Kangwon-Kyungki Math. Jour. 11 (2003), No. 2, pp. 147 153 ON RIGHT(LEFT) DUO PO-SEMIGROUPS S. K. L and K. Y. Park Abstract. W invstigat som proprtis on right(rsp. lft) duo po-smigroups. 1. Introduction Khayopulu([6]) prov that vry idal of an N -class of an ordrd smigroup dos not contain propr prim idals. As a consqunc, ach prim idal of an ordrd smigroup is dcomposabl into its N - classs. In this papr, w giv th rlation btwn th lft(rsp. right) filtrs and th prim lft(rsp. right) idals. W dfin a smilattic congrunc N l (rsp. N r ) gnratd by th lft(rsp. right) filtr on a right(rsp. lft) duo po-smigroup and invstigat som proprtis on th right(rsp. lft) duo po-smigroups. Also w prov that vry lft(rsp. right) idal of N l -classrsp. N r -class) of a right(rsp. lft) duo po-smigroup dos not contain th propr prim lft(rsp. right) idals. As a consqunc, ach prim lft(rsp. right) idal of a right(rsp. lft) duo po-smigroup is dcomposabl into its N l - classs(rsp. N r -classs). A po-smigroup(: ordrd smigroup) is an ordrd st S at th sam tim a smigroup such that a b = xa xb and ax bx for all x S. Lt S b a po-smigroup. A nonmpty subst A of S is calld a lft(rsp. right) idal of S if (1) SA A(rsp. SA A), (2) a A and b a for b S = b A([3,5]). A is calld an idal of S if it is a lft and right idal of S. Rcivd July 14, 2003. 2000 Mathmatics Subjct Classification: 03G25, 06F35. Ky words and phrass: po-smigroup, idal, lft(right) idal, right(lft) duo, prim, prim lft(right) idal, lft(right) filtr, lft(right) congrunc, congrunc, smilattic congrunc.
148 S. K. L and K. Y. Park A po-smigroup S is said to b right(rsp. lft) duo if vry right(rsp. lft) idal is a lft(rsp. right) idal([4,5]). A non-mpty subst T of a po-smigroup S is said to b prim if AB T = A T or B T for substs A, B of S([8]). Equivalnt Dfinition: For lmnts a, b in a subst T ab T a T or b T. T is calld a prim lft(rsp. right) idal if T is prim as a lft(rsp. right) idal([2]). A non-mpty subsmigroup F of a po-smigroup S is calld a lft(rsp. right) filtr of S if (1) ab F for a, b S = b F (rsp. a F ), (2) a F, a c for c S = c F ([9]). A subsmigroup F of S is calld a filtr of S if F is a lft and right filtr ([2,4,5]). An quivalnc rlation σ on S is calld a lft congrunc(rsp. right congrunc) on S if (a, b) σ = (ac, bc) σ(rsp. (ca, cb) σ) for all c S. An quivalnc rlation σ on S is calld a congrunc it is a lft and right congrunc. A rlation σ is calld a smilattic congrunc on S if σ is a congrunc such that (x 2, x) σ and (xy, yx) σ([1,2,4]). Notation. For a smilattic congrunc σ, (z) σ is a class of th smilattic congrunc σ containing an lmnt z in a po-smigroup S. 2. Main Rsults. Lmma ([9]). Lt S b a po-smigroup and F a nonmpty subst of S. Th following ar quivalnt: 1) F is a lft(rsp. right) filtr of S. 2) S \ F = or S \ F is a prim lft(rsp. right) idal of S. From Lmma, w gt th following corollary. Corollary 1([2]). Lt S b a po-smigroup and F a nonmpty subst of S. Th following ar quivalnt: 1) F is a filtr of S. 2) S \ F = or S \ F is a prim idal of S. Proposition 1. A po-smigroup S dos not contain propr lft(rsp. right) filtrs if and only if S dos not contain propr prim lft(rsp. right) idals. Proof.. Assum that S contains a propr prim lft idal L of S. Thn = S \ L S. Sinc S \ (S \ L) = L, w not that S \ (S \ L)
