Solvability by radicals of a separable field of characteristic $p$ Preliminary Definitions
A field $K$ is said to be solvable by proper radicals over $F$ if there exist fields $F_0,F_1,\dots,F_m$ such that $F=F_0\subseteq F_1\subseteq F_2\subseteq\dots\subseteq F_m$ and $K\subseteq F_m$ and for all $i$ we wave $F_i=F_{i-1}(a_i)$ with $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$.
A field $N$ is said to be metacyclic over $F$ if $N$ is Galois over $F$ and its Galois group is solvable.
For every $n$ not multiple of the characteristic of the field $F$, let $C_n$ be the $n$-th cyclotomic field.
My Question
I know that a Galois field $N$ over $F$ of characteristic $0$ is solvable by proper radicals if and only if its Galois group is solvable.
I would like to know what would be an equivalent condition about the Galois group of a Galois field $N$ over $F$ of characteristic $p$ for $N$ to be solvable by proper radicals over $F$, not including Artin-Schreier polynomials.
I found only a partial answer to this, namely:
Theorem 1: Let $P$ be a set of primes different from $p$ such that for every $r\in P$ then all prime factors of $r-1$ are in $P$. If $N$ is metacyclic of degree $n$ over $F$ of characteristic $p$ and every prime divisor of $n$ is in $P$, then $N$ is solvable by proper radicals over $F$.
Theorem 2: If a Galois field $N$ of degree $n$ over $F$ of characteristic $p$ is solvable by proper radicals over $F$, then $N$ is metacyclic over $F$ and $p\nmid n$.
My Attempt
I tried to prove some facts and theorems, but I do not know how to improve them to complete the problem. For example, the set $P$ in Theorem 1 looks so ill-behaved or broken in pieces, an example of such a set being the set of all primes less than $p$.
Remark: In the definition of solvability by proper radicals, we can suppose without loss of generality that $[F_i:F_{i-1}]$ is prime.
Indeed, if $K=F(a)$ and $a^{mn}\in F$ and $[K:F]=mn$, then let $L=F(a^n)$, then $K=L(a)$ and $L=F(a^n)$ and $a^n\in L$ and $(a^n)^m=a^{mn}\in F$, so that $[K:L]\leq n$ and $[L:F]\leq m$, but $[K:L][L:F]=[K:F]=mn$, so $[K:L]=n$ and $[L:F]=m$, so $K=L(a)$ and $a^{[K:L]}\in L$ and $L=F(a^n)$ and $(a^n)^{[L:F]}\in F$.
Fact 1: Let $F$ be a field containing a primitive $n$-th root of $1$. Let $E=F(a)$ where $a^n\in F$ and no smaller power of $a$ is in $F$. Then $E$ is a Galois extension of $F$ with cyclic Galois group of order $n$. Conversely, if $n$ is prime and $E$ is a cyclic extension of $F$ of degree n, then $E=F(a)$ for some $a$ with $a^n\in F$.
a) Suppose that $F$ has a primitive $n$-root $\zeta$ of unity. Let $K=F(a)$ e suppose that $n$ is the least such that $a^n\in F$. Then $x^n-a^n\in F[x]$ and $x^n-a^n=(x-a)(x-\zeta a)\dots(x-\zeta^{n-1}a)$ is separable and all its roots are in $K$, so $K$ is Galois over $F$. Let $G=\mathrm{Aut}(K:F)$. For every $\sigma \in G$, then $\sigma(a^n)=a^n$, so $(\sigma(a)/a)^n=1$, so let $\zeta_\sigma=\sigma(a)/a$, then $\zeta_\sigma\in F$. For $\sigma,\theta\in G$, then $\sigma\tau(a)=\sigma(\zeta_\tau a)=\zeta_\tau\sigma(a)=\zeta_\tau\zeta_\sigma a$, so $\zeta_{\sigma\tau}=\zeta_\tau\zeta_\sigma$. Hence $f:\sigma\mapsto\zeta_\sigma$ is an homomorphism from $G$ to the group $H$ of the $n$-roots of unity. If $\zeta_\sigma=1$, then $\sigma(a)=a$, so $\sigma=I$; then $f$ is injective. If $d=|\mathrm{Im}(f)|$, then for all $\sigma\in G$ we have $\zeta_\sigma^d=1$, so $(\sigma(a)/a)^d=1$, so $\sigma(a^d)=a^d$, so $a^d\in F$, so $d=n$, therefore $[K:F]=|G|=n$.
