Smarandache-Zero Divisors in Group Rings

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1 Smarandache-Zero Dvsors n Group Rngs W.B. Vasantha and Moon K. Chetry Department of Mathematcs I.I.T Madras, Chenna The study of zero-dvsors n group rngs had become nterestng problem snce 1940 wth the famous zero-dvsor conjecture proposed by G.Hgman [2]. Snce then several researchers [1, 2, 3] have gven partal solutons to ths conjecture. Tll date the problem remans unsolved. Now we ntroduce the notons of Smarandache zero dvsors (S-zero dvsors) and Smarandache week zero dvsors (S-weak zero dvsors) n group rngs and study them. Both S-zero dvsors and S- weak zero dvsors are zero dvsors but all zero dvsors are not S-zero dvsors or S-weak zero Even here we can modfy the zero dvsor conjecture and suggest the S-zero dvsor conjecture and S-weak zero dvsor conjecture for group rngs. Thus the study has ts own mportance. Unlke n case of group rngs of fnte groups over feld of characterstc zero where one s always gurranteed the zero dvsors, we can not establsh the same n case of S-zero dvsors or S-weak zero In ths paper we study the condtons on Z n, the rng of nteger modulo n to have S- zero dvsors and S-weak zero We have proved f n s a composte number of the form n = p 1 p 2 p 3 or (n = p α ) where p 1, p 2 and p 3 are dstnct prmes or (p a prme, α 3), then Z n has S-zero We further obtan condtons for Z n to have S-weak zero We prove all group rngs Z 2 G where G a cyclc group of non prme order and G 6 have S-zero The result s extended to any prme p.e. we generalze the result for Z p G where p any prme. Condtons for group rngs ZG and KG (Z the rng of nteger and K any feld of characterstc zero) to have S-zero dvsors s derved. Ths paper s organzed nto 5 sectons. In secton 1, we recall the basc defntons of S-zero dvsors and S-weak dvsors and some of ts propertes. In secton 2, we fnd condtons for the fnte rng Z n to have S-zero dvsors and S-weak zero We gve examples to ths effect. In secton 3, we analze the group rngs Z 2 G where G s a fnte group of order greater than or equal to 6 and G a cyclc group of non-prme order. In secton 4 we fnd condtons for ZG to have S-zero We prove ZS n and ZD 2n have S-zero Further f K s a feld of characterstc zero then KS n and KD 2n have S-zero When G s any group of order n and n a composte number, then ZG and KG have S-zero In secton 5 we gve the concluson based on our study.

2 1. Some Basc Defntons: Here we just recall some basc results about S-zero dvsors and S-weak zero Defnton 1.1 [5]: Let R be a rng. An element a R \ {0} s sad to be a S-zero dvsor f a.b = 0 for some b 0 n R and there exsts x, y R \ {0, a, b} such that. a.x = 0 or x.a = 0. b.y = 0 or y.b = 0. x.y 0 or y.x 0 Note that n case of commutatve rngs we just need. a.x = 0,. b.y = 0 and. x. y 0 Theorem 1.1 [5]: Let R be a rng. Every S-zero dvsor s a zero dvsor but a zero dvsor n general s not a S-zero dvsor. Example 1.1: In Z 4 = {0, 1, 2, 3} the rng of ntegers modulo 4, 2 s a zero dvsor but t s not a S-zero dvsor. Defnton 1.2 [5]: Let R be a rng. An element a R \ {0} s a S-weak zero dvsor f there exsts b R \ {0, a} such that a.b = 0 satsfyng the followng condtons: There exsts x, y R \ {0, a, b} such that. a.x = 0 or x.a = 0. b.y = 0 or y.b = 0. x.y = 0 or y.x = 0 Example 1.2: Let Z 12 = {0, 1, 2,,11} be the rng of nteger modulo (mod 12), (mod12), and (mod 12) also (mod12) So 3 and 4 are S-weak zero dvsors n Z 12. We can check that Z 12 contans S-zero dvsor also. For (mod 12) and (mod12) (mod12) but (mod (12)) Thus Z 12 has both S-zero dvsor and S-weak zero

