Internatonal Electronc Journal of Algebra Volume 3 2008 7-24 P.P. PROPERTIES OF GROUP RINGS Lbo Zan and Janlong Chen Receved: May 2007; Revsed: 24 October 2007 Communcated by John Clark Abstract. A rng s called left p.p. f the left annhlator of each element of R s generated by an dempotent. We prove that for a rng R and a group G, f the group rng RG s left p.p. then so s RH for every subgroup H of G; f n addton G s fnte then R. Counterexamples are gven to answer the queston whether the group rng RG s left p.p. f R s left p.p. and G s a fnte group wth R. Let G be a group actng on R as automorphsms. Some suffcent condtons are gven for the fxed rng R G to be left p.p. Mathematcs Subject Classfcaton 2000: 6D50, 6P70 Keywords: p.p. rng, Baer rng, group rng. Introducton Throughout ths paper all rngs are assocatve wth dentty. A rng R s called Baer f the left annhlator of every nonempty subset of R s generated by an dempotent. The concept of a Baer rng was ntroduced by Kaplansky to abstract propertes of rngs of operators on a Hlbert space n hs 965 book [9]. The defnton of Baer s ndeed left-rght symmetrc by [9]. Closely related to Baer rngs are p.p. rngs. A rng R s called a left p.p. rng f each prncpal left deal of R s projectve, or equvalently, f the left annhlator of each element of R s generated by an dempotent. Smlarly, rght p.p. rngs can be defned. A rng s called a p.p. rng f t s both a left and a rght p.p. rng. The concept of a p.p. rng s not left-rght symmetrc by Chase [2]. A left p.p. rng R s Baer so p.p. when R s orthogonally fnte by Small [] and a left p.p. rng s p.p. when R s Abelan by Endo [5]. For more detals on left p.p. rngs, see [3,7,8]. Baer rngs are clearly p.p. rngs, and von Neumann regular rngs are p.p. rngs by Goodearl [6]. The second author was supported by the Natonal Natural Scence Foundaton of Chna No.057026, the Natural Scence Foundaton of Jangsu Provnce No.BK2005207, and the Specalzed Research Fund for the Doctoral Program of Hgher Educaton 20060286006.
8 LIBO ZAN AND JIANLONG CHEN Gven a rng R and a group G, we wll denote the group rng of G over R by RG. Wrte R G for the augmentaton deal of RG generated by { g : g G}. If H s a fnte subgroup of G, we let Ĥ = h H h. If g G has fnte order, we defne ĝ = Ĥ where H = g. We wrte C n for the cyclc group of order n, Z for the rng of ntegers and Z n for the rng of ntegers modulo n. As usual, Q s the feld of ratonals and C s the feld of complex numbers. The magnary unt s denoted by. For a subset X of R, l R X denotes the left annhlator of X n R. In [3], Z. Y and Q. Y. Zhou studed Baer propertes of group rngs. Motvated by them, we dscuss the p.p. propertes of group rngs. Some methods and proofs are smlar to those n [3].. Necessary Condtons Theorem.. Let R be a subrng of a rng S both wth the same dentty. Suppose that S s a free left R-module wth a bass G such that G and ag = ga for all a R and all g G. If S s left p.p., then so s R. Proof. For a R, snce S s left p.p., l S a = Se where e 2 = e S. Wrte e = e 0 g 0 + + e n g n where g 0 =, g G are dstnct and e R. Then 0 = ea = e 0 g 0 + + e n g n a = e 0 ag 0 + + e n ag n, and so e a = 0 for = 0,..., n. Thus e l S a = Se, mplyng that e = e e. Then e 0 g 0 = e 0 = e 0 e = e 0 e 0 g 0 + + e n g n = e 2 0g 0 + e 0 e g + + e 0 e n g n, whence e 0 = e 2 0 R. Because e 0 a = 0, we have Re 0 l R a. For r l R a l S a = Se, we have r = re = re 0 g 0 + + e n g n = re 0 g 0 + + re n g n. So r = re 0 Re 0. Hence l R a = Re 0 and R s left p.p. Corollary.2. Let R be a rng and G be a group. If RG s left p.p., then so s R. Proof. Note that S = RG = g G Rg s a free left R-module wth a bass G satsfyng the assumptons of Theorem.. Corollary.3. If R[x] or R[x, x ] s left p.p., then so s R. Proof. Note that R[x] and R[x, x ] are free R-modules wth bases {x : = 0,,...} and {x : = 0, ±,...} satsfyng the assumptons of Theorem.. Corollary.4. If R[x]/x n + a x n + + a n s left p.p., where a,, a n R and n s a postve nteger, then R s left p.p. Proof. Note that S = R[x]/x n + a x n +... + a n = n =0 Rx s a free left R- module wth a bass {, x,..., x n } satsfyng the assumptons of Theorem..
