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A note on a conjecture by Ulas on polynomial substitutions
Journal of Number Theory 205 (2019) 122–123
Contents lists available at ScienceDirect
Journal of Number Theory www.elsevier.com/locate/jnt
General Section
A note on a conjecture by Ulas on polynomial substitutions Peter Müller Institut für Mathematik, Universität Würzburg, 97074 Würzburg, Germany
a r t i c l e
i n f o
Article history: Received 28 May 2019 Accepted 24 June 2019 Available online 18 July 2019 Communicated by S.J. Miller
a b s t r a c t We prove a recent conjecture by Ulas on reducible polynomial substitutions. © 2019 Elsevier Inc. All rights reserved.
Keywords: Univariate polynomials Reducibility
We prove the following result, which was conjectured by Ulas in [2] and proven for d ≤ 4 there. Theorem. Let f (X) be an irreducible polynomial of degree d ≥ 3 over a field K. Then there is a polynomial h(X) ∈ K[X] of degree ≤ d − 1 such that f (h(X)) is reducible over K. Proof. Let α be a root of f (X) in some extension of K, and g(X) be the minimal polynomial of α1 over K. (So g(X) is, up to a non-zero factor from K, the reciprocal of f (X).) Since K( α1 ) = K(α), and α1 has degree d over K, we have α = h( α1 ) for some h(X) ∈ K[X] of degree at most d − 1.
E-mail address: [email protected]. https://doi.org/10.1016/j.jnt.2019.06.004 0022-314X/© 2019 Elsevier Inc. All rights reserved.
P. Müller / Journal of Number Theory 205 (2019) 122–123
123
We obtain f (h( α1 )) = f (α) = 0. So f (h(X)) shares the root α1 with the irreducible polynomial g(X), thus g(X) divides f (h(X)). The degree of h is at least 2, for otherwise α would have degree at most 2 over K, contrary to d ≥ 3. The assertion now follows from deg f (h(X)) ≥ 2d and deg g(X) = d. 2 Remark. (a) Of course one can formulate the proof without reference to the algebraic element α. Suppose without loss that f (X) is monic. Set g(X) = X d f (1/X) and h(X) = (1 − g(X))/X. Then h(X) is a polynomial, and g(X) divides f (h(X)). This example is similar to the one by Schinzel in [1, Lemma 10], which, as pointed out by Ulas [2, page 59], proves the theorem provided that d − 1 does not divide the characteristic of K. (b) The answer to [2, Question 5.5], a kind of converse to the above theorem, is negative if K = R by lack of irreducible polynomials of degree ≥ 3. References [1] Andrzej Schinzel, On two theorems of Gelfond and some of their applications, Acta Arith. 13 (1967/1968) 177–236. [2] Maciej Ulas, Is every irreducible polynomial reducible after a polynomial substitution? J. Number Theory 202 (2019) 37–59.
Contents lists available at ScienceDirect
Journal of Number Theory www.elsevier.com/locate/jnt
General Section
A note on a conjecture by Ulas on polynomial substitutions Peter Müller Institut für Mathematik, Universität Würzburg, 97074 Würzburg, Germany
a r t i c l e
i n f o
Article history: Received 28 May 2019 Accepted 24 June 2019 Available online 18 July 2019 Communicated by S.J. Miller
a b s t r a c t We prove a recent conjecture by Ulas on reducible polynomial substitutions. © 2019 Elsevier Inc. All rights reserved.
Keywords: Univariate polynomials Reducibility
We prove the following result, which was conjectured by Ulas in [2] and proven for d ≤ 4 there. Theorem. Let f (X) be an irreducible polynomial of degree d ≥ 3 over a field K. Then there is a polynomial h(X) ∈ K[X] of degree ≤ d − 1 such that f (h(X)) is reducible over K. Proof. Let α be a root of f (X) in some extension of K, and g(X) be the minimal polynomial of α1 over K. (So g(X) is, up to a non-zero factor from K, the reciprocal of f (X).) Since K( α1 ) = K(α), and α1 has degree d over K, we have α = h( α1 ) for some h(X) ∈ K[X] of degree at most d − 1.
E-mail address: [email protected]. https://doi.org/10.1016/j.jnt.2019.06.004 0022-314X/© 2019 Elsevier Inc. All rights reserved.
P. Müller / Journal of Number Theory 205 (2019) 122–123
123
We obtain f (h( α1 )) = f (α) = 0. So f (h(X)) shares the root α1 with the irreducible polynomial g(X), thus g(X) divides f (h(X)). The degree of h is at least 2, for otherwise α would have degree at most 2 over K, contrary to d ≥ 3. The assertion now follows from deg f (h(X)) ≥ 2d and deg g(X) = d. 2 Remark. (a) Of course one can formulate the proof without reference to the algebraic element α. Suppose without loss that f (X) is monic. Set g(X) = X d f (1/X) and h(X) = (1 − g(X))/X. Then h(X) is a polynomial, and g(X) divides f (h(X)). This example is similar to the one by Schinzel in [1, Lemma 10], which, as pointed out by Ulas [2, page 59], proves the theorem provided that d − 1 does not divide the characteristic of K. (b) The answer to [2, Question 5.5], a kind of converse to the above theorem, is negative if K = R by lack of irreducible polynomials of degree ≥ 3. References [1] Andrzej Schinzel, On two theorems of Gelfond and some of their applications, Acta Arith. 13 (1967/1968) 177–236. [2] Maciej Ulas, Is every irreducible polynomial reducible after a polynomial substitution? J. Number Theory 202 (2019) 37–59.