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# Algebra III MATH 7753

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Math 753 Algebra lll Christopher Drupieski Fall 2005 Contents Introduction ii 1 WedderburniArtin Theory 1 11 Basic Terminology and Examples 1 Exercises for 1 6 12 Semisirnplicity 9 Exercises for 2 12 13 Structure of Semisirnple Rings 12 Exercises for 3 17 2 Jacobson Radical Theory 20 24 The Jacobson Radical 20 Exercises for 4 26 25 Jacobson Radical Under Change of Rings 28 Exercises for 5 30 26 Group Rings and the J Sernisirnplicity Problem 31 Exercises for 6 32 3 Introduction to Representation Theory 35 37 Modules over Finite Dimensional Algebras 35 Exercises for 7 41 38 Representations of Groups 43 Exercises for 8 56 7 Local Rings Semilocal Rings and Idempotents 58 719 Local Rings 58 721 The Theory of Idempotents 60 Introduction This document represents my notes from Mr Wangfs Fall 2005 section of Math 753 Algebra Ill The chapter and section numbers of this document have been set up to correspond to the chapter and section numbers of the textbook we used7 TY Lam7s A First Course in Noncommutatz39ve Rings In places where my notes were ambiguous or incomplete7 l have relied on Lam7s book to complete this document lncluded also in this document are my solutions to various exercises from Lam7s book Chapter 1 Wedderburn Artin Theory 11 Basic Terminology and Examples We assume that all rings are unital7 though not necessarily commutative We assume that subrings share the same identity element as their parent ring7 and that all ring homomor phisms R a S necessarily map 1R gt gt 15 De nition An element a E R is a left zero divisor if 3 0 31 b E R such that ab 0 An element a E R is a right zero divisor if 3 0 31 b E R such that ba 0 De nition We call a ring reduced if it has no nonzero nilpotent ideals De nition An element a E R is right invertible if 3b 6 R such that ab 1 An element a E R is left invertible if 3 b E R such that ba 1 De nition Let k be a eld7 R klt1 7xmy17 yngt the polynomial ring in noncom muting determinates xhyi 1 S 239 S 71 Let F ziyj 7 yixj 7 617 We call RF Ank the n th Weyl Algebra De nition Given a ring k and a semi group G with multiplicative group operation7 de ne the group ring kG as a set to be the set of all k linear combinations of elements of G7 kG 960 k g with multiplication de ned by 2960 0499 Shea hh EM ag hgh Note that k k1G Q kG7 G 1k G Q kG7 k and G commute in the multiplication of kG7 and kG is commutative if and only if k and G are each commutative De nition Given a ring k de ne the ring of formal power series in commuting indetermi nates Z39E I by 2 6 1 F f0 f1 39 39 39 lfn homogeneous polynomial in of degree n De ne the ring of formal Laurent series in one indeterminate x by ajzj0ltNltoo 77N Note that U F f0 6 L1007 and U F E the lowest coef cient of F is in k De nition Given a polynomial ring and a ring endomorphism a k 7 k de ne the left skew polynomial ring Mam as a set to be the polynomial ring klz with new skew product determined by xi b Uibi for all b E k Remark H De ne right skew polynomial rings and rings of skew formal power series analogously D Given a polynomial ring and an endomorphism a E Endk7 the corresponding left and right skew polynomial rings are not necessarily identical eg7 in one ring z may be a left zero divisor but not a right zero divisor if a is not injective C40 lf 0 is injective and k is a domain7 then Ma 0 is a domain and U klzgal r If a E Autk7 we can de ne the ring of skew formal Laurent series kza7 and U k0 U WW De nition Call a map 6 k 7 k a derivation of k if 6a b 6a 6b and 6ab a6b 6ab for all ab 6 k De nition Given a derivation 6 of the ring k de ne the differential polynomial ring Ma 6 as a set to be the polynomial ring klz with product z a ax 6a for all a E k Remark Let 6 be an inner derivation of the ring k say 6a ca ca 7 ac Then in klzg l x 7 ca ax 6a 7 ca ax ca 7 ac 7 ca aa 7 ac az 7 c7 ie7 a and x 7 c commute Now p Magi 7 Mt mapping x 7 c gt gt t and a gt gt a for a E k is a ring isomorphism Example Let k0 