b) If $n$ is prime and $E=F(\xi)$ is cyclic of degree $n$ over $F$, then designate the roots of the minimum function of $\xi$ by $\xi_i=\sigma^{i}(\xi)$. These elements are in $E$ and so are:
$$a_i=\xi_0+\zeta^i\xi_1+\dots+\zeta^{i(n-1)}\xi_{n-1},\quad\quad(i=0,\dots,n-1).$$
The determinant:
$$d=\mathrm{det}(\xi^{ij}),\quad\quad(i,j=0,\dots,n-1)$$
is the Vandermonde determinant $\prod_{i<j}(\xi^i-\xi^j)\neq 0$, so that if the $a_i$ are all in $F$ then $\xi_i$ would be all in $F$, contrary to the hypothesis $E=F(\xi_0)\neq F$. So some $a_i$ is not in $F$ and this is not $a_0$ which is $\xi_0+\dots+\xi_{n-1}\in F$. So $\sigma(a_i)=\zeta^{-i}a_i$ and then the conjugates $\sigma^k(a_i)$, $k=0,\dots,n-1$ are all distinct by primeness of $n$, so the minimal polynomial of $a_i$ over $F$ has degree at least $n$, so $E=F(a)$ and $a^n\in F$ because $\sigma(a_i^n)=\zeta^{-ni}a_i^n=a_i^n$.
Fact 2: Let $N$ be metacyclic over $F$ of characteristic $p$ of degree $n=p_1\dots p_r$ not multiple of $p$ over $F$, and let $F$ fave primitive $p_i$-roots of unity for primes $p_i$, $i=1,\dots,r$. Then there are fields $F=F_0\subseteq\dots\subseteq F_m=N$ such that for all $i$ we wave $F_i=F_{i-1}(a_i)$ with $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$.
For $N$ has the form:
$$F=F_0\subseteq F_1\subseteq \dots\subseteq F_m=N$$
where $F_i$ is cyclic of prime degree $p_i\neq p$ over $F_{i-1}$, so by Fact 1 we have $F_i=F_{i-1}(a_i)$ with $a_i^{p_i}\in F_{i-1}$.
Fact 3: Let $p_1,\dots,p_r$ be primes different from $p$ and $C_{p_1},\dots,C_{p_r}$ the corresponding cyclotomic fields over $F$ over characteristic $p$. Then their composite $C$ is metacyclic over $F$.
The result is evident when $r=1$ since $C_{p_1}$ is cyclic over $F$. Assuming that the composite $D$ of $C_{p_1},\dots,C_{p_{r-1}}$ is metacyclic over $F$, then $D$ and $C_{p_r}$ are Galois over $F$, so their composite $C$ is Galois over $F$ and $\mathrm{Aut}(C:D)$ is isomorphic to a subgroup of $\mathrm{Aut}(C_{p_r}:F)$, which is isomorphic to a subgroup of the cyclic group $\mathbb{Z}_{p_r}^*$, so $\mathrm{Aut}(C:D)$ is cyclic. Then $C$ is metacyclic over $D$ and, since $D$ is metacyclic over $F$, the field is metacyclic over $F$.
Fact 4: Let $F$ be a field of characteristic $p$ and $P$ be a set of primes different from $p$ such that for every $r\in P$ then all prime factors of $r-1$ are in $P$. For every $r\in P$, if $C$ is the composite of the $C_q$ with $q\in P$ and $q\leq r$, then there are fields $F=F_0\subseteq \dots\subseteq F_m=C$ such that for all $i$ we have $F_i=F_{i-1}(a_i)$ and $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$.
Let $r\in P$, $C$ be the composite of $C_q$ with $q\in P$ and $q<r$ and $D$ the composite of $C$ and $C_r$. As $C_r$ is Galois over $F$, then $D$ is Galois over $C$ and $\mathrm{Aut}(D:C)$ is isomorphic to a subgroup of $\mathrm{Aut}(C_r:F)$, which is isomorphic to a subgroup of the cyclic group $\mathbb{Z}_r^*$, so $\mathrm{Aut}(D:C)$ is cyclic of order $d\mid r-1$, but then for every $q$ prime divisor of $d$, then $q\in P$ and $q<r$, so that $C$ has a primitive $q$-root of unity; so by Fact 2 we have $C=K_0\subseteq\dots\subseteq K_n=D$ such that for all $j$ we wave $K_j=K_{j-1}(b_j)$ with $b_j^{[K_j:K_{j-1}]}\in K_{j-1}$. By induction hypothesis, there are $F=F_0\subseteq \dots\subseteq F_m=C$ such that for all $i$ we have $F_i=F_{i-1}(a_i)$ and $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$.
