a property of the radical closure of a field Definition 1: Let $K$ be a field. If $\alpha$ is an algebraic element over K, such that $\alpha^n \in K $ and such that $x^n-\alpha^n \in K[x] $is also irreducible over K. Then we call $ K(\alpha)$ a simple radical and irreducible extension.
Definition 2:
Let $K$ be a field with $char(K)=0$ , let's define $ K^{rad}\subset \overline{K}$ by:
$$
K^{rad}  = \bigcup\limits_{i \geqslant 1} {K_i } 
$$
where $$
K_1  = K
$$
$$
K_{i + 1}  = K_i \left( {\alpha  \in \overline K ;\exists n;\alpha ^n  \in K{}_i} \right)
$$
Problem:
Given $x\in K^{rad}$ , prove that there exist $n$ and a tower of fields $$ K=E_0 \subset E_1 \subset .... \subset E_n \subset \overline{K}$$
such that $ x\in E_n$ and $ E_{i-1} \subset E_i $ is a simple radical and irreducible extension.
 A: Let $K$ be a field of characteristic $0$.
Let $p$ be a prime number.
Let $\alpha$ be an element of an algebraic closure of $K$ such that $\alpha^p \in K$ and $X^p - \alpha^p$ is irreducible over $K$. Then we call $K(\alpha)/K$ a simple prime radical extension.
Let $K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ be a tower of fields.
If $K_i/K_{i-1}$ is a simple prime radical extension for $i =1, \dots, n$, we call $L/K$ is a prime radical extension.
Let $L/K$ be a field extension.
If there exists a prime radical extension $E/K$ such that $L \subset E$, we call $L/K$ a prime radically solvable extension.
Lemma 1
Let $K$ be a field.
Let $p$ be a prime number such that $char(K) \neq p$.
Let $a \in K$.
Suppose $X^p - a$ has no roots in $K$.
Then $X^p - a$ is irreducible over $K$.
Proof:
Let $\bar K$ be an algebraic closure of $K$.
Let $\zeta$ be a $p$-th root of unity in $\bar K$.
Let $\alpha$ be a root of $X^p - a$ in $\bar K$.
Then $\alpha, \alpha\zeta, \dots, \alpha\zeta^{p-1}$ are distinct roots of $X^p - a$.
Suppose $X^p - a$ is not irreducible over $K$.
Then $X^p - a = g(X)h(X)$, where $g(X)$ and $h(x)$ are monic irreducible polynomials of degree $> 0$.
Suppose deg $g(X) = k$ and $b$ is the constant term of $g(X)$.
Then $b = \pm\alpha^k\zeta^m$ for some integer $m$.
Hence $b^p = \pm a^k$ if $p$ is odd, and $b^p = a^k$ if $p = 2$.
Hence there exists $c \in K$ such that $c^p = a^k$.
Let $K^*$ be the multiplicative group of $K$.
Let $(K^*)^p = \{x^p| x \in K^*\}$.
Let $\gamma$ be the image of $a$ by the canonical homomorphism $K^* \rightarrow K^*/(K^*)^p$.
Then $\gamma \neq 1$ and $\gamma^p = 1$.
Hence the order of $\gamma$ is $p$.
Since $c^p = a^k$, $\gamma^k = 1$.
This is a contradiction.
QED
Lemma 2
Let $K$ be a field.
Let $p$ be a prime number such that $char(K) \neq p$.
Let $\alpha$ be an element of an algebraic closure of $K$ such that $\alpha^p \in K$.
Let $\alpha^p = a$.
Then one of the following three cases occurs.
1) $X^p - a$ is irreducible over $K$.
2) $K(\alpha) = K(\zeta)$, where $\zeta$ is a primitive $p$-th root of unity.
3) $K(\alpha) = K$.
Proof:
Let $\bar K$ be an algebraic closure containing $\alpha$.
Let $\zeta$ be a primitive $p$-th root of unity in $\bar K$.
Suppose $X^p - a$ is not irreducible.
By Lemma 1, there exists $b \in K$ such that $b^p = a$.
Since $b, b\zeta,\dots,b\zeta^{p-1}$ are all the roots of $X^p - a$,
$\alpha = b\zeta^i, 0 \le i \le p -1$.
