Revisiting the Factorization of x n +1 over Finite Fields
aa r X i v : . [ m a t h . N T ] S e p Revisiting the Factorization of x n + 1 over FiniteFields Arunwan Boripan and Somphong Jitman
Abstract
The polynomial x n + 1 over finite fields has been of interest due to itsapplications in the study of negacyclic codes over finite fields. In this paper,a rigorous treatment of the factorization of x n + 1 over finite fields is givenas well as its applications. Explicit and recursive methods for factorizing x n + 1 over finite fields are provided together with the enumeration formula.As applications, some families of negacyclic codes are revisited with moreclear and simpler forms. keywords : Factorization, Enumeration, Polynomials, Negacyclic Codes Mathematics Subject Classification : 11T71, 11T60, 12Y05
In coding theory, the polynomial x n + 1 over finite fields plays an important role inthe study of negacyclic codes (see [1], [2], [5], [9], and references therein). Precisely,a negacyclic code of length n over F q can be uniquely determined by an ideal inthe principal ring F q [ x ] / h x n + 1 i generated by a monic divisor of x n + 1. A briefdiscussion on the factorization of x n + 1 over finite fields F q has been given in [5]and [9]. In the case where the characteristic of F q is even, the factorization of x n + 1 = x n − F q has been given and applied in the study of cyclic codesover finite fields in [7]. In [3] and [6], an explicit form of the factorization of x i + 1over finite fields of odd characteristic has been established. This research was supported by the Thailand Research Fund under Research GrantRSA6280042.A. Boripan is with the Department of Mathematics, Faculty of Science, RamkhamhaengUniversity, Bangkok 10240, Thailand (email: [email protected])S. Jitman (Corresponding Author) is with the Department of Mathematics, Faculty of Sci-ence, Silpakorn University, Nakhon Pathom 73000, Thailand (email: [email protected]).
1n this paper, we focus on the factorization of x n + 1 over finite fields F q forarbitrary positive integers n and all odd prime powers q . If the characteristic of F q is p , we have x p s n + 1 = ( x n + 1) p s for all integers n ≥ s ≥
0. It is therefore sufficient to study the factorizationof x n + 1 over F q such that n is co-prime to q . Here, we write n = 2 i n ′ for someinteger i ≥ n ′ such that gcd( n ′ , q ) = 1.Before proceed to the general results, we consider a pattern on the factorizationof x i + 1 over F . We have x · + 1 = f ( x ) f ( x ) f ( x ) f ( x ) f ( x ) f ( x ) x · + 1 = f ( x ) f ( x ) f ( x ) f ( x ) f ( x ) f ( x )... x i · + 1 = f ( x i − ) f ( x i − ) f ( x i − ) f ( x i − ) f ( x i − ) f ( x i − )for all i ≥
1, where f ( x ) = x + 2, f ( x ) = x + 3, f ( x ) = x + x + x + 2 x + x + 2, f ( x ) = x + 2 x + x + 2 x + 3 x + 2, f ( x ) = x + 3 x + x + 3 x + 3 x + 3 and f ( x ) = x + 4 x + x + 3 x + x + 3. It is easily seen that the factorization can bedetermined recursively on the exponent i of 2 and the number of monic irreduciblefactors of x i + 1 is a constant independent of i ≥ x i n ′ + 1 over F q is given. Precisely, we prove that there exists a positive integer k such that the number of monic irreducible factors of x i n ′ + 1 over F q becomesa constant for all positive integers i ≥ k . In the cases where ord n ′ ( q ) is odd, acomplete recursive factorization of x i n ′ + 1 over F q is provided together with arecursive formula for the number of its monic irreducible factors for all positiveintegers i . In the cases where ord n ′ ( q ) is even, a recursive factorization of x i n ′ + 1over F q is given for all positive integers i ≥ k . As applications, constructions andenumerations of some negacyclic codes of lengths 2 i n ′ over F q are given based onthe above results.The paper is organized as follows. Preliminary concepts and results on thefactorization of x n + 1 over finite fields are recalled in Section 2. In Section 3,the number theoretical results and properties of q -cyclotomic cosets required inthe study of the factorization of x i n ′ + 1 are established. Recursive methodsfor factorizing x i n ′ + 1 and enumerating its monic irreducible factors are givenin Section 4. Applications in the study of negacyclic codes over finite fields arerevisited in Section 5. 2 Preliminary
