Ideals Of The Ring Of Higher Dimensional Dual Numbers

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1 Journal of Advances in Algebra (AA). ISSN Volume 9, Number 1 (2016), pp. 1 8 Research India Publications Ideals Of The Ring Of Higher Dimensional Dual Numbers Basem Alkhamaiseh Yarmouk University. basem.m@yu.edu.jo Abstract Let R be a commutative ring with unity and let R[α 1,α 2,...,α n ] be the ring of the higher dimensional dual numbers ring. The purpose of this article is to characterize the ideal structure of the ring R[α 1,α 2,...,α n ] and its relation to the ideal structure of the ring R. In particular, a describtion of the maximal ideals, prime ideals, finitely generated ideals and pure ideals are also given. AMS subject classification: 13A15, 13A25. Keywords: Polynomial ring, Dual numbers ring, Maximal ideal, Prime ideal, Principal ideal ring, PM-ring. 1. Introduction Throghout this article R is a commutative ring with a nonzero identity 1. Let R[x 1, x 2,...,x n ] be the polynomial ring over R with indetemenants x 1,x 2,...,x n. Then the factor ring of R[x 1,x 2,...,x n ] modulo the ideal n x i x j, which is the ideal of 1 i,j R[x 1,x 2,...,x n ] generated by the set {x i x j } n 1 i,j, has the form R[x 1,x 2,...,x n ]/ x i x j n 1 i,j Suppose the set R[α 1,α 2,...,α n ]={a 0 + a 1 α 1 + a 2 α a n α n : a i R and α i α j = 0 for 1 i, j n}. Then R[α 1,α 2,...,α n ] over the operations, addition pointwise and multiplication defined by (a 0 + a 1 α 1 + a 2 α a n α n )(b 0 + b 1 α 1 + b 2 α b n α n ) = a 0 b 0 +(a 0 b 1 +a 1 b 0 )α 1 +(a 0 b 2 +a 2 b 0 )α (a 0 b n +a n b 0 )α n, forms a commutative ring with unity 1. Furthermore, there is a ring isomorphism φ between R[x 1,x 2,...,x n ]/ n x i x j 1 i,j and R[α 1,α 2,...,α n ] which is given by φ(a 0 + a 1 x 1 + a 2 x a n x n + n x i x j 1 i,j ) = a 0 + a 1 α 1 + a 2 α a n α n. As a special

2 2 Basem Alkhamaiseh case, If n = 1, then the ring R[α 1 ] = R[x 1 ]/ x1 2 is precisely the ring of dual numbers. Hence, Vasantha et.al in [7] have named R[α 1,α 2,...,α n ] as the higher dimensional dual numbers ring. There are connections between some of the results in this paper and the earlier work of Asfahani et.al. [5], where some results in this paper are somewhat similar to those in [5], but there are some key differences since we use different rings. Throughout the paper, U(R)denotes the group of units of R, and Idem(R), Nil(R) denote the set of idemptents of R and the set of nilpotents of R respectively. Any undefined notation or terminology is standard, as in [1]. For more about the ring R[x 1,x 2,...,x n ] consult [6]. 2. Ideal Structure of the ring R[α 1,α 2,...,α n ] In this section, we shall describe the ideals in the ring R[α 1,α 2,...,α n ] in terms of those in R. We start this section with the following lemma which is necessary to construct the ideals of R[α 1,α 2,...,α n ]. Lemma 2.1. Let R be a ring and n a positive integer. Then J ={a 0 + a 1 α 1 + a 2 α 2 + +a n α n : a i I i, where I i ideals of R with I 0 I i for 1 i n} is an ideal of R. Moreover, we denote such ideal J as J = I 0 + I 1 α 1 + I 2 α 2 + +I n α n. The next lemma shows that if I is an ideal of R[α 1,α 2,...,α n ], then we can construct a family of ideals of R depending on I. Lemma 2.2. Let R be a ring, and I is an ideal of