Exercise 7.15

Proof. From exercise 7.12, we know that $x^n-1$ splits into linear factors in some extension field $K$, with $[K:F]<\infty$ :

Then $u^{\prime }(x)\wedge u(x)=n{x}^{n-1}\wedge (x^n-1)=1$, since $x(n{x}^{n-1})-n(x^n-1)=n$, and $n\ne 0$ in the field $F$, because we know from the hypothesis $q\wedge n=1$ that the characteristic $p$ doesn’t divide $n$. So the $n$ roots of $x^n-1$ are distinct.

The set $G=\{x\in K|x^n=1\}$ is a subgroup of $K^{\text{∗}}$, thus $G$ is cyclic of order $n$. Let $\zeta$ a generator of $G$. Then

Let $p(x)$ be the minimal polynomial of $\zeta$ on $F$, and $f$ the degree of $p$ :

Thus $\mathrm{Card}F[\zeta ]=q^f$, and since $\zeta \in F{[\zeta ]}^{\text{∗}}$, ${\zeta }^{q^f-1}-1=0$. As the order of $\zeta$ in the group $G$ is $n$, $n\mid q^f-1$, namely $q^f\equiv 1(\mathit{mod}n)$.

Let $k$ be any positive integer such that $q^k\equiv 1(\mathit{mod}n)$.

Then $n\mid q^k-1$, thus ${\zeta }^{q^k-1}-1=0$, and ${\zeta }^{q^k}-\zeta =0$. Let $L$ be any extension of $K$ such that $x^{q^k}-x$ splits in linear factors in $L$, and let $M=\{\alpha \in L\mid {\alpha }^{q^k}=\alpha \}$. We know that $M$ is a subfield of $L$ (see Ex. 13), with cardinality $q^k$, so that $[M:F]=k$. As ${\zeta }^{q^k}-\zeta =0$, $\zeta$ belongs to $M$. Therefore $F[\zeta ]\subset M$, thus $f=[F[\zeta ]:F]\le k=[M:F]$.

So $f=[F[\zeta ]:F]$ is the smallest $k\in {\mathbb{N}}^{\text{∗}}$ such that $q^k\equiv 1(\mathit{mod}n)$.

If $K$ is any extension of $F$ containing the roots of $x^n-1$, then $K\supset F[\zeta ]$, where $\zeta$ is a primitive root of unity, so $[K:F]\ge [F[\zeta ]:F]=f$.

Conclusion: the minimal degree of a extension $K\supset F$ containing the roots of $x^n-1$, with $n\wedge q=1$, is the smallest positive integer $f$ such that $q^f\equiv 1(\mathit{mod}n)$, the order of $q$ modulo $n$. □