Fact 5: If $K$ is a separable extenstion of $F$ of characteristic $p$ and is solvable by proper radicals, then there are fields $F=F_0\subseteq F_1\subseteq\dots\subseteq F_M$ such that $K\subseteq F_M$ and for all $i$ we have $F_i=F_{i-1}(a_i)$ and $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$ and $[F_i:F_{i-1}]$ prime different from $p$. In particular, $p\nmid[K:F]$.
If $F$ is a field of characteristic $p$ and $K=F(a)$ with $a^p\in F$ and $[K:F]=p$ and $L=K(b)$ with $b^q\in K$ and $[L:K]=q$ prime with $q\neq p$, then the minimal polynomial of $b$ over $K$ is $x^q-b^q$, which is separable over $K$, so the minimal polynomial $f$ of $b$ over $K(b^p)$ is also separable, but it also divides $(x-b)^p=x^p-b^p\in K(b^p)[x]$, so $f=x-b$ and $b\in K(b^p)$, therefore $K(b^p)=K(b)$. For all $t\in K$, we have $t=\sum_kc_ka^k$ with $c_k\in F$, so $t^p=\sum_kc_k^pa^{pk}\in F$. So $(b^p)^q=(b^q)^p\in F$. As $L=K(b^p)=F(b^p,a)$ and $a^p\in F\subseteq F(b^p)$, then $[L:F(b^p)]\leq p$, and also $(b^p)^q\in F$ so $[F(b^p):F]\leq q$, but $[L:F(b^p)][F(b^p):F]=[L:F]=[L:K][K:F]=qp$, then $[L:F(b^p)]=p$ and $a^p\in F(b^p)$ and $[F(b^p):F]=q$ and $(b^p)^q\in F$.
Therefore, if $F$ is a field of characteristic $p$ and $K$ is separable over $F$ and solvable by proper radicals over $F$, then there are $F=F_0\subseteq F_1\subseteq\dots\subseteq F_m$ such that $K\subseteq F_m$ and for all $i$ we have $F_i=F_{i-1}(a_i)$ with $a_i\in{[F_i:F_{i-1}]}\in F_{i-1}$ and $[F_i:F_{i-1}]$ prime, so by the preceding paragraph we can suppose without loss of generality that there is $M$ such that for all $i\leq M$ we have $[F_i:F_{i-1}]\neq p$ and for all $i>M$ we have $[F_i:F_{i-1}]=p$, so for all $\xi\in K$ then $\xi$ is separable over $F$, so $\xi$ is separable over $F_M$, but as $\xi\in F_m$, we have $\xi^p\in F_{m-1}$ and by induction we have $\xi^{p^{m-M}}\in F_M$, so the minimal polynomial $f$ of $\xi$ over $F_M$ divides $(x-\xi)^{p^{m-M}}=x^{p^{m-M}}-\xi^{p^{m-M}}\in F_M[x]$, and $f$ is separable over $F_M$, so $f=x-\xi$ and $\xi\in F_M$. Therefore $K\subseteq F_M$.
Theorem 1: Let $P$ be a set of primes different from $p$ such that for every $r\in P$ then all prime factors of $r-1$ are in $P$. If $N$ is metacyclic of degree $n$ over $F$ of characteristic $p$ and every prime divisor of $n$ is in $P$, then $N$ is solvable by proper radicals over $F$.
For let $N$ be metacyclic of degree $n$ over $F$ of characteristic $p$ such that every prime divisor of $n$ is in $P$, and let $r$ be the greatest prime divisor of $n$, and $C$ be the composite of the $C_q$ with $q\in P$ and $q\leq r$. The composite $D$ of $N$ and $C$ is Galois over $C$ and its Galois group is isomorphic to a subgroup $H$ of the group $G$ of $N$ over $F$. The group $G$ is solvable and so is $H$. Therefore $D$ is metacyclic over $C$ and its degree $m$ is a divisor of $n$, so that for every $q$ prime divisor of $m$ then $C$ has a primitive $q$-root of unity. By Fact 2, there are fields $C=K_0\subseteq\dots\subseteq K_m=N$ such that for all $j$ we wave $K_j=K_{j-1}(b_j)$ with $b_j^{[K_j:K_{j-1}]}\in K_{j-1}$, and by Fact 4 there are fields $F=F_0\subseteq \dots\subseteq F_m=C$ such that for all $i$ we have $F_i=F_{i-1}(a_i)$ and $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$. Therefore, $D$ is solvable by proper radicals over $F$.
