The Degree and Order of Polynomials in a Contracted Length Semigroup Ring R[S]

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1 International Journal of Algebra, Vol. 2, 2008, no. 18, The Degree and Order of Polynomials in a Contracted Length Semigroup Ring R[S] Ronnason Chinram and Kittima Pattamavilai Department of Mathematics, Faculty of Science Prince of Songkla University, Hat Yai Songkhla 90112, Thailand ronnason.c@psu.ac.th Abstract In this paper, we generalize some properties of the degree and the order of polynomials in R[x]. Mathematics Subject Classification: 12Y05, 16R50 Keywords: Polynomials, degree, order 1 Introduction and Preliminaries Let R be a ring. f is said to be a polynomial in x with coefficients in R if f is the form of sum f = a n x n where a n R for all n N {0} n=0 such that a n = 0 for all but a finite number of indices n. Let R[x] be the set of all polynomials in x with coefficients in R. We have that R[x] is a ring under usual addition and multiplication of polynomials. A ring R[x] is called the ring of polynomials in x with coefficients in R or the polynomial ring (see [4]). Let f R[x] where f 0. The degree of f is max{n a n 0}. The degree of f is denoted by deg f. The order of f is min{n a n 0}. The order of f is denoted by ord f. The following theorem is well-known. Theorem 1.1 Let R be a ring and f,g R[x] {0}. The following statements are true.

2 882 R. Chinram and K. Pattamavilai (i) fg =0or deg (fg) deg f + deg g. (ii) f + g =0or deg (f + g) max{deg f, deg g}. (iii) If deg f deg g, then deg(f + g) =max{deg f, deg g}. (iv) fg =0or ord (fg) ord f + ord g. (v) f + g =0or ord (f + g) min{ord f, ord g}. (vi) If ord f ord g, then ord(f + g) =min{ord f, ord g}. Our aim of this paper is to generalize Theorem 1.1. Let R be a ring and S be a semigroup. f is said to be a polynomial on S with coefficients in R if f is the form of finite sum f = a s s where s S and a s R. Let R[S] be the set of all polynomials on S with coefficients in R. For any f = a s s and g = s s, define binary operations + and on R[S] by b f + g = a s s + b s s = (a s + b s )s and f g =( a s s)( b s s)= (a s b s )ss. s S Then R[S] is a ring under these two binary operations. A ring R[S] is called a contracted semigroup ring. In [1], Chinram has given some properties of the degree and the order of polynomials in R[F 1 A ] where F A is a free semigroup generated by a nonempty set A and F 1 A = F A {1} (can see the definition of free semigroups in [3]). A semigroup S is called a length semigroup if there exists a function L : S N {0} such that L(xy) =L(x) +L(y) for all x, y S. The following theorem holds. Theorem 1.2 Let S be a length semigroup. The following statements hold. (i) If S has a zero 0, then L(x) =0for all x S. (ii) If e is an idempotent of S, then L(e) =0. (iii) If S has an identity 1, then L(1) = 0. Proof. (i) We have that L(0) = L(0 x) =L(0) + L(x) for all x S. Thus L(x) = 0 for all x S. (ii) Let e be an idempotent of S. Then L(e) = L(e 2 ) = L(e) +L(e). Therefore L(e) = 0. (iii) It follows by (ii). In this paper, we give some properties of the degree and the order of polynomials in R[S] where S is a length semigroup. In the remainder of this paper, let R be a ring and S be a length semigroup.

3 The degree and order of polynomials The degree of polynomials in R[S] Let f R[S] {0}. The degree of f is max{l(s) a s 0}. The degree of f is denoted by deg f. The two following theorems hold. Theorem 2.1 Let f,g R[S] {0}. Then fg =0or deg fg deg f + deg g. Proof. Let f = a s s and g = s s be nonzero polynomials in R[S]. b Then fg = (a s b s )ss. Assume fg 0. Thus s S deg fg = max{l(ss ) a s b s 0} = max{l(s)+l(s ) a s b s 0} max{l(s)+l(s ) a s 0 and b s 0} max{l(s) a s 0} + max{l(s ) b s 0} = deg f + deg g. Therefore fg = 0 or deg fg deg f + deg g. Theorem 2.2 Let f,g R[S] {0}. Then f + g =0or deg (f + g) max{deg f, deg g}. Proof. Let f = a s s and g = s s be nonzero polynomials in R[S]. b Assume that deg f = n and deg g = m. Case 1 : m>n. We have that f + g = (a s + b s )s + b s s.,l(s) n,l(s)>n Then deg(f+g) = max{l(s) b s 0} = m = max{n, m} = max{deg f,deg g}. Case 2 : m<n. We have that f + g = (a s + b s )s + a s s.,l(s) m,l(s)>m So deg(f + g) = max{l(s) a s 0} = n = max{n, m} = max{deg f,deg g}. Case 3 : m = n. We have that f + g = (a s + b s )s. Case 3.1 : f = g. Then f + g =0. Case 3.2 : f g. Then deg(f + g) max{deg f,deg g}. Therefore f + g = 0 or deg(f + g) max{deg f,deg g}, as required. The following corollary follows by the proof of Case 1 and Case 2 of Theorem 2.2. Corollary 2.3 Let f,g R[S] {0}. Ifdegf deg g, then deg (f + g) = max{deg f, deg g}.

