MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $q$ be a power of a prime. It's well-known that the function $B(n, q) = \frac{1}{n} \sum_{d | n} \mu \left( \frac{n}{d} \right) q^d$ counts both the number of irreducible polynomials of degree $n$ over $\mathbb{F}_q$ and the number of Lyndon words of length $n$ over an alphabet of size $q$. Does there exist an explicit bijection between the two sets?

share|cite|improve this question
up vote 24 down vote accepted

In Reutenauer's "Free Lie Algebras", section 7.6.2:

A direct bijection between primitive necklaces of length n over F and the set of irreducible polynomials of degree n in F[x] may be described as follows: let K be the field with qn elements; it is a vector space of dimension n over F, so there exists in K an element θ such that the set {θ, θq, ..., θqn-1} is a linear basis of K over F.

With each word w = of length n on the alphabet F, associate the element β of K given by β = a0θ + a1θq + ... + an-1 θqn-1. It is easily shown that to conjugate words w, w' correspond conjugate elements β, β' in the field extension K/F, and that w \mapsto β is a bijection. Hence, to a primitive conjugation class corresponds a conjugation class of cardinality n in K; to the latter corresponds a unique irreducible polynomial of degree n in F[x]. This gives the desired bijection.

share|cite|improve this answer

See section 38.10 "Generating irreducible polynomials from Lyndon words" of

share|cite|improve this answer

I believe such a bijection is presented in

S. Golomb. Irreducible polynomials, synchronizing codes, primitive necklaces and cyclotomic algebra. In Proc. Conf Combinatorial Math. and Its Appl., pages 358– 370, Chapel Hill, 1969. Univ. of North Carolina Press.

but I don't have immediate access to this paper - I'm pretty sure it's in there though.

share|cite|improve this answer
I don't seem to have access to it either, but at least one other paper (<a href="">Berstel and Reutenauer</a>) suggests that this is an open problem. Indeed I have essentially the same motivation as them for asking this question, so I suppose I should've read this paper more carefully. – Qiaochu Yuan Oct 16 '09 at 18:43
A tiny bit of additional evidence (still not conclusive): "As there exists a bijection between Lyndon words over an alphabet of cardinality k and irreducible polynomials over Fk [10]..." where the reference [10] is to Golomb's paper. – Alon Amit Oct 16 '09 at 19:12

The correspondence invented by Golomb relies on the choice of a primitive element a in the field of order q^n. Then, to each Lyndon word L=(l_0,l_1,...,l_{n-1}) one assigns the primitive polynomial having as a root the element a^{m(L)} where m(L) is the integer sum of l_i*q^i over i=0,1,...,n-1.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.