Research Article Characterization and Enumeration of Good Punctured Polynomials over Finite Fields

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1 International Mathematics and Mathematical Sciences Volume 016, Article ID , 7 pages Research Article Characterization and Enumeration of Good Punctured Polynomials over Finite Fields Somphong Jitman, 1 Aunyarut Bunyawat, Supanut Meesawat, Arithat Thanakulitthirat, and Napat Thumwanit 1 Department of Mathematics, Faculty of Science, Silpakorn University, Nakhon Pathom 73000, Thailand Department of Mathematics, Mahidol Wittayanusorn School, Nakhon Pathom 73170, Thailand Correspondence should be addressed to Somphong Jitman; sjitman@gmail.com Received 5 November 015; Revised 3 March 016; Accepted 6 March 016 Academic Editor: Shyam L. Kalla Copyright 016 Somphong Jitman et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. A family of good punctured polynomials is introduced. The complete characterization and enumeration of such polynomials are given over the binary field F. Over a nonbinary finite field F q, the set of good punctured polynomials of degree less than or equal to are completely determined. For n 3, constructive lower bounds of the number of good punctured polynomials of degree n over F q are given. 1. Introduction From the fundamental theorem of algebra, every polynomial over the rational numbers Q (orovertherealnumbersr)has arootinc. However,itisnotguaranteedthatapolynomial has a root in Q or in R. Therefore, for a given polynomial over Q (resp., R), it is of natural interest to determine whether it has a root in Q (resp., R). In general, determining whether a given polynomial has a root in a nonalgebraically closed field is an interesting problem and has been extensively studied (see, e.g., [1 4]). In this paper, we introduce punctured forms of a polynomial f(x) over a field F (see the definition below) and focus on determining whether the punctured parts of f(x) have a root in F. Due to the rich algebraic structures and various applications of polynomials over finite fields (see [5 9] and references therein), their properties such as factorization, root finding, and irreducibility have extensively been studied (see [10 13]). In this paper, we mainly focus on punctured polynomials over a finite field which is not an algebraically closed field. The readers may refer to [5] for more details on finitefieldsandpolynomialsoverfinitefields. Let q be a prime power and let F q denotethefinitefieldof q elements. Denote by F q the set of nonzero elements in F q. Let F q [x] m = a i x i a i F q,1,,...,m, m N 0}} be the set of all polynomials with indeterminate x over F q.let F q [x] n =f(x) F q [x] deg (f (x)) n}, F q [x] n =f(x) F q [x] deg (f (x)) =n} be the set of all polynomials of degree less than or equal to n over F q andthesetofallpolynomialsofdegreen over F q, respectively. (1) ()

2 International Mathematics and Mathematical Sciences Given a polynomial f(x) = m a ix i F q [x] of degree m,foreachj 0,1,,...,m},thejth punctured polynomial of f(x) isdefinedtobe f (j) (x) = j 1 a i x i + m i=j+1 a i x i 1. (3) For convenience, by abuse of notation, the degree of zero polynomial is defined to be 0.Hence,wecanwritef (0) (x) = 0 for all constant polynomials f(x) = a F q. Apolynomialf(x) F q [x] of degree m is said to be good punctured if f (j) (x) has a root in F q for all j 0,1,,...,m}. Otherwise, f(x) is said to be bad