On right(lft) duo po-smigroups 149 is a prim lft idal of S. By Lmma 1, S \ L is a propr lft filtr of S. It is impossibl. Hnc S dos not contain propr prim lft idals.. Suppos that F is a propr lft filtr of S. Thn S \ F. By Lmma 1, S \ F is a propr prim lft idal of S. It is impossibl. Hnc S dos not contain propr prim lft filtrs. By Proposition 1, w hav th following corollary. Corollary 2([6, Rmark 2]). A po-smigroup S dos not contain propr filtrs if and only if S dos not contain propr prim idals. Now w dfin a rlation N l on a po-smigroup S as follows: N l := {(x, y) N l (x) = N l (y)}, N r := {(x, y) N r (x) = N r (y)} whr N l (x) (rsp. N r (x)) is th lft (rsp. right) filtr of S gnratd by x S. Proposition 2. N l (rsp. N r ) is a smilattic congrunc on a right(rsp. lft) duo po-smigroup S. Proof. It is asy to chck that N l is an guivalnc rlation on S. Lt (x, y) N l. Thn N l (x) = N l (y). Sinc xz N l (xz) for all z S and N l (xz) is a lft filtr, w gt x N l (xz) and z N l (xz). Thus N l (x) N l (xz) and so y N l (y) = N l (x) N l (xz). Sinc y, z N l (xz) and N l (xz) is a subsmigroup of S, w gt yz N l (xz). Thrfor N l (yz) N l (xz). By symmtry, w gt N l (xz) N l (yz). Hnc N l (xz) = N l (yz). Thrfor N l is a right congrunc. Now w shall show that (x 2, x) N l. Lt x S. Sinc x 2 N l (x 2 ) and N l (x 2 ) is a lft filtr, w gt x N l (x 2 ). Thus N l (x) N l (x 2 ). Sinc x N l (x) and N l (x) is a subsmigroup of S, w gt x 2 N l (x). Hnc N l (x 2 ) N l (x). Thrfor N l (x 2 ) = N l (x), and so (x 2, x) N l. Nxt w shall show that (xy, yx) N l. Lt x, y S. Sinc xy N l (xy) and N l (xy) is a lft filtr, w hav x N l (xy). Suppos that y / N l (xy). Thn y S \ N l (xy). Sinc S \ N l (xy) is a prim right idal and S is a right duo, xy S(S \ N l (xy)) S \ N l (xy). It is impossibl. Thus y N l (xy). Sinc N l (xy) is a filtr, yx N l (xy). Thus N l (yx) N l (xy). By symmtry, N l (xy) N l (yx). Thrfor N l (xy) = N l (yx) and so (xy, yx) N l.
150 S. K. L and K. Y. Park Finally, w shall show that N l is a lft congrunc. Lt (x, y) N l, and z S. Thn N l (zx) = N l (xz) = N l (yz) = N l (zy). Thrfor N l is a lft congrunc. It follows that N l is a smilattic congrunc. Proposition 3. Lt S b a po-smigroup. If F is a lft filtr of S and F (z) Nl for z S, thn (z) Nl F. Proof. Assum that F is a lft filtr of S and a F (z) Nl for z S. If y (z) Nl thn (y) Nl = (z) Nl = (a) Nl. Thus (y, a) N l, and so N l (y) = N l (a). Sinc F is a lft filtr of S and a F, w hav N l (a) F. Thus y N l (y) = N l (a) F. Hnc (z) Nl F. Proposition 4. and (a, ab) N r. For a po-smigroup S, a b implis (a, ba) N l Proof. Suppos that a b. Sinc a N l (a) and N l (a) is a lft filtr, w gt b N l (a). Thus ba N l (a), and so N l (ba) N l (a). Sinc ba N l (ba) and N l (ba) is a lft filtr, w hav a N l (ba). Thus N l (a) N l (ba). Hnc N l (a) = N l (ba), and so (a, ba) N l. By symmtry, w can prov that (a, ab) N r. Proposition 5. Lt S b a right duo po-smigroup. If L is a lft idal of (z) Nl for z S thn L dos not contain propr prim lft idals. Proof. From Proposition 1, it is sufficint to prov that L dos not contain propr lft filtrs (of L). Lt F b a lft filtr of L and a F. Now w dfin T := {x S a 2 x F }. Thn T is a nonmpty st, sinc a 2 a = a 3 F. Now w show that F = T L. If y F, thn a 2 y F. Thus y T. Sinc F is a lft filtr of L, F L. Hnc y T L, and so F T L. Convrsly, if y T L, thn a 2 y F. Sinc F is a lft filtr of L, w gt y F. Thrfor F = T L. Nxt w show that T is a lft filtr of L. If x T and y T, thn a 2 x, a 2 y F. Sinc F is a lft filtr, w hav x, y F. Sinc a F, a 2 xy F. Thus xy T. If xy T for x, y L, thn (a 2 x)y = a 2 (xy) F. Sinc F is a lft filtr of L, w gt y F. If x T and x y for y L, thn a 2 x F. Sinc x y, w gt a 2 x a 2 y. Sinc F is a lft filtr, a 2 y F. Thus y T. Thrfor T is a lft filtr of L.