In this section, basic concepts and tools used in the study of the factorization of x n + 1 over finite fields and the enumeration of its monic irreducible factors arerecalled.For a positive integer a and an integer s , the notation 2 s || a is used whenever s is the largest integer such that a is divisible by 2 s , or equivalently, 2 s | a but 2 s +1 ∤ a .For an integer a and a positive integer n , denote by Θ n ( a ) the additive order of a modulo n . In the case where gcd( a, n ) = 1, denote by ord n ( a ) the multiplicativeorder of a modulo n . By abuse of notation, we write ord ( a ) = 1.For a prime power q , a positive integer n co-prime to q , and an integer 0 ≤ a Let q be an odd prime power and let n ′ be an odd positive integer suchthat gcd( q, n ′ ) = 1 . Then | Cl q, i +1 n ′ ( a ) | = | Cl q, i +1 n ′ ( a + 2 i n ′ ) | all odd integers a and for all positive integers i .Proof. Let a be an odd integer and let i be a positive integer. ThenΘ i +1 n ′ ( a ) = 2 i +1 n ′ gcd(2 i +1 n ′ , a ) = 2 i +1 n ′ gcd( n ′ , a ) = 2 i +1 n ′ gcd( n ′ , a + 2 i n ′ )= 2 i +1 n ′ gcd(2 i +1 n ′ , a + 2 i n ′ ) = Θ i +1 n ′ ( a + 2 i n ′ ) . | Cl q, i +1 n ′ ( a ) | = ord Θ i +1 n ′ ( a ) ( q ) = ord Θ i +1 n ′ ( a +2 i n ′ ) ( q ) = | Cl q, i +1 n ′ ( a + 2 i n ′ ) | as desired.Properties of q -cyclotomic cosets with q ≡ q ≡ q ≡ In this subsection, we focus on properties of q -cyclotomic cosets in the case where q ≡ i ( q ) for all odd prime powers q ≡ i . Lemma 4. Let q be an odd prime power and let β be the positive integer such that β || ( q − . Let i be a positive integer. If q ≡ , then ord i ( q ) = if i = 1 , if ≤ i ≤ β, i − β +1 if i ≥ β + 1 . Proof. Assume that q ≡ || ( q − 1) and 2 i | ( q − 1) for all 2 ≤ i ≤ β .Since q − q − q + q + 1) and q + q + 1 is odd, we have 2 || ( q − ( q ) = 1 and ord i ( q ) = 2 for all 2 ≤ i ≤ β .Assume that i ≥ β + 1. Since q ≡ q j ≡ j ≥ 1. Hence, 2 || ( q j + 1) for all j ≥ 1. Since ( q i − β − q i − β + 1) = q i − β +1 − q − i − β Q j =1 ( q j + 1), we have 2 i || ( q β − i +1 − 1) and 2 i ∤ ( q t − 1) for all t ≤ β + i . Hence, ord i ( q ) = 2 i − β +1 for all i ≥ β + 1.Properties of q -cyclotomic cosets modulo 2 i +1 n ′ containing odd integers areestablished in the next proposition. Proposition 1. Let q be a prime power such that q ≡ and let n ′ bean odd positive integer such that gcd( q, n ′ ) = 1 . Let λ ≥ be the integer suchthat λ || ord n ′ ( q ) and let β be the positive integer such that β || ( q − . Then thefollowing statements hold. i ) If λ = 0 , then the following statements hold. a ) Cl q, i +1 n ′ ( a ) = Cl q, i +1 n ′ ( a + 2 i n ′ ) all odd integers a and integers ≤ i ≤ β − . ) Cl q, i +1 n ′ ( a ) = Cl q, i +1 n ′ ( a + 2 i n ′ ) = Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) + 2 i n ′ ) for allodd integers a and integers i = 1 or i ≥ β . ii ) If λ > , then the following statements hold. a ) Cl q, λ + β − n ′ (1) = Cl q, λ + β − n ′ (1 + 2 λ + β − n ′ ) . b ) Cl q, i +1 n ′ ( a ) = Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) + 2 i n ′ ) for all odd integers a andintegers i ≥ λ + β − .Proof. First, we observe that β ≥ 3, 2 || ( q − 1) and 2 β − || ( q + 1).To prove i ), assume that λ = 0. In this case, ord n ′ ( q ) is odd which implies thatord Θ n ′ ( a ) ( q ) is odd for all odd positive integers a .To prove a ), let a be an odd integer and let