punctured. Theconstant polynomials f(x) = a F q are always good punctured and referred to as trivial good punctured polynomials. A good punctured polynomial is called nontrivial if it is not a trivial good punctured polynomial. Example 1. Let f(x) = x +x+be a polynomial in F 3 [x]. Then f () (x)=x+, f (1) (x) = x +,andf (0) (x) = x +.It is not difficult to see that f(x) is good punctured. Given a positive integer n and a prime power q, lett (q,n) and T (q,n) denote the set of good punctured polynomials of degree less than or equal to n over F q and the set of all good punctured polynomials of degree n over F q,respectively. Precisely, T (q,n) =f(x) F q [x] n f (j) (x) has a root in F q 0 j deg (f (x))}, T (q,n) =f(x) F q [x] n f (j) (x) has a root in F q 0 j n}. By convention, since deg(0) = 0,wehaveT (q,0) = T (q,0) = F q and T (q,0) = T (q,0) =q. Remark. From the definitions of T (q,n) and T (q,n),wehave the following facts: (1) T (q,n) is a subset of T (q,n). (4) () T (q,n) =T (q,n 1) T (q,n) is a disjoint union for all n. (3) T (q,n) = T (q,n 1) + T (q,n) for all n. Example 3. Over the finite fields F,wehave T (,) = 0, 1, x,x +x,x +x+1}, T (,) =x,x +x,x +x+1}. Hence, T (,) =5and T (,) =3,respectively. In this paper, we focus on the characterization and enumeration of the good punctured polynomials of degree n over F q. The complete characterization and enumeration of (5) good punctured polynomials over the binary field F are given in Section. In Section 3, good punctured polynomials of degree n over F q,whereq >, are studied. The good punctured polynomials of degree less than or equal to over fields F q are completely determined. Lower bounds ofthesizeofthesetofgoodpuncturedpolynomialsof degree greater than areprovidedaswell.conclusionand some discussions about future researches on punctured polynomials are provided in Section 4.. Good Punctured Polynomials over the Binary Field In this section, we focus on good punctured polynomials over the finite field F. The characterization and enumeration of such polynomials are completely determined. First, we determine the set T (,n) of good punctured polynomials of degree less than or equal to n over the binary field F.ItisnotdifficulttoseethatT (,1) = F.Forn,theset T (,n) is given as follows. Theorem 4. Let n be a positive integer. Then T (,n) =A B C, (6) where A= k xi 0 k n/}, B=xf(x) f(x) F [x] n 1 and f(1) = 0},andC=x f(x) f(x) F [x] n }. Proof. First, we prove that A B C T (,n).letg(x) A B C. We distinguish the proof into three cases. Case 1 (g(x) A). Then g(x) = k xi for some 0 k n/. It follows that 1 is a root of g (j) (x) = k 1 x i for all 0 j k.hence,g(x) T (,n). Case (g(x) B). Then g(x) = xf(x) for some f(x) F [x] n 1.Wehaveg (0) (1) = f(1) = 0 and g (i) (0) = g(0) = 0 for all 0<i deg(g(x)).hence,g (i) (x) has a root in F for all 0 i deg(g(x)). Therefore, g(x) T (,n). Case 3 (g(x) C). Then g(x) = x h(x) for some h(x) F [x] n.itfollowsthat0 is a root of g (0) (x) = g (1) (x) = xh(x) and g (i) (x) = x h (i ) (x) for all i deg(g(x)). Therefore, f (i) (x) has a root in F for all 0 i deg(g(x)). As desired, g(x) T (,n). On the other hand, let g(x) T (,n).writeg(x) = deg(g(x)) g i x i and consider the following two cases. Case 1 (g 0 =0) Case 1.1 (g 1 =0). Then g(x) C. Case 1. (g 1 = 1). Then deg(g(x)) 1. Sinceg (0) (x) = deg(g(x)) 1 g i+1 x i,wehaveg (0) (0) = g 1 =1.Itfollowsthat 0=g (0) (1) = deg(g(x)) 1 g i+1 = g(1).hence,g(x) B.

