Irreducible compositions in an exceptional unicritical family
Henry Zweiman
Abstract
We construct irreducible compositions in a positive-shift subfamily left exceptional by a recent theorem on unicritically generated semigroups. Let be even and let be an integer which is not a th power for any odd prime , nor a fourth power when . Put and . We prove that is irreducible over for every composition of integral unicritical polynomials whose degrees have prime divisors among those of . For a finite equal-degree generating set containing and , this gives an explicit positive lower density. Within the exceptional four-constant alphabet, the shorter prefix suffices. The result includes infinitely many pairs in every degree divisible by four for which is reducible and fixes a square. The proof combines one real embedding, a field norm, and gaps between consecutive prime powers.
1. Introduction and statements
Let be a finite set of polynomials of degree at least two in , and let be the semigroup they generate under composition. A natural question is whether the presence of one irreducible polynomial forces a positive proportion of the elements of to be irreducible over , when polynomials are counted by degree. For generators of common degree , write
The prime-degree case was treated in [2], following the quadratic work [3]. For general common degree, Bhardwaj, Boyer-Paulet, Hindes, Qiu, and Sun [1, Theorem 1.1] establish the desired positive-proportion conclusion outside explicit exceptional families. In even degree, the exceptional constants are contained in
They expect the positive-proportion conclusion to extend to the exceptional cases. We treat a positive-shift subfamily of (1).
Throughout, superscripts on integers denote ordinary powers. We write compositions explicitly, including rather than , to distinguish iteration from multiplication. The empty composition is the identity polynomial.
Theorem 1.1.
Let be even, and let be an integer satisfying
for every odd prime ;
if , then .
Define
Then is irreducible over for every finite composition of polynomials with , , and . This includes the empty composition.
Corollary 1.2.
Under the hypotheses of Theorem 1.1, let be generated by distinct polynomials , with , including and . Then
If instead the generating set is a subset of
containing and , then is irreducible for every and for the empty composition, and
In particular, .
The main new instances occur when and . If is odd and nonzero, the hypotheses hold for every such : the valuation excludes every odd perfect power and every fourth power. For , , every nonzero integer is allowed. In these instances and , while
is reducible. Thus an irreducible positive-shift generator is not available to obtain the conclusion by the existing norm argument. The first example is
The two-letter prefix gives the lower bound in the semigroup generated by this pair.
The argument in [1, Proposition 3.8] excludes the positive-shift partner in (2). Its norm sufficient condition [1, Proposition 3.1] gives no conclusion at the first transition here, since is a square in the displayed subfamily. Our proof deals with that transition using a negative real root of and then separates the fourth-power obstruction by its norm. All later transitions follow from elementary gaps between prime powers. We do not resolve the full exceptional locus (1).
2. Algebraic irreducibility criteria
We use the classical binomial criterion: for a field of characteristic zero and , the polynomial is irreducible over if and only if
Let be irreducible, let be a root of , and let be nonconstant. Then is irreducible over if and only if is irreducible over .
Proof.
If is a root of , then and . The tower law gives
The right side equals precisely when is irreducible over . Since is a root of , the equality is equivalent to irreducibility of that polynomial.
The following familiar norm consequence is included to specify the parity and fourth-power issue exactly; compare [1, Proposition 3.1].
Lemma 2.2.
Let be monic and irreducible of even degree , and let with and . If for every prime , then is irreducible over .
Proof.
Suppose otherwise, and put for a root of . Lemma 2.1 and (4) imply either for a prime , or and , with . In the first case its norm is a rational th power. In the second case its norm is
which is a rational square because is even. In both cases this contradicts the assumptions, since
3. The first mixed composition
Lemma 3.1.
Let be even and be prime. If is an integer, then
Proof.
For an integer , factor the difference of powers to obtain
Apply this with . Since is even, .
Proposition 3.2.
Under the hypotheses of Theorem 1.1, both and are irreducible over .
Thus , hence . Here we use the elementary fact that an integer which is a rational th power is an integer th power. Since , it also does not belong to . Criterion (4) proves irreducible.
Choose its negative real root and regard as a subfield of . The element is negative in this embedding, so it is not a square in . Also
For an odd prime , equation (5) and the hypothesis exclude .
It remains to address the exceptional clause when . If for some , then, since ,
This would make a rational fourth power, and therefore an integer fourth power, contrary to hypothesis. Thus (4) proves irreducible over . Lemma 2.1 proves irreducible over .
We can therefore append the factors of one at a time. Suppose that is already irreducible, and append as in the theorem. Then , so
by (7). The polynomial remains monic and has even degree. Lemma 2.2 proves irreducible and completes the induction.
Proposition 4.1.
Under the hypotheses of Corollary 1.2 for the alphabet (2), is irreducible for every finite word in that alphabet, including the empty word.
Proof.
Put . All four generator constants belong to . The first three maps in (2) take into integers at least , because . For the last map, an input of magnitude maps to . For all other inputs in , integrality gives , and
Thus its output is greater than . Consequently every word in the alphabet preserves .
For we have , so (6) implies that is not a rational th power for any prime . Starting from irreducible by Proposition 3.2, append letters as before. Each required norm value has the form , where the generator constant belongs to and . Lemma 2.2 proves irreducibility at every step.
5. Counting by degree
We include the elementary same-degree freeness argument to pass from words to actual polynomials; compare [1, Proposition 3.9].
Lemma 5.1.
In characteristic zero, distinct polynomials , with common degree , generate a free semigroup under composition.
Proof.
Degree determines word length. Suppose two words of equal length greater than one define the same polynomial. Removing their outer letters leaves monic polynomials of equal positive degree , with
If , then
since has leading coefficient and degree . This contradicts . Hence and . Induction recovers all letters.
By Lemma 5.1, there are exactly distinct polynomials of word length , each of degree . Every word of length with prefix is irreducible by Theorem 1.1; there are such words. Therefore, for ,
Since , we have , and the right side tends to . For the restricted alphabet, Proposition 4.1 supplies the prefix of length two, giving the analogous limit .
Remark 5.2.
The restrictions on in Theorem 1.1 are sufficient hypotheses for the real-place and norm proof; no necessity assertion is intended. The result does not settle exceptional pairs with the negative shift in place of , when the positive-shift generator is absent. The mixed-degree assertion concerns irreducibility of the indicated compositions, not a density theorem for a mixed-degree semigroup.
References
[1] A. Bhardwaj, A. Boyer-Paulet, W. Hindes, E. Qiu, and A. Sun, Prime-powered images and irreducible polynomials in dynamical semigroups, arXiv:2510.10310v1 (2025), https://arxiv.org/abs/2510.10310.
[2] W. Hindes, R. Jacobs, B. Keller, A. Kim, P. Ye, and A. Zhou, On the proportion of irreducible polynomials in unicritically generated semigroups, J. Algebra 677 (2025), 394–429.
[3] W. Hindes, R. Jacobs, and P. Ye, Irreducible polynomials in quadratic semigroups, J. Number Theory 248 (2023), 208–241.
[4] S. Lang, Algebra, third revised edition, Graduate Texts in Mathematics, vol. 211, Springer, New York, 2002.