Exercise 8.5.1

Proof. The characteristic of $F$ is 0 in this section.

Let $L_1,L_2$ two splitting fields of $f\in F[x]$ over $F$. Then there exists an isomorphism $\phi :L_1\to L_2$ which is the identity on $F$.

Suppose that $F\subset L_1$ is a solvable extension.

As $L_1$ is a splitting field of $f$ over $F$, $F\subset L_1$ is a normal extension, and as the characteristic of $F$ is 0, this is a separable extension, so $F\subset L_1$ is a Galois extension.

Let $\zeta$ be a $m$th primitive root of unity, where $m=[L_1:F]$. Write $M_1=L_1(\zeta )$. As $F\subset L_1$ is a solvable Galois extension, by Corollary 8.3.4, $F\subset M_1$ is a radical extension, so there exist fields $F_i,i=1,\cdots ,m$ such that

and $F_i=F_{i-1}({\gamma }_i),{\gamma }_i\in F_i,{\gamma }_i^{m_i}\in F_{i-1},m_i>0(i=1,\cdots ,m)$.

Moreover, $M_1=L_1(\zeta )$ is the splitting field of $x^m-1$ over $L_1$. Let $M_2=L_2({\zeta }^{\prime })$ the splitting field of $x^m-1$ over $L_2$. By theorem 5.1.6, there exists an isomorphism $\bar{\phi }:M_1\to M_2$ such that $\phi =\bar{\phi }{\text{|}}_{L_1}$. Then $M_2$ is such that $F\subset L_2\subset M_2$.

Write $F_i^{\prime }=\bar{\phi }(F_i)$. Then

and $F_i^{\prime }=F_{i-1}^{\prime }({\gamma }_i^{\prime })$, where ${\gamma }_i^{\prime }=\bar{\phi }({\gamma }_i)\in F_i^{\prime }$ satisfies ${{\gamma }_i^{\prime }}^{m_i}=\bar{\phi }{({\gamma }_i)}^{m_i}=\bar{\phi }({\gamma }_i^{m_i})\in \bar{\phi }(F_{i-1})=F_{i-1}^{\prime }$.

So $F\subset M_2$ is radical, and $F\subset L_2$ is solvable.

By exchanging $L_1,L_2$, we show similarly that $F\subset L_2$ is solvable implies $F\subset L_1$ is solvable, so

$F\subset L_1$ is solvable if and only if $F\subset L_2$ is solvable. □