Irreducibility of cyclotomic polynomials

For every integer $n\geq 1$, the $n$th cyclotomic polynomial $\Phi_n$ is the monic polynomial whose complex roots are the primitive $n$th roots of unity. A priori, this is a polynomial with complex coefficients, but since every $n$th root of unity is a primitive $d$th root of unity, for a unique divisor $d$ of $n$, one has the relation
\[ T^n-1 = \prod_{d\mid n} \Phi_d(T), \]
which implies, by induction and euclidean divisions, that $\Phi_n \in \mathbf Z[T]$ for every $n$.
The degree of the polynomial $\Phi_n$ is $\phi(n)$, the Euler indicator, number of units in $\mathbf Z/n\mathbf Z$, or number of integers in $\{0,1,\dots,n-1\}$ which are prime to $n$.

The goal of this note is to explain a few proofs that these polynomials are irreducible in $\mathbf Q[T]$ — or equivalently, in view of Gauss's lemma, in $\mathbf Z[T]$. This also amounts to saying that $\deg(\Phi_n)=\phi(n)$ or that the cyclotomic extension has degree $\phi(n)$, or that the canonical group homomorphism from the Galois group of $\mathbf Q(\zeta_n)$ to $(\mathbf Z/n\mathbf Z)^\times$ is an isomorphism.

1. The case where $n=p$ is a prime number.

One has $T^p-1=(T-1)(T^{p+1}+\dots+1)$, hence $\Phi_p=T^{p-1}+\dots+1$. If one reduces it modulo $p$, one finds $\Phi_p(T)\equiv (T-1)^{p-1}$, because $(T-1)\Phi_p(T)=T^p-1\equiv (T-1)^p$. Moreover, $\Phi_p(1)=p$ is not a multiple of $p^2$. By the Eisenstein criterion (after a change of variables $T=1+U$, if one prefers), the polynomial $\Phi_p$ is irreducible.

This argument also works when $n=p^e$ is a power of a prime number. Indeed, since a complex number $\alpha$ is a primitive $p^e$th root of unity if and only if $\alpha^{p^{e-1}}$ is a primitive $p$th root of unity, one has $\Phi_{p^e}= \Phi_p(T^{p^{e-1}})$. Then the Eisenstein criterion gives the result.

Comment. — From the point of view of algebraic number theory, this proof makes use of the fact that the cyclotomic extension $\mathbf Q(\zeta_p)$ is totally ramified at $p$, of ramification index $p-1$.
Consequently, it must have degree $p-1$. More generally, it will prove that $\Phi_p$ is irreducible over the field $\mathbf Q_p$ of $p$-adic numbers, or even over any unramified extension of it, or even over any algebraic extension of $\mathbf Q_p$ for which the ramification index is prime to $p-1$.

2. The classical proof

Let us explain a proof that works for all integer $n$. Let $\alpha$ be a primitive $n$th root of unity, and let $P\in\mathbf Z[T]$ be its minimal polynomial — one has $P\mid \Phi_n$ in $\mathbf Z[T]$. Let (A priori, the divisibility is in $\mathbf Q[T]$, but Gauss's lemma implies that it holds in $\mathbf Z[T]$ as well.) Fix a polynomial $Q\in\mathbf Z[T]$ such that $\Phi_n=PQ$.

By euclidean division, one sees that the set $\mathbf Z[\alpha]$ of complex numbers of the form $S(\alpha)$, for $S\in\mathbf Z[T]$, is a free abelian group of rank $\deg(P)$, with basis $1,\alpha,\dots,\alpha^{\deg(P)-1}$.

Let $p$ be a prime number which does not divide $n$. By Fermat's little theorem, one has $P(T)^p \equiv P(T^p) \pmod p$, so that there exists $P_1\in\mathbf Z[T]$ such that $P(X)^p-P(X^p)=pP_1(T)$. This implies that $P(\alpha^p)=p P_1(\alpha)\in p\mathbf Z[\alpha]$.

Since $p$ is prime to $n$, $\alpha^p$ is a primitive $n$th root of unity, hence $\Phi_n(\alpha^p)=0$. Assume that $P(\alpha^p)\neq 0$. Then one has $Q(\alpha^p)=0$. Differentiating the equality $\Phi_n=PQ$, one gets $nT^{n}=T\Phi'_n(T)=TP'Q+TPQ'$; let us evaluate this at $\alpha_p$, we obtain $n=\alpha^p P(\alpha_p) Q'(\alpha^p)=p \alpha^p P^1(\alpha^p)Q'(\alpha^p)$. In other words, $n\in p\mathbf Z[\alpha]$, which is absurd because $n$ does not divide $p$. Consequently, $P(\alpha^p)=0$, and $P$ is also the minimal polynomial of $\alpha^p$.

