In Algebra I by Falko Lorenz, page 7. After introducing
Definition 2. Let $E$ be an extension of the filed $K$
- We say that $E$ arises from $K$ by ADJOINING A SQUARE ROOT if there exists $w \in E$ such that $w^2 \in K$ and $E = K(w)$. We call $w$ a SQUARE ROOT of the element $v$
- We say that $E$ arises from $K$ by SUCCESSIVELY ADJOINING A SQUARE ROOTs if there is a chain $K = K_0 \subseteq K_1 \subseteq \cdots \subseteq K_m = E$ of subfields $K_i$ of $E$ where each $K_i$ is obtained from $K_{i-1}$ by adding a square root.
The book shows an example.
$E = \mathbb{Q}(z)$, where $z = e^{2 \pi i /3}$ is obtained from $\mathbb{Q}$ by adjoining a square root. For since $z = -\frac{1}{2} + \frac{1}{2} i \sqrt{3}$, we have $\mathbb{Q}(z) = \mathbb{Q}(\sqrt{-3})$.
I don't understand how we can conclude that $\mathbb{Q}(z) = \mathbb{Q}(\sqrt{-3})$.
Because $\mathbb Q(z)$ and $\mathbb Q(\sqrt{-3})$ are the smallest field extensions containing $z$ and $\sqrt{-3}$ respectively, it suffices to show that $\sqrt{-3} \in \mathbb Q(z)$ and $z \in \mathbb Q(\sqrt{-3})$. This is shown easily: $$ z = −\frac{1}{2}+\frac{1}{2} \sqrt{-3} \in \mathbb Q(\sqrt{-3}), \quad \sqrt{-3} = 2z +1 \in \mathbb Q(z). $$