Exercise 8.6.4

Proof.

(a)

We show that in the situation of diagram (8.24), where $M\subset L$ is Galois, and $[L:M]=n<\infty$, then $M(\gamma )\subset L(\gamma )$ is also a Galois extension.

By the Theorem of the Primitive Element (Theorem 5.4.1), as $M\subset L$ is Galois, there exists a separable element $\delta \in L$ such that $L=M(\delta )$. Let $f$ be the minimal polynomial of $\delta$ over $M$. Then $f\in M[x]$ is a separable polynomial, and as $M\subset L$ is normal, $f$ splits completely over $L$. Write ${\delta }_1=\delta ,{\delta }_2,\dots ,{\delta }_n$ the roots of $f$ in $L$.

Then

$$L(\gamma )=M(\delta ,\gamma )=M(\gamma )(\delta ).$$

Moreover, as ${\delta }_i\in L\subset L(\gamma )=M(\gamma )(\delta )$, with $\delta ={\delta }_1$, then $M(\gamma )(\delta )=M(\gamma )({\delta }_1,\dots {\delta }_n)$, so

$$M(\gamma )\subset L(\gamma )=M(\gamma )({\delta }_1,\dots {\delta }_n)$$

is the splitting field of the separable polynomial $f\in M[x]\subset M(\gamma )[x]$ over $M(\gamma )$.

This implies that $M(\gamma )\subset L(\gamma )$ is a Galois extension.

(b)

We show that $L$ lies in no radical extension of $M$, as in the proof of Proposition 8.6.4. and Exercise 1.

As the extension $M\subset K$ is radical, there exist ${\gamma }_1,\dots ,{\gamma }_n\in K$ 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 $L_0=L$ and $L_i=L({\gamma }_1,\dots ,{\gamma }_i)$. Then $M_0=M\subset L_0=L$ and $M_i\subset L_i$.

$[L_0:M_0]=[L:M]=p$. Reasoning by induction, 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 part (a) shows that $M_i\subset L_i$ is Galois, and the proof of Proposition 8.6.10 shows that

$$[L_i:M_i]=[L_{i-1}({\gamma }_i):M_{i-1}({\gamma }_i)]=p,i=1,\dots ,n,$$

therefore $[L({\gamma }_1,\dots ,{\gamma }_n):K]=[L_n:M_n]=[L_0:M_0]=[L:M]$, so

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

Moreover, $M\subset L$, and $K=M({\gamma }_1,\cdots ,{\gamma }_n)$, hence $\mathit{KL}=L({\gamma }_1,\cdots ,{\gamma }_n)$ in a fixed extension $\mathrm{\Omega }$ of $K$ and $L$. Indeed $L({\gamma }_1,\cdots ,{\gamma }_n)$ is the smallest subfield of $\mathrm{\Omega }$ containing $K$ and $L$, so $\mathit{KL}=L({\gamma }_1,\dots ,{\gamma }_n)$.

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

It follows that $\mathit{KL}\ne K$, so $L\not\subset K$ (see Exercise 1). Since $M\subset K$ is an arbitrary real radical extension of $M$, we conclude that $L$ cannot lie in a radical extension, so the extension $M\subset L$ is not solvable