Over a field whose characteristic does not divide $n$, the roots of ${\mathrm{\Phi }}_n$ are exactly the primitive roots of unity

By [L1] the polynomial $t^n-1$ is separable over $K$ and ${\mu }_n(E)$ is cyclic of order $n$; so $t^n-1$ has $n$ distinct roots in $E$, namely the elements of ${\mu }_n(E)$, and $t^n-1={\prod }_{\zeta \in {\mu }_n(E)}(t-\zeta )$ by [L3].

Each $\zeta \in {\mu }_n(E)$ has order dividing $n$ by [L5], and for a positive divisor $m$ of $n$ the condition ${\zeta }^m=1$ says exactly $\mathrm{ord}(\zeta )\mid m$; so ${\mu }_m(E)$ is the disjoint union of the $S_d$ over positive divisors $d$ of $m$.

For every positive divisor $m$ of $n$ the polynomial $t^m-1$ divides $t^n-1$ in $\mathbb{Z}[t]$, since $t^n-1=(t^m-1)(t^{n-m}+t^{n-2m}+\cdots +1)$; hence it splits over $E$ with distinct roots, which are the elements of ${\mu }_m(E)$, and $t^m-1={\prod }_{\zeta \in {\mu }_m(E)}(t-\zeta )$ with $\mid {\mu }_m(E)\mid =m$.

Consequently ${\prod }_{d\mid m}{\mathrm{\Psi }}_d={\prod }_{\zeta \in {\mu }_m(E)}(t-\zeta )=t^m-1$ for every positive divisor $m$ of $n$.

For every positive divisor $d$ of $n$ the image of ${\mathrm{\Phi }}_d$ in $E[t]$ is ${\mathrm{\Psi }}_d$, by induction on $d$ through the divisors of $n$: at $d=1$ both are $t-1$, since $S_1=\{1\}$; and if the claim holds for every positive divisor $e$ of $d$ with $e<d$, then [L2] and step 3.1 give $\left({\prod }_{e\mid d,e<d}{\mathrm{\Psi }}_e\right){\mathrm{\Phi }}_d=t^d-1=\left({\prod }_{e\mid d,e<d}{\mathrm{\Psi }}_e\right){\mathrm{\Psi }}_d$, and cancelling the nonzero left factor in the integral domain $E[t]$ ([L6]) gives ${\mathrm{\Phi }}_d={\mathrm{\Psi }}_d$.

Taking $d=n$: the image of ${\mathrm{\Phi }}_n$ in $E[t]$ is ${\prod }_{\zeta \in S_n}(t-\zeta )$, so ${\mathrm{\Phi }}_n$ splits over $E$ and its roots there are exactly the elements of $S_n$, which are the elements of order $n$ in ${\mu }_n(E)$, that is the primitive $n$-th roots of unity in $E$; there are $\phi (n)$ of them by [L1], in agreement with $\mathrm{deg}{\mathrm{\Phi }}_n=\phi (n)$ from [L2].

${\mathrm{\Phi }}_n$ is separable over $K$. The product identity [L2] shows that the image of ${\mathrm{\Phi }}_n$ divides $t^n-1$ in $K[t]$. If an extension field $L\mathrm{/}K$ contained an $a$ for which $(t-a{)}^2$ divided the image of ${\mathrm{\Phi }}_n$, then the same square would divide the image of $t^n-1$ in $L[t]$, making $a$ a repeated root of $t^n-1$. This contradicts the separability of $t^n-1$ supplied by [L1]. Thus [L4] applies.