3 2. S-zero dvsors and S-weak zero dvsors n Z n In ths secton we fnd condtons for Z n to have S-zero dvsors and S-weak zero dvsors and prove when n s of the form n = p α (α 3) or n = p 1 p 2 p 3 where p 1, p 2, p 3 are dstnct prmes, then Z n has S-zero dvsor. Proposton 2.1: Z p = {0, 1, 2,, p-1}, the prme feld of characterstc p where p s a prme has no S-zero dvsors or S-weak zero Proof: Let Z p = {0, 1, 2,.., p-1}, where p s a prme. Z p has no zero Hence Z p can not have S-zero dvsor as every S-zero dvsor s a zero dvsor. Further Z p cannot have S-weak zero dvsor as every S-weak zero dvsor s also a zero dvsor. Hence the clam Example 2.1: Let Z 6 = {0, 1, 2, 3, 4, 5} be the rng of ntegers modulo 6. Here 0 (mod 6), (mod 6) are the only zero dvsors of Z 6. So Z 6 has no S-zero dvsor and S-weak zero Example 2.2: Z 8 = {0, 1, 2, 3, 4, 5, 6, 7}, the rng of ntegers modulo 8. Here (mod ) and (mod 8), (mod 8) but (mod8). So Z has 4 as S-zero dvsor, but has no S-weak zero Theorem 2.1: Z n has no S-weak zero dvsors when n = 2p, p a prme. Proof: To get any zero dvsors a, b Z n \ {0} such that a.b 0 (mod n) one of a or b must be p and the other an even number. So we cannot get x, y Z n \ {p} such that a.x 0 (mod n), b.y 0 (mod n) and x.y 0 (modn). Hence the clam. Theorem 2.2: Z n has S-weak zero dvsors when n = p 1 p 2 (orp 2 ) where p 1 and p 2 are dstnct odd prmes wth both of them greater than 3 (or p an odd prme greater than 3). Proof: Take a = p 1, b = p 2, then a.b 0 (mod n). Agan take x = 3p 2 and y 2p 1 then a.x 0 (mod n) and b.y 0 (mod n), also x.y 0 (mod n).

4 In case of = p 2 a smlar proof holds good. Hence the clam. Theorem 2.3: Z n has no S-zero dvsors f n = p 1 p 2 where p 1, p 2 are prmes. Proof: Let n = p 1 p 2. Suppose a.b 0 (mod n), a, b Z n \ {0} then p 1 s factor of a and p 2 s a factor of b or vce-versa. Suppose p 1 s a factor of a and p 2 s a factor of b. Now to fnd x, y Z n \ {0, a, b} such that a.x 0 (mod n) p 2 s a factor of x, and b.y 0 (mod n) p 1 s a factor of y. Ths forcng x.y 0 (mod n). Thus f n = p 1 p 2, Z n has no S-zero Corollary 2.1: Z n has no S-zero dvsors when n = p 2. Theorem 2.4: Z n has S-zero dvsors f n = α p 1 where p 1 s a prme and α 3. Proof: Take α 1 α 1 a = p1, b = p1 then a.b 0 (mon n). Agan take x = p 1 and y = p 1 p 2 where p 2 s the prme next to p 1. Then clearly a.x 0 (mod n), b.y 0 (mod n) but x.y 0 (mod n). Hence the clam. Example 2.3: In Z 27 = {0, 1, 2,, 26}; rng of ntegers modulo 27, we have (mod 27), (mod 27) also (mod 27) but (mod 27). So 3 2 = 9 s a S-zero dvsor n Z 27. Theorem 2.5: Z n has S-zero dvsor when n s of the form n = p 1 p 2 p 3, where p 1, p 2, p 3 are prmes. Proof: Take a = p 1 p 2, b = p 2 p 3 then a.b 0 (mod n). agan take x = p 1 p 3 and y = p 1 p 4 where p 4 s the prme next to p 3. Then a.x 0 (mod n) and b.y 0 (mod n). But x.y 0 (mod n). Hence the clam Example 2.4: Let Z 30 = {0, 1, 2,, 29} be the rng of ntegers modulo 30. Here (mod 30), (mod 30), (mod 30) But (mod 30), so 6 and 15 are S-zero dvsor n Z 30. We can generalze the Theorem 2.5 as follows: Theorem 2.6: If n = p 1 p 2 p 3..p n then Z n has S-zero