i be an integer such that 2 ≤ i ≤ β − 1. By Lemma 4, it follows that ord i ( q ) = 2 = ord i +1 ( q ). Sinceord Θ n ′ ( a ) ( q ) is odd, it can be deduced that ord Θ i +1 n ′ ( a ) ( q ) = ord i +1 Θ n ′ ( a ) ( q ) =lcm(ord i +1 ( q ) , ord Θ n ′ ( a ) ( q )) = lcm(ord i ( q ) , ord Θ n ′ ( a ) ( q )) = ord Θ in ′ ( a ) ( q ). Supposethat Cl q, i +1 n ′ ( a ) = Cl q, i +1 n ′ ( a + 2 i n ′ ) . Since a a + 2 i n ′ (mod 2 i +1 n ′ ), there ex-ists 0 < j < ord Θ i +1 n ′ ( a ) ( q ) such that a + 2 i n ′ ≡ aq j (mod 2 i +1 n ′ ) . Hence, wehave a ≡ aq j (mod 2 i n ′ ) which implies that ord Θ in ′ ( a ) ( q ) ≤ j < ord Θ i +1 n ′ ( a ) ( q ) =ord Θ in ′ ( a ) ( q ), a contradiction. Therefore, Cl q, i +1 n ′ ( a ) = Cl q, i +1 n ′ ( a + 2 i n ′ ) as de-sired.To prove b ), let a be an odd integer and let i be an integer such that i = 1 or i ≥ β . By Lemma 4, we have ord i +1 ( q ) = 2ord i ( q ). Since ord n ′ ( q ) is odd, we haveord i +1 n ′ ( q ) = lcm(ord i +1 ( q ) , ord n ′ ( q )) = lcm(2ord i ( q ) , ord n ′ ( q )) = 2ord i n ′ ( q )which implies that aq ord in ′ ( q ) a (mod 2 i +1 n ′ ). Since aq ord in ′ ( q ) ≡ a (mod 2 i n ′ ),we have aq ord in ′ ( q ) ≡ a + 2 i n ′ (mod 2 i +1 n ′ ). Hence, a + 2 i n ′ ∈ Cl q, i +1 n ′ ( a ) whichimplies that Cl q, i +1 n ′ ( a ) = Cl q, i +1 n ′ ( a + 2 i n ′ ). This proves the first equality.For the second equality, let b ∈ Cl q, i +1 n ′ ( a ). Then b ≡ aq j (mod 2 i +1 n ′ ) forsome 0 ≤ j < ord Θ i +1 n ′ ( a ) ( q ). It follows that b ≡ aq j (mod 2 i n ′ ). If b < i n ′ ,then b ∈ Cl q, i n ′ ( a ). Otherwise, b − i n ′ ∈ Cl q, i n ′ ( a ) which implies that b ∈ Cl q, i n ′ ( a ) + 2 i n ′ . Hence, Cl q, i +1 n ′ ( a ) ⊆ Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) + 2 i n ′ ). Since Cl q, i n ′ ( a ) and Cl q, i n ′ ( a ) + 2 n ′ are disjoint sets of the same size ord Θ in ′ ( a ) ( q ), wehave | Cl q, i +1 n ′ ( a ) | = ord Θ i +1 n ′ ( a ) ( q ) = 2ord Θ in ′ ( a ) ( q ) = | Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) +2 i n ′ ) | . Therefore, Cl q, i +1 n ′ ( a ) = Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) + 2 i n ′ ) as desired.To prove ii ), assume that λ > 0. For a ), suppose that 1 ∈ Cl q, λ + β − n ′ (1 +2 λ + β − n ′ ). If λ = 1, then λ + β − β , we have ord λ + β − ( q ) = 2 = ord λ + β − ( q ) byLemma 4. Since 2 || ord n ′ ( q ), we have ord n ′ ( q )2 is odd and it follows that ord λ + β − n ′ ( q ) =lcm(ord λ + β − ( q ) , ord n ′ ( q )) = lcm(ord λ + β − ( q ) , ord Θ n ′ ( a ) ( q )) = ord Θ λ + β − n ′ ( a ) ( q ).Assume that λ ≥ 2. Since λ + β − ≥ β + 1, we have ord λ + β − ( q ) = 2 λ and6rd λ + β − ( q ) = 2 λ − by Lemma 4. Since 2 λ || ord n ′ ( q ), it follows thatord λ + β − n ′ ( q ) = lcm(ord λ + β − ( q ) , ord n ′ ( q ))= lcm(2 λ , ord n ′ ( q ))= lcm(2 λ − , ord n ′ ( q ))= lcm(ord λ + β − ( q ) , ord n ′ ( q ))= ord λ + β − n ′ ( q ) . Since 2 λ || ord n ′ ( q ), ord n ′ ( q )2 λ is odd. Hence, ord λ + β − n ′ ( q ) = lcm(ord λ + β − ( q ) , ord n ′ ( q ))= lcm(ord λ + β − ( q ) , ord n ′ ( q )) = ord λ + β − n ′ ( q ). Since 1+2 λ + β − n ′ λ + β − n ′ ),we have 1 + 2 λ + β − n ′ ≡ q j (mod 2 λ + β − n ′ ) for some 0 < j < ord Θ λ + β − n ′ (1) ( q ) =ord λ + β − n ′ ( q ). It follows that 1 ≡ q j (mod 2 λ + β − n ′ ) which implies that ord λ + β − n ′ ( q ) ≤ j < ord λ + β − n ′ ( q ) = ord λ + β − n ′ ( q ), a contradiction. Therefore, Cl q, λ + β − n ′ (1) = Cl q, λ + β − n ′ (1 + 2 λ + β − n ′ ) as desired. .To prove b ), let a be an odd integer and let i be an integer such that i ≥ λ + β − i ≥ β which implies that ord i +1 ( q ) = 2ord i ( q ) and ord i ( q ) = 2 i − β +1 ≥ λ by Lemma 4. Since 2 λ || ord n ′ ( q ), ord n ′ ( q )2 λ is odd andord i +1 n ′ ( q ) = lcm(ord i +1 ( q ) , ord n ′ ( q ))= lcm(2ord i ( q ) , ord n ′ ( q ))= lcm(2ord i ( q ) , ord n ′ ( q )2 