be a eld7 k kdy and 6 Then ax za if a E k and xy ya 1 We have kolyllz 6 g k0ltx7ygtzy 7 ya 7 1 the Weyl algebra A1k0 De nition Let RS be rings7 M an R7 S bimodule We de ne the triangular ring A R 69 M 69 S as the set of all matrices of the form RM 7 m Ailt0 S i0 8 r Rs SmEM with the usual matrix operations Lemma In the triangular ring A R 69 M 69 S 1 R is a left ideal 2 S is a right ideal 3 M is an ideal with M2 0 4 R 69 M and M 69 S are ideals Proposition Let A be the triangular ring A R 69 M 69 S 1 The left ideals of A are of the form I1 69 I27 where I2 Q S is a left ideal of S and I1 is a left R subrnodule of R 69 M such that MI2 Q I1 2 The right ideals of A are of the form J1 69 J2 where J1 Q R is a right ideal and J2 is a right S subrnodule of M 69 S such that J1M Q J2 3 The ideals of A are of the form K1 BKO EB K27 where K1 31 R7 Kg 31 S7 and K0 is an R7 S sub birnodule of M such that K0 2 K1M MK2 Proof 1 If I1 and I2 satisfy the conditions of 17 then I1 69 I2 is a left ideal of A Conversely7 let I be a left ideal of A7 z TEn mE12 SE22 6 I Then E22 5E22 E I and Ens rEn mE12 6 J7 so I I1 EBIZ7 I2 I S Q S a left ideal of S7 I1 Q REBM REBM I aleft Rsubrnodule ofREBM And MI2 MI S Q I MQI R MI1 E0 If J1 and J2 satisfy the conditions of 27 then J1 69 J2 is a right ideal of A Conversely7 let J be a right ideal of A7 z rEu mE12 SE22 6 J Then zEn rEn 6 J7 E22 mElz i SEgg E J SO J J1 J27 J1 Q R J1 ideal7 J2 M EDS J a right S subrnodule of M EDS And J1M J RM Q J MQJ M BSJ2 9 lf K K1 BKOEBKZ satis es the conditions of 37 then K is an ideal of A Conversely7 let K be an ideal of A7 z TE11 mE12 SE22 6 K Then E22 SE22 6 K7 zEn rEll 6 K7 hence also mE12 E K So K K1 EBKO BKZ7 where K1 K R lR K2 K S S7 and K0 K M Since M and K are ideals ofA7 K1MMK2QK MK0 El Theorem Let A REBM BS be a triangular ring Then A is left resp right noetherian iff RS are left resp right noetherian and RM resp M5 is noetherian Proof We prove the left noetherian case i By the lernrna7 R E AM EB S7 S E AREB M7 so A noetherian implies that RS are left noetherian Let M1 Q M2 Q be a chain of R subrnodules of M Then M1 Q M2 Q as A subrnodules7 embedded in the northeast corner7 so it must stablize since A is left noetherian Hence stablizes Let I97 be an increasing chain of left ideals of A By the proposition7 I 10gt m R ea M ea 10gt m S Let If 10gt m R ea M 1 10gt m S Then 19 is an increasing chain of left ideals in S7 and I9 is an increasing chain of left R subrnodules of R 69 M7 both of which are left noetherian by assumption Hence these chains rnust stablize7 and we conclude that A is left noetherian De nition A left or right R module M is called artinian if the family of submodules of M satis es the descending chain condition7 that is7 if every descending chain of submodules stabilizes Example 1 Z is noetherian but not artinian 2 Let R be a noetherian domain7 r E R an element with no left or right inverses Then r 2 r2 2 r3 2 is an in nite descending chain of R submodules which does not stabilize Conclude R is not artinian Proposition Let N be an R submodule of M Then M is noetherian resp artinian if and only if N and MN are noetherian resp artinian Proof If M is artinian7 then N and MN are artinian since any descending chain of sub modules of N is also a descending chain of submodules of M7 and any descending chain of submodules in MN lifts to a descending chain of submodules in M containing N Conversely7 suppose N and MN are artinian7 and let M1 2 M2 2 M3 2 be a descending chain of submodules of M Since MN is artinian7 the descending chain M1 NN 2 M2 NN Q of submodules of MN stabilizes7 say at M NN The descending chain M1 N 2 M2 N 2 of submodules of N must also stabilize7 say at Mk N Let n maxj7k Given m E Mn7 m 6 Mn since MnNN Mn1 NN7 mim E N Mn But Mn NMnH N7 so mim E Mn N m m m7n 6 Mn Conclude Mn Q Mn ie7 Mn Mn and the chain stabilizes at M D Corollary Finite direct sums of artinian modules are artinian Proof lt suf ces to consider the direct sum of two artinian modules M7 N By the previous proposition7 M 69 N is artinian if and only if N and M