Exercise 8.2.6

Proof.

Let $\sigma \in \mathrm{Gal}(M∕F)$.

There exists an ascending series ${(F_i)}_{0\le i\le n}$ of subfields of $L$ such that

where ${\gamma }_i^{m_i}\in F_{i-1}$.

Write $F_i^{\prime }=\sigma F_i,i=0,\cdots ,n$, and ${\gamma }_i^{\prime }=\sigma ({\gamma }_i)$. Then $F_0^{\prime }=\sigma F_0=\mathit{\sigma F}=F$ and $F_n^{\prime }=\mathit{\sigma L}$.

As $F_i=F_{i-1}({\gamma }_i),i=1,\cdots ,n$, then

Therefore

Moreover, ${({\gamma }_i^{\prime })}^{m_i}=\sigma {({\gamma }_i)}^{m_i}=\sigma ({\gamma }_i^{m_i})\in \sigma (F_{i-1})=F_{i-1}^{\prime }$.

Consequently $F\subset \mathit{\sigma L}$ is a radical extension. □