If $i = 0, K(\alpha) = K$.
If $i > 0$, $\zeta^i$ is a primitive $p$-th root of unity.
Hence $K(\alpha) = K(b\zeta^i) = K(\zeta^i) = K(\zeta)$.
QED
Lemma 3
Let $K$ be a field.
Let $n > 0$ be an integer which is not divisible by $char(K)$.
Suppose $K$ contains a primitive $n$-th root $\zeta$ of unity.
Let $a \in K$.
Let $\alpha$ be a root of $X^n - a$ in an algebraic closure of $K$.
Then $K(\alpha)/K$ is a cyclic extension and $(K(\alpha)\colon K)$ is a divisor of $n$.
Proof:
If $a = 0$, the assertion is trivial. Hence we assume $a \neq 0$.
Hence $\alpha \neq 0$.
Hence $\alpha, \alpha\zeta,\dots,\alpha\zeta^{n-1}$ are the distincts roots of $X^n - a$.
Since $\zeta \in K, K(\alpha, \alpha\zeta,\dots,\alpha\zeta^{n-1}) = K(\alpha)$.
Hence $K(\alpha)/K$ is a Galois extension.
Let $G$ be the Galois group of $K(\alpha)/K$.
Let $\Gamma$ be the set of $n$-th roots of unity in $K$.
$\Gamma$ is a cyclic group of order $n$.
Let $\sigma \in G$.
Since $\sigma(\alpha)$ is a root of $X^n - a$, there exists a unique $\omega \in \Gamma$ such that $\sigma(\alpha) = \alpha\omega$.
Hence we get a map $\psi\colon G \rightarrow \Gamma$ such that $\sigma(\alpha) = \alpha\psi(\alpha)$.It is easy to see that $\psi$ is an injective homomorphism.
QED
Lemma 4
Let $\Omega$ be an algebraically closed field of characteristic $0$.
Let $p$ be a prime number.
Let $L/K$ be a prime radical extension of degree $p$, where $L$ is a subfield of $\Omega$.
Let $\zeta$ be a primitive $p$-th root of unity in $\Omega$.
Let $F$ be a subfield of $\Omega$ containing $\zeta$.
Then $FL/FK$ is a trivial extension or a simple prime radical extension of degree $p$.
Proof:
Since $L/K$ is a prime radical extension of degree $p$, there exists $\alpha \in L$ such that $L = K(\alpha)$ and $\alpha^p \in K$.
Then $FL = FK(\alpha)$ and $\alpha^p \in K \subset FK$.
By Lemma 3, $FL/FK$ is a trivial extension or a cyclic extension of degree $p$.
Suppose $FL/FK$ is a cyclic extension of degree $p$.
Then $X^p - a$ is the minimal polynomial of $\alpha$ over $FK$, where $a = \alpha^p$.
Hence $X^p - a$ is irreducible over $FK$.
Hence $FL/FK$ is a simple prime radical extension of degree $p$.
QED
Lemma 5
Let $\Omega$ be an algebraically closed field of characteristic $0$.
Let $L/K$ be a prime radical extension of degree $n$, where $L$ is a subfield of $\Omega$.
Let $p_1,\dots,p_r$ be all the distinct prime factors of $n$.
Let $m = \prod_i p_i$.
Let $\zeta$ be a primitive $m$-th root of unity in $\Omega$.
Let $F$ be a subfield of $\Omega$ containing $\zeta$.
Then $FL/FK$ is a prime radical extension and $(FL \colon FK)$ is a divisor of $n$.
Proof:
Since $L/K$ is a prime radical extension, there exists a tower of subfields of L:
$K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ such that each $K_i/K_{i-1}$ is a simple radical extension.
Since $n = \prod_i (K_i\colon K_{i-1})$, each $(K_i\colon K_{i-1})$ is one of the primes $p_1, \dots,p_r$.
Hence $F$ contains a primitive $p_i$-th unity for each $i$.
By Lemma 4, $FK_i/FK_{i-1}$ is trivial or a simple prime radical extension.
Hence the assertions follow.
