Suppose $x^n-1\in F[x]$ for $F$ a finite field of order $q$. I read that if $(q,n)=1$, then the least degree of a field over which $x^n-1$ splits is the smallest integer $f$ such that $q^f\equiv 1\pmod{n}$.
I wrote out the following, but never used the fact that $(q,n)=1$, so I think I am making a mistake.
I take $K$ to be the splitting field of least degree for $x^n-1$, with degree $f$. Since $|K|=q^f$, every nonzero element of $K$ satisfies $x^{q^f-1}-1$. Since $x^n-1$ splits in $K$, all its roots are in $K$, so $x^n-1\mid x^{q^f-1}-1$. Thus $n\mid q^f-1$, so $q^f\equiv 1\pmod{n}$.
Conversely, suppose $m<f$. If $L$ is a an extension of degree $m$, then $L$ is not a splitting field of $x^n-1$. If $q^m\equiv 1\pmod{n}$, then $n\mid q^m-1$, hence $x^n-1\mid x^{q^m-1}-1$ in $L[x]$. So every root of $x^n-1$ is a root of $x^{q^m-1}-1$, and since the roots of $x^{q^n-1}-1$ are precisely the elements of $L^*$, it follows that $x^n-1$ splits over $L$, a contradiction.
But I don't see if I used $(q,n)=1$ anywhere. Is there a mistake I am overlooking? Thanks.
Here you are implicitly assuming that $x^n-1$ has distinct roots - if there are repeated roots, the divisibility cannot be concluded, since $x^{q^{f}-1} - 1$ is divisible by each linear factor only once.