ED NN E M are arinian7 both of which are true by assumption D De nition A composition series for a module M is a chain of submodules 0 M0 C M1 C M2 C C Mn M such that the quotients MiMi1 are simple for all 1 S 239 S 71 If M has a composition series7 we call M n the lengh of M7 and we call the simple quotients MiMi1 the composition factors of M Theorem JordaniHo39lder If M has a composition series7 then any two composition series for M have the same length7 and the composition factors of M are unique up to isomorphism and ordering Proof Let0 M0 C M1 C Mand0 No C N1 C Nk Mbetwocomposition series for M We argue by induction on minnk7 the case minnk 1 iff M is simple being clear Given a submodule S of M7 consider the inclusion S Mi1 gt MiMi1 Since the composition factors are by assumption simple7 the image of this map is either 0 01 all Of Then either S m S m Miil OI Miil g MiMi1 is simple Then 0 S M0 Q S M1 Q Q S Mn S M S becomes a composition series for S after deleting redundant terms as necessary 4 lf Mn1 Nk17 we can apply the induction hypothesis with S Mn1 Suppose Mn1 31 Nk1 Without loss of generality7 assume Nk1 g Mn1 Then M Mn1 Nk1 since Mn1 Mn1 Nk1 and MMn1 is simple Set S Mn1 Nk1 Then Mn1S E MNk1 is simple and Nk1S MMn1 is simple We now have S C Mn1 C Mn M and S C Nk1 C Nk M Moreover7 S has a composition series of length lS lt minnk7 so the induction hypothesis applies to S The result now follows for M by again applying the induction hypothesis D Proposition An R module M has a nite composition series if and only if M is both noetherian and artinian Proof Let M be an R module with nite composition series Argue that M is notherian and artinian by induction on lM7 the case M 1 ltgt M is simple being clear By assumption7 M has a composition series 0 M0 C M1 C C Mn1 C Mn M Since lMn1 n 7 1 lt 717 Mn1 is notherian and artinian by the induction hypothesis But MMn1 is simple7 hence noetherian and artinian7 which implies by previous results that M must be both noetherian and artinian Conversely7 let M 31 0 be an R module that is both noetheiran and artinian Since M is artinian7 it contains a minimal nonzero submodule M17 which is simple by minimality lf M1 M7 we are done Otherwise7 MM1 is artinian7 hence contains a minimal nonzero submodule lifting to a submodule M1 Q M2 of M7 and M2M1 is simple by minimality Since M is noetherian this process must eventually terminate7 and the resulting series will be by construction a composition series for M Remark If M is a nitely generated left module over a left noetherian resp artinian ring7 then M is noetherian resp artinian7 since R is notherian resp artinian if R is noetherian resp artinian7 and M is isomorphic to a quotient of R for some n E N Theorem Let A R 69 M 69 S be a triangular ring Then A is left resp right artinian iff RS are left resp right artinian and RM resp M5 is artinian Corollary Let S be a commutative noetherian domain7 R its fraction eld Assume S 31 R Then the triangular ring R R 69 R 69 S is left noetherian but not right noetherian7 and is neither left nor right artinian Proof Follows from the theorem and the fact that S is not artinian because by assumption it contains a nonzero non unit and R5 is not nitely generated it it were nitely generated7 we could nd a common denominator t E S for all fractions in R7 in which case 1t2 st 1t s E S i R S7 hence is not noetherian Corollary Let S Q R be elds7 ding 00 Then the triangular ring A R 69 R 69 S is left noetherian and left artinian7 but neither right noetherian nor right artinian Proof AA has a composition series7 0 C R C R 69 S C A7 so by the above proposition A is both left noetherian and left artinian But R5 is neither notherian nor artinian7 so by the above theorem A is niether right notherian nor right artinian D Example Let k be a division ring7 a an endomorphism of k that is injective but not surjective Then R Hz a is left noetherian but not right noetherian 5 Proof We have za oax for all a E k Let I be a left ideal of R Choose f E I monic of minimal degree Then I Rf So every left ideal of R