Exercise 9.1.12

Proof.

(a)

$n=p_1^{{\nu }_1}{p}_2^{{\nu }_2}\cdots p_r^{{\nu }_r}$. Write $m=p_1\cdots p_r$. Then

$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_n(x))=\varphi (n)=p_1^{{\nu }_1-1}{p}_2^{{\nu }_2-1}\cdots p_r^{{\nu }_r-1}(p_1-1)(p_2-1)\cdots (p_r-1)$.

$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_m(x)))=\varphi (p_1{p}_2\cdots p_n)=(p_1-1)(p_2-1)\cdots (p_r-1)$, therefore

$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_m(x^{n∕m}))=p_1^{{\nu }_1-1}{p}_2^{{\nu }_2-1}\cdots p_r^{{\nu }_r-1}(p_1-1)(p_2-1)\cdots (p_r-1)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_n(x))$.

Moreover these polynomials are monic and ${\Phi }_n$ is separable. It remains to show that every root $\zeta$ of ${\Phi }_n(x)$ is a root of ${\Phi }_m(x^{n∕m})$.

Such a root $\zeta$ has order $n=p_1^{{\nu }_1}{p}_2^{{\nu }_2}\cdots p_r^{{\nu }_r}$ in the group ${ℂ}^{\text{∗}}$.

Write $\xi ={\zeta }^{n∕m}={\zeta }^{p_1^{{\nu }_1-1}{p}_2^{{\nu }_2-1}\cdots p_r^{{\nu }_r-1}}$.

Then the order of $\xi$ is $m=p_1{p}_2\cdots p_r$. Indeed, for all $k\in \mathbb{Z}$,

${\xi }^k=1⟺{\zeta }^{k{p}_1^{{\nu }_1-1}{p}_2^{{\nu }_2-1}\cdots p_r^{{\nu }_r-1}}=1⟺p_1^{{\nu }_1}{p}_2^{{\nu }_2}\cdots p_r^{{\nu }_r}\mid k{p}_1^{{\nu }_1-1}{p}_2^{{\nu }_2-1}\cdots p_r^{{\nu }_r-1}$

$⟺p_1{p}_2\cdots p_r\mid k$.

Therefore, by definition of ${\Phi }_m$, ${\Phi }_m({\zeta }^{n∕m})={\Phi }_m(\xi )=0$.

The hypotheses of the lemma are satisfied, thus

$${\Phi }_n(x)={\Phi }_m(x^{n∕m})$$

(b)

We show that ${\Phi }_{2n}(x)={\Phi }_n(-x)$ ( $n>1,n$ odd, so $n\ge 3$).

Note first that $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_{2n}(x))=\varphi (2n)=\varphi (2)\varphi (n)=\varphi (n)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_n(-x))$.

If $n>2$, then $\varphi (n)$ is even. Indeed, we can group in pairs the elements of ${(\mathbb{Z}∕\mathit{n\mathbb{Z}})}^{\text{∗}}$, with the pairs $\{[d],-[d]\}$, where $d\wedge n=1$ and $[d]\ne -[d]$ since $(n\mid 2d,d\wedge n=1)\Rightarrow n\mid 2$, which is impossible if $n>2$. Hence

$${(-1)}^{\varphi (n)}=1(n>2).$$

${\Phi }_{2n}(x)$ is monic by definition, and the leading coefficient of ${\Phi }_n(-x)$ is ${(-1)}^{\varphi (n)}=1$, so ${\Phi }_n(-x)$ is also monic.

Let $\alpha$ be any root of ${\Phi }_n(-x)$. Then $\alpha =-\zeta$, where $\zeta$ is a $n$th primitive root of unity, so $\zeta$ is an element of order $n$ in the group ${ℂ}^{\text{∗}}$.

Then the order of $\alpha =-\zeta$ is $2n$. Indeed, for all $k\in \mathbb{Z}$,

${(-\zeta )}^k=1$, that is ${(-1)}^k{\zeta }^k=1$, implies ${\zeta }^{2k}=1$, thus $n\mid 2k$, so $n\mid k$ (since $n$ is odd), therefore ${\zeta }^k=1,{(-1)}^k=1$ and so $2\mid k$.

As $n\wedge 2=1,2n\mid k$.

Conversely, if $2n\mid k,{(-\zeta )}^{2n}={[{(-1)}^2]}^n{[{\zeta }^n]}^2=1$.

Conclusion: ${(-\zeta )}^k=1⟺2n\mid k$, so the order of $\alpha =-\zeta$ is $2n$, hence $x=-\zeta$ is a root of ${\Phi }_{2n}$.

Every root of ${\Phi }_n(-x)$ in $ℂ$ is a root of ${\Phi }_{2n}(x)$. Moreover ${\Phi }_n(-x)$ is a separable polynomial, and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_{2n}(x))=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_n(-x))$. Then the lemma gives the conclusion, for all odd $n$, $n>1$,

$${\Phi }_{2n}(x)={\Phi }_n(-x)$$

(c)

We show first that ${\Phi }_n(x)$ divides ${\Phi }_n(x^p)$. As ${\Phi }_n(x)$ is separable, it is sufficient to verify that every root $\zeta$ of ${\Phi }_n(x)$ is a root of ${\Phi }_n(x^p)$. Such a root $\zeta$ is a $n$th primitive root of unity, so its order is $n$. Then the order of ${\zeta }^p$ is also $n$. Indeed, for all $k\in \mathbb{Z}$, as $n\wedge p=1$,

$${({\zeta }^p)}^k=1⟺{\zeta }^{\mathit{pk}}=1⟺n\mid \mathit{pk}⟺n\mid k.$$

Therefore ${\zeta }^p$ is a root of ${\Phi }_n$, so ${\Phi }_n({\zeta }^p)=0$ and $\zeta$ is a root of ${\Phi }_n(x^p)$.

$${\Phi }_n(x)\mid {\Phi }_n(x^p)(p\nmid n).$$

We compare the degrees:

$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_{\mathit{pn}}(x))=\varphi (\mathit{pn})=\varphi (p)\varphi (n)=(p-1)\varphi (n)$,

$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_n(x^p)∕{\Phi }_n(x))=\mathit{p\varphi }(n)-\varphi (n)=(p-1)\varphi (n)$, thus

$$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_n(x^p)∕{\Phi }_n(x))=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_{\mathit{pn}}(x)).$$

Moreover, these two polynomials are monic, and ${\Phi }_{\mathit{pn}}$ is separable.

We show that every root $\zeta$ of ${\Phi }_{\mathit{pn}}(x)$ is a root of ${\Phi }_n(x^p)∕{\Phi }_n(x)$.

If $\zeta$ is a root of ${\Phi }_{\mathit{pn}}(x)$, then $o(\zeta )=\mathit{pn}$, therefore $o({\zeta }^p)=n$

(indeed, for all $k\in \mathbb{Z}$, ${({\zeta }^p)}^k=1⟺{\zeta }^{\mathit{pk}}=1⟺\mathit{pn}\mid \mathit{pk}⟺n\mid k$).

So ${\zeta }^p$ is a root of ${\Phi }_n(x)$, which is equivalent to $\zeta$ is a root of ${\Phi }_n(x^p)$.

As $o(\zeta )=\mathit{pn}$, ${\zeta }^n\ne 1$, ${\Phi }_n(\zeta )\ne 0$, therefore $\zeta$ is a root of ${\Phi }_n(x^p)∕{\Phi }_n(x)$.

The hypotheses of the lemma are so satisfied, so

$${\Phi }_{\mathit{pn}}(x)={\Phi }_n(x^p)∕{\Phi }_n(x)(p\nmid n).$$