R[α 1,α 2,...,α n ]. Then I i ={r R : there are r 0,r 1,...,r i 1,r i+1,...,r n R with r 0 + r 1 α 1 + +r i 1 α i 1 + rα i + r i+1 α i+1 +,r n α n I} is an ideal of R for 0 i n. Moreover, I 0 I i for 1 i n. Proof. Let a I i, and r R. So there are a 0,a 1,...,a i 1,a i+1,...,a n R with x = a 0 + a 1 α 1 + +a i 1 α i 1 + aα i + a i+1 α i+1 + +a n α n I. Since I is an ideal of R[α 1,α 2,...,α n ],rx = ra 0 + ra 1 α 1 + +ra i 1 α i 1 + raα i + ra i+1 α i+1 + +ra n α n I. Hence, ra I i. Now, let b I 0. So there are a 1,a 2,...,a n R with y = b + a 1 α 1 + a 2 α 2 + +a n α n I. Then since I is an ideal of R[α 1,α 2,...,α n ], yα i = (b + a 1 α 1 + a 2 α 2 + +a n α n )α i = bα i I. Thus, b I i, and I 0 I i,for 1 i n. Now we use Lemma 2.1 and Lemma 2.2 to describe the ideals of R[α 1,α 2,...,α n ]. Theorem 2.3. Let R be a ring. Then each ideal I of R[α 1,α 2,...,α n ] has the form I = I J, where J = I 0 + I 1 α 1 + I 2 α 2 + +I n α n, where I i is an ideal of R defined as in Lemma 2.2 for 0 i n. Proof. Let I be an ideal of R[α 1,α 2,...,α n ]. Then by Lemma 2.1 and Lemma 2.2,

3 Ideals Of The Ring Of Higher Dimensional Dual Numbers 3 J = I 0 + I 1 α 1 + I 2 α 2, +I n α n is an ideal of R[α 1,α 2,...,α n ]. It is also clear that I J. Therefore, I = I J. Corollary 2.4. Let R be a ring. Then any ideal I = I J of R[α 1,α 2,...,α n ] where J = I 0 + I 1 α 1 + I 2 α 2 + +I n α n is uniquely determined by I 0,I 1,I 2,...,I n. The following example shows that, for an ideal I of R[α 1,α 2,...,α n ], it is unnecessary that J = I 0 + I 1 α 1 + I 2 α 2 + +I n α n I. That means, an ideal I of R[α 1,α 2,...,α n ], need not have the form I = I 0 + I 1 α 1,I 2 α 2,...,I n α n. Example 2.5. Let I = α 1 + 2α 2 be an ideal of Z[α 1,α 2 ] generated by α 1 + 2α 2.It is clear that I 0 = 0,I 1 = Z,I 2 = 2Z. Hence I Zα 1 + 2Zα 2. The other inclusion is not correct since α 1 Zα 1 +2Zα 2 and α 1 / I otherwise α 1 = (α 1 +2α 2 )(a+bα 1 +cα 2 ) = aα 1 + 2aα 2. Hence a = 1, and 2a = 0, which is a contradiction. Also the following example shows that if I = I (I 0 + I 1 α 1 + I 2 α 2 + +I n α n ) is an ideal of R[α 1,α 2,...,α n ], then it is not true in general that I 0 I. Example 2.6. Let I = 2+α 1 be an ideal of Z[α 1 ] generated by 2+α 1. Then it is easy to show that I has the form I = 2 + α 1 (2Z + Zα 1 ). Suppose that 2Z 2 + α 1, then 2 = (2 + α 1 )(a + bα 1 ) for some a + bα 1 Z[α 1 ]. Thus we have 2a = 2 and 2b + a = 0. Solving these two equations produces a = 1 and 2b = 1 which is a contradiction. Hence 2Z 2 + α 1. By adding some suitable assumptions on the ideals of R[α 1,α 2,...,α n ]. one can prove the next result easily. Lemma 2.7. Let R be a ring, and let I = I (I 0 + I 1 α 1 + I 2 α 2 + +I n α n ) be an ideal of R[α 1,α 2,...,α n ]. If α i I, for all 1 i n, then 1. I 0 I, and 2. I = I 0 + Rα 1 + Rα 2 + +Rα n. 3. Prime and Maximal ideals of the ring R[α 1,α 2,...,α n ] In this section, we will describe the prime and maximal ideals of the ring R[α 1,α 2,...,α n ]. We also prove some elementary results concerning the Nilradical and the Jacobson radical of R[α 1,α 2,...,α n ]. Lemma 3.1. Let R be a ring, and let I 0 be an ideal of R. Then R[α 1,α 2,...,α n ]/(I 0 + Rα 1 + Rα 2 + +Rα n ) = R/I 0 Proof. Define the mapping θ : R[α 1,α 2,...,α n ] R/I 0