5 Proof: Take a = p 1 p 2 p n-1, b = p 2 p 3 p n-1. and x = p 1 p 3 p n y = p 1 p n+1 whre p n+1 s the prme next to p n. Then t s easy to see that a.b 0 (mod n), a.x y 0 (mod n), b.y 0 (mod n), But x.y 0 (mod n). So a and b are S-zero dvsors n Z n. Fnally we can characterze Z n for havng S-zero dvsors as follows. Theorem 2.7: Z n has S-zero dvsors f and only f n s the product of atleast three prmes. 3. S-zero dvsors n the group rng Z 2 G Here we prove that the group rng Z 2 G, where G s a fnte cyclc group of non prme order and G 6 has S-zero We llustrate by certan examples the non-exstence of S-zero dvsors before we prove our clam. Example 3.1: Consder the group rng Z 2 G of the group G = {g/g 2 = 1} over Z 2. Clearly (1 + g) 2 = 0 s the only zero dvsor, so t can not have S-zero dvsors or S-weak zero Example 3.2: Let G = {g / g 3 = 1} be the cyclc group of order 3. Consder the group rng Z 2 G of the group G over Z 2. Clearly (1 + g + g 2 ) (g + g 2 ) = 0 (1 + g + g 2 ) (1 + g) = 0 (1 + g + g 2 ) (1 + g 2 ) = 0 are the only zero dvsors n Z 2 G. We see none of these are S-zero dvsors or S-weak zero Example 3.3: Consder the group rng Z 2 G. where G = { g / g 4 = 1} s the cyclc group of order 4. Then (1 + g) (1 + g + g 2 + g 3 ) = 0 (1 + g 2 ) (1 + g + g 2 + g 3 ) = 0 (1 + g 3 ) (1 + g + g 2 + g 3 ) = 0 (g + g 2 ) (1 + g + g 2 + g 3 ) = 0 (g + g 3 )(1 + g + g 2 + g 3 ) = 0 (g 2 + g 3 ) (1 + g + g 2 + g 3 ) = 0 are some of the zero dvsors n Z 2 G. So t has S-zero dvsors and no S-weak zero Theorem 3.1: Let Z 2 G be the group rng where G s a cyclc group of prme order p. Then the group rng Z 2 G has no S-zero dvsors or S-weak zero

6 Proof: The only possble zero dvsors n Z 2 G are (1 + g + +g p-1 ) Y = 0 where Y Z 2 G and support of T s even But we can not fnd A Z 2 G \ (1 + g + + g p-1 ) Such that AY = 0, supp A < p. Hence Z 2 G can not have S-zero dvsors or S-weak zero Theorem 3.2: Let Z 2 G be the group rng of a fnte cyclc group of composte α α order n 6 (n = p 2 1 p ) then Z 2 G has S-zero 1 1 Proof: Let G be a cyclc group of order n where G has atleast one subgroup H. Let H be generated by g t, t > 1. Now (1 + (g t ) r ) (1 + g + g g n-1 ) = 0; (g t ) r 1 agan (1 + g) (1 + g + g g n-1 ) = 0 also (1 + (g t ) r ) (1 + g t + (g t ) 2 + +(g t ) s+1 ) = 0, ( (g t ) s+1 = 1) clearly (1 + g) (1 + g t + (g t ) (g t ) s+1 ) 0. Thus Z 2 G has S-zero Theorem 3.3: Let Z 2 S n be the group rng of the symmetrc group S n over Z 2. then Z 2 S n has S-zero Proof: Let A = 1 + p 1 + p 2 + p 3 + p 4 + p 5 and B = p 4 + p 5 p 1 = n n, p 2 = n n p 3 = n n, p 4 = n n n p 5 = n And 1 s the dentty permutaton. Clearly AB = 0 Take X = 1 + p 1, and Y = 1 + p 4 + p 5 Then