3 International Mathematics and Mathematical Sciences 3 Case (g 0 =1). Since g (0) (0) = g 0 =1,wehave0=g (0) (1) = deg(g(x)) i=1 g i. Suppose that there exists 1 j<deg(g(x)) such that g j =0.Sinceg (j) (0) = g 0 =1and g (deg(g(x))) (0) = g 0 =1, we have 0=g (j) (1) = j 1 g i + 0=g (deg(g(x))) (1) = deg(g(x)) i=j+1 g i, deg(g(x)) 1 It follows that 0=g j =g deg(g(x)) =1, a contradiction. Hence, g i = 1 for all 0 i deg(g(x)). Since0 = g (0) (1) = deg(g(x)) i=1 g i,thedegreeofg(x) mustbeeven.weconclude that g(x) A. From the two cases, we have g(x) A B C,and,hence, T (,n) A B C. Therefore, T (,n) =A B C Corollary 5. If n is a positive integer, then g i. (7) T (,n) = if n = 1, 3 n + n +1 ifn. (8) Proof. By direct calculation, we have T (,1) = 0, 1} and T (,1) =. Next, assume that n.bytheorem4,wehave T (,n) = A B C, (9) where A= k xi 0 k n/}, B=xf(x) f(x) F [x] n 1 and f(1) = 0},andC=x f(x) f(x) F [x] n }. Since A and B Care disjoint, by the inclusion-exclusion principle, we have T (,n) = A B C = A + B + C B C. (10) Clearly, A = n/ + 1 and C = n 1.Observethatxf(x) = x n 1 f ix i Bif and only if 0 i n 1 f i =1} is even. Hence, It is not difficult to see that B = n 1. (11) B C=x f (x) f(x) F [x] n, f(1) =0}, (1) and, hence, Therefore, B C = n. (13) T (,n) = n +1+n 1 + n 1 n =3 n + n +1 (14) Next, we determine the set T (,n) of good punctured polynomials of degree n over the binary field F.SinceT (,1) = F, we have T (,1) =0.Forn,theset T (,n) can be determined as follows. Theorem 6. If n is a positive integer, then T (,n) = A B C B C if n is even, if n is odd, (15) where A = n xi }, B = x(f(x) + x n 1 ) f(x) F [x] n and f(1) = 1},and C =x (f(x) + x n ) f(x) F [x] n 3 }. Proof. We prove the statement by determining the elements in T (,n) of degree n. LetA= k xi 0 k n/}, B= xf(x) f(x) F [x] n 1 and f(1) = 0}, andc=x f(x) f(x) F [x] n } bedefinedasintheorem4. ItisnotdifficulttoseethatthesetofelementsinB (resp., C) of degree n is B (resp., C). If n is even, then the set of elements in A of degree n is A. In the case where n is odd, the set of elements in A of degree n is empty. By Theorem 4, the result, therefore, follows. Corollary 7. If n is a positive integer, then 0 if n = 1, T (,n) = 3 if n =, 3 n 3 if n 3 is odd, 3 n 3 +1 ifn 4iseVen. Proof. By direct calculation, we have T (,1) =0and (16) T (,) =x,x +x,x +x+1}. (17) Hence, we have T (,1) =0and T (,) =3. Next, assume that n 3.ByTheorem6,wehave T (,n) = A B C B C if n is even, if n is odd, (18) where A, B, and C aredefinedasintheorem6.since A and B C are disjoint, by the inclusion-exclusion principle, we have T (,n) = A B C = A + B + C B C. (19) We note that A = 1 and C = n. Since x(f(x)+x n 1 )=x( n f ix i +x n 1 ) B if and only if is odd, we have i 0 i n, f i =1} (0) B =n. (1)

4 4 International Mathematics and Mathematical Sciences Table 1: Punctured polynomials over F. n T (,n) T (,n) It is not difficult to see that Theorem 10. Let q>be a prime power. Then T (q,) =a x +a 1 x+a 0 (a,a 1,a 0 ) F q F q F q} a x a F q } F q. (5) B C =x (f (x) +x n ) f(x) F [x] n 3, f(1) =1}, and, hence, () B C =n 3. (3) Therefore, by (18), we have T (,n) = 1+ n + n n 3 if n is even, n + n n 3 if n is odd = 3 n n 3 if n is even, if n is odd (4) Table 1 presents the numbers T (,n) and T (,n) for n= 1,,...,13.Therelation T (,n) = T (,n 1) + T (,n) in Remark can be easily seen. 3. Punctured Polynomials over Nonbinary Finite Fields In this section, we focus on punctured polynomials