Theorem 2: If a Galois field $N$ of degree $n$ over $F$ of characteristic $p$ is solvable by proper radicals over $F$, then $N$ is metacyclic over $F$ and $p\nmid n$.
Let $N$ be a Galois field over $F$ with automorphism group $G$ and assume that $N$ is solvable by proper radicals over $F$. Then, by Fact 5, there are $F=F_0\subseteq F_1\subseteq\dots\subseteq F_M$ such that $K\subseteq F_M$ and for all $i$ we have $F_i=F_{i-1}(a_i)$ and $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$ and $[F_i:F_{i-1}]=p_i$ prime different from $p$ and let $C$ be the composite of the cyclotomic fields $C_{p_i}$. The group $G'=\mathrm{Aut}(C:F)$ is solvable by Fact 3. If $D=N\cap C$ and $H=\mathrm{Aut}(N:D)$ and $H'=\mathrm{Aut}(C:D)$, then $D$ is Galois over $F$ and $\mathrm{Aut}(D:F)$ is isomorphic to both $G'/H'$ and $G/H$. But $G'/H'$ is solvable, and so $G/H$ is solvable. Also, if $N_0$ is the composite of $N$ and $C$, then $H_0=\mathrm{Aut}(N_0,C)$ is isomorphic to $H$. We define $L_t=C(a_1,\dots,a_M)$, then $L_i=L_{i-1}(a_i)$ and $L_{i-1}$ contains a primitive $p_i$-root of unity and $a_i^{p_i}\in L_{i-1}$, so that by primeness of $p_i$ and Fact 1 either $L_i$ is cyclic of prime degree $p_i$ over $L_{i-1}$ or $L_i=L_{i-1}$. Is $N_1$ is the composite of $N_0$ and $L_1$, then $H_1=\mathrm{Aut}(N_1,L_1)$ is isomorphic either to $H_0$ or a normal subgroup of index $p_1$ of $H_0$. In fact, $[H_0:H_1]=[N_0\cap L_1:C]$ and, since $[L_1:C]$ is $p_1$ or $1$, then $[N_0\cap L_1:C]$ is $p_1$ or $1$. We define $N_i$ as the composite of $N_{i-1}$ and $L_i$ and obtain a sequence of groups:
$$H_0\supseteq H_1\supseteq\dots\supseteq H_M,$$
where $H_i=\mathrm{Aut}(N_i:L_i)$ and either $H_i$ is $H_{i-1}$ or a normal subgroup of prime index $p_i$ of $H_{i-1}$. But $N_M$ is the composite of $N_0$ and $L_M$ and since $N\subseteq F_M$ we have $N_0\subseteq L_M$ and $N_M=L_M$, $H_M$ must be the identity group. Each $H_i\subset H_{i-1}$ is evidently a maximal normal divisor of $H_{i-1}$ and if we delete the repeated groups from the sequence $H_0,H_1,\dots,H_M$ we obtain a composition series of $H_0$. Then $H_0$ is solvable, and so is $H$. Therefore the group $G$ is solvable.
 A: Radical means a tower of extensions of the form $M(a)/M$ with $a^d\in M$ for some $d$.
For a finite characteristic $p$ extension $F/K$, let $L$ be the normal closure, $F/K$ is contained in a radical extension iff $Aut(L/K)$ is solvable and with
$L_n=L^{p^{n!}}K(\zeta_{n!}), K_n=K(\zeta_{n!})$, for some thus all $n$ large enough
$$p\nmid [L_n:K_n]$$

*

*This latter condition is because otherwise there is a subfield $E\subset L_m$ such that $L_m/E$ is cyclic of degree $p$ and $[L_n:EK_n]=p$ for all $n$,
$F/K$ contained in a radical extension implies that $L_n/EK_n$ is contained in a separable radical extension, for $n$ large enough we have the roots of unity to say it must be $EK_n(a)$ with $x^d-a^d\in EK_n[x]$ irreducible, which is inseparable since $p=[L_n:EK_n]$ must divide $d$, a contradiction.


*Conversely if $Aut(L/K)$ is solvable and $p\nmid [L_n:K_n]$ then for $n$ large enough: $L_n/K_n$ is a tower of cyclic extensions of degree coprime with $p$ and we have the roots of unity to use the same argument as in characteristic $0$, that $\sum_j \zeta_d^j \sigma^j(a)$ is a $d$-th root.
