Irreducible cyclotomic polynomial

Question

I want to know if there is a way to decide if a cyclotomic polynomial is irreducible over a field $\mathbb{F}_q$?

Answer 1

Yes there is.

Let $p$ be the characteristic, so $q=p^m$ for some positive integer $m$.

Assuming $\gcd(q,n)=1,$ the $n^{th}$ cyclotomic polynomial $\Phi_n(x)\in\mathbb{Z}[x]$ will remain irreducible (after reduction mod $p$) in $\mathbb{F}_q[x]$ if and only if the residue class of $q$ generates the multiplicative group $\mathbb{Z}_n^*$ of residue classes coprime to $n$.

This is because if $z$ is a root of $\Phi_n(x)$ in an extension of $\mathbb{F}_q$, then its conjugates are $z^q, z^{q^2},$ et cetera. If you get the same number of conjugates as you would get over $\mathbb{Q}$, then you are done. But over $\mathbb{Q}$ the conjugates are exactly $z^a, \gcd(a,n)=1, 1\le a<n$.

---

More details. Let $z$ be a primitive $n^{th}$ root of unity in an extension $\mathbb{F}_q$. Let $\mathbb{F}_q[z]=\mathbb{F}_{q^k}$. Because the multiplicative group of $\mathbb{F}_{q^k}$ is cyclic of order $q^k-1$, we know that $k$ is the smallest positive integer with the property that $n\mid q^k-1$. By the Galois theory of finite fields the minimal polynomial of $z$ is $$ m(x)=(x-z)(x-z^q)(x-z^{q^2})\cdots(x-z^{q^{k-1}}). $$ This will always be a factor of the cyclotomic polynomial $\Phi_n(x)$. The roots of the latter are $z^a, 1\le a<n, \gcd(a,n)=1$. The polynomial $\Phi_n(x)$ is thus irreducible precisely when the two sets of roots are the same.

Here $z^{q^i}=z^a$ if and only if $q^\ell\equiv a\pmod{n}$. Therefore all the primitive roots $z^a$ are zeros of $m(x)$ only, if all the exponents $a$ are congruent to a power of $q$ modulo $n$.

---

All of the above assumed that $\gcd(n,q)=1$. Let us next consider the case where that is not true. Here $q$ is the order of a finite field, so it is a power of a prime number $p$. Therefore $\gcd(n,p)>1$ if and only if $p\mid n$, so we can write $n=mp^\ell$ for some integer $\ell\ge1$, $m$ coprime to $p$. Then we have in the ring $\mathbb{F}_p[x]$ the factorization $$ x^n-1=(x^m-1)^{p^a} $$ as a consequence of Freshman's dream: $$ (a+b)^p=a^p+b^p. $$ Therefore all the roots of $\Phi_n(x)$ in $\overline{\mathbb{F}_q}$ are actually roots of $x^m-1$ as well. Hence any one of them has at most $\phi(m)<\phi(n)$ conjugates. Therefore $\Phi_n(x)$ cannot be irreducible in $\mathbb{F}_q[x]$. (the part in italics is incorrect, see below)

Edit: As pointed out by Yecabel, the last claim is a touch too sweeping. We do see that any zero of $\Phi_n(x)$ has at most $\phi(m)$ conjugates. But, it is possible that $\phi(m)=\phi(n)$. As $n=mp^\ell$, $p\nmid m$, we have $\phi(n)=\phi(m) p^{\ell-1}(p-1)$. So for $\phi(n)$ to be equal to $\phi(m)$ we need that $p=2$ and $\ell=1$. Leaving the special case of $q$ even, $n=2m$, $m$ odd, to deal with. We always have $\Phi_{2m}(x)=\Phi_m(-x)$. And in characteristic two $\Phi_m(-x)=\Phi_m(x)$, so $\Phi_n(x)$ is irreducible if and only if $\Phi_m(x)$ is. The conclusion is thus

If $\gcd(q,n)>1$ then $\Phi_n(x)\in\Bbb{Z}[x]$ stays irreducible in $\Bbb{F}_q[x]$ only, if $q$ is a power of two, $n=2m$, $2\nmid m$, and $\Phi_m(x)$ stays irreducible (see the result in the main case).