over nonbinary finite fields. Given a prime power q>, the characterization and enumeration of good punctured polynomials of degree less than or equal to over F q are completely determined. For n 3,weconstructsubsetsofT (q,n) and T (q,n) which lead to lower bounds of the cardinalities of T (q,n) and T (q,n),respectively. Theorem 8. If q>is a prime power, then T (q,1) = F q. Proof. By the definition, F q T (q,1).letf(x) = ax+b T (q,1). Since f (0) (x) = a has a root in F q,wehavea = 0.Hence, f(x) = b F q The next corollary follows immediately from Theorem 8. Proof. Let A=a x +a 1 x+a 0 (a,a 1,a 0 ) F q F q F q} and B=a x a F q }. Let f(x) A B F q.wewritef(x) = a x +a 1 x+a 0 and consider the proof as two cases. Case 1 (f(x) A). We have f (0) (x) = a x+a 1, f (1) (x) = a x+a 0,andf () (x) = a 1 x+a 0.Sincea and a 1 are nonzero, it follows that a 1 a 1, a 0a 1,and a 0a 1 1 are roots of f (0) (x), f (1) (x),andf () (x),respectively.hence,f(x) T (q,). Case (f(x) B). Then f(x) = a x for some a F q.it follows that 0 is a root of f (0) (x) = a x, f (1) (x) = a x,and f () (x) = 0. Therefore, f(x) T (q,). Case 3 (f(x) F q ). Then, by the definition, f(x) T (q,). On the other hand, let f(x) = a x +a 1 x+a 0 T (q,).if a =0,thenf(x) = a 1 x+a 0 T (q,1),and,hence,f(x) = a 0 F q by Theorem 8. Assume that a =0.Thenf (0) (x) = a x+a 1, f (1) (x) = a x+a 0,andf () (x) = a 1 x+a 0 have a root in F q. Case 1 (a 1 =0). We have that f () (x) = a 0 has a root in F q which implies that a 0 =0. Therefore, f(x) = a x B. Case (a 1 =0). Since f (0) (x) = a x+a 1 has a root in F q,we have a =0.Hence,f(x) A. From the two cases, it can be concluded that f(x) A B F q. As desired, we have T (q,) =A B F q. Corollary 11. If q>is a prime power, then T (q,) =q(q q+3) 1. (6) Proof. Let A=a x +a 1 x+a 0 (a,a 1,a 0 ) F q F q F q } and B = a x a F q } be defined as in the proof of Theorem 10. It is not difficult to see that A, B, andf q are disjoint. By Theorem 10, we have Corollary 9. If q>is a prime power, then the following statements hold: (i) T (q,1) =q. (ii) T (q,1) =0. (iii) T (q,1) =0. T (q,) = A B F q = A + B + F q =(q 1)(q 1)q+(q 1)+q =q(q q+3) 1 (7)

5 International Mathematics and Mathematical Sciences 5 Corollary 1. If q is a prime power, then T (q,) =a x +a 1 x+a 0 (a,a 1,a 0 ) F q F q F q} (8) First, we show that S T (q,3).letg(x) = x(a x +a 1 x+ a 0 ) S,wherea F q and a 1,a 0 F q.theng (0) (x) = a x + a 1 x+a 0 has a root in F q and 0 is a root of g (1) (x) =a x +a 1 x, a x a F q }. g () (x) =a x +a 0 x, (34) Proof. From Theorem 10, it is not difficult to see that the polynomials of degree less than in T (q,) are f(x) = a F q. Hence, the result follows. Corollary 13. If q is a prime power, then T (q,) = (q 1) (q q+1). (9) Proof. From Corollaries 11 and 1, it follows that T (q,) = T (q,) q=q(q q+3) 1 q =(q 1)(q q+1). (30) In the case where n 3, determining the sets T (q,n) and T (q,n) is more tedious and complicated. For these cases, we give constructive lower bounds of T (q,n) and T (q,n). The following results are important tools in constructing lower bounds of T (q,n) and T (q,n). Theorem 14 (see [14, Page 588]). Let n be a positive integer and let q be a prime power. Then the number of monic irreducible polynomials of degree n in F q [x] is g (3) (x) =a 1 x +a 0 x. Hence, g(x) T (q,3) is good punctured. By Theorem 14, the number of monic irreducible polynomials of degree over F q is L(q,)= 1 d μ (d) q /d = 1 (μ (1) q +μ() q 1 ) = 