4 4 Basem Alkhamaiseh As θ(a 0 + a 1 α 1 + a 2 α a n α n ) = a 0 + I 0. It is clear that θ is an onto ring homomorphism with kernel, ker(θ) = I 0 + Rα 1 + Rα 2 + +Rα n. Then the result is obtained by applying the first isomorphism theorem for rings. Theorem 3.2. Let R be a ring. Then 1. P is prime ideal of R[α 1,α 2,...,α n ], if and only if P = P 0 + Rα 1 + Rα 2 + +Rα n for some prime ideal P 0 of R. 2. M is maximal ideal of R[α 1,α 2,...,α n ], if and only if M = M 0 + Rα 1 + Rα 2 + +Rα n for some maximal ideal M 0 of R. Proof. (1) Suppose that P is prime ideal of R[α 1,α 2,...,α n ]. By Theorem 2.3, P = P (P 0 + Rα 1 + Rα 2 + +Rα n ). Now since α 2 i = 0 P,for 1 i n, we have α i P. Thus by Lemma 2.7, P = P 0 + Rα 1 + Rα 2 + +Rα n. The rest of the proof obtained by applying Lemma 3.1, and using the fact R/I is an integral domain if and only if I is a prime ideal. (2) Similar to part (1), but we use the fact that R/I is a field if and only if I is a maximal ideal. In commutative ring theory, the Jacobson radical (Nilradical) of the ring R is the intersection of all maximal (prime) ideals of R (see [1]). Corollary 3.3. Let R be a ring. Then 1. J(R[α 1,α 2,...,α n ]) = J(R)+ Rα 1 + Rα 2 + +Rα n 2. Nil(R[α 1,α 2,...,α n ]) = Nil(R)+ Rα 1 + Rα 2 + +Rα n. Proof. 1. J(R[α 1,α 2,...,α n ]) = Intersection of all maximal ideals of R[α 1,α 2,...,α n ] = 1 + Rα 2 + +Rα n ) M Max(R) = ( M Max(R) M )+ Rα 1 + Rα 2 + +Rα n = J(R)+ Rα 1 + Rα 2 + +Rα n

5 Ideals Of The Ring Of Higher Dimensional Dual Numbers 5 2. Nil(R[α 1,α 2,...,α n ]) = Intersection of all prime ideals of R[α 1,α 2,...,α n ] = 1 + Rα 2 + +Rα n ) P Spec(R) = ( P Spec(R) P )+ Rα 1 + Rα 2 + +Rα n = Nil(R)+ Rα 1 + Rα 2 + +Rα n Corollary 3.4. Let R be a ring. Then R is local ring if and only if R[α 1,α 2,...,α n ] is so. Recall that a ring R is called a PM-ring if for each prime ideal in R is contained in a unique maximal ideal, for more see [2]. Theorem 3.5. Let R be a ring. Then R is PM-ring if and only if R[α 1,α 2,...,α n ] is PM-ring. Proof. Suppose that R is PM-ring and let P be a prime ideal of R[α 1,α 2,...,α n ] then by Theorem 3.2, P = P 0 +Rα 1 +Rα 2 + +Rα n for some P 0 a prime ideal of R. Since R is Gelfand ring, P 0 is contained in a unique maximal ideal, say M 0 Also, again by Theorem 3.2, M = M 0 + Rα 1 + Rα 2 + +Rα n is maximal ideal of R[α 1,α 2,...,α n ]. Note that, P is contained in M. Now, If there exists a maximal ideal N of R[α 1,α 2,...,α n ] that contains P, then by Theorem 3.2, N = N 0 + Rα 1 + Rα 2 + +Rα n for some maximal ideal N 0 of R such that P 0 + Rα 1 + Rα 2 + +Rα n = P N = N 0 + Rα 1 + Rα 2 + +Rα n It is clear that P 0 is contained in N 0 which is a contradiction except if N 0 = M 0. Hence, N 0 = M 0.Therefore, R[α 1,α 2,...,α n ] is PM-ring. By the same argument we can prove the converse direction. 4. Finitely Generated Ideals of R[α 1,α 2,...,α n ] In this section, we will study and describe the finitely generated ideals of the ring R[α 1,α 2,...,α n ]. Then we will characterize when R[α 1,α 2,...,α n ] is a principal ideal ring. Theorem 4.1. Let R be a ring, and let a 0 +a 1 α 1 +a 2 α 2 + +a n α n R[α 1,α 2,...,α n ]. Then the ideal generated by a 0 + a 1 α 1 + a 2 α 2 + +a n α n has the form I = a 0 + a 1 α 1 + a 2 α 2 + +a n α n = I (I 0 + I 1 α 1 + I 2 α 2 + +I n α n ), where I 0 = a 0, and I i = a 0,a i for 1 i n.