4 884 R. Chinram and K. Pattamavilai 3 The order of polynomials in R[S] For f R[S] {0}, let the order of f be min{l(s) a s 0}. The order of f is denoted by ord f. The two following theorems are true. Theorem 3.1 Let f,g R[S] {0}. Then fg =0or ord fg ord f + ord g. Proof. Let f = a s s and g = s s be nonzero polynomials in R[S]. b Then fg = (a s b s )ss. Assume fg 0. Therefore s S ord fg = min{l(ss ) a s b s 0} = min{l(s)+l(s ) a s b s 0} min{l(s)+l(s ) a s 0 and b s 0} min{l(s) a s 0} + min{l(s ) b s 0} = ord f + ord g. Hence fg = 0 or ord fg ord f + ord g. Theorem 3.2 Let f,g R[S] {0}. Then f + g =0or ord (f + g) min{ord f, ord g}. Proof. Let f = a s s and g = s s be nonzero polynomials in R[S]. b Assume that ord f = n and ord g = m. Case 1 : m>n. We have that f + g = (a s + b s )s + a s s.,l(s) m,l(s)<m So ord (f + g) = min{l(s) a s 0} = n = min{n, m} = min{ord f,ord g}. Case 2 : m<n. We have that f + g = (a s + b s )s + b s s.,l(s) n,l(s)<n Thus ord (f+g) = min{l(s) b s 0} = m = min{n, m} = min{ord f,ord g}. Case 3 : m = n. We have that f + g = (a s + b s )s. Case 3.1 : f = g. Then f + g =0. Case 3.2 : f g. Then ord (f + g) min{ord f,ord g}. Therefore f + g = 0 or ord (f + g) min{ord f,ord g}, as required. The following corollary follows by the proof of Case 1 and Case 2 of Theorem 3.2. Corollary 3.3 Let f,g R[S] {0}. Ifordf ord g, then ord (f + g) = min{ord f, ord g}.

5 The degree and order of polynomials Remarks In this section, we give some interesting remarks. Remark 4.1 Let S = {x n n N {0}} be a semigroup under multiplication. Define L : S N {0} by L(x n )=n for all n N {0}. It is easy to see that R[S] is R[x]. Moreover, the degree and the order of polynomials in R[S] is the degree and the order of polynomials in R[x]. By Remark 4.1, we know that the degree and the order of polynomials in R[S] generalize the degree and the order of polynomials in R[x]. Remark 4.2 Let F A be a free semigroup generated by a nonempty set A, FA 1 = F A {1} and R be an integral domain. Let L : FA 1 N {0} defined by L(1) = 0 and for s = a 1 a 2 a n F A where a i A, L(s) =n. The following theorem is well-known (can see the proof in [1]). Theorem 4.1 [1] Let R be an integral domain and f,g R[FA 1 ]{0}. The following statements are true. (i) deg (fg)=deg f + deg g. (ii) If deg f deg g, then deg (f + g) =max{deg f, deg g}. (iii) ord (fg)=ord f + ord g. (iv) If ord f ord g, then ord (f + g) =min{ord f, ord g}. Moreover, we have that the ring R[FA 1 ] has no zero divisor. To generalize this, let R(F 1 A)={ f g f,g R[F 1 A] and g 0}. It is easy to prove that R(FA 1 ) is a ring under usual addition and multiplication of rational polynomials. Let f g R(F A 1) {0}. Let the degree of f be deg f deg g and the order g of f g be ord f ord g. The degree and order of f g are denoted by deg(f g ) and ord ( f ), respectively. The following theorems are true. g Theorem 4.2 Let R be an integral domain and f 1 R(FA 1 ) {0}. Then deg ( f 1 g 1 f2 g 2 )=deg f 1 g 1 + deg f 2 g 2.

6 886 R. Chinram and K. Pattamavilai Proof. We have deg( f 1 f2 ) = deg( f 1f 2 ) = deg(f 1 f 2 ) deg( )= deg(f 1 ) + deg(f 2 ) deg(g 1 ) deg(g 2 ) = (deg(f 1 ) deg(g 1 )) + (deg(f 2 ) deg(g 2 )) = deg f 1 + deg f 2. Theorem 4.3 Let R be an integral domain and f 1 R(FA 1 ) {0}. If deg (f 1 g 2 ) deg (g 1 f 2 ), then deg ( f 1 + f 2 )=max{deg f 1, deg f 2 }. Proof. We have deg ( f 1 + f 2 ) = deg( f 1g 2 + g 1 f 2 ) = deg(f 1 g 2 + g 1 f 2 ) deg( ) = max{deg(f 1 g 2 ), deg(g 1 f 2 )} deg( ) = max{deg(f 1 g 2 ) deg( ), deg(g 1 f 2 ) deg( )} = max{deg(f 1 ) deg(g 1 ), deg(f 2 ) deg(g 2 )} = max{deg f 1 g 1, deg f 2 g 2 }. Therefore the theorem is proved. Theorem 4.4 Let R be an integral domain and f 1 R(FA 1 ) {0}. Then ord ( f 1 f2 )=ord f 1 + ord f 2. Proof. It is similar to the proof of Theorem 4.2. Theorem 4.5 Let R be an integral domain and f 1 R(FA 1 ) {0}. If ord (f 1 g 2 ) ord (g 1 f 2 ), then ord ( f 1 + f 2 )=min{ord f 1,ord f 2 }. Proof. It is similar to the proof of Theorem 4.3. References [1] R. Chinram, The degree and the order of polynomials in the ring R[F 1 A ], Songklanakarin Journal of Science and Technology 29(2007), [2] P. A. Grillet, Algebra, John Wiley & Sons, Inc., New York, 1999.

7 The degree and order of polynomials 887 [3] J. M. Howie, An introduction to semigroup theory, Academic Press, London, [4] T. W. Hungerford, Algebra, Springer-Verlag, New York, Received: May 22, 2008