QED
Lemma 5.5
Let $\Omega$ be an algebraically closed field.
Let p be a prime number such that $p \neq char(\Omega)$.
Let $L/K$ be a cyclic extension of degree $p$, where $L$ is a subfield of $\Omega$.
Let $\zeta$ be a primitive $p$-th root of unity in $\Omega$.
Then $L(\zeta)/K(\zeta)$ is a trivial or a simple prime radical extension.
Proof:
Since $L(\zeta) = LK(\zeta)$, $Gal(L(\zeta)/K(\zeta)$) is isomorphic to a subgroup of $Gal(L/K)$.
Hence $|Gal(L(\zeta)/K(\zeta))| = 1$ or $p$.
Suppose $|Gal(L(\zeta)/K(\zeta))| = p$.
Then by this theorem, $L(\zeta)/K(\zeta)$ is a simple prime radical extension.
QED
Lemma 6
Let $K$ be a field of characteristic $0$.
Let $\zeta$ be a primitive $n$-th root of unity in an algebraic closure $\bar K$ of $K$.
Then $K(\zeta)/K$ is a prime radically solvable extension.
Proof:
We use induction on $n$.
Since the assertion of the lemma is clear if $n = 1$, we assume that $n > 1$.
Let $G$ be the Galois group of $L/K$, where $L = K(\zeta)$.
$G$ is isomorphic to a subgroup of $(\mathbb{Z}/n\mathbb{Z})^\times$.
Hence, by Galois theory, there exists a tower: $K = K_0 \subset K_1 \subset \cdots \subset K_r = L$
such that each $K_i/K_{i-1}$ is a cyclic extension of a prime degree.
Let $p_1,\dots,p_s$ be all the distinct prime divisors of $|G|$.
Since $|G| = \prod_i (K_i \colon K_{i-1})$, each $(K_i \colon K_{i-1})$ is one of $p_1,\dots,p_s$.
Let $m = \prod_j p_j$.
Let $\eta$ be a $m$-th primitive root of unity in $\bar K$.
By Lemma 5.5, $K_i(\eta)/K_{i-1}(\eta)$ is trivial or a simple prime radical extension.
Hence, $L(\eta)/K(\eta)$ is a prime radical extension.
Since $m \le |G| \le |(\mathbb{Z}/n\mathbb{Z})^*| < n$, by the induction hypothesis, there exists a prime radical extension $E/K$ such that $K(\eta) \subset E$.
By Lemma 5, $EL(\eta)/EK(\eta) = EL/EK = EL/E$ is a prime radical extension.
Hence $EL/K$ is a prime radical extension.
Since $L \subset EL$, $L/K$ is a prime radically solvable extension.
QED
Lemma 7
Let $\Omega$ be a field of characteristic $0$.
Let $K$ be a subfield of $\Omega$.
Let $L/K$ be a simple prime radical subextension of $\Omega/K$.
Let $F/K$ be a subextension of $\Omega/K$.
Then $FL/F$ is a prime radically solvable extension.
Proof:
Since $L/K$ is a simple prime radical extension, there exist a prime number $p$ and $\alpha \in L$ such that $L = K(\alpha)$ and $\alpha^p \in K$.
Then $FL = F(\alpha)$.
By Lemma 2 and Lemma 6, the assertion follows.
QED
Lemma 8
Let $\Omega$ be an algebraically closed field of characteristic $0$.
Let $K \subset M \subset L$ be a tower of subfields of $\Omega$.
Suppose $M/K$ is a prime radically solvable extension and $L/M$ is a simple prime radical extension.
Then $L/K$ is a prime radically solvable extension.
Proof:
There exists a tower $K \subset M \subset F$ of subfields of $\Omega$ such that $F/K$ is a prime radical extension. By Lemma 7, $FL/F$ is a prime radically solvable.
Hence $FL/K$ is also a prime radically solvable extension.
Hence the assertion follows.
QED
Lemma 9
Let $K \subset M \subset L$ be a tower of fields.
Suppose $M/K$ is a prime radically solvable extension and $L/M$ is a prime radical extension.
Then $L/K$ is a prime radically solvable extension.
Proof:
This follows immediately from Lemma 8.