P.P. PROPERTIES OF GROUP RINGS 9 Theorem.5. If RG s left p.p., then so s RH for every subgroup H of G. Proof. For x RH, because RG s left p.p. and RH RG, we have l RG x = RGe, where e 2 = e RG. Wrte e = h H a hh + g / H b gg. Then 0 = ex = h H a hhx + g / H b ggx. Note that f h H and g / H then hg / H. Ths shows that the support of g / H b ggx s contaned n G\H. So by the above equalty that α := h H a hh l RH x l RG x = RGe, and hence h H a hh = h H a hhe = h H a hh 2 + h H a hh g / H b gg. Therefore, α 2 = α and RHα l RH x. If y l RH x, then yx = 0. So y = ye = y h H a hh + y g / H b gg, showng that y = y h H a hh = yα. Hence RHα = l RH x and RH s left p.p. Theorem.6. If G s a fnte group and RG s left p.p., then R. Proof. It s well-known that l RG Ĝ = RG. Snce RG s left p.p., we have R G = l RG Ĝ = RGe where e2 = e RG. Then R G s a drect summand of RG. By [0, Lemma 3.4.6], s nvertble n R. Example.7. ZG s not left p.p. for any nontrval fnte group G. Example.8. Let G be a fnte group and n be an nteger wth n >. Then the followng are equvalent: Z n G s Baer; Z n G s left p.p.; gcdn, = and n s square-free. Proof. clearly mples. Suppose that holds. Wrte n = p s ps k k and s > 0. Then Z n = Zp s Z s p k, and Z n G k = Z s p where all p are prme numbers G Z s p k G. It follows k s left p.p. and p s by Theorem.6. s left p.p. then s =. from that each Z s p G s p.p. So Z s p Clam. If Z p s Proof. Assume that s >. Snce Z s p e 2 = e Z s p. Because Z s p l Zp s s left p.p., l Zp s p = Z s p e, where p = 0 or s local, ether e = 0 or e =. Then l Zp s, a contradcton. Thus s = and p. Hence holds. p = Z s p If holds, then Z n G s a semsmple rng by Maschke s Theorem, hence holds.
20 LIBO ZAN AND JIANLONG CHEN Proposton.9. Let R be a von Neumann regular rng and G be a locally fnte group. Then the followng are equvalent: RG s left p.p.; the order of every fnte subgroup of G s a unt n R. Proof. Suppose that holds. Snce RG s left p.p., by Theorem.5 we have RH s left p.p for every fnte subgroup H of G. So we have H R by Theorem.6. Hence holds. Suppose holds. By [], RG s von Neumann regular, so RG s left p.p. In the followng, S 3 denotes the symmetrc group of order 6. Lemma.0. [4, Lemma 4.7 ] If 6 R, then RS 3 = R R M2 R. By [8, Proposton 9], f R s a left p.p. rng then so s ere for e 2 = e R. Thus f M 2 R s left p.p. then R s left p.p. So we have Corollary.. If 6 R, then RS 3 s left p.p. f and only f M 2 R s left p.p. 2. Group Rngs of Fnte Cyclc Groups Let R be a rng and G be a fnte group. If the group rng RG s left p.p. then R s left p.p. and R by Corollary.2 and Theorem.6. Thus t s natural to ask whether the converse holds. In ths secton, counterexamples to ths queston are gven. Proposton 2.. RC 2 s left p.p. f and only f R s left p.p. and 2 R. Proof. By [3, Lemma 2.], f 2 R then RC 2 = R R. Thus the result follows from Corollary.2 and Theorem.6. Proposton 2.2. RC 4 s left p.p. f and only f R[x]/x 2 + s left p.p. and 2 R. Proof. By [3, Lemma 2.3], f 2 R then RC 4 = R R R[x]/x 2 +. Thus the result follows from Corollary.4 and Theorem.6. Proposton 2.3. If R C, then RC 3 s left p.p. f and only f R[x]/x 2 + x + s left p.p. and 3 R. Proof. By [3, Lemma 2.5], f R C and 3 R then RC 3 = R R[x]/x 2 + x +. Thus the result follows from Corollary.4 and Theorem.6.
