightarrow a^{\mathit{qk}}{b}^{\mathit{qk}}=e\Rightarrow a^{\mathit{qk}}=e\Rightarrow p\mid \mathit{qk}\Rightarrow p\mid k.$$

Similarly

$${(\mathit{ab})}^k=e\Rightarrow {(\mathit{ab})}^{\mathit{pk}}=e\Rightarrow a^{\mathit{pk}}{b}^{\mathit{pk}}=e\Rightarrow b^{\mathit{pk}}=e\Rightarrow q\mid \mathit{pk}\Rightarrow q\mid k.$$

Consequently, using again $p\wedge q=1$,

$${(\mathit{ab})}^k=e\Rightarrow (p\mid k\mathrm{and}q\mid k)\Rightarrow \mathit{pq}\mid k.$$

To conclude,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},{(\mathit{ab})}^k=e⟺\mathit{pq}\mid k,$$

The order of $\mathit{ab}$ is thus $\mathit{pq}$. □

Lemma 2. If $G$is an Abelian group and $c_1,\cdots ,c_r\in G$are of respective order $p_1,\cdots ,p_r$, where $p_1,\cdots ,p_r$are pairwise relatively prime, then the order of $c=c_1\cdots c_r$is $p_1\cdots p_r$.

Proof of Lemma 2: Using the induction hypothesis $|c_1\cdots c_k|=p_1\cdots p_k,k<r$, and applying Lemma 1 to $a=c_1\cdots c_k$ of order $p_1\cdots p_k$, and $b=c_{k+1}$ of order $p_{k+1}$, where $p_1\cdots p_k\wedge p_{k+1}=1$, then $|c_1\cdots c_{k+1}|=p_1\cdots p_{k+1}$. □

$\text{∙}$ Suppose that $F$ contains a root of unity $c$ of order $n=p_1\cdots p_r$. Write $c_i=c^{n∕p_i},i=1,\cdots r$. Then $c_i\in F$ and Exercise 6 proves that $c_i$ is of order $p_i$.

$\text{∙}$ Conversely, suppose that $F$ contains some elements $c_i$ of order $p_i$, for all $i,1\le i\le r$. The $p_i$ are distinct prime numbers, so are pairwise relatively prime.

Let $c=c_1\cdots c_r$. Lemma 2 applied in the Abelian group $G=F^{\text{∗}}$ shows that the order of $c_1\cdots c_r$ is $p_1\cdots p_r$.

Conclusion: if $p_1,\cdots ,p_r$ are distinct prime numbers, $F$ contains a primitive $(p_1\cdots p_r)$th of unity if and only if it contains primitive $p_i$th roots of unity for all $i=1,\cdots ,r$.

(b)

Suppose that $F$ contains a primitive $p_1\cdots p_r$th root of unity ${\zeta }_{p_1\cdots p_r}$. By part (a), $F$ contains also $p_i$th primitive roots of unity ${\zeta }_{p_i}$ for $i=1,\cdots ,r$.

So the condition (8.12) is satisfied: $F$ contains a $p$th primitive root of unity for all $p$ dividing $\mathrm{Gal}(L∕F)=[L:F]$. The part $(b)\Rightarrow (a)$ (special case) in the proof of Theorem 8.3.3 shows that $F\subset L$ is a radical extension.

(c)

Let $\zeta$ a $p_1\cdots p_r$th primitive root of unity.

As proven in the text, there exists an injective group homomorphism (8.13)

$$\mathrm{Gal}(L(\zeta )∕F(\zeta ))\to \mathrm{Gal}(L∕F)$$

thus $|\mathrm{Gal}(L∕F)|$ is a multiple of $|\mathrm{Gal}(L(\zeta )∕F(\zeta ))|$.

So $F(\zeta )$ contains a $p$th primitive root of unity for all $p$ dividing $|\mathrm{Gal}(L(\zeta )∕F(\zeta ))|$, and then the part (b) proves that $F(\zeta )\subset L(\zeta )$ is a radical extension. As $F\subset F(\zeta )$ is radical, by Lemma 8.2.7(a), $F\subset L(\zeta )$ is a radical extension.