Lemma 10
Let $K \subset M \subset L$ be a tower of fields.
Suppose $M/K$ and $L/M$ are prime radically solvable extensions.
Then $L/K$ is also a prime radically solvable extension.
Proof:
This follows immediately from Lemma 9.
Lemma 11
Let $K$ be a field of characteristic $0$.
Let $n > 1$ be an integer.
Let $\alpha$ be an element of an algebraic closure of $K$ such that $\alpha^n \in K$.
Then $K(\alpha)/K$ is a prime radically solvable extension.
Proof:
We use induction on $n$.
If $n = 2$, the assertion is clear.
We assume $n > 2$.
Let $p$ be a prime divisor of $n$.
Let $m = \frac{n}{p}$.
Since $(\alpha^p)^m = \alpha^n \in K$, by the induction hypothesis, $K(\alpha^p)/K$ is a prime radically solvable extension.
By Lemma 2 and Lemma 6, $K(\alpha)/K(\alpha^p)$ is a prime radically solvable extension.
Hence, by Lemma 10, $K(\alpha)/K$ is a prime radically solvable extension.
QED
Let $K$ be a field of characteristic $0$.
Let $L/K$ be an extension.
Let $n > 0$ be an integer.
Suppose there exists an element $\alpha$ of $L$ such that $L = K(\alpha)$ and $\alpha^n \in K$.
Then we call $L/K$ is a simple radical extention.
Let $K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ be a tower of fields.
If $K_i/K_{i-1}$ is a simple radical extension for for $i =1, \dots, n$, we call $L/K$ is a radical extension.
Let $M/K$ be an extention.
If there esists a radical extension $L/K$ such that $M$ is a subfield of $L$, $M/K$ is called a radically sovable extension.
Lemma 12
Let $\Omega/K$ be an extension of a field $K$ of characteristic $0$.
Let $L/K$ be a radically solvable subextension of $\Omega/K$.
Let $F/K$ be a subextension of $\Omega/K$.
Then $FL/F$ is radically solvable.
Proof:
There exists a tower:
$K = K_0 \subset K_1 \subset \cdots \subset K_n = E$ such that $L \subset E$, where $K_i/K_{i-1}$ is a simple radical extension for for $i =1, \dots, n$
Then $F = FK_0 \subset FK_1 \subset \cdots \subset FK_n = FE$.
Since $FK_i/FK_{i-1}$ is a simple radical extension, $FL/F$ is radically solvable.
QED
Lemma 13
Let $K$ be a field of characteristic $0$.
Let $K \subset M \subset L$ be a tower of fields.
Suppose $M/K$ and $L/M$ are radically solvable.
Then $L/K$ is also radically solvable.
Proof:
There exists a tower:
$K = K_0 \subset K_1 \subset \cdots \subset K_n = F$ such that $M \subset F$, where $K_i/K_{i-1}$ is a simple radical extension for for $i =1, \dots, n$
By Lemma 12, $FL/F$ is radically solvable.
Hence $L/K$ is radically solvable.
QED
Lemma 14
Let $K$ be a field of characteristic $0$.
Let $x \in K^{rad}$.
Then $K(x)/K$ is radically solvable.
Proof:
For each integer $i \ge 1$, let $K_i$ be the subfield of $\bar K$ as defined by the title question.
Let $r$ be the least integer such that $x \in K_r$.
We use induction on $r$.
If $r = 1$, the assertion is clear.
Suppose $r > 1$.
Then there exists a positive integer $n$ such that $x^n \in K_{r-1}$.
By the induction hypothesis, $K(x^n)/K$ is radically solvable.
Clearly $K(x)/K(x^n)$ is radically solvable.
Hence, by Lemma 13, $K(x)/K$ is radically solvable.
QED
Lemma 15
A radically solvable extension $L/K$ is a prime radically solvable extension.
Proof:
This follows immediately from Lemma 10 and Lemma 11.
Proposition
Let $K$ be a field of characteristic $0$.
Let $x \in K^{rad}$.
Then $K(x)/K$ is prime radically solvable.
Proof:
This follows immediately from Lemma 14 and Lemma 15.
