# Galois extension of prime degree

I don't understand a step in the proof of corollary 96 in the book "Galois Theory" by J. Rotman. The full book can be found here https://epdf.pub/galois-theory-second-edition-universitext.html

Corollary 96: Let $$E / F$$ be a Galois extension of prime degree $$p$$. If $$F$$ has a primitive $$p$$th root of unity, then $$E = F(\beta)$$, where $$\beta^p \in F$$, and so $$E / F$$ is a pure extension.

Here is the beginning of the proof:

Proof: If $$\omega$$ is a primitive $$p$$th root of unity, then $$N(\omega) = \omega^p = 1$$, because $$\omega \in F$$. Now $$Gal(E/F) \simeq \mathbb{Z}_p$$, by Corollary 71, hence is cyclic; let $$\sigma$$ be a generator. ...

($$N$$ denotes the norm.) And Corollary 71 says:

Corollary 71: Let $$p$$ be a prime, let $$F$$ be a field containing a primitive $$p$$th root of unity, and let $$f(x) = x^p - c \in F[x]$$ have splitting field $$E$$. Then either $$f(x)$$ splits and $$Gal(E/F) = 1$$ or it is irreducible and $$Gal(E/F) \simeq \mathbb{Z}_p$$.

My question: I do not see how $$Gal(E/F) \simeq \mathbb{Z}_p$$ follows from Corollary 71. Isn't it true that every group of prime order $$p$$ is cyclic and isomorph to $$\mathbb{Z}_p$$? Have I overseen something?

## 1 Answer

You are correct, Corollary 71 is not relevant and you only have to observe that every group of order $$p$$ is cyclic.