is nitely generated hence R is left noetherian Let b E kok Claim 20 be is a direct sum of right ideals Suppose 0 snbzf x xnmbzfnm is a nontrivial dependence relation Since z is not a left zero divisor we have bzfnx 7 bfn1z smbxfnmx So bzfnx xgz 31 0 Comparing highest power terms mam Marx 0 boaTH oczTH 0 b oca1 E ok Then 20 11be is a direct sum of right ideals hence R is not right noetherian 1 Exercises for 1 114 True or false lf ab is a unit then ab are units 1 If a 6 Pi then a 6 Pi 2 If a is left invertible and not a right zero divisor then a 6 Pi 3 If R is a domain then R is Dedekind nite Proof False Let k be a eld V a k vector space with countable basis 6162 Let R EndkV Let f E R be determined by fe 611 and let 9 E R be determined by 961 0 96139 614 Then 9 P but 9 o f idV 6 Pi H The proof is routine E0 Suppose ba 1 Nowcaar 0for all 07 c E B so 0 a7aaba7a ab71a ab 1 Conclude a 6 Pi 9 Suppose ab 1 for some ab E R Then by b ba 1 Conclude R is Dedekind nite 1 116 Let ab be elements in a ring R If 1 7 ba is left invertible resp invertible show that 1 7 ob is left invertible resp invertible and construct a left inverse resp inverse for it explicitly Proof Suppose 1 7 ba is left invertible Then R1 7 ab 2 Rb1 7 ob R1 7 bab Rb In particular ab 6 R1 7 ob so 1 7 ob ab 1 E R1 7 ab and 1 7 ob is left invertible lf 1 7 ba is invertible then 1 7 abR Q 1 7 abaR a1 7 baR 1R In particular ab 6 17abR so 17abab 1 E 17abR and 17ab is right invertible hence invertible Suppose 1 7 ba has left inverse resp inverse u E R Then 1 aub1 7 ob 17abau17bab17abab1 D 117 Let B17Bn be left ideals resp ideals in a ring R Show R B1 69 69 En iff there exist idempotents resp central idempotents 61 6 with sum 13 such that eiej 0 for 2 79739 and B R6 for all 2 Suppose the Bs are ideals If R B1 69 69 En then each B is a ring with identity 6139 and R 2 B1 gtlt gtlt B Show that any isomorphism of R with a nite direct product of rings arises in this way Proof Suppose there exist idempotents 61 6 E R such that 1 61 6 eiej 0 for 279739 and B Rei for all 1 32 72 Thenfor all 6 R77 T61T6n E BNLunLBn7 so R Q B1 Bn Suppose 6161 anon 0 for some 017cn E R Then for each 1 S 2 S n7 0 6161 onenki cm Conclude Bl Bj 0 for 2 31 j hence RBl Bn Suppose R B169 Bn for left ideals B1 7B Then there exist 61 6 B1 7 on 6 En suchthat 1 616n Thenfor each 1 23 727 0 6176 ei61en761 606 21 616739 7 21 6739 6139 Since eiej E B and 21 6739 e 6 Bi conclude by the directness of the sum R B1 69 69 En that eiej 0 for all 2 31 j Now 6 6i61 on 6 ie7 each 6 is idempotent Let b 6 Bi Then bl bi61 on and 0 blel f b 21 biog Again conclude by the directness of the sum R B1 69 69 En that biej 0 for all 2 79739 and b biei E Roi Then Bl R6 for all 1 S 2 S n 1 119 Show that for any ring R7 the center of the matrix ring MAR consists ofthe diagonal matrices TI 7 E ZR 1112 A left R module is said to be hop an if any surjective R endomorphism of M is an automorphism 1 Show that any noetherian module M is hop an 2 Show that the left regular module RR is hop an iff R is Dedekind nite 3 Deduce that any left noetherian ring R is Dedekind nite Proof 1 Let f E EndRM be surjective Then fk M a M is also surjective for all k E N We have kerf Q ker f2 Q Say the chain stabilizes at ker fm Let g E ker f Then 3h 6 M such that g fmg Now 9 E kerf Q kerfm i 0 fmg f2mh i h E ker f2 ker f i g fmh 0 So ker f Conclude f is an automorphism E0 Suppose RR is hop an Let ab E R and suppose ab 1 Then the map p R a R given by s gt gt 5b is a surjective R endomorphism7 hence an automorphism by Now 27092171 ba71b bab 7b bib 0 Conclude ba 17 so R is Dedekind nite Suppose R id Dedekind nite Let p R a R be a surjective R endomorphism Let x E kerltp Since p is surjective7 3y 6 R such that y 1 Now 1 My yltp1 1y 1 since R is Dedekind nite Now z 21 ltpy xyltp1 xltp1y zw 0 Then ker p 0 and p is injective Conclude that p is an automorphism and RR is hop an 3 Let R be a noetherian ring Then R is noetherian as a module over itself i R is hop an by 1 i R is Dedekind nite by D 1113 Let A be an algebra over a eld k such that every element of