# 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.

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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.

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Thanks for your very complete answer!! I have a question for you, what book I can use to study more of radical extensions? Some basic book . – Daniel Nov 18 '12 at 2:18
@Daniel Radical extensions appear in connection with Galois theory. I learned Galois theory from some textbooks on abstract algebra. For example, Van der Waerden, Lang. Van der Waerden is very good, but you may think it's too old. Lang is also very good, but you may think it's too terse. I heard Dummit-Foote is good. I also heard Rotman's Galois theory is good. There's a Milne's online course note on Galois theory. – Makoto Kato Nov 18 '12 at 3:16
Neither on Dummit nor Rotman talk about the radical closure :/ I'm actually using these texts – Daniel Nov 18 '12 at 23:30
@Daniel But you said radical extensions. I'm sure they talk about them. – Makoto Kato Nov 19 '12 at 0:08
But also about radical closure, and all that stuff :( – Daniel Nov 19 '12 at 0:26