Exercise 8.2.5

Proof.

(a)

By hypothesis, $F\subset K_2$, so $F_0=F\subset K_2=F_0^{\prime }$.

Using induction, if we suppose that $F_{i-1}\subset F_{i-1}^{\prime }$ for some $i,1\le i<n$, then $F_{i-1}({\gamma }_i)\subset F_{i-1}^{\prime }({\gamma }_i)$, therefore $F_i\subset F_i^{\prime }$.

Conclusion: $\forall <!--\; FUNCTION\; APPLICATION\; -->i,0\le i\le n,F_i\subset F_i^{\prime }$.

Consequently, ${\gamma }_i^{m_i}\in F_{i-1}\subset F_{i-1}^{\prime },i=1,\cdots ,n$, so $K_2\subset F_n^{\prime }$ is a radical extension.

(b)

We show that $K_n^{\prime }=K_1{K}_2$.

$K_1=F({\gamma }_1,\cdots ,{\gamma }_n)$, and $F\subset K_2$, therefore $K_1{K}_2=K_2({\gamma }_1,\cdots ,{\gamma }_n)=F_n^{\prime }$.

Indeed, $F_n^{\prime }=K_2({\gamma }_1,\cdots ,{\gamma }_n)$ by construction, and

$\text{∙}$

$$\begin{array}{llll}\hfill & K_2({\gamma }_1,\cdots ,{\gamma }_n)\supset K_2,& \hfill & \\ \hfill & K_2({\gamma }_1,\cdots ,{\gamma }_n)\supset F({\gamma }_1,\cdots ,{\gamma }_n)=K_1,& \hfill & \end{array}$$

therefore $F_n^{\prime }\supset K_1,F_n^{\prime }\supset K_2$.

$\text{∙}$

If $K$ is any subfield of $L$ which contains $K_1,K_2$, $K\supset K_1=F({\gamma }_1,\cdots ,{\gamma }_n)$, so

$K\supset \{{\gamma }_1,\cdots ,{\gamma }_n\}$ and $K\supset K_2,$

therefore $K\supset K_2({\gamma }_1,\cdots ,{\gamma }_n)=F_n^{\prime }$.

So $F_n^{\prime }$ is the smallest subfield of $L$ which contains $K_1$ and $K_2$,

$$K_1{K}_2=F_n^{\prime }.$$

We conclude that $K_1{K}_2$ is a radical extension of $K_2$.