7 But Hence Z 2 S n has S-zero dvsor. AX = 0 and BY = 0 XY 0. Example 3.4: The group rng Z 2 S 4 where S 4 s the symmetrc group of order 4 has s-zero Let a =, b =, c = S Put A = (1 + a + b + c) Let g =, then g = 1 = : Now. Let B = 1 + g + g 2 + g 3, Clearly (1 + g) s = 0 (1 + g). B = 0 also, s S4 A s = 0, s S4 A.B = 0. Hence (1 + g) and s S s 4 are S-zero dvsors n Z 2 S 4. Theorem 3.4: Let Z 2 D 2n be the group rng of the Dhedral group D 2n, over Z 2 where n s a composte number, then Z 2 D 2n has S-zero Proof: Snce n s a composte number, by the theorem 3.2, we see Z 2 D 2n has S- zero Example 3.5: Let Z 2 D 8 be the group rng, where D 8 s the Dhedral group of order 8. then Z 2 D 8 has s-zero Here D 8 = {a, b / a 2 = b 4 = 1, bab = a}. Let x = 1 + b + b 2 + b 3, y = g = (1 + a + ab + ab 2 + ab 3 + b + b 2 + b 3 ) g D8 then x.y = 0 Suppose A = (1 + b 2 ) and B = (1 + ab). Then clearly A.x = 0 also B.y = 0 But

8 A.B 0. Hence x and y are S-zero dvsors n Z 2 D. Example 3.6: Z 2 G where G = {g / g 6 = 1} be the group rng. Z 2 G has S-zero Let X = (1 + g + g 2 + g 3 + g 4 + g 5 ), and y = (1 + g 2 + g 4 ) Then x.y = 0. Let a = 1 + g 3 and b = 1 + g 2 Clearly a.x = 0 and b.y = 0 but a. b 0. Hence Z 2 G gas S-zero 4. S-zero dvsors of group rngs over rngs of characterstc zero. In ths secton we prove ZG, the group rng of a fnte group G over Z,. the rng of ntegers, has non-trval S-zero Further we prove ZS n and ZD 2n has nontrval S-zero Snce Z K for any feld K of characterstc zero, we can say KG has non-trval S-zero Example 4.1: ZG the group rng of the group G over Z; where G = {g / g 8 = 1}, we have for (1-g 4 ), (1 + g + g 2 + +g 7 ) ZG (1-g 4 ). (1 + g + g 2 + +g 7 ) = 0 also (1-g). (1 + g + g 2 + +g 7 ) = 0 (1-g 4 ). (1 + g 2 + g 4 +g 6 ) = 0 but (1-g). (1 + g 2 + g 4 +g 6 ) 0. Hence the clam Theorem 4.1: Let G be a fnte group where G = n, n s a composte number ; then the group rng ZG has S-zero Proof: Let a G such that a m =1, m < n. Now for (1-a), (1 + g g n-1 ) ZG We have (1-a) (1 + g 1 + g g n-1 ) = 0, g G (1 a) (1 + a + a a m-1 ) = 0, a G and a m = 1 (1-h) (1 + g 1 + g g n-1 ) = 0, h G and h a, I = 1, 2,, m but (1-h) (1 + a + a a m-1 ) 0, m < n. The complete the proof.

