Exercise 6.5.3

Proof. We show that the Galois group $G$ of an Abelian equation is Abelian.

Let $L=F({\alpha }_1,\dots ,{\alpha }_n)$ a splitting field of $f\in F[x]$, and $\alpha ={\alpha }_1$.

By definition of an Abelian equation, there exists ${𝜃}_i\in F(x)$ tels que ${\alpha }_i={𝜃}_i(\alpha )(i=1,\dots ,n)$, so $L=F(\alpha )$ (see Exercise 1).

$\text{∙}$ $\sigma \in \mathrm{Gal}(L∕F)$, and $f\in F[x]$, thus $\sigma (\alpha )$ is also a root ${\alpha }_i,1\le i\le n$ of $f$:

$\sigma (\alpha )={\alpha }_i={𝜃}_i(\alpha )$. Similarly $\tau (\alpha )={𝜃}_j(\alpha ),1\le j\le n$.

$\text{∙}$ if $\mathit{\sigma \tau }=\mathit{\tau \sigma }$, then $\sigma (\tau (\alpha ))=(\mathit{\sigma \tau })(\alpha )=(\mathit{\tau \sigma })(\alpha )=\tau (\sigma (\alpha ))$.

Conversely, if $\sigma (\tau (\alpha ))=\tau (\sigma (\alpha ))$, then $(\mathit{\sigma \tau })(\alpha )=(\mathit{\tau \sigma })(\alpha )$.

As $L=F(\alpha )$, and as $\mathit{\sigma \tau }$ and $\mathit{\tau \sigma }$ are identity over $F$ and send $\alpha$ on the same element, $\mathit{\sigma \tau }=\mathit{\tau \sigma }$.

$\text{∙}$ $\sigma (\tau (\alpha ))=\sigma ({𝜃}_j(\alpha ))$. Moreover $\sigma$ is a $F$-automorphism of fields, and ${𝜃}_j\in F(x)$ a polynomial, thus $\sigma ({𝜃}_j(\alpha ))={𝜃}_j(\sigma (\alpha ))={𝜃}_j({𝜃}_i(\alpha ))$. Therefore $\sigma (\tau (\alpha ))={𝜃}_j({𝜃}_i(\alpha ))$. Similarly $\tau (\sigma (\alpha ))={𝜃}_i({𝜃}_j(\alpha ))$.

The equation $f=0$ being Abelian, ${𝜃}_j({𝜃}_i(\alpha ))={𝜃}_i({𝜃}_j(\alpha ))$, thus $\sigma (\tau (\alpha ))=\tau (\sigma (\alpha ))$, so $\mathit{\sigma \tau }=\mathit{\tau \sigma }$, and this is true for all $\sigma ,\tau \in \mathrm{Gal}(L∕F)$: $\mathrm{Gal}(L∕F)$ is Abelian.

Conclusion: the Galois group of an Abelian equation is Abelian. □