## Tuesday, July 2, 2019

### 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. Using 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 suffices to prove that these extensions are linearly disjoint, which is the object of the following lemma.

Lemma. — 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 $(\mathbf Z/d\mathbf Z)^\times$.