I do not quite understand why there may not be a $n$-th primitive root in splitting fields of $x^n-1$ of characteristic non- $0$. May I please ask for the reason (I tried to understand it and it takes me some time. Now I think I need some explicit explaination now...).
If possible, may I ask for a specific example of a splitting field of $x^n-1$ with some positive characteristic where some $n$-th primitive root does not exist in the field? Thanks!
On
$p$ is prime.
By uniqueness of finite fields and since $\mathbb{F}_{p^m}^\times$ is cyclic of order $p^m-1$ then $$F_{p}(\zeta_n) =\mathbb{F}_{p^m}$$ where $m$ is the least natural number such that $n|p^m-1$, ie. the order of $p \bmod n$.
And hence $\mathbb{F}_p(\zeta_{n})$ doesn't exist whenever $p | n$. Assume the opposite, since $\zeta_n$ is in every case a root of $X^n-1$ then $[\mathbb{F}_p(\zeta_{n}):\mathbb{F}_p] \le n$ in particular it is finite, but this is impossible since $\mathbb{F}_{p^m}^\times$ is cyclic of order $p^m-1$ so it doesn't contain an element of order $n$
The problem is if $p\mid n$, where $p$ is the characteristic. Notice that if $n=pm$ then $(x^n-1)=(x^m-1)^p$, so any root of $x^n-1$ is also a root of $x^m-1$. It follows that no field of characteristic $p$ can have a primitive $n$th root of unity.
For a very concrete example, take the field $\mathbb{F}_p$ and $n=p$. Note that $x^p-1$ factors as $(x-1)^p$, so it already splits in $\mathbb{F}_p$. But the only root of it is $x=1$, which is not a primitive $p$th root of unity.