λ )= 2lcm(ord i ( q ) , ord n ′ ( q )2 λ )= 2lcm(ord i ( q ) , ord n ′ ( q ))= 2ord i n ′ ( q )which implies that aq ord in ′ ( q ) a (mod 2 i +1 n ′ ). Since aq ord in ′ ( q ) ≡ a (mod 2 i n ′ ),we have aq ord in ′ ( q ) ≡ a + 2 i n ′ (mod 2 i +1 n ′ ). Hence, a + 2 i n ′ ∈ Cl q, i +1 n ′ ( a ) whichimplies that Cl q, i +1 n ′ ( a ) = Cl q, i +1 n ′ ( a + 2 n ′ ). The first equality holds.For the second equality, let b ∈ Cl q, i +1 n ′ ( a ). Then b ≡ aq j (mod 2 i +1 n ′ ) forsome 0 ≤ j < ord Θ i +1 n ′ ( a ) ( q ). It follows that b ≡ aq j (mod 2 i n ′ ). If b < i n ′ ,then b ∈ Cl q, i n ′ ( a ). Otherwise, b − i n ′ ∈ Cl q, i n ′ ( a ) which implies that b ∈ Cl q, i n ′ ( a ) + 2 i n ′ . Hence, Cl q, i +1 n ′ ( a ) ⊆ Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) + 2 i n ′ ). Since Cl q, i n ′ ( a ) and Cl q, i n ′ ( a ) + 2 i n ′ are disjoint sets of the same size ord Θ in ′ ( a ) ( q ), wehave | Cl q, i +1 n ′ ( a ) | = ord Θ i +1 n ′ ( a ) ( q ) = 2ord Θ in ′ ( a ) ( q ) = | Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) +2 i n ′ ) | . Therefore, Cl q, i +1 n ′ ( a ) = Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) + 2 i n ′ ) as desired. q ≡ Here, we investigate properties of q -cyclotomic cosets in the case where q ≡ i ( q ).7 emma 5. Let q be an odd prime power and let β be the positive integer such that β || ( q − . Let i be a positive integer. If q ≡ , then ord i ( q ) = if ≤ i ≤ β − , i − β +1 if i ≥ β. Proof. Assume that q ≡ β − || ( q − 1) which implies thatord i ( q ) = 1 for all 1 ≤ i ≤ β − 1. Next, assume that i ≥ β . Since q ≡ q j ≡ j ≥ 0. Hence, 2 || ( q j + 1) for all j ≥ q i − β − q i − β + 1) = q i − β +1 − q − i − β Q j =0 ( q j + 1), it can be concludedthat 2 i || ( q β − i +1 − 1) and 2 i ∤ ( q t − 1) for all t ≤ β + i . As desired, we haveord i ( q ) = 2 i − β +1 for all i ≥ β . Proposition 2. Let q be a prime power such that q ≡ and let n ′ be anodd positive integer such that gcd( q, n ′ ) = 1 . Let λ ≥ be the integer such that λ || ord n ′ ( q ) and let β be the positive integer such that β || ( q − . Then followingstatements hold. i ) If λ = 0 , then a ) Cl q, i +1 n ′ ( a ) = Cl q, i +1 n ′ ( a + 2 i n ′ ) for all odd integers a and integers ≤ i ≤ β − .b ) Cl q, i +1 n ′ ( a ) = Cl q, i +1 n ′ ( a + 2 i n ′ ) = Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) + 2 i n ′ ) for allodd integers a and integers i ≥ β − . ii ) If λ > , then(a) Cl q, λ + β − n ′ (1) = Cl q, iλ + β − n ′ (1 + 2 λ + β − n ′ ) .(b) Cl q, i +1 n ′ ( a ) = Cl q, i +1 n ′ ( a + 2 i n ′ ) = Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) + 2 i n ′ ) forodd integers a and integers i ≥ λ + β − .Proof. First, we observe that β ≥ 3, 2 || ( q + 1) and 2 β − || ( q − λ = 0, then ord Θ i +1 n ′ ( a ) ( q ) = ord Θ in ′ ( a ) ( q ) for all odd integers a and integers1 ≤ i ≤ β − 2, and ord i +1 n ′ ( q ) = 2ord i n ′ ( q ) for all integers i ≥ β − λ > 0, then ord λ + β − n ′ ( q ) = ord λ + β − n ′ ( q ), and ord i +1 n ′ ( q ) = 2ord i n ′ ( q ) forall integers i ≥ λ + β − Factorization of x n + 1 over Finite Fields In this section, the factorization of x i n ′ + 1 over F q is established. First, we provethat there exists a positive integer k such that the number of monic irreduciblefactors of x i n ′ + 1 over F q becomes a constant for all integers i ≥ k . In the casewhere ord n ′ ( q ) is odd, a complete recursive factorization of x i n ′ + 1 over F q is giventogether with a recursive formula for the number of its monic irreducible factorsfor all positive integers i in Subsection 4.1. In the case where ord n ′ ( q ) is even, arecursive factorization of x i n ′ + 1 over F q is given as well as a recursive formulafor the number of its monic irreducible factors for all integers i ≥ k in Subsection4.2. x n + 1 over F q with Odd ord n ′ ( q ) In this