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.
Best Answer
Here are some remarks :
For those who haven't Dummit and Foote (lemma 40, Sec. 13.6 in the third edition), let me recall that they define $\Phi_n(X)$ as the product of the $X-\zeta$ with $\zeta \in \mu_n$ going through the primitive $n$-th roots of unity.
In their proof, they say : $f$ divides $X^n-1$ in $\Bbb Q(\zeta_n)[X]$ and also in $\Bbb Q[X]$ by the division algorithm. Then, $X^n-1=f(X)\Phi_n(X)$ in $\Bbb Q[X]$ and thanks to Gauss' lemma ($f$ being monic in $\Bbb Z[X]$), one has $f \in \Bbb Z[X]$ and $\Phi_n \in \Bbb Z[X]$. In this version, they need to use Gauss' lemma.
But I think your version is also right. Actually, since $f$ is assumed to be a monic polynomial in $\Bbb Z[X]$, the division algorithm allows us to write $X^n-1 = q(X)f(X)+r(X)$ in $\Bbb Z[X]$ as you did. In particular, you don't need to use Gauss' lemma here. Then the conclusion follows by your argument, which seems correct to me.
Sometimes, the $n$-th cyclotomic polynomial is defined as the minimal polynomial of $\zeta_n$ over $\mathbb Q$. Then $\Phi_n$ divides $X^n-1$ in $\mathbb Q[X]$, since it is a minimal polynomial. By Gauss' lemma, we get : $\Phi_n$ divides $X^n-1$ in $\mathbb Z[X]$, in particular $\Phi_n \in \Bbb Z[X]$.