The normal closure of a radical extension is again radical

We induct on the length $r$ of the radical tower. For $r=0$, the extension is $F\mathrm{/}F$, whose normal closure is itself.

Assume $r>0$, let $N\mathrm{/}F$ be the normal closure of $F_{r-1}\mathrm{/}F$, and write $F_r=F_{r-1}(\alpha ),{\alpha }^m=a\in F_{r-1}\mathrm{.}$

If $\alpha =0$, then $F_r=F_{r-1}$, so the normal closure of $F_r\mathrm{/}F$ is just $N$.

Assume instead that $\alpha \mathrm{\ne }0$. Because $N\mathrm{/}F$ is normal and contains $F_{r-1}$, every $F$-conjugate of $a$ lies in $N$. Let $a_1,\dots ,a_s\in N$ be the distinct conjugates of $a$ over $F$, and choose roots ${\alpha }_j$ with ${\alpha }_j^m=a_j$. Every $F$-conjugate of $\alpha$ is then a nonzero root of some polynomial $x^m-a_j$, so [L1] says it has the form $\zeta {\alpha }_j$ for some $\zeta \in {\mu }_m$. Therefore the normal closure $M$ of $F_r\mathrm{/}F$ is exactly $M=N({\alpha }_1,\dots ,{\alpha }_s,{\mu }_m)\mathrm{.}$

By the induction hypothesis, $N\mathrm{/}F$ is radical.

In the case of step 2.2, the field $M$ is radical over $N$: adjoin the finitely many ${\alpha }_j$ one at a time, each by one equation $x^m-a_j$, and then adjoin generators of ${\mu }_m$ by roots of $x^m-1$. Concatenating that tower with the radical tower for $N\mathrm{/}F$ from step 2.3 shows that $M\mathrm{/}F$ is radical.

Step 2.1 handles the case $\alpha =0$. Otherwise step 2.2 identifies the normal closure as $M$, and step 3.1 shows that $M\mathrm{/}F$ is radical. Thus the induction closes, so the normal closure of every finite radical extension is radical.