subsection, we established a complete recursive factorization of x i n ′ + 1over F q in the case where ord n ′ ( q ) is odd. Subsequently, a formula for the numberof monic irreducible factors of x i n ′ + 1 over F q is given recursively on i . q ≡ q -cyclotomic cosets and their induced poly-nomials for the case q ≡ Lemma 6. Let q be a prime power such that q ≡ and let n ′ be an oddpositive integer such that gcd( q, n ′ ) = 1 and ord n ′ ( q ) is odd. Let β be the positiveinteger such that β || ( q − . Let i be a positive integer and let a be an odd integer.Then one of the following statements holds. i ) Cl q, i +1 n ′ ( a ) and Cl q, i +1 n ′ ( a + 2 i n ′ ) induce distinct monic irreducible polyno-mials of degree | Cl q, i n ′ ( a ) | for all ≤ i ≤ β − .ii ) For each i = 1 or i ≥ β , if f ( x ) is induced by Cl q, i n ′ ( a ) , then Cl q, i +1 n ′ ( a ) induces f ( x ) .Proof. To prove i ), assume that 2 ≤ i ≤ β − 1. By Proposition 1 i.a ), we have Cl q, i +1 n ′ ( a ) = Cl q, i +1 n ′ ( a + 2 i n ′ ). From Lemma 3, it follows that | Cl q, i +1 n ′ ( a ) | = | Cl q, i +1 n ′ ( a + 2 i n ′ ) | which equals to | Cl q, i n ′ ( a ) | by the proof of Proposition 1 i.a ). Hence, Cl q, i +1 n ′ ( a ) and Cl q, i +1 n ′ ( a + 2 i n ′ ) induce distinct monic irreduciblepolynomials of degree | Cl q, i n ′ ( a ) | .To proof ii ), assume that i = 1 or i ≥ β . Assume that f ( x ) is induced by Cl q, i n ′ ( a ). Let α be a 2 i +1 n ′ th root of unity. Then α is a 2 i n ′ th root of unity9nd f ( x ) = Q j ∈ Cl q, in ′ ( a ) ( x − ( α ) j ). From Proposition 1 i.b ), we have Cl q, i +1 n ′ ( a ) = Cl q, i n ′ ( a ) ∪ ( Cl q, i n ′ ( a ) + 2 i n ′ ). It follows that Y j ∈ Cl q, i +1 n ′ ( a ) ( x − α j ) = Y j ∈ Cl q, in ′ ( a ) ∪{ Cl q, in ′ ( a )+2 i n ′ } ( x − α j )= Y j ∈ Cl q, in ′ ( a ) ( x − α j ) × Y j ∈{ Cl q, in ′ ( a )+2 i n ′ } ( x − α j )= Y j ∈ Cl q, in ′ ( a ) ( x − α j )( x − α j +2 i n ′ )= Y j ∈ Cl q, in ′ ( a ) ( x − α j )( x + α j )= Y j ∈ Cl q, in ′ ( a ) ( x − α j )= f ( x ) . Therefore, Cl q, i +1 n ′ ( a ) induces f ( x ) as desired.The next corollary can be deduced directly from the above lemma. Corollary 1. Assume the notations as in Lemma 6 with i ≥ β . If f ( x ) is inducedby Cl q, i n ′ ( a ) , then f ( x j ) is irreducible for all j ≥ β − i . In order to simplify the notations in the next theorem, let α and γ be 2 i n ′ thand 2 i +1 n ′ th roots of unity, respectively. For each a ∈ SO q (2 i n ′ ), let f a ( x ) = Y j ∈ Cl q, in ′ ( a ) ( x − α j ) and g j ( x ) = Y j ∈ Cl q, i +1 n ′ ( a ) ( x − γ j ) (4.1)be the irreducible polynomials induced by Cl q, i n ′ ( a ) and Cl q, i +1 n ′ ( a ), respectively.Using these notations, a recursive factorization of x i n ′ + 1 is given as follows. Theorem 1. Let q be a prime power such that q ≡ and let n ′ be an oddpositive integer such that gcd( q, n ′ ) = 1 and ord n ′ ( q ) is odd. Let β be the positiveinteger such that β || ( q − . Then the following statements hold. i ) If i = 0 , then x i n ′ + 1 = x n ′ + 1 = Y a ∈ SO q (2 n ′ ) f a ( x ) (4.2) ii ) If i ≥ , then x i n ′ + 1 = Q a ∈ SO q (2 i n ′ ) f a ( x ) if i = 1 or i ≥ β, Q a ∈ SO q (2 i n ′ ) g a ( x ) g a +2 i n ′ ( x ) if ≤ i ≤ β − , (4.3) where f a ( x ) and g a ( x ) are given in (4.1) . n this case, we have x β − i n ′ + 1 = Y a ∈ SO q (2 β n ′ ) f a ( x i ) for all i ≥ .Proof. From (2.3), we note that x i n ′ + 1 = Y a ∈ SO q (2 i +1 n ′ ) f a ( x ) . (4.4)The first statement is the special case where i = 0. From Proposition 1 i ), it canbe deduced that SO q (2 i +1 n ′ ) = SO q (2 i n ′ ) if i = 1 or i ≥ β,SO q (2 i n ′ ) ∪ ( SO q (2 i n ′ ) + 2 i n ′ ) if 2 ≤ i ≤ β − , where the union is disjoint. The results therefore follow from Lemma 6.A recursive formula for the number of monic irreducible factors of x i n ′ + 1 over F q follows immediately from the