I don't understand why Gauss's lemma is invoked in the proof in Dummit and Foote that $\Phi_n(x)$ (the $n$th cyclotomic polynomial) belongs to $\mathbb{Z}[x]$. I'm an analyst and I wanted to remind myself about cyclotomic polynomials. The following is the proof as I've written it, and because it's been a while since I've done algebra I want to know if there is something I'm taking for granted.
It is a fact that if $R$ is a unital commutative ring, $f \in R[x]$ is a monic polynomial and $g \in R[x]$ is a polynomial, then there are $q,r \in R[x]$ with $g = qf+r$, $r=0$ or $\deg r < \deg f$.
First, $\Phi_1(x)=x-1 \in \mathbb{Z}[x]$. For $n>1$, assume that $\Phi_d(x) \in \mathbb{Z}[x]$ for $1 \leq d < n$. Then let $f = \prod_{d \mid n, d<n} \Phi_d$, which by hypothesis belongs to $\mathbb{Z}[x]$; and since each $\Phi_d$ is monic, so is $f$.
On the one hand, since $g(x)=x^n-1 \in \mathbb{Z}[x]$, there are $q,r \in \mathbb{Z}[x]$ with $g = q f + r$ and $r=0$ or $\deg r < \deg f$. On the other hand, using $x^n-1= \prod_{d \mid n} \Phi_d(x)$ we have $g = \Phi_n f \in \mathbb{C}[x]$.
Thus $\Phi_n f = qf+r \in \mathbb{C}[x]$, so $r = f \cdot (\Phi_n - q) \in \mathbb{C}[x]$. If $\Phi_n \neq q$ then $\deg r = \deg f + \deg (\Phi_n-q) \geq \deg f$, contradicting that $r=0$ or $\deg r < \deg f$. Therefore $\Phi_n = q \in \mathbb{C}[x]$, and because $q \in \mathbb{Z}[x]$ this means that $\Phi_n \in \mathbb{Z}[x]$.
I don't see any tacit assumptions, like for example degree meaning two different things in $\mathbb{C}[x]$ and $\mathbb{Z}[x]$, but if I'm not assuming anything then I don't see why Gauss's lemma is being used in the proofs I've come across.