6 6 Basem Alkhamaiseh Proof. Suppose that I is an ideal of R[α 1,α 2,...,α n ] generated by a 0 + a 1 α 1 + a 2 α 2 + +a n α n. Then by Theorem 2.3, I = I (I 0 +I 1 α 1 +I 2 α 2 + +I n α n ). Now let x I 0. So there exist b 1,b 2,...,b n R such that (x +b 1 α 1 + +b n α n ) = (a 0 +a 1 α a n α n )(c 0 + c 1 α 1 + +c n α n ) for some c 0 + c 1 α 1 + +c n α n R[α 1,α 2,...,α n ]. Thus, x = a 0 c 0 and hence I 0 a 0. Also since a 0 + a 1 α 1 + a 2 α 2 + +a n α n I we have a 0 I 0. To prove that I i = a 0,a i for 1 i n. Let x I i. Then there exist b 0, b 1,...,b i 1, b i+1,...,b n R such that (b 0 + b 1 α 1 + +xα i b n α n ) = (a 0 + a 1 α 1 + a 2 α 2 + +a n α n )(c 0 + c 1 α 1 + c 2 α 2 + +c n α n ) for some c 0 + c 1 α 1 + c 2 α 2 + +c n α n R[α 1,α 2,...,α n ]. Hence we have x = a 0 c i + a i c 0 a 0,a i. That is I i a 0,a i. To prove the other inclusion, let x a 0,a i. So x = a 0 t + a i s for some s, t R. Thus, there exist a 0 s, a 1 s, a 2 s,...,a i 1 s, a i+1 s,...,a n s R with a 0 s + a 1 sα 1 + a 2 sα 2 + +xα i + +a n sα n I. Therefore, a 0,a i I i. Corollary 4.2. Let R be a ring. Then R is principal ideal ring if and only if R[α 1, α 2,...,α n ] is so. Proof. Let I be an ideal of R[α 1,α 2,...,α n ]. Then by Theorem 2.3, I = I (I 0 + I 1 α 1 + I 2 α 2 + +I n α n ). Since R is principal ideal ring, I i = a i for 1 i n. Also since I 0 I i for 1 i n we have b R such that a i divides b for 1 i n. Thus, I = I ( b + a 1 α 1 + a 2 α a n α n ) = I ( b + a 1,b α 1 + a 2,b α a n,b α n ) = b + a 1 α 1 + a 2 α 2 + +a n α n Therefore, R[α 1,α 2,...,α n ] is a principal ideal ring. The other direction is obvious. The following corollary describes the form of the finitely generated ideal of the ring R[α 1,α 2,...,α n ]. { n } Corollary 4.3. Let R be a ring, and let x j = a ij α i be a family of elements of R[α 1,α 2,...,α n ]. Then the ideal generated by the finite set {x j } m j=1 has the form I = x 1,x 2,...x m = I (I 0 + I 1 α 1 + I 2 α 2 + +I n α n ), where I 0 = {a 0j } m j=1, and I i = {a 0j,a ij } m j=1 for 1 i n. i=0 5. The Pure ideals of the ring R[α 1,α 2,...,α n ] In this section, we will exhibit the relationship between the pure ideals of R and the pure ideals of R[α 1,α 2,...,α n ]. Recalling that an deal I of R is said to be pure if for every element x I, there exits y I such that xy = x.

