From Galois Theory (Rotman):
If F is a finite field, then $F^*$ [which is the multiplicative group] is cyclic and $F=\Bbb{Z}_p(\alpha)$ for some $\alpha$.
Proof If $|F|=q$, take $\alpha$ to be a primitive (q-1)st root of unity.
I find this proof a bit confusing…
1) How do we know that F has a (q-1)st primitive root of unity, or any root of unity for that matter?
2) How do we know that $F=\Bbb{Z}_p(\alpha)$? I know that every finite field must have an isomorphic copy of $\Bbb{Z}_p$, but I don't understand how a finite field is equal $\Bbb{Z}_p(\alpha)$.
Thank you in advance
Best Answer
Recall that every element of $\mathbb{F}_q$ is fixed by the Frobenius map $x\mapsto x^q$. But this means every element satisfies the polynomial equation $x^q-x=0$. If we throw out $0$, consider $x(x^{q-1}-1)=0$ and we get that every non-zero element is a $(q-1)$st root of unity. Conversely, since there are $q-1$ distinct non-zero elements we see that the groups are the same.
So $\mathbb{F}_q^*$ is the group of $(q-1)$st roots of unity which is cyclic and generated by a primitive root of unity. Depending on the author, $\mathbb{F}_q$ is the splitting field of $x^q-x$ over $\mathbb{F}_p$. By the previous part of the argument, if you adjoin a generator for $\mathbb{F}_q^*$, then you get all roots by taking powers. Thus $\mathbb{F}_p(\alpha)=\mathbb{F}_q$.