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?


1 Answer 1


Welcome to MSE!

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?).

I hope this helps ^_^

  • 1
    $\begingroup$ Thank you @HallaSurvivor, I think that makes sense. Keyword being roots of 'unity' not necessarily roots of a. $\endgroup$
    – Oderus
    Commented May 17, 2021 at 5:38
  • $\begingroup$ Happy to help! And yeah, that's it exactly ^_^ $\endgroup$ Commented May 17, 2021 at 5:39

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