Answer 2

I know this thread is very old but I have been thinking about this problem recently and I had problems understanding Jyrki's proof. Maybe my argument will be clearer for future readers. I mean, all the main ideas are in the previous post; but I found it hard to make the connections. I will prove a more general assertion present in an article on Wikipedia:

Suppose $f(x)$ is the $n$-th cyclotomic polynomial with coefficients in the finite field $\mathbb{F}_q$, with $q=p^m$ for some prime number $p$. Also, assume $\gcd (n,p)=1$. Then, $f(x)$ can be factored into $\frac{\varphi(n)}{d}$ different irreducible polynomials (all of them with order $d$), where $\varphi(n)=\deg(f)$ is Euler's totient function and $d$ is the multiplicative order of $q\!\!\!\!\mod\!\!n$ in $\mathbb{Z}_n^{\times}$ (the multiplicative group of units in the ring of integers modulo $n$).

Remark: The original question is a particular case when $d=\varphi(n)$.

Proof: Let's divide the proof into different items:

1. Since $\gcd(n,p)=1$ (then $\gcd(n,q)=1$), we have that $f(x)$ is separable over $\mathbb{F}_q$ and that $(q\!\!\!\!\mod\!\!n)\in \mathbb{Z}_n^{\times}$. $f(x)$ is separable because it's a factor of $x^m-1$ and this polynomial has no repeated roots (use the formal derivative criterion). $q$ modulo $p$ is a unit in $\mathbb{Z}_n$ because of Bézout's identity: $xq+yn=1$; this establishes that $q\!\!\!\!\mod\!\!n$ is in $\mathbb{Z}_n^{\times}$ and so its multiplicative order there is well defined.

2. We have a root $\zeta$ of $f(x)$ in a field extension $\mathbb{F}_{q^s}/\mathbb{F_q}$ (including degree 1 extensions) iff $n \mid (q^s-1)$. To see this, notice that the field $\mathbb{F}_{q^s}$ has a cyclic multiplicative group. Let's call $w$ a generator of that multiplicative group. Then the desired root would be $\zeta=w^{\frac{q^s-1}{n}}$. Notice the roots of $f(x)$ are precisely the primitive $n$-roots of unity, i.e., elements with multiplicative order $n$. This characterization of the roots of $f(x)$ is possible thanks to the separability of $x^n-1$, because we have $n$ different roots, and the roots form a cyclic multiplicative group.

3. A monic polynomial $g(x)$ with root $\zeta$ is irreducible over a field iff $g(x)$ is the minimal polynomial of $\zeta$. For a proof, see for example theorem 3.113 of Rotman's Advanced Modern Algebra. Define $d$ as the minimal positive integer $s$ for which $n \mid (q^s-1)$. Attending to what we said in the second item, the polynomial of minimal degree having $\zeta$ as a root has degree $d$, i.e., $\deg g =d$. Now, $g(x)\mid f(x)$, and this fact can also be consulted in the same theorem 3.113. Since this argument can be made to any root of $f(x)$, all irreducible polynomial factors of $f(x)$ have degree $d$. Since $\deg f =\varphi(n)$, the number of irreducible polynomial factors is $\frac{\varphi(n)}{d}$. To complete the proof, we just need to notice that $d$ can also be seen as the minimal integer for which $q^s \!\!\!\!\mod\!\! n =1$, i.e., $d$ is the multiplicative order of $q \!\!\!\!\mod\!\! n$ in $\mathbb{Z}_n^{\times}$.

Examples:

In order to illustrate my previous result I will offer examples of the three main possibilities:

1. $d=1$. For example, if $n=6$ and $q=7$. Here, $f(x)=x^2-x+1$ and it can be factored as $f(x)=(x-3)(x-5)$ (remember we are working in $\mathbb{Z}_7[x]$). In this case, all members of the group are sixth roots of unity, $3$ and $5$ are the primitive roots.
2. $1<d<\varphi(n)$. For example, $n=10$ and $q=19$. Here $f(x)=x^4-x^3+x^2-x+1$ and it can be factored as $f(x)=(x^2-5x+1)(x^2+4x+1)$. Both quadratic factors are irreducible over $\mathbb{F}_{19}$.
3. $d=\varphi(n)$. For example, $n=6$ and $q=11$. Here, $f(x)=x^2-x+1$ is truly irreducible.