Research preprint

Divisors and unique factorizations of Carlitz binomial polynomials

Henry Zweiman

September 8, 2026

Abstract

Let . We classify the divisors in of every finite product of the basic Carlitz binomial polynomials . Each divisor is, up to a unit, a subproduct of the given basic factors. Thus the multiplicative monoid generated by these factors is divisor-closed and factorial. In particular, every power of every Carlitz binomial polynomial has a unique factorization into irreducibles, answering a question of Tichy and Windisch. The proof combines a simultaneous-minimum identity for weighted products over the sets of polynomials of bounded degree with an extraction argument at a prime of prescribed degree.

1. Introduction and main theorem

Fix a prime power , and write and . The ring of integer-valued polynomials on is

Here and are independent indeterminates. Carlitz constructed a regular -module basis of this ring that is analogous to the ordinary binomial basis of ; we follow the notation and normalization in Tichy–Windisch [2].

Set , , and . For , let

We also put . The convention is used. If is the base- expansion of a nonnegative integer, then its Carlitz binomial polynomial is

In particular, .

Tichy and Windisch proved that the are irreducible in and that each is absolutely irreducible: every power of an individual factors uniquely into irreducibles [2, Theorem 1.2]. Their Question 3.1 asks whether this uniqueness persists for powers of the general polynomial (2). We prove the following stronger divisor statement.

Theorem 1.1.

Let , and put

An element divides in if and only if

The unit and the exponent vector are unique.

Corollary 1.2.

For every and , the polynomial has a unique factorization into irreducibles in , up to order and associates. If , this factorization is

The mixed-product assertion is stronger than absolute irreducibility of the separate . The proof must show that extracting a highest-level basic factor from an arbitrary divisor preserves integer-valuedness at every prime. Section 2 provides the required fixed-divisor identity; Section 3 performs the extraction and proves the theorem. The argument also recovers the known irreducibility of the .

2. A simultaneous-minimum identity

For a monic irreducible polynomial , let be the corresponding valuation on , normalized by , with . If is nonzero, write

This minimum is finite: is infinite, and has only finitely many roots. It is the -adic valuation of the fixed divisor of .

The residue counts for and the formula for used below are classical; see Wagner [3, p. 51]. We require a point that simultaneously minimizes the valuation of and of an arbitrary product of its linear factors with nonnegative multiplicities.

Lemma 2.1.

Let , let be monic irreducible of degree , and put

Let

Then, for every ,

Moreover,

Proof.

Write , where . For , the reduction map

is surjective, and each fiber has elements. Indeed, division by the monic polynomial expresses a polynomial of degree less than uniquely as a remainder of degree less than plus times a polynomial of degree less than . Consequently, for every ,

Empty sums have value zero, so this also holds for .

The residue class contains exactly elements of . For this says simply that . These elements occupy at most of the residue classes modulo lying above modulo . Since , choose in one of the unoccupied classes. Then , and for every ,

For valuations less than , this follows from the congruence between and ; for valuations at least , it follows from the choice of the class of modulo .

Equation (6) now gives

It also shows that for all . Start with an attaining . The resulting attains this same minimum and also minimizes . Hence

Taking and proves the first equality in (5).

The polynomial divides exactly when , and then its multiplicity is one. This follows from the description of the roots of as and from its derivative . Applying to the product defining gives the second equality in (5).

Corollary 2.2.

Every belongs to . More generally, if , , , and is as in Lemma 2.1, then

Proof.

An element belongs to if and only if for every monic irreducible . Therefore, for a nonzero polynomial ,

Lemma 2.1 makes these inequalities identical for the two polynomials in (8). It also proves . The case is immediate. All primes in the criterion are finite primes of ; no condition at the infinite place is imposed.

3. Extraction and divisor classification

Lemma 3.1.

Let , let with , and suppose satisfy

There is an integer with such that

Both quotients split over with all roots in .

Proof.

Unique factorization in gives representations

where , , and is the multiplicity of in the numerator of . In particular,

Choose a monic irreducible polynomial of degree . Such a polynomial exists over every finite field in every positive degree. The set is a complete system of representatives modulo , and Lemma 2.1 gives

Set . Comparing leading coefficients in the identity for gives

For each , evaluate at . Then , while for every . The integer-valuedness of and yields

For in the nonempty set , equations (9) and (10) force . Since on this set, we have .

For all , the multiplicities and are nonnegative. They are zero on . We can consequently write

where has nonnegative root multiplicities and all roots in . Regard as a weighted product over with zero weights on the outer set. Corollary 2.2 gives

The same argument applied to proves the second assertion.

Proof of Theorem 1.1.

The units of are exactly . Indeed, a polynomial unit and its inverse must both have -degree zero, and the constant elements of are exactly .

Suppose with . We induct on the largest index such that , omitting trailing zero exponents. If all exponents are zero, , and the assertion follows from the description of the units. If , unique factorization in gives

Evaluation at shows , so .

Now let . Apply Lemma 3.1. The two resulting integer-valued quotients have product

The induction hypothesis gives

Taking proves (3). Conversely, a polynomial of that form has an integer-valued complementary product, so it divides .

For uniqueness, compare the multiplicity of any root in in two expressions of the form (3). This determines . Cancel this power of and descend. At the final level, the multiplicity of zero determines . Equality of the remaining constants then determines .

Proof of Corollary 1.2.

The irreducibility of each was proved in [2]; it also follows by applying Theorem 1.1 to . If an irreducible polynomial divides a product in that theorem, its expression (3) must contain exactly one basic factor, counted with multiplicity. Hence every irreducible divisor of is associate to some . Uniqueness of the exponent vector proves the assertion after taking .

4. Consequences and examples

Corollary 4.1.

The submonoid

of is divisor-closed. Its quotient by its units is the free commutative monoid on the classes of .

Proof.

Divisor closure is exactly Theorem 1.1. The uniqueness statement in that theorem shows that the exponent-vector map identifies with the direct sum .

In particular, if , then the divisor classes of form a product of finite chains:

This is also an isomorphism of partially ordered sets, with divisibility on the left and coordinatewise order on the right. Thus

where denotes the set of factorization lengths. The uniqueness of factorization is stronger than the singleton assertion for lengths.

Example 4.2.

For every , the index gives

Its th power has exactly divisor classes, represented by with . Its unique factorization has length . These mixed-index families have unbounded degree and are not covered by the prime-power assertion alone.

Remark 4.3.

The analogue of Corollary 4.1 for the ordinary binomial polynomials fails. Put . Then

The numerator of is divisible by , as it contains three consecutive integers. It is also divisible by : for odd the square suffices, while for even the product suffices. However, is not a product of ordinary binomial polynomials, up to a unit. Any such product with roots contained in has weakly decreasing root multiplicities in the order , whereas those of are . Thus products of the ordinary binomial polynomials need not form a divisor-closed monoid, even though each is absolutely irreducible [1].

References

[1] R. Rissner and D. Windisch, Absolute irreducibility of the binomial polynomials, Journal of Algebra 578 (2021), 92–114. doi:10.1016/j.jalgebra.2021.03.007.

[2] R. Tichy and D. Windisch, Irreducibility properties of Carlitz’ binomial coefficients for algebraic function fields, Finite Fields and Their Applications 96 (2024), article 102413. doi:10.1016/j.ffa.2024.102413. Preprint: arXiv:2310.02061.

[3] C. G. Wagner, On the factorization of some polynomial analogues of binomial coefficients, Archiv der Mathematik 24 (1973), 50–52. Author’s copy.