Exercise 6.5.6

Proof. Suppose that $F\subset L$ is normal and separable and that $G=\mathrm{Gal}(L∕F)$ is an Abelian group.

(a)

As $F\subset L$ is separable, the Theorem of the Primitive Element shows that there exists a separable element $\alpha \in L$ such that $L=F(\alpha )$.

(b)

Let $f$ be the minimal polynomial of $\alpha$ over $F$. Then $f$ is irreducible and separable. As $F\subset L$ is normal, the roots ${\alpha }_1=\alpha ,\dots ,{\alpha }_n$ of $f$ are all in $L$, so $L=F(\alpha )=F({\alpha }_1,\dots ,{\alpha }_n)$ is the splitting field of $f$. By Exercise 1, there exist polynomials ${𝜃}_i\in F[x]$ such that ${\alpha }_i={𝜃}_i(\alpha ),i=1,\dots ,n$.

Let $1\le i,j\le n$. As $f$ is separable and irreducible, by Proposition 6.3.7, the Galois group $G$ acts transitively on the set of the roots of $f$, so there exists $\sigma ,\tau \in G$ such that ${𝜃}_i(\alpha )=\sigma (\alpha )$ and ${𝜃}_j(\alpha )=\tau (\alpha )$.

Exercise 3 shows that $(\mathit{\sigma \tau })(\alpha )={𝜃}_j({𝜃}_i(\alpha ))$ and $(\mathit{\tau \sigma })(\alpha )={𝜃}_i({𝜃}_j(\alpha ))$. As $G$ is Abelian by hypothesis, $\mathit{\sigma \tau }=\mathit{\tau \sigma }$, so

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

The equation $f=0$ is Abelian.

Conclusion: If the finite extension $F\subset L$ is normal and separable and has an Abelian Galois group, and if $f$ is the minimal polynomial of a primitive element $\alpha$, then $f=0$ is an Abelian equation.