Exercise 8.3.1

Proof.

(a)

Let $f=x^m-1,m\in {\mathbb{N}}^{\text{∗}}$. Then $f^{\prime }=m{x}^{m-1}$ is relatively prime with $f$. Indeed $m\ne 0$ in $L$ since the characteristic of $L$ is $0$, and $-f+m^{-1}x{f}^{\prime }=-x^m+1+x^m=1$ is a Bézout’s relation between $f$ and $f^{\prime }$. Therefore (Prop. 5.3.2), $f$ is a separable polynomial, so $x^m-1$ has $m$ distinct roots in $M$, the splitting field of $f$ over $L$.

(b)

We show that ${𝕌}_m=\{\alpha \in M|{\alpha }^m=1\}$, the set of the $m$ roots of $f$ in $M$, is a subgroup of $M^{\text{∗}}$.

$\text{∙}$

$1^m=1$, therefore $1\in {𝕌}_m\ne \varnothing$.

$\text{∙}$

If $\alpha ,\beta \in {𝕌}_m$, then ${(\alpha {\beta }^{-1})}^m={\alpha }^m{({\beta }^m)}^{-1}=1$, therefore $\alpha {\beta }^{-1}\in {𝕌}_m$.

The roots of $x^m-1$ in $M$ form a group under multiplication, subgroup of the multiplicative group of a field, so it is cyclic.