On right(lft) duo po-smigroups 151 W not that a F = T L L (z) Nl, and so T (z) Nl. Sinc T is a lft filtr of L, w hav (z) Nl T by Proposition 3. Thus L = (z) Nl L T L = F L, and so F = L. Hnc L dos not contain propr lft filtrs (of L). Thrfor by Proposition 1, L dos not contain propr prim right idals. Proposition 6. Lt S b a right duo po-smigroup and L a prim lft idal of S. Thn L = {(x) Nl x L}. Proof. Lt t (x) Nl for som x L. Sinc (x) Nl is a lft idal of (x) Nl, (x) Nl dos not contain propr prim lft idals by Proposition 5. If w prov that (x) Nl L is a prim lft idal of (x) Nl thn (x) Nl L = (x) Nl. W first show that (x) Nl L is a lft idal of (x) Nl. W not that (x) Nl L sinc x (x) Nl L. And (x) Nl ((x) Nl L) = (x) 2 Nl (x) Nl L (x) Nl SL (x) Nl L. Lt a (x) Nl L and b a for b (x) Nl. Sinc L is a lft idal of S, b is containd in L. Thus b (x) Nl L. Hnc (x) Nl L is a lft idal of (x) Nl. Finally, w show that (x) Nl L is prim in (x) Nl. Lt yz (x) Nl L for y, z (x) Nl. Sinc yz L and L is a prim lft idal of S, y is containd in L or z is containd in L. Hnc y (x) Nl L or z (x) Nl L. Thrfor (x) Nl L is a prim lft idal of (x) Nl. It follows that L {(x) Nl x L = {(x) Nl L x L} L. Thrfor L = {(x) Nl x L}. B. y similar mthods of Proposition 3, 5 and 6, w hav th followings: (1) If F is a right filtr of a po-smigroup S and F (z) Nr for z S, thn (z) Nr F. (2) If R is a right idal of (z) Nr of lft duo po-smigroups thn R dos not contain propr prim right idals. (3) If R is a prim right idal of lft duo po-smigroups, thn R = {(x) Nr x R}.
152 S. K. L and K. Y. Park 3. Exampls Now w giv an xampl of a lft filtr which is not a right filtr in po-smigroups and an xampl of a lft and right filtr in a posmigroup. Exampl 1([7]). Lt S := {a, b, c, d,, f} b a po-smigroup with Cayly tabl and Hass diagram on S as follows: a b c d f a b c d f b c d d d d c d d d d d d d d d d d d d d d d d f f f f f f f. d a b c Th st A := {, f} is a lft filtr, but not a right filtr of S. Thus A is not a filtr of S. Exampl 2([8]). Lt S := {a, b, c, d,, f} b a po-smigroup with Cayly tabl (Tabl 2) and Hass diagram (Figur 2) on S as follows: a b c d f a b c d f a b b d f b b b b b b b b b b b b d b b d f f f f f f f f f f d f. a b c Tabl 2 Figur 2 Th st B := {a, d, } is a lft and right filtr of S, and so B is a filtr of S.
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