No field of characteristic $p > 0$ contains a primitive $p^{th}$ root of unity.
I spent the whole day yesterday on this problem, still no progress!
I reworded the question and came up with this,
If $F$ is a field of characteristic $p$, then $g(x)=x^{p-1}+x^{p-2}+\cdots+1$ is irreducible $\forall x \in F$.
Clearly, $1$ is a root, but I guess it has to be different from $1$.
I tried using this $$x^p-1 = (x-1)(x^{p-1}+x^{p-2}+\cdots+1)$$
If $F$ does not have a $p^{th}$ root of unity then $g(x)=x^{p-1}+x^{p-2}+\cdots+1$ is irreducible in $F$ so if I adjoin root $\alpha$ of $g$ to $F$ then $[F(\alpha):F]=p-1$.
Not sure if this leads to something.
I also used freshman's dream to get
$$f = x^p – 1 = x^p-1^p=(x-1)^p.$$
So any factors of $f$ have to have the form $(x-1)^k $, $1 \leq k \leq p$,
but still nothing, any hints?
Best Answer
You've almost got it. If $x^p-1=(x-1)^p=0$ then $x=1$ because fields are integral domains.