Exercise 9.1.8

Proof. As in the proof of Theorem 9.1.9, the Gauss’s Lemma in the form of Corollary 4.2.1 allows us to assume that there exists a polynomial $f(x)\in \mathbb{Z}[x]$ of ${\Phi }_n(x)$ such that ${\Phi }_n(x)=f(x)g(x),f(x),g(x)\in \mathbb{Z}[x]$, where $f$ is monic and irreducible over $\mathbb{Q}$. Let $p$ be a prime number such that $p\nmid n$.

Reasoning by contradiction, we suppose that $\zeta$ is a root of $f$ such that $f({\zeta }^p)\ne 0$.

(a)

As $\zeta$ is the root of $f$, where $f$ divides ${\Phi }_n$, $\zeta$ is a $n$th primitive root of unity. Since $p\nmid n,p\wedge n=1$, hence ${\zeta }^p$ is also a $n$th primitive root of unity by Exercise 7(a), therefore $0=\Phi ({\zeta }^p)=f({\zeta }^p)g({\zeta }^p)$. As $f({\zeta }^p)\ne 0$, $g({\zeta }^p)=0$, so

$\zeta$ is a root of $g(x^p)$.

As $f$ is irreducible, $f$ is the minimal polynomial of $\zeta$ over $\mathbb{Q}$, and $\zeta$ is a root of $g(x^p)\in \mathbb{Q}[x]$, hence

$$f(x)\mid g(x^p).$$

(b)

As $f$ is monic, the refined division algorithm of Exercise 4 show that the quotient $q(x)$ of $g(x^p)$ by $f(x)$ lies in $\mathbb{Z}[x]$, so $f(x)$ divides $g(x^p)$ in $\mathbb{Z}[x]$.

The projection homomorphism on ${𝔽}_p[x]$ gives $\bar{g}(x^p)=\bar{f}(x)\bar{q}(x)$, thus $\bar{f}(x)$ divides $\bar{g}(x^p)$ in ${𝔽}_p[x]$.

(c)

As the characteristic of ${𝔽}_p(x)$ is $p$, writing $\bar{g}(x)=\underset{i=0}{\overset{r}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}_i{x}^i\in {𝔽}_p[x]$, then (as in Exercise 7)

$$\bar{g}{(x)}^p={\left(\underset{i=0}{\overset{r}{\sum }}{a}_i{x}^i\right)}^p=\underset{i=0}{\overset{r}{\sum }}{a}_i^p{x}^{\mathit{ip}}=\underset{i=0}{\overset{r}{\sum }}{a}_i{x}^{\mathit{ip}}=\bar{g}(x^p).$$

Therefore $\bar{f}(x)$ divides $\bar{g}{(x)}^p$ in ${𝔽}_p[x]$.

(d)

Let $h(x)\in {𝔽}_p[x]$ an irreducible factor of $\bar{f}(x)$. Then $h(x)\mid \bar{g}{(x)}^p$. Since $h$ is irreducible (hence prime) in ${𝔽}_p[x]$, then $h\mid \bar{g}$.

$h(x)\mid \bar{f}(x),h(x)\mid \bar{g}(x)$, so $h{(x)}^2\mid \bar{f}(x)\bar{g}(x)$.

(e)

Therefore $h^2\mid {\bar{\Phi }}_n$, and ${\bar{\Phi }}_n\mid x^n-1$, thus $h^2\mid x^n-1\in F_p[x]$.

(f)

As $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)>1$, every root of $h$ in the splitting root of $x^n-1\in {𝔽}_p[x]$ is not a simple root, thus $x^n-1$ would not be separable.

But $n$ is relatively prime to $p$, so ${(x^n-1)}^{\prime }=n{x}^{n-1}$ is relatively prime to $x^n-1$, and so $x^n-1\in {𝔽}_p[x]$ is separable: this is a contradiction, therefore.

$$f(\zeta )=0\Rightarrow f({\zeta }^p)=0.$$

We conclude that ${\Phi }_n$ is irreducible as in the conclusion of the proof of Theorem 9.1.9.