P.P. PROPERTIES OF GROUP RINGS 2 The proof of the next theorem s smlar to that of [3, Theorem 2.6]. Theorem 2.4. Let R be a subrng of C and let QR denote the quotent feld of R. Consder the polynomal x 2 + a x + a 2 R[x] wth a 2 4a 2 0. Let α be a soluton of x 2 + a x + a 2 = 0 n C. Then R[x]/x 2 + a x + a 2 s left p.p. f and only f ether α R or Rα R = 0.e., α / QR. Proof. Let T denote the rng R[x]/x 2 +a x+a 2 and x 2 +a x+a 2 = x αx β where α, β C. By hypothess, α β. Frst suppose α / QR. Then T s a doman. In partcular T s p.p. Next suppose α QR. Then β QR. Defne the map ϕ : R[x] QR QR by ϕfx = fα, fβ. Then the kernel of ϕ s x 2 + a x + a 2. Hence T can be regarded as a subrng of QR QR. It s clear that T s not a doman. Clam. T s left p.p. f and only f T contans the dempotent 0, QR QR. Proof. Snce T s not a doman, f T s left p.p. then T contans the nontrval dempotents of QR QR. The nontrval dempotents of QR QR are exactly, 0 and 0,. So 0, T. Assume 0, T. Then, 0 T. Consder any 0, 0 a, b T, where a, b QR. If a 0, b 0, l T a, b = 0; f a = 0, b 0, l T a, b = T, 0; f a 0, b = 0, l T a, b = T 0,. So T s left p.p. Moreover, 0, T f and only f there exsts ax+b R[x] such that aα+b = 0 and aβ + b =. Snce x 2 + a x + a 2 = x αx β, we have that a a b = [ α + βa ]b = [ 2b ]b = 2bb = 2 aα aβ = 2a 2 a 2. Hence b = aa b 2aa 2. So α = b a R. Example 2.5. Let R 0 = {n/2 k : n, k Z, k 0}. Then R 0 s a subrng of Q. Set R = {a + pb : a, b R 0 } where p > 2, p s a prme. Then R s a subrng of C wth 2 R. Because R s a doman, t s certanly p.p. Clearly / R. Moreover, for r = p and s = p, we have s = p R R. So, by Theorem 2.4, R[x]/x 2 + s not left p.p. Hence RC 4 s not left p.p. by Proposton 2.2. Example 2.6. [3, Example 2.8] Let R 0 = {n/3 k : n, k Z, k 0}. Then R 0 s a subrng of Q. Set R = {a + 3b : a, b R 0 }.
