P.P. PROPERTIES OF GROUP RINGS. Libo Zan and Jianlong Chen

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1 Internatonal Electronc Journal of Algebra Volume 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 , the Natural Scence Foundaton of Jangsu Provnce No.BK , and the Specalzed Research Fund for the Doctoral Program of Hgher Educaton

2 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 e n g n where g 0 =, g G are dstnct and e R. Then 0 = ea = e 0 g e n g n a = e 0 ag 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 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 e n g n = re 0 g 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..

3 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.

4 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.

5 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 }.

6 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 = / R. Let r = 2 3, s = Then s = rα Rα R. Hence RC 3 s not left p.p. by Proposton 2.3 and Theorem 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 =

7 P.P. PROPERTIES OF GROUP RINGS 23 { } { } a b 0 b 0 : a, b K. So JR G = : b K. If x =, 0 a then l R Gx = JR G. Because JR G can not be generated by an dempotent, R G 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, [2] S. U. Chase, A generalzaton of the rng of trangular matrces, Nagoya Math. J [3] A. W. Chatters and W. M. Xue, On rght duo p.p. rngs, Glasgow Math. J [4] J. L. Chen, Y. L. L and Y. Q. Zhou, Morphc group rngs, J. Pure Appl. Alg., , [5] S. Endo, Note on p.p. rngs, Nagoya Math. J [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., [8] C. Huh, H. K. Km, and Y. Lee, p.p. rngs and generalzed p.p. rngs, J. Pure Appl. Alg., [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, [] L. W. Small, Semheredtary rngs, Bull. Amer. Math. Soc [2] Z. Y, Homologcal dmenson of skew group rngs and crossed products, J. Algebra, , 0 23.

8 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