I am trying to understand this proof for the nth cyclotomic polynomial $\Phi_n(x)$ having integer coefficients, given in Fraleigh’s 7th edition A First Course in Abstract Algebra. The proof assumes $K$ is the splitting field of $x^n-1 \in F[x]$ where $\mathrm{char} F$ does not divide $n$.
“Since an automorphism of the Galois group $G(K/F)$ must permute the primitive nth roots of unity, we see that $\Phi_n(x)$ is left fixed under every element of $G(K/F)$ regarded as extended in the natural way to $K[x]$. Thus $\Phi_n(x) \in F[x]$. In particular, for $F = \mathbf{Q}$, $\Phi_n(x) \in \mathbf{Q}[x]$, and $\Phi_n(x)$ is a divisor of $x^n - 1$. Thus over $\mathbf{Q}$, we must actually have $\Phi_n(x) \in \mathbf{Z}[x]$, by Theorem 23.11.”
Theorem 23.11 asserts the following:
“If $f(x) \in \mathbf{Z}[x]$, then $f(x)$ factors into a product of two polynomials of lower degrees $r$ and $s$ in $\mathbf{Q}[x]$ if and only if it has such a factorization with polynomials of the same degrees $r$ and $s$ in $\mathbf{Z}[x]$.”
Why must it be that $\Phi_n(x) \in \mathbf{Z}[x]$? I understand why $\Phi_n(x)$ is in $\mathbf{Q}[x]$, since the automorphism take the set of multiplicative order $n$ elements to itself. All I understand from this is that Theorem 23.11 shows there exists a polynomial in $\mathbf{Z}[x]$ that divides $x^n - 1$ of degree $\phi(n)$. For example, $\frac{1}{2}(x-1)$ divides $(x-1)(x-2)$ in $\mathbf{Q}[x]$ but obviously $\frac{1}{2}(x-1) \notin \mathbf{Z}[x]$.
Any help would be greatly appreciated!