22 LIBO ZAN AND JIANLONG CHEN Then R s a subrng of C wth 3 R. Because R s a doman, t s certanly p.p. Clearly α = + 3 2 / R. Let r = 2 3, s = 3 + 3. Then s = rα Rα R. Hence RC 3 s not left p.p. by Proposton 2.3 and Theorem 2.4. 3. Fxed Rngs Let G be a group actng on R as automorphsms and let R G be the fxed rng of G actng on R. Here we study the condtons under whch R G becomes left p.p. Theorem 3.. Let R be a rng and G be a group actng on R as automorphsms such that ether ee g = e g e for all e 2 = e R and all g G or G s fnte wth R. If R s left p.p., so s R G. Proof. For any a R G, snce R s left p.p., we have l R a = Re where e 2 = e R. For g G, It follows that Re g = R g e g = Re g = l R a g = l R ga g = l R a = Re. e g = e g e and e = ee g for all g G. 3. Suppose that holds. It follows that e = e g for all g G, so e R G. Snce ea = 0, we have that R G e l R Ga. For r l R Ga, we have ra = 0, so r l R a = Re. Thus r = re R G e. Hence l R Ga = R G e. Suppose that holds. Let f = g G eg. Note that, for all g, h G, 3. mples e h e g = e h ee g = e h ee g = e h e = e h. Ths shows that f 2 = h G eh = 2 h G g G eg g G eh = = h G g G 2 eh e g h G eh = f. g G eg = Moreover, f g = f for all g G. So f R G. Because ea = 0 and f = g G eg e Re by 3., we have R G f l R Ga. Note that l R Ga l R a = Re g for all g G. Thus, for r l R Ga, r = re g for all g G. Hence r = r = p.p. g G reg = rf R G f, so l R Ga = R G f. Therefore, R G s left The assumptons and n the prevous theorem are necessary by the next example. Example 3.2. [2, Example 6.4] Let K be a feld wth chark = p > 0. Let R = M 2 K and G = g where g : R R, r u ru, wth u =. 0 Then R s left p.p. smple Artnan ndeed. Drect calculatons show that R G =
P.P. PROPERTIES OF GROUP RINGS 23 { } { } a b 0 b 0 : a, b K. So JR G = : b K. If x =, 0 a 0 0 0 0 then l R Gx = JR G. Because JR G can not be generated by an dempotent, R G 0 0 0 s not left p.p. If e = R, then e 2 = e and e g =. It s clear 0 0 that ee g = e e g = e g e. Moreover, = p s zero n R. The next example shows that R beng left p.p. s not necessary for R G to be left p.p. Example { 3.3. [3, Example } 3.3] Let K be a feld wth 2 K and R be the rng a b a b a b : a, b K. Let g : R R be gven by, 0 a 0 a 0 a { } a 0 and G = g. Then R G = : a K = K. So R G s p.p., but R s not 0 a left p.p. by Example 3.2. References [] M. Auslander, On regular group rngs, Proc. Amer. Math. Soc. 8957, 658 664. [2] S. U. Chase, A generalzaton of the rng of trangular matrces, Nagoya Math. J. 8 96 3 25. [3] A. W. Chatters and W. M. Xue, On rght duo p.p. rngs, Glasgow Math. J. 32 990 22 225. [4] J. L. Chen, Y. L. L and Y. Q. Zhou, Morphc group rngs, J. Pure Appl. Alg., 205 2006, 62 639. [5] S. Endo, Note on p.p. rngs, Nagoya Math. J. 7 960 67 70. [6] K. R. Goodearl, Von Neumann Regular Rngs, Ptman, London, 979. [7] C. Y. Hong, N. K. Km, and T. K. Kwak, Ore extensons of Baer and p.p. rngs, J. Pure Appl. Alg., 5 2000 25 226. [8] C. Huh, H. K. Km, and Y. Lee, p.p. rngs and generalzed p.p. rngs, J. Pure Appl. Alg., 67 2002 37 52. [9] I. Kaplansky, Rngs of Operators, Benjamn, New York, 965. [0] C. P. Mles and S. K. Sehgal, An Introducton to Group Rngs,Kluwer Academc Publshers, Dordrecht, 2002. [] L. W. Small, Semheredtary rngs, Bull. Amer. Math. Soc. 73 967 656 658. [2] Z. Y, Homologcal dmenson of skew group rngs and crossed products, J. Algebra, 64 994, 0 23.
24 LIBO ZAN AND JIANLONG CHEN [3] Z. Y and Y. Q. Zhou, Baer and quas-baer propertes of group rngs, J. Austral. Math. Soc., n press. Lbo Zan College of Math & Physcs Nanjng Unversty of Informaton Scence & Technology, Nanjng, Chna E-mal: zanlbo@yahoo.com.cn Janlong Chen Department of Mathematcs Southeast Unversty, Nanjng, Chna E-mal: jlchen@seu.edu.cn