3.10 Radical Extensions

Definition

We say $F/K$ is a radical extension, if $F=K(u_1,\cdots ,u_n)$ such that $u_1^{m_1}\in K$ for some $m_1⩾1$, and for any $i⩾2$, $u_i^{m_i}\in K(u_1,\cdots ,u_{i-1})$ for some $m_i⩾1$.

Facts.

• $F_1/K$ and $F_2/F_1$ are radical yield that $F_2/K$ radical.
• $F_1/K$ and $F_2/K$ are radical yield that $F_1{F}_2/K$ radical.

Definition

We say $f(T)\in K[T]$ is solvable by radical if there exists radical $F/K$ such that $f(T)$ splitting in $F$.

In this section, we aim to show a polynomial equation $f(T)$ is solvable by radical iff Galois group of $f(T)$ is solvable as a group, which is stated in ^7b35db.

Normal Closure

Lemma

Let $E/K$ be an algebraic extension. Then there exists an extension $N/E$ such that

• $N/K$ is normal;
• no proper subfield of $N$ containing $E$ is normal over $K$.

We call $N$ the normal closure of $E/K$.

Lemma

Let $F/K$ be an radical extension, and let $N$ be a normal closure of $F/K$. Then $N/K$ is also radical.

\begin{proof} Let $F=K(u_1,\cdots ,u_n)$. Note that $N$ is a splitting field over $K$ of $S=\{\mathrm{Irr}(u_i,K)=f_i\}$. Let $v$ be a root of one of $f_i$. Then we have an field automorphism ${\tau }_v:K(u_i)\stackrel{\sim }{\to }K(v)$. By ^28f933, there exists $\widetilde{{\tau }_v}:N\stackrel{\sim }{\to }N$. This deduces that $F\simeq \widetilde{{\tau }_v}(F)$ and $v\in \widetilde{{\tau }_v}(F)$. Note that $N$ is a composite of $\widetilde{{\tau }_v}(F)$ with $v$ running through all roots of $S$. But $\widetilde{{\tau }_v}(F)/K$ is radical because $F/K$ is radical. Hence, $N/K$ is radical. \end{proof}

Remark. Recall that we always assume that $\mathrm{char}F=0$, then $N/K$ is normal and radical yields that $N/K$ is Galois.

Theorem

Suppose that $\mathrm{char}K=0$. Let $F/K$ be a finite Galois extension such that $F/K$ is a radical extension. Then $\mathrm{Gal}(F/K)$ is solvable.

\begin{proof} Write $F=K(u_1,\cdots ,u_n)$, with $u_1^{m_1}\in K,\cdots ,u_n^{m_n}\in K(u_1,\cdots ,u_{n-1})$.

Let $m=m_1\cdots m_n$, and let $\zeta$ is a primitive $m$-th root of unity. Note that $\mathrm{Gal}(F/K)$ is solvable iff $\mathrm{Gal}(F(\zeta )/K)$ is solvable iff $\mathrm{Gal}(F(\zeta )/K(\zeta ))$ is solvable.

Let $E_0=K(\zeta )$, and let $E_i=K(\zeta ,u_1,\cdots ,u_i)$ for all $1⩽i⩽n$. Consider the following subfields and subgroups

$$\begin{array}{rl}& F(\zeta )=E_n\supseteq \cdots \supseteq E_0=K(\zeta ),\\ & 1=H_n<\cdots <H_i=\mathrm{Gal}(F(\zeta )/E_i)<\cdots <H_0=\mathrm{Gal}(F(\zeta )/K(\zeta )).\end{array}$$

We claim that the group series is a solvable series. Note that $u_i^{m_i}\in E_{i-1}$, $E_i=E_{i-1}(u_i)$ and ${\zeta }^{m/m_i}\in E_{i-1}$, by ^e267b2 we have $E_i/E_{i-1}$ is Galois and furthermore is a cyclic extension. It deduces that $H_i⊲H_{i-1}$ and $H_{i-1}/H_i$ is solvable. Therefore, $H_0=\mathrm{Gal}(F(\zeta )/K(\zeta ))$ is solvable and so for $\mathrm{Gal}(F/K)$. \end{proof}

Remark. The process involves first adjoining roots of unity, whose Galois group (a subgroup of $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$) is abelian and thus solvable; then, with these roots of unity present in the base field, all the “radical” elements are adjoined through a series of cyclic extensions, which are also solvable.

Corollary

Suppose $K\subseteq E\subseteq F$ is a finite extension with $F/K$ is radical. Then ${\mathrm{Aut}}_{K}E$ is solvable.

\begin{proof} Let $K_0=E^{{\mathrm{Aut}}_{K}E}$. Then $E/K_0$ is Galois, and ${\mathrm{Aut}}_{K}E={\mathrm{Aut}}_{K_0}E$. Still $F/K_0$ is still Galois. So WLOG assume $K=K_0$. Let $N$ be the normal extension of $F/K$. Then by ^bb62fa $N/K$ is also radical. So WLOG assume $F=N$. In summary, we assume $K\subseteq E\subseteq F$ with $E/K,F/E,F/K$ Galois. It deduces that $\mathrm{Gal}(E/K)=\mathrm{Gal}(F/K)/\mathrm{Gal}(F/E)$ is solvable. \end{proof}

Definition

We say $f(x)\in K[x]$ is solvable by radical, if there exists radical extension $R/K$ such that $f(x)$ splits over $R$.

Corollary

Suppose that $\mathrm{char}K=0$. If $f(x)\in K[x]$ is solvable by radical. Then its Galois group is solvable.

\begin{proof} Suppose $R/K$ is as in above definition. Then $R$ contains a splitting field $E$ of $f(x)$, that is, $K\subseteq E\subseteq R$ and $R/K$ is radical. Then by ^5c0107, ${\mathrm{Aut}}_{K}E$ is solvable. \end{proof}

abel

Suppose $\mathrm{char}k=0$. Consider the general polynomial $p_n(x)=x^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$ of degree $n$ over $K=k(a_0,\cdots ,a_{n-1})$. Then it is NOT solvable by radicals, if $n⩾5$.

\begin{proof} Let $E$ be a splitting field of $p_n(x)$ over $K$. One can prove that $\mathrm{Gal}(k(x_1,\cdots ,x_n)/k(f_1,\cdots ,f_n))\simeq \mathrm{Gal}(E/K)$ and so $\mathrm{Gal}(E/K)\simeq S_n$ by symmetric polynomial. Then it deduces that $p_n(x)$ with $n⩾5$ is not solvable by radicals.

For more details, see here. \end{proof}

Theorem

A polynomial equation $f(T)$ is solvable by radical iff Galois group of $f(T)$ is solvable as a group.