Exercise 7.9

Proof. Let $k=[K:F]$. From hypothesis, $k\wedge n=1$, so there exist integers $u,v$ such that $\mathit{uk}+\mathit{vn}=1$.

As $n\mid q-1$, $n\wedge (q-1)=n$, the hypothesis " $x^n=\alpha$ is not solvable in $F$" implies that ${\alpha }^{(q-1)∕n}\ne 1$ (Prop. 7.1.2).

Write $\omega ={\alpha }^{(q-1)∕n}$, so $\omega \ne 1$ and ${\omega }^n=1$.

As $n\mid q-1$, $n\mid q^k-1$ and

Moreover $1+q+\cdots +q^{k-1}\equiv k(\mathit{mod}n)$, and ${\omega }^n=1$, so ${\alpha }^{(q^k-1)∕n}={\omega }^k$.

If ${\omega }^k=1$, then $\omega ={\omega }^{\mathit{uk}+\mathit{vn}}={({\omega }^k)}^u{({\omega }^n)}^v=1$, which is in contradiction with $\omega ={\alpha }^{(q-1)∕n}\ne 1$.

Thus ${\alpha }^{(q^k-1)∕n}={\omega }^k\ne 1$. This proves that the equation $x^n=\alpha$ has no solution in $K$. □