By induction, one has $P(\alpha^m)=0$ for every integer $m$ which is prime to $n$. All primitive $n$th roots of unity are roots of $P$ and $\deg(P)=\phi(n)=\deg(\Phi_n)$. This shows that $P=\Phi_n$.

Comment. — Since this proof considers prime numbers $p$ which do not divide $n$, it makes implicit use of the fact that the cyclotomic extension is unramified away from primes dividing $n$. The differentiation that appears in the proof is a way of proving this non-ramification: if $P(\alpha^p)$ is zero modulo $p$, it must be zero.

3. Landau's proof

A 1929 paper by Landau gives a variant of this classical proof which I just learnt from Milne's notes on Galois theory and which I find significantly easier.

We start as previously, $\alpha$ being a primitive $n$th root of unity and $P\in\mathbf Z[T]$ being its minimal polynomial.

Let us consider, when $k$ varies, the elements $P(\alpha^k)$ of $\mathbf Z[\alpha]$. There are finitely many of them, since this sequence is $n$-periodic, so that they can be written as finitely polynomials of degree $<\deg(P)$ in $\alpha$. Let $A$ be an upper-bound for their coefficients. If $p$ is a prime number, we have $P(\alpha^p) \in p\mathbf Z[\alpha]$ (by an already given argument). This implies $P(\alpha^p)=0$ if $p>A$.

By induction, one has $P(\alpha^m)=0$ for any integer $m$ whose prime factors are all $>A$.

One the other hand, if $m$ is an integer prime to $n$ and $P$ is the product of all prime number $p$ such that $p\leq A$ and $p$ does not divide $m$, then $m'=m+nP $ is another integer all of which prime factors are $>A$. (Indeed, if $p\leq A$, then either $p\mid m$ in which case $p\nmid nP$ so that then $p\nmid m'$, or $p\nmid m$ in which case $p\mid nP$ so that $p\nmid m'$.) Since $m'\equiv m \pmod n$, one has $P(\alpha^{m})=P(\alpha^{m'})=0$.

This shows that all primitive $n$th roots of unity are roots of $P$, hence $P=\Phi_n$.

Comment. —This proof is quite of a mysterious nature to me.

4. Can one use Galois theory to pass from local information to global information?

The cyclotomic extension $K_n$ contains, as subextension, the cyclotomic extensions $K_{p^e}$, where $n=\prod p_i^{e_i}$ is the decomposition of $n$ has a product of powers of prime numbers. By the first case, $K_{p^e}$ has degree $\phi(p^e)=p^{e-1}(p-1)$ over $\mathbf Q$. To prove that $\Phi_n$ is irreducible, it would be sufficient to prove that these extensions are linearly disjoint.

This is what I had claimed in a first version of this blog post. Unfortunately, as Olivier Fouquet made me observe (September 26, 2020), this cannot be true without any additional argument that uses the fact that the ground field is the field of rational numbers.

Assume, for example, that the ground field $K=\mathbf Q(\sqrt 3)$. Over this field, the polynomials  $\Phi_4=T^2+1$ and $\Phi_3=T^2+T+1$ are both irreducible (because they have degree 2 and no root). However $\Phi_{12}=T^4-T^2+1=(T^2+1)^2-3T^2=(T^2-\sqrt 3T+1)(T^2+\sqrt 3T+1)$ is not irreducible.

 
(For consistency, I leave below the lemma that I had used. I have not yet thought about where the mistake lies, nor to how I could correct the statement.)

Unlemma. — Let $m$ and $n$ be integers and let $d$ be their gcd. Then $K_m\cap K_n=K_d$.

This is an application of Galois theory (and the result holds for every ground field as soon as its characteristic does not divide $m$ and $n$). Let $M$ be the least common multiple of $m$ and $n$. One has $K_N=K_m\cdot K_n$, and the cyclotomic character furnishes a group morphism $\operatorname{Gal}(K_N/\mathbf Q)\to (\mathbf Z/N\mathbf Z)^\times$. The Galois groups $\operatorname{Gal}(K_N/K_m)$ and $\operatorname{Gal}(K_N/K_n)$ corresponding to the subfields $K_m$ and $K_n$ are the kernels of the composition of the cyclotomic character with the projections to $(\mathbf Z/m\mathbf Z)^\times$ and $(\mathbf Z/n\mathbf Z)^\times$, and their intersection to the subgroup generated by these two kernels, which is none but the kernel of the composition of the cyclotomic character with the projection to $(\mathbf Z/d\mathbf Z)^\times$.