1 (q q). (35) Hence, the number of irreducible polynomials of degree over F q is 1 (q 1)(q q). (36) By Theorem 15, the number of polynomials of degree over F q having a root in F q is S =(q 1)q 1 (q 1) (q q) = 1 (37) (q 1) q (q + 1). Since S T (q,3),wehave where L(n,q)= 1 n d nμ (d) q n/d, (31) T (q,3) S = 1 (q 1) q (q + 1) (38) μ (n) 1 if n = 1, = 0 if n contains a repeated prime factor, ( 1) r if n is a product of r distinct primes is themöbius function. (3) Theorem 15 (see [3, Section 4., Theorem 1]). Let f(x) be a polynomial of degree or 3 in F q [x]. Thenf(x) is reducible if and only if f(x) has a root in F q [x]. Theorem 16. If q>is a prime power, then Corollary 17. If q>is a prime power, then T (q,3) 3q3 4q +5q. (39) Proof. By Corollary 11 and Theorem 16, we have T (q,) =q(q q+3) 1, T (q,3) 1 (q 1) q (q + 1). (40) Hence, by Remark, we have the relation T (q,3) = T (q,) + T (q,3) T (q,3) 1 (q 1) q (q + 1). (33) Proof. Let S fl xf(x) f(x) F q [x] and f(x) has a root in F q }. q(q q+3) 1+ 1 (q 1) q (q + 1) = 3q3 4q +5q. (41)

6 6 International Mathematics and Mathematical Sciences Theorem 18. If q>is a prime power, then From Theorem 18, we have T (q,4) 1 (q 1) q (q +1). (4) Proof. Let S fl xf(x) f(x) F q [x] 3 and f(x) has a root in F q }. First, we show that S T (q,4).letg(x) = x(a 3 x 3 +a x + a 1 x+a 0 ) S,wherea 3 F q and a,a 1,a 0 F q.theng (0) (x) = a 3 x 3 +a x +a 1 x+a 0 has a root in F q and 0 is a root of g (1) (x) =a 3 x 3 +a x +a 1 x, g () (x) =a 3 x 3 +a x +a 0 x, g (3) (x) =a 3 x 3 +a 1 x +a 0 x, g (4) (x) =a x 3 +a 1 x +a 0 x. (43) Therefore, g(x) T (q,4) is good punctured By Theorem 14, the number of monic irreducible polynomials of degree 3 over F q is L(q,3)= 1 3 d 3μ (d) q 3/d = 1 (μ (1) q3 +μ(3) q 1 ) = 1 (q3 q). (44) Hence, the number of irreducible polynomials of degree 3 over F q is T (q,4) 1 (q 1) q (q +1). (50) Hence, by Remark, we have the relation T (q,4) = T (q,3) + T (q,4) 3q3 4q +5q = q4 +q 3 3q +4q. + 1 (q 1) q (q +1) (51) Theorem 0. Let q>be a prime power and let n 5be an integer. Then T (q,n) qn 1 +q 3 q +q. (5) Proof. Let A=x f(x) f(x) F q [x] n }. First, we show that A T (q,) T (q,n). Clearly, T (q,) T (q,n).letg(x) A. Then0 is a root of g (i) (x) for all i = 0, 1,..., deg(g(x)), and,hence,g(x) T (q,n).notethata T (q,) = ax a F q }. By Corollary 11, we have Therefore, consider T (q,) =q(q q+). (53) 1 (q 1) (q3 q). (45) By Theorem 15, the number of polynomials of degree 3 over F q having a root in F q is T (q,n) A + T (q,) A T (q,) =q n 1 +q(q q+3) 1 (q 1) =q n 1 +q 3 q +q. (54) Since S T (q,4),wehave S =(q 1)q 3 1 (q 1) (q3 q) = 1 (q 1) q (q +1). (46) Corollary 1. Let q>be a prime power and let n 5be an integer. Then T (q,n) (q 1)q n. Proof. The set of elements in A of degree n in the proof of Theorem 0 is A = x f(x) f(x) F q [x] n }.By Theorem 0, we have T (q,4) S = 1 (q 1) q (q +1) (47) T (q,n) A (q 1)qn. (55) Corollary 19. If q>is a prime power, then T (q,4) q4 +q 3 3q +4q. (48) Proof. By Corollary 17, we have T (q,3) 3q3 4q +5q. (49) 4. Conclusion and Open Problems The concepts of punctured polynomials and good punctured polynomials are introduced. Over the finite field F,thecomplete characterization and enumeration of such polynomials are given. Over nonbinary finite fields, the good punctured polynomials of degree less than or equal to are completely determined. For n 3, constructive lower bounds of the number of good punctured polynomials of degree n are given.

