A Note on Stable Quadratic Polynomials over Fields of Characteristic Two
aa r X i v : . [ m a t h . N T ] O c t A Note on Stable Quadratic Polynomials overFields of Characteristic Two
Omran Ahmadi
Claude Shannon InstituteUniversity College DublinDublin 4, Ireland [email protected]
June 13, 2018
Abstract
In this note, first we show that there is no stable quadratic poly-nomial over finite fields of characteristic two and then show that thereexist stable quadratic polynomials over function fields of characteristictwo.
Let K be a field and K [ x ] be the polynomial ring over K . A polynomial f ( x ) ∈ K [ x ] is called stable if f ( x ) , f f ( x ) , f f f ( x ) , . . . , f ( n ) ( x ) , . . . is a se-quence of irreducible polynomials in K [ x ]. Recently there has been an inter-est in the study of stable polynomials .(see [1, 2, 3, 5, 7, 8, 9])In [9], Ostafe and Shparlinski studied stable quadratic polynomials overfinite fields and obtained an upper bound on the length of the critical or-bits of stable quadratic polynomials over finite fields with odd characteristic.They also posed the question of estimating the number of stable quadraticpolynomials over finite fields. Here we show that there is no stable quadraticpolynomial over finite fields of characteristic two by proving the followingtheorem. 1 heorem 1. Let q = 2 m , and let f ( x ) = cx + ax + b ∈ IF q [ x ] where IF q denotes the field with q elements and a, b ∈ IF ∗ q . Then f f f ( x ) cannot be anirreducible polynomial over IF q . For simplicity we assume that c = 1 and hence f ( x ) is a monic polynomial.The same proof can be used for the non-monic polynomials. We also notethat in what follows Tr | K ||| L | ( · ) denotes the trace map from field K with | K | elements to its subfield L with | L | elements. Our proof is based on thefollowing two well-known lemmas. Lemma 2. [4, Corollary 3.6] Let q = 2 m , and let f ( x ) = x + ax + b ∈ IF q [ x ] where IF q denotes the field with q elements and a, b ∈ IF ∗ q . Then f ( x ) isirreducible over IF q if and only if Tr q | ( ba ) = 1 . The following lemma is known as Capelli’s lemma and can be found in [6]too.
Lemma 3.
Let f ( x ) be a degree n irreducible polynomial over F q , and let g ( x ) ∈ F q [ x ] . Then p ( x ) = f ( g ( x )) is irreducible over F q if and only if forsome root α of f ( x ) in F q n , g ( x ) − α is an irreducible polynomial over F q n . Now suppose that both f ( x ) = x + ax + b and f f ( x ) are irreducibleover IF q . We show that f f f ( x ) cannot be an irreducible polynomial over IF q .Suppose for the contrary that f f f ( x ) is irreducible over IF q . Using Lemma 3, f f f ( x ) is irreducible over IF q if and only if h ( x ) = f ( x ) − α is irreducible overIF q for some root α of f f ( x ) in IF q . Applying Lemma 2, h ( x ) is irreducibleover IF q if and only if Tr q | ( b − αa ) = 1. Using the properties of the trace mapin one hand Tr q | (cid:18) b − αa (cid:19) = Tr q | (cid:18) ba (cid:19) − Tr q | (cid:16) αa (cid:17) (1)and on the other handTr q | (cid:18) ba (cid:19) = 4Tr q | (cid:18) ba (cid:19) = 0 . q | (cid:0) αa (cid:1) = 1. ButTr q | (cid:16) αa (cid:17) = Tr q | (cid:0) Tr q | q (cid:0) αa (cid:1)(cid:1) = Tr q | (cid:18) Tr q | q ( α ) a (cid:19) . Now suppose that f f ( x ) = x + c x + c x + c x + c . Then it is easy tosee that c = 0. Finally, since α is a root of f f ( x ) in IF q , we deduce thatTr q | q ( α ) = c = 0, and hence Tr q | (cid:0) αa (cid:1) = 0 which is a contradiction. Let IF ( t ) be the rational function field in t over IF , where t is transcendentalover IF , and let IF ( t )[ x ] be the polynomial ring over IF ( t ) in the variable x . One expects that a theorem similar to Theorem 1 to hold for quadraticpolynomials in IF ( t )[ x ] but this is not true as the following example shows.Suppose f ( x ) = x + t ∈ IF ( t )[ x ]. Then it is easy to see that f ( n ) ( x ) = x n + t n − + t n − + · · · + t + t. Now from Eisenstein’s criterion for function fields [10, Proposition III.1.14]it is easy see that for every n ≥ f ( n ) ( x ) is irreducible over IF ( t ) and hence f ( x ) is a stable polynomial in IF ( t )[ x ]. The author would like to thank Igor Shparlinski for helpful comments on anearlier version of this note.
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