A is algebraic over k 1 Show that A is Dedekind nite 2 Show that a left zero divisor of A is also a right zero divisor of A 3 Show that a non zero element of A is a unit iff it is not a zero divisor 4 Let B be a subalgebra of A7 and b E B Show I E B iff b E A 1115 Let A xgo where 0 denotes complex conjugation 1 Show ZA Rx2 2 Show Azz 1 g H 3 Show Az4 1 M2C 1116 Let K be a division ring with center k 1 Show that the center of the polynomial ring R is 2 For any a E Kk7 show that the ideal generated by z 7 a in is the unit ideal 3 Show that any ideal I Q R has the form Rh for some h E 1117 Let my be elements in a ring R such that Pm By Show that there exists a right R module isomorphism f zR 7 yR such that f y Proof Say y ax z by De ne f zR 7 yR by fzr yr This map is a well de ned right R module homomorphism7 since if x7 x5 then yr ax axs ys Similarly7 de ne g yR 7 zR by 9yr mquot This map is a well de ned right R module homomorphism7 since if yr go then x7 byr bys x5 Now fog 11R and gof 1213 Conclude that f is a right R module isomorphism 1119 Let R be a domain If R has a minimal left ideal7 show that R is a division ring In particular7 a left artinian domain must be a division ring Proof Suppose R has a nonzero minimal left ideal I Let 0 31 1 E I Then 0 31 12 Q a Q I7 in which case 12 a I by the minimality of I Then 37 E R such that a raz Then 0 a 7 raz a17 m But a 31 0 and R is a domain7 so 17 m 07 ie7 7711 ThenI1 R Nowgiven07r b R7 Q 1 I b 1 Iby minimality of I i b 6 Pi Conclude R is a division ring D 1120 Let E EndRM be the ring of endomorphisms of an R module M7 and let 71M denote the direct sum of 71 copies of M Show that EndRnM E Proof Say M is a right R module7 and write the endomorphisms on the left Let 6739 M a nM be the j th inclusion7 and m nM a M the 24th projection For any F E EndMnM7 let fij be the composition mFej E EndMnM De ne Oz EndMnM a by aF flj this map is an isomorphism D 1122 For any ring k let A and let R resp S denote the ring of n gtlt 71 upper resp lower triangular matrices over k 1 Show R S 2 Suppose k has an anti automorphism resp involution Show the same is true for A7 R and S 3 Under the assumption of 27 show that R7 S R0 7 8quot are all isomorphic 1126 For any right ideal in a ring R7 the idealizer of A is de ned to be llMA r E A TA Q A 1 Show llMA is the largest subring of R that contains A as an ideal 2 The ring EMA llMAA is known as the eigenring of the right ideal A Show EMA E EndMRA as rings 1127 Let R where k is a ring7 A the right ideal of R consisting of all matrices whose rst 7 rows are zero Compute llMA and EMA 12 Semisimplicity De nition An R module M is semisimple if for every submodule N of M7 there exists a submodule N of M such that M N 69 N Remark A simple R module is not necessarily semisimple Lemma Submodules and quotients of semisimple modules are semisimple Proof Let M be a semisimple R module7 N a submodule Let W be a submodule of N By semisimplicity of M7 M WEBW for some submodule W of M Then N W BN W A submodule of MN lifts to a submodule V containing N By semisimplicity of M7 M VEBV for some submodule V of M Then MN VNEB V NN D Lemma Any nonzero semisimple R module M contains a simple submodule Proof Let 0 31 m E M It suf ces to nd a simple submodule of Rm Let f be the collection of submodules of Rm not containing m By the usual Zorn7s Lemma argument7 f contains a maximal element N7 N g Rm Now M semisimple Rm semisimple i 3 N S Rm such that Rm NED N Now N must be simple If 0 31 N S N 7 then by maximality of N7 N N RmNEBN N N D Theorem The following are equivalent for a left R module RM 9 1 M is semisimple 2 M is a direct sum of simple modules 3 M is a sum of simple modules Proof 1 i 2 Let f S S M Sis a direct sum of simple modules Then f 31 Q by the previous lemma By the usual Zorn7s Lemma argument7 f contains a maximal element M S M Claim M M lf not7 M M 69 M for a nonzero semisimple submodule M Then HS 3 M simple7 so M M EBS E f 2 i 3 Clear 3 i 1 Take N S M Let M 2 Mi Ml simple Consider id f JCLZMi MiN Mi 0 ieJ ieJ ieJ Then f 31 Q if N 31 M7 so by