7 Ideals Of The Ring Of Higher Dimensional Dual Numbers 7 Lemma 5.1. Let R be a ring, and let I = I (I 0 + I 1 α 1 + I 2 α 2 + +I n α n ) be an ideal of R[α 1,α 2,...,α n ]. If I 0 = I 1 = I 2 = =I n, then 1. I 0 I. and 2. I = I 0 + I 0 α 1 + I 0 α 2 + +I 0 α n. Proof. It is clear that I (I 0 + I 0 α 1 + I 0 α I 0 α n ). Now let a 0 + a 1 α 1 + a 2 α a n α n (I 0 + I 0 α 1 + I 0 α I 0 α n ). Hence by the assumption a i I 0 for all 0 i n. So there exist b 1,i,b 2,i,...,b n,i R such that a i + b 1,i α 1 + b 2,i α 2 + +b n,i α n I. Since I is an ideal of R[α 1,α 2,...,α n ], we have a i α i = (a i + b 1,i α 1 + b 2,i α b n,i α n )α i I for all 1 i n. Using the same argument and since I (I 0 + I 0 α 1 + I 0 α 2 + +I 0 α n ), we deduce b t,0 α t I for all 1 t n. Thus, a 0 I. Therefore, a 0 + a 1 α 1 + a 2 α 2 + +a n α n I, and (I 0 + I 0 α 1 + I 0 α 2 + +I 0 α n ) I. This proves (2). Lemma 5.2. Let R be a ring, and let I = I (I 0 + I 1 α 1 + I 2 α 2 + +I n α n ) be an ideal of R[α 1,α 2,...,α n ]. If I is a pure ideal, then 1. I 0 = I i for all 1 i n. 2. I 0 is a pure ideal of R. Proof. Suppose that I = I (I 0 +I 1 α 1 +I 2 α 2 + +I n α n ) is an ideal of R[α 1,α 2,...,α n ]. Now 1. By the costruction of the ideal I in Theorem 2.3, we have I 0 I i for all i n. Hence, it is enough to prove the other inclusion. Let a I k for any 1 k n. So there are b 0,b 1,...,b i 1,b i+1,...,b n R such that b 0 + b 1 α aα k + +b n α n I. Since I is a pure ideal of R[α 1,α 2,...,α n ], there exists c 0 +c 1 α 1 + +c k α k + +c n α n I such that b 0 +b 1 α 1 + +aα k + +b n α n = (b 0 + b 1 α 1 + +aα k + +b n α n )(c 0 + c 1 α 1 + +c k α k + +c n α n ). That is, a = b 0 c k + ac 0 I 0. Hence, I 0 = I k. 2. Let x I 0. So there exist y 1,,y n R such that x + y 1 α 1 + +y n α n I. Hence, there is z 0 + z 1 α 1 + +z n α n I such that x + y 1 α 1 + +y n α n = (x + y 1 α 1 + +y n α n )(z 0 + z 1 α 1 + +z n α n ), So x = xz 0 and this ends the proof since z 0 I. De Marco [3] has studied purity and projectivity of ideals. Among many results he has proved that is if A is an ideal of the ring R, and if {a 1,a 2,...,a n } is a finite subset of A, then there exists b A such that a i = a i b for all 1 i n. This result will be used in the following lemma. Lemma 5.3. Let R be a ring, and let I = I 0 + I o α 1 + I o α 2 + +I 0 α n be an ideal of R[α 1,α 2,...,α n ]. If I 0 is a pure ideal of R, then I is pure ideal of R[α 1,α 2,...,α n ].

8 8 Basem Alkhamaiseh Proof. Let I 0 be a pure ideal of R. If a 0 + a 1 α 1 + a 2 α 2 + +a n α n I, then exists b I 0 I such that a i = a i b for all 0 i n. Hence, a 0 +a 1 α 1 +a 2 α 2 + +a n α n = (a 0 + a 1 α 1 + a 2 α 2 + +a n α n )b which completes the proof. Gathering the results in Lemma 5.1, Lemma 5.2 and Lemma 5.3 produces the following theorem. Theorem 5.4. Let R be a ring, and let I be an ideal of R[α 1,α 2,...,α n ]. Then I is pure ideal of R[α 1,α 2,...,α n ] if and only if I = I 0 + I 0 α 1 + I 0 α 2 + +I 0 α n with I 0 a pure ideal of R. The following result is a direct consequence of Theorem 3.2 and Theorem 5.4. Corollary 5.5. Let R be a ring, and let I be a pure ideal of R[α 1,α 2,...,α n ].Then I never be maximal nor prime ideal. References [1] Atiyah MF, MacDonald IG. Introduction to Commutative Algebra. Boston, MA, USA: Addison-Wesley Publishing Company, [2] G. De Marco, A. Orsatti, Commutative rings in which every prime ideal is contained in a unique maximal ideal, Proc. Amer. Math. Soc., 1971, 30 (3): [3] G. De Marco. Projectivity of pure ideals, Rend. Sem. Mat. Univ. Padova,1983, 68: [4] Kai Long, Qichuan Wang, Lianggui Feng. Morphic property of a quotient ring over polynomial ring. Bulletin of the Korean Mathematical Society 2013, 50: [5] Nasr-Isfahani A., Moussavi, A. On a quotient of polynomial rings, Commun Algebra 2010; 38: [6] Thomas Hungerford, Algebra, Graduate Texts in Mathematics, vol. 73, Springer- Verlag, New York-Berlin, [7] Vasantha Kandasamy W.B., Smarandache F., Dual Numbers, Zip publishing, Ohio,2012.