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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.

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up vote 4 down vote accepted

Since $x^n-1$ splits in $K$, all its roots are in $K$, so $x^n-1\mid x^{q^f-1}-1$

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.

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Thanks Sanchez. Otherwise, is the proof sound? – Noomi Holloway Jan 27 '13 at 0:28
@NoomiHolloway, I think so :) – user27126 Jan 27 '13 at 0:30

When you say, "all roots of $x^n-1$ are in $K$, so $x^n-1 | x^{q^f-1}-1$."

If $(q,n)>1$, then $x^n-1$ has mulitple roots, as its derivative is $0$.

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Thanks Berci. So other than that, is the proof correct? – Noomi Holloway Jan 27 '13 at 0:28
Yes, seems perfect. – Berci Jan 27 '13 at 0:31

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