theorem. Corollary 2. Let q be a prime power such that q ≡ and let n ′ be anodd positive integer such that gcd( q, n ′ ) = 1 and ord n ′ ( q ) is odd. Let i ≥ be aninteger and let β be the positive integer such that β || ( q − . Then N q ( n ′ ) = X d | n ′ φ (2 d )ord d ( q ) (4.5) and N q (2 i n ′ ) = N q ( n ′ ) if i = 1 , N q (2 i − n ′ ) = 2 i − N q ( n ′ ) if ≤ i ≤ β − ,N q (2 β − n ′ ) = 2 β − N q ( n ′ ) if i ≥ β. (4.6) Proof. Equation (4.5) is a special case of (2.4). Equation (4.6) follows immediatelyfrom Theorem 1. q ≡ q ≡ q -cyclotomic coset Cl q, i +1 n ′ ( a ) and its induced polynomial are established.11 emma 7. Let q be a prime power such that q ≡ and let n ′ be an oddpositive integer such that gcd( q, n ′ ) = 1 and ord n ′ ( q ) is odd. Let β be the positiveinteger such that β || ( q − . Let i be a positive integer and let a be an odd integer.Then one of the following statements holds. i ) Cl q, i +1 n ′ ( a ) and Cl q, i +1 n ′ ( a + 2 i n ′ ) induce distinct monic irreducible polyno-mials of the same degree for all ≤ i ≤ β − .ii ) For each i ≥ β − , if f ( x ) is induced by Cl q, i n ′ ( a ) , then Cl q, i +1 n ′ ( a ) induce f ( x ) .Proof. The proof can be obtained using the arguments similar to those in the proofof Lemma 6 while Proposition 2 i ) is applied instead of Proposition 1 i ). Corollary 3. Assume the notations as in Lemma 7 with i ≥ β − . If f ( x ) isinduced by Cl q, i n ′ ( a ) , then f ( x j ) is irreducible for all j ≥ β − i − . The factorization of x i n ′ + 1 is given in the next theorem. Theorem 2. Let q be a prime power such that q ≡ and let n ′ be an oddpositive integer such that gcd( q, n ′ ) = 1 and ord n ′ ( q ) is odd. Let β be the positiveinteger such that β || ( q − . Then the following statements hold. i ) If i = 0 , then x i n ′ + 1 = x n ′ + 1 = Y a ∈ SO q (2 n ′ ) f a ( x ) (4.7) ii ) If i ≥ , then x i n ′ + 1 = Q a ∈ SO q (2 i n ′ ) g a ( x ) g a +2 i n ′ ( x ) if ≤ i ≤ β − , Q a ∈ SO q (2 i n ′ ) f a ( x ) if i ≥ β − , (4.8) where f a ( x ) and g a ( x ) are given in (4.1) .In this case, we have x β − i n ′ + 1 = Y a ∈ SO q (2 β − n ′ ) f a ( x i ) for all i ≥ .Proof. The proof can be obtained using the arguments similar to those in theproof of Theorem 1 while Proposition 2 i ) and Lemma 7 are applied instead ofProposition 1 i ) and Lemma 6 . 12rom the theorem, the enumeration of monic irreducible factor of x i n ′ − F q can be concluded in the next corollary. Corollary 4. Let q be a prime power such that q ≡ and let n ′ be anodd positive integer such that gcd( q, n ′ ) = 1 and ord n ′ ( q ) is odd. Let i ≥ be aninteger and let β be the positive integer such that β || ( q − . Then N q ( n ′ ) = X d | n ′ φ (2 d )ord d ( q ) (4.9) and N q (2 i n ′ ) = N q (2 i − n ′ ) = 2 i N q ( n ′ ) if ≤ i ≤ β − ,N q (2 β − n ′ ) = 2 β − N q ( n ′ ) if i ≥ β − . (4.10) Proof. Equation (4.9) is given in (2.4). Equation 4.10 follow immediately fromTheorem 2. x n + 1 over F q with Even ord n ′ ( q ) In this subsection, we focus on the case where ord n ′ ( q ) is even, i.e., 2 λ || ord n ′ ( q )for some positive integer λ . The results are not strong as the previous subsection.Precisely, a recursive factorization of x i n ′ +1 over F q is given only for all sufficientlylarge positive integers i .In general, the factorization of x i n ′ +1 over F q is given in (2.3). For i ≥ λ + β − Theorem 3. Let q be an odd prime power and let n ′ be an odd positive integersuch that gcd( q, n ′ ) = 1 . Let λ be the positive integer such that λ || ord n ′ ( q ) and let β be the positive integer such that β || ( q − . Then x λ + β − j n ′ + 1 = Y a ∈ SO q (2 λ + β n ′ ) f a ( x j ) for all j ≥ .Proof. The proof can be obtained using the arguments similar to those in the proofof Theorem 1 while Proposition 2 ii ) and Proposition 1 ii ) are applied instead ofProposition 