7 International Mathematics and Mathematical Sciences 7 In general, the following related problems are also interesting: (1) Determine the sets T (q,n) and T (q,n),whereq>is a prime power and n 3is an integer. [13] A. Knopfmacher and J. Knopfmacher, Counting polynomials with a given number of zeros in a finite field, Linear and Multilinear Algebra,vol.6,no.4,pp.87 9,1990. [14] D. S. Dummit and R. M. Foote, Abstract Algebra,JohnWiley& Sons, New York, NY, USA, 3rd edition, 003. () Determine the exact values of T (q,n) and T (q,n), where q>is a prime power and n 3is an integer. (3) Improve lower bounds of T (q,n) and T (q,n),where q>is a prime power and n 3is an integer. (4) Characterize and enumerate the good punctured polynomials of degree n over the real numbers R or over the rational numbers Q. Competing Interests The authors declare that there are no competing interests regarding the publication of this paper. Acknowledgments This research is supported by the Thailand Research Fund under Research Grant TRG References [1] B. Kalantari, Polynomial Root-Finding and Polynomiography, World Scientific Publishing, River Edge, NJ, USA, 008. [] V. Y. Pan and A.-L. Zheng, New progress in real and complex polynomial root-finding, Computers & Mathematics with Applications, vol. 61, no. 5, pp , 011. [3] W.K.Nicholson,Introduction to Abstract Algebra, JohnWiley & Sons, New York, NY, USA, 01. [4] C. C. Pinter, ABookofAbstractAlgebra,Dover,Mineola,NY, USA, nd edition, 010. [5] R. Lidl and H. Niederreiter, Finite Fields, Cambridge University Press, Cambridge, UK, [6] E. Sangwisut, S. Jitman, S. Ling, and P. Udomkavanich, Hulls of cyclic and negacyclic codes over finite fields, Finite Fields and their Applications, vol. 33, pp. 3 57, 015. [7] A. Akbary and Q. Wang, On some permutation polynomials over finite fields, International Mathematics and Mathematical Sciences,vol.005,no.16,pp ,005. [8] R. A. Mollin and C. Small, On permutation polynomials over finite fields, International Mathematics and Mathematical Sciences,vol.10,no.3,pp ,1987. [9] A. Sahni and P. T. Sehgal, Hermitian self-orthogonal constacyclic codes over finite fields, Discrete Mathematics, vol. 014, Article ID , 7 pages, 014. [10] E. R. Berlekamp, Factoring polynomials over finite fields, The Bell System Technical Journal,vol.46,pp ,1967. [11] M. C. R. Butler, On the reducibility of polynomials over a finite field, The Quarterly Mathematics, vol. 5, pp , [1] J.-M. Couveignes and R. Lercier, Fast construction of irreducible polynomials over finite fields, Israel Mathematics,vol.194,no.1,pp ,013.

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