9 Theorem 4.2: The group rng ZS n has S-zero dvsors, where S n s the symmetrc group of degree n. Proof: For (1-p 1 ), (1 + p 1 + p 2 + p 3 + p 4 + p 5 ) ZS n We have (1 p 1 ) (1 + p 1 + p 2 + p 3 + p 4 + p 5 ) = 0 where p are as n theorem 3.3 agan (1-p 1 ) (1 + p 1 ) = 0 (1 + p 2 ) (1 + p 1 + p 2 + p 3 + p 4 + p 5 ) = 0 but (1+p 1 ) (1 + p 2 ) 0. Hence the clam Theorem 4.3: The group rng ZD 2n has S-zero dvsors, where D 2n s the Dhedral group of order 2n, n s a composte number. Proof: As n s a composte number, result wll follow from the theorem 4.1 Theorem 4.4: Let KG be the group rng of G over K, where K s a feld of characterstc zero and G any group of composte order, then KG has S-zero Proof: As Z K, so ZG KG and as ZG has S-zero dvsors so KG has nontrval S-zero Example 4.2: Let Z 3 G be the group rng of G over Z 3 where G = {g / g 6 = }. For (1 + g + g 2 + g 3 + g 4 + g 5 ), (1 + g + g 4 ) Z 3 G, we have (1 + g + g 2 + g 3 + g 4 + g 5 ) (1 + g 2 + g 4 ) = 0 (g ) (1 + g 2 + g 4 ) = 0 also (1 + g + g 3 ) (1 + g + g 2 + g 3 + g 4 + g 5 ) = 0 but (g 2 + 2) (1 + g + g 3 ) 0. Hence Z 3 G has s-zero Theorem 4.5: The group rng Z p G, where G = {g / g n = 1} and p/n, has S-zero Proof: Here (1 + g + g 2 + +g n-1 ) (1 + g p + g 2p + +g rn-1 ) = 0 where rp = n agan (1 + g + g 2 + +g n-1 ) ((p-1) + g)) = 0, (where (s, p) = 1) ((p-1) + g tp ) (1 + g p + g 2p + +g rp-1 ) = 0, t < r but ((p-1) + g 8 ) ((p-1) + g tp ) 0. Hence the clam

10 Example 4.3: The group rng Z 3 S 3 has S-zero Let (1 + p 4 + p 5 ), (1 + p 1 + p 2 + p 3 + p 4 + p 5 ) Z 3 S 3 Now (1 + p 4 + p 5 ), (1 + p 1 + p 2 + p 3 + p 4 + p 5 ) = 0 (2 + p 1 ) (1 + p 1 + p 2 + p 3 + p 4 + p 5 ) = 0 also ( 2 + p 5 ) (1 + p 4 + p 5 ) = 0 but ( 2 + p 5 ) (2 + p 1 ) 0. Theorem 4.6: The group rng Z p S n when p n! (.e p < n) has S-zero Proof: As p n! S n has a cyclc subgroup H of order p. We have for (1 + h + h h p-1 ), s Z S, s Sn p n (1 + h + h h p-1 ) s = 0, g D8 h H and ((p-1) + s) s = 0, s H, s S n g D8 and ((p-1) + h r ) (1 + h + h 2 + +h p-1 ) = 0, r < p 1 but ((p-1) + s) ((p-1) + h r ) 0. So Z p S n has S-zero Theorem 4.7: Let Z p D 2n be the group rng of the Dhedral group D 2n over Z p, p a prme such that p/n then Z p D 2n has S-zero Proof: Gven Z p D 2n s the group rng of D 2n over Z p such that p s a prme and p/n. Now p/n D 2n has a cyclc subgroup of order p, say H = < t >. For (1 + t + t t p-1 ), g Z D We have Hence the clam. Concluson g D2n p 2n t p = 1, t D 2n (1 + t + t t p-1 ). g 0 = g D2n ((p-1) t r ) (1 + t + t t p-1 ) = 0 r < p - 1 ((p-1) g r ) g 0 gr t r for r = 1, 2, 3,, p 1. = g D2n But ((p-1)-t r ) ((p-1)-g r ) 0. In the frst place we proved that Z p S n has S-zero dvsors provded p n!, p s a prme such that p < n. Further we proved ZS n and ZD 2n has S-zero dvsors only when n s a composte number.

11 Z p D 2n has S-zero dvsors for p n. We are not n a poston to obtan any nce algebrac structure for the collecton of S-zero dvsors or S-weak zero dvsors n group rngs. Also another nterestng problem would be to fnd the number of S- zero dvsors and s-weak zero dvsors n case of Z p S n (p n!, p < n) and Z p D 2n (p n). The soluton even n case of Z p G where G s a cyclc group of order n such that p/n, p a prme s not complete. References [1]. CONNEL I.G., On the group rng, Can.J. Math 15 (1963), [2]. HIGMAN G., The unts of group rngs, Proc., London math. Soc. 2, 46 (1940) [3] PASSMAN D.S., The algebrac structure of group rngs, Wley nterscence (1977). [4] PASSMAN D.S. What s a group rng? Amer. Math Monthly 83 (1976), [5] VASANTHA KANDASAMY, W.B. Smarandache Zero dvsors, (2001)

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