2 i ) and Proposition 1 i ).The next corollary follows immediately. Corollary 5. Let q be an odd prime power and let n ′ be an odd positive integersuch that gcd( q, n ′ ) = 1 . Let λ be the positive integer such that λ || ord n ′ ( q ) and let β be the positive integer such that β || ( q − . Then N q (2 i n ′ ) = N q (2 λ + β − n ′ ) for all i ≥ λ + β − . .3 Algorithm and Examples In this subsection, the above results are summarized as an algorithm for factorizing x i n ′ + 1 over F q . Some illustrative examples are given as well.An algorithm for the factorization of x i n ′ + 1 over F q is given in Algorithm 1.Algorithm 1: Algorithm for the Factorization of x i n ′ + 1 over F q Input: odd prime power q , odd integer n ′ with gcd( q, n ′ ) = 1 , and integer i ≥ .1) Compute the positive integer β such that β || ( q − .2) Compute ord n ′ ( q ) and the integer λ such that λ || ord n ′ ( q ) .3) Consider the following cases:I) λ = 0 .i) q ≡ .a) i = 0 . Compute x i n ′ + 1 = x n ′ + 1 = Q a ∈ SO q (2 n ′ ) f a ( x ) . b) ≤ i ≤ β − . Compute x i n ′ + 1 = Y a ∈ SO q (2 i n ′ ) g a ( x ) g a +2 i n ′ ( x ) and SO q (2 i +1 n ′ ) = SO q (2 i n ′ ) ∪ ( SO q (2 i n ′ ) + 2 i n ′ ) .c) i ≥ β − . Compute x i n ′ + 1 = Y a ∈ SO q (2 β − n ′ ) f a ( x i − β +2 ) . ii) q ≡ .a) ≤ i ≤ . Compute x i n ′ + 1 = Q a ∈ SO q (2 n ′ ) f a ( x i ) . b) ≤ i ≤ β − . Compute x i n ′ + 1 = Y a ∈ SO q (2 i n ′ ) g a ( x ) g a +2 i n ′ ( x ) and SO q (2 i +1 n ′ ) = SO q (2 i n ′ ) ∪ ( SO q (2 i n ′ ) + 2 i n ′ ) .c) i ≥ β . Compute x i n ′ + 1 = Y a ∈ SO q (2 β n ′ ) f a ( x i − β +1 ) . II) λ ≥ .i) ≤ i ≤ λ + β − . Compute x i n ′ + 1 directly using (2.3) ii) i ≥ λ + β − . Compute x i n ′ + 1 = Y a ∈ SO q (2 λ + β n ′ ) f a ( x i − λ − β +1 ) . Note that where f a ( x ) and g a ( x ) are given in (4.1) . For the enumeration of monic irreducible factors of x i n ′ + 1 over F q , it can be14alculated using (2.4). With more information on n ′ , i , and q , the formula can besimplified using Corollaries 2, 4, and 5 of the form N q (2 i n ′ ) = i N q ( n ′ ) if λ = 0 , ≤ i ≤ β − q ≡ , β − N q ( n ′ ) if λ = 0 , i ≥ β − q ≡ N q ( n ′ ) if λ = 0 , i = 1 and q ≡ , i − N q ( n ′ ) if λ = 0 , ≤ i ≤ β − q ≡ , β − N q ( n ′ ) if λ = 0 , i ≥ β and q ≡ ,N q (2 λ + β − n ′ ) if λ ≥ i ≥ λ + β − , (4.11)where λ is the positive integer such that 2 λ || ord n ′ ( q ), β is the positive integer suchthat 2 β || ( q − N q ( n ′ ) = X d | n ′ φ (2 d )ord d ( q ) . From (4.11), the number N q (2 i n ′ ) of monic irreducible factors of x i n ′ +1 over F q becomes a constant independent of i for all i ≥ λ + β − λ = 0 and q ≡ i ≥ λ + β − N q (2 i n ′ )of monic irreducible factors of x i n ′ + 1 over F q with odd ord n ′ ( q ) and even ord n ′ ( q )are given in Table 1 and Table 2, respectively.In Table 1, the results for the q ∈ { , } and q ∈ { , } are obtained fromCorollary 2 and Corollary 4, respectively.In Table 2, the last row of each n ′ is obtained from Corollary 5. Otherwise, itis computed using (2.4). In this section, the factorization of x i n ′ + 1 over F q obtained in Section 4 areapplied in the study of negacyclic codes. Some known results are revisited insimpler forms.A linear code of length n over F q is defined to be a subspace of the the F q -vectorspace F nq . The dual of a linear code C of length n over F q is defined to be C ⊥ = { ( v , v , . . . , v n − ) ∈ F nq | n − X i =0 c i v i = 0 for all ( c , c , . . . , c n − ) ∈ C } . A linear code C is said to be self-dual if C = C ⊥ and it is said to be complementarydual if C ∩ C ⊥ = { } . 