Exercise 8.6.1

Proof.

To achieve the induction in part (a), it is necessary, in the situation of diagram 8.24, to prove that $M(\gamma )\subset L(\gamma )$ is a Galois extension, knowing that the extension $L\subset M$ is a Galois extension. This is done in Exercise 4.

(a)

As the extension $M\subset K$ is radical, there exist ${\gamma }_1,\dots ,{\gamma }_n\in K\subset ℝ$ such that the subfields $M_k=M({\gamma }_1,\dots ,{\gamma }_k),k=0,\dots ,n$, of $K$ satisfy $$M_0=M\subset M_1\subset \cdots \subset M_n=K,$$

and $M_i=M_{i-1}({\gamma }_i),{\gamma }_i\in M_i,{\gamma }_i^{m_i}\in M_{i-1},m_i>0$, with $m_i$ prime (Lemma 8.6.2).

Let $L_0=L,L_i=L({\gamma }_1,\dots ,{\gamma }_i)$. Then $M_0=M\subset L_0=L$ and $M_i\subset L_i$.

$[L_0:M_0]=[L:M]=p$. Reasoning by induction for $i=1,\dots ,n$, we suppose that $M_{i-1}\subset L_{i-1}$ is a Galois extension and $[L_{i-1}:M_{i-1}]=p$ (where $p$ is the odd prime $[L:M]$). As $L_i=L_{i-1}({\gamma }_i),M_i=M_{i-1}({\gamma }_i),i=1,\cdots ,n$, the Exercise 8.6.4(a) shows that $M_i\subset L_i$ is Galois, and the proof of Proposition 8.6.4 shows that $[L_i:M_i]=[L_{i-1}({\gamma }_i):M_{i-1}({\gamma }_i)]=p$, and the induction is done. Therefore $[L({\gamma }_1,\dots ,{\gamma }_n):K]=[L_n:M_n]=[L_0:M_0]=[L:M]$, so

$$[L({\gamma }_1,\dots ,{\gamma }_n):K]=[L:M]=p.$$

Moreover, $M\subset L$, and $K=M({\gamma }_1,\cdots ,{\gamma }_n)$, hence $\mathit{KL}=L({\gamma }_1,\dots ,{\gamma }_n)$. Indeed $ℝ$ is a field which contains $K$ and $L$, and $L({\gamma }_1,\cdots ,{\gamma }_n)$ is the smallest subfield of $ℝ$ containing $K$ and $L$, so $\mathit{KL}=L({\gamma }_1,\dots ,{\gamma }_n)$.

$$[\mathit{KL}:K]=[L:M]=p.$$

(b)

We show that $$\mathit{KL}=K⟺L\subset K.$$

( $\Leftarrow$) If $L\subset K$, $K$ is the smallest subfield of $ℝ$ containing $L$ and $K$, so $\mathit{KL}=K$.

$\text{(⇒)}$ If $\mathit{KL}=K$, then $L\subset \mathit{KL}=K$, so $L\subset K$.