Exercise 6.5.7

Proof. Suppose that the implication (a) $\Rightarrow$(b) of Theorem 6.5.5 is true.

Let $f\in \mathbb{Q}[x]$ such that the equation $f=0$ is Abelian. Then $f$ has a root $\alpha$ such that $L=F(\alpha )$ is the splitting field of $F$, so the extension $F\subset L$ is normal. By Theorem 6.5.3 (and Exercise 3), as the equation $f=0$ is Abelian, $\mathrm{Gal}(L∕\mathbb{Q})$ is an Abelian group. The hypothesis (a) is so satisfied, and (b) follows : $L\subset \mathbb{Q}({\zeta }_n)$, where ${\zeta }_n=e^{2\mathit{i\pi }∕n}$. As the roots of $f$ are in $L$, these roots can be expressed rationally in terms of a root of unity.

Conversely, suppose that the roots ${\alpha }_1,\dots ,{\alpha }_n$ of any Abelian equation $f=0$ in the splitting field of $f$ can be expressed rationally in terms of a root of unity ${\zeta }_n$, and suppose also (a) : $\mathbb{Q}\subset L\subset ℂ$, the extension $\mathbb{Q}\subset L$ is normal, and $\mathrm{Gal}(L∕\mathbb{Q})$ is an Abellan group.

As the characteristic of $\mathbb{Q}$ is 0, $\mathbb{Q}\subset L$ is also separable, and there exists a primitive element $\alpha$ for the extension $\mathbb{Q}\subset L$. Let $f$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$. By Exercise 6, since $\mathbb{Q}\subset L$ is normal and separable, the equation $f=0$ is Abelian. By hypothesis, the roots ${\alpha }_1=\alpha ,\dots ,{\alpha }_n$ of $f$ can be expressed rationally in terms of a root of unity ${\zeta }_n$, therefore ${\alpha }_i\in \mathbb{Q}({\zeta }_n),1\le i\le n$. In particular $\alpha \in \mathbb{Q}({\zeta }_n)$, thus $L=\mathbb{Q}(\alpha )\subset \mathbb{Q}({\zeta }_n)$. (b) is so proved under the hypothesis (a).

Conclusion : (a) $\Rightarrow$(b) is equivalent to the assertion of Kronecker : the roots of an Abelian equation over $\mathbb{Q}$ can be expressed rationally in terms of a root of unity. □