15able 1: N q (2 i n ′ ) of monic irreducible factors of x i n ′ + 1 over F q with odd ord n ′ ( q ) q n ′ ord n ′ ( q ) λ β i N q (2 i n ′ )3 1 1 0 3 0 11 1 ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ A linear code C of length n over F q is said to be negacyclic if it is closed underthe negacyclic shift. Precisely, ( − c n − , c , c , . . . , c n − ) ∈ C for every ( c , c , . . . , c n − , c n − ) ∈ C . Under the map π : F nq → F q [ x ] / h x n + 1 i defined by( c , c , . . . , c n − , c n − ) c + c x + c x + · · · + c n − x n − , it is well known (see [9]) that a linear code C of length n over F q is negacyclic ifand only if π ( C ) is an ideal in the principal ideal ring F q [ x ] / h x n + 1 i . The map π induces a one-to-one correspondence between negacyclic codes of length n over F q and ideas in F q [ x ] / h x n + 1 i . In this case, π ( C ) is uniquely generated by themonic divisor of x n + 1 of minimal degree in π ( C ). The such polynomial is call the generator polynomial of C .Let q be an odd prime power and let n ′ be an odd positive integer such that16able 2: N q (2 i n ′ ) of monic irreducible factors of x i n ′ + 1 over F q with even ord n ′ ( q ) q n ′ ord n ′ ( q ) λ β i N q (2 i n ′ )3 5 4 2 3 0 21 32 6 ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ gcd( q, n ′ ) = 1. Let λ be the positive integer such that 2 λ || ord n ′ ( q ) and let β be thepositive integer such that 2 β || ( q − k = λ + β − λ = 0 and q ≡ ,λ + β − . In general, negacyclic codes have been studied in [4], [5], and [9]. Here, we focuson negacyclic codes of length n = p s i n ′ with i ≥ k , where p is the characteristic17f F q . The construction and enumeration of such negacyclic codes are simplifiedusing the results from Section 4.From (2.3), we have x k n ′ + 1 = N q (2 k n ′ ) Y j =1 r j ( x ) . (5.1)Based on Theorem 1, Theorem 2, and Theorem 3, it follows that x p s i n ′ + 1 = ( x i n ′ + 1) p s = N q (2 k n ′ ) Y j =1 ( r j ( x i − k )) p s (5.2)and r j ( x i − k ) is irreducible for all i ≥ k .The following characterization and enumeration of negacyclic codes of length n = p s i n ′ with i ≥ k are straightforward. The proof is committed. Theorem 4. Assume the notations above. The the following statements hold.1. The map T : F q [ x ] / h x p s k n ′ + 1 i → F q [ x ] / h x p s i n ′ + 1 i defined by f ( x ) f ( x i − k ) is a ring isomorphism for all integers i ≥ k .2. For each integer i ≥ k , g ( x ) is the generator polynomial of a negacyclic codeof length p s k n ′ over F q if and only if g ( x i − k ) is the generator polynomial ofa negacyclic code of length p s i n ′ over F q 3. The number of negacyclic codes of length p s i n ′ over F q is ( p s + 1) N q (2 k n ′ ) for all i ≥ k . From the theorem, all negacyclic codes of length p s i n ′ over F q with i ≥ k canbe determined using the negacyclic codes of length p s k n ′ over F q . References [1] G. K. Bakshi, M. Raka, Self-dual and self-orthogonal negacyclic codes oflength 2 p n over a finite field, Finite Fields and Their Applications , (2013)39–54.[2] T. Blackford, Negacyclic duadic codes, Finite Fields and Their Applications , (2008) 930–943. 183] I. F. Blake, S. Gao, R. C. Mullin, Explicit factorization of x k + 1 over F p withprime p ≡ , (1993) 89–94.[4] A. Boripan, S. Jitman, SRIM and SCRIM factors of x n + 1 over finite fieldsand their applications, preprint, https://arxiv.org/abs/1909.03826.[5] S. Jitman, S. Prugsapitak, M. Raka, Some generalizations of good integersand their applications in the study of self-dual negacyclic codes, Advances inMathematics of Communications , (2020) 35–51.[6] H. Meyn, Factorization of the cyclotomic polynomial x n + 1 over finite fields,Finite Fields and Their Applications , (1996) 439442.[7] Y. Jia, S. Ling, C. Xing, On self-dual cyclic codes over finite fields, IEEETransaction on Information Theory , (2011) 2243–2251.[8] S. Ling, C. Xing, Coding Theory : A First Course , Cambridge UniversityPress, 2004.[9] E. Sangwisut, S. Jitman, S. Ling, P. Udomkavanich, Hulls of cyclic and nega-cyclic codes over finite fields, Finite Fields and Their Applications33