Proving that cyclotomic polynomials have integer coefficients

Question

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.

Answer 1

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 $\Phi_n \in \Bbb Z[X]$.

Answer 2

This proof seems to be more elementary. We begin by contending that if $$(x^n -1) =({\sum}^p_{i=1}{a_i{x^i}})({\sum}^q_{j=1}{b_j{x^i}}),$$ where ${\sum}^p_{i=1}{a_i{x^i}}\in{\bf{Z}[x]},$ then every coefficient $b_j$, must be an integer. To prove this, we argue by backward induction. Since $a_p{b_q} = 1,$ and since $a_p\in{\bf Z},$ we conclude that $a_p=b_q = \pm{1}$ and $b_q$ is therefore an integer. Now, assume as our induction hypothesis, that $b_q,b_{q-1},\cdots b_{r+1}$ are all integers. Then, the coefficient of $x^{p+r}$ is $${a_p}{b_r}+{a_{p-1}}{b_{r+1}}+\cdots +{a_{(p+r)-q}}{b_q}=0.$$ Since ${a_p}=\pm1$, the above identity shows that $${b_r} = \pm({a_{p-1}}{b_{r+1}}+\cdots +{a_{(p+r)-q}}{b_q})\in{\bf{Z}}.$$ Thus, we conclude that every coefficient $b_j$, must be an integer by mathematical induction. Now, we use the identity $$(x^n - 1) = {\prod}_{d|n}{\Phi}_d(x).$$ We claim that $\Phi_n(x)\in{\bf{Z}}[x]$ by induction on $n$. The assertion is clear for $n=1$ since $\Phi_1(x)=(x-1).$ Assume now, as our induction hypothesis, that $\Phi_d(x)\in{\bf{Z}}[x]$ for all positive integers $d<n$. Then, writing $$(x^n - 1) = {{\prod}_{d|n,d<n}}{\Phi}_d(x)\cdot{\Phi_n(x)},$$ since ${{\prod}_{d|n,d<n}}{\Phi}_d(x)\in {\bf{Z}}[x]$, it follows by our argument above that ${\Phi_n(x)}\in {\bf{Z}}[x].$