# Proposition 36 of Chapter 14, Dummit and Foote

Proposition 36 of Chapter 14 of Dummit and Foote states

"Let $$F$$ be a field of characteristic not dividing $$n$$ which contains the $$n$$-th roots of unity. Then the extension $$F(\sqrt[n]{a})$$ for $$a \in F$$ is cyclic over $$F$$ of degree dividing $$n$$."

My question is that if the field $$F$$ already contains all of the $$n$$-th roots of unity, then is $$F(\sqrt[n]{a})$$ even an extension? It seems like if you already have the $$n$$-th roots of unity then you already have all of the elements that could be obtained by the extension $$(\sqrt[n]{a})$$, but maybe I am interpreting the statement of the Proposition wrong?

Notice $$\mathbb{Q}$$ already contains all the square roots of unity (that is, $$1$$ and $$-1$$), and yet that does not mean that $$\mathbb{Q}$$ contains all square roots.
Perhaps less trivially, if $$\omega$$ is a primitive cube root of unity, than $$\mathbb{Q}(\omega)$$ is a field containing the cube roots of unity, and yet $$\mathbb{Q}(\omega, \sqrt[3]{2})$$ is a nontrivial extension. In particular, $$\sqrt[3]{2} \not \in \mathbb{Q}(\omega)$$ (do you see why?).