Zorn7s Lemma7 f has a maximal element J Let M NEBGBiEJMi Claim M M If M M7 then HMZO g M because M Mic M 0 because Mic is simple7 M Mic N 69 NHO Mi so J U 2390 6 f a contradiction to the maximality of J D Theorem For a ring R7 the following are equivalent 1 All left R modules are semisimple 2 All nitely generated left R modules are semisimple 3 All cyclic left R modules are semisimple 4 RR is semisimple A ring with these properties is called a left semisimple ring Proof The implications 1 i 2 i 3 i 4 are clear For the implication 4 i 17 let M be an R module Trivially7 M ZmeM Rm Now Rm R Annm a quotient of the semisimple module RR Then for each m 6 M7 Rm and hence all of M is a sum of simple modules7 which implies that M is semisimple by the previous theorem Remark Let R be a semisimple ring Then RR 161 Ai Ai minimal left ideals7 and 1B 2161 almost all 11 04116 Ai Then 7 2161 ml for all r E R7 and 1 lt oo Corollary If R is semisimple7 then R has a composition series with composition factors Ai 239 E I In this case R is also left noetherian and left artinian De nition 1 A left R module P is called projective if for any exact sequence M L N a 0 of module homomorphisms and module homomorphism g P a N7 there exists a module homomorphism h P a M such that f o h g 10 2 A left R module I is called injective if for any exact sequence 0 a M L N of mod ule homomorphisms and module homomorphism g M a I7 there exists a module homomorphism h N a I such that h o f 9 Lemma For a left R module P7 the following are equivalent 1 P is projective 2 P g a direct summand of a free R module F 3 Every exact sequence of R modules M a P a 0 splits Proof a i 0 Suppose P is projective Then given idp P a P7 there exists h P a M such that f o h idp Then M L P a 0 splits c i b Let F pepR 10 the free R module with basis the elements of P Now F a P a 0 is exact7 so it splits by Then 0 holds b i a Let M L N a 0 be exact Assume F is free and F P 69 Q Given 9 P a N7 we can extend it to a map 9 F a N by mapping Q to zero Say F has generators Find ml 6 M such that gipossible since f is surjective De ne H F a M by mi h Hlp Then f oh g Conclude P is projective D Theorem For a ring R the following are equivalent 1 R is left semisimple 2 All left R modules are projective 3 All nitely generated left R modules are projective 4 All cylic left R modules are projective Proof The implications 2 i 3 i 4 are clear For the implication 1 i 27 if R is left semisimple7 then every exact sequence of left R modules M a P a 0 splits because P is a direct sum of simple PL submodules7 each of which must be either contained in or disjoint from the image of the map M a P7 which is equivalent to P being projective 4 i 1 Let I be a submodule of RR Now RI is cyclic7 so by 4 the exact sequence 0 a I a R a RI a 0 splits Then R E I BRI Conclude R is semisimple 1 Remark A left R module I is injective if and only if any exact sequence of left R modules 0 a I a M splits Exercises for 2 123 What are the semisimple Z modules A simple Z module is a simple abelian group7 and the simple abelian groups are precisely the cyclic groups of prime order Thus a semisimple Z module is a direct sum of cyclic groups of prime order 124 Let R be the commutative ring of all real valued continuous functions on 01 ls R a semisimple ring No For n 2 17 let In f ER f 0V E OJ71 Then 1 C 2 C 3 C is an in nite strictly increasing chain of ideals in R Conclude that R is not noetherian7 hence not semisimple since all semisimple rings are noetherian 126 Let R be a right semisimple ring For my 6 R7 show that Pm Ry if and only if z uy for some u 6 Pi Proof If x uy for some u 6 Pi then clearly Pm Ry Suppose Pm By By Exercise 11177 there exists a right R module homomorphism f xR a yR such that f y Since R is right semisimple7 we can write R zR 69 N yR 69 N for some complimentary right ideals NN of R Furthermore7 since zR g yR we must have N E N as right R submodules of RR Let A1 7A be a complete list of pairwise nonisomorphic right ideals of R Say RR 2 1niAi7 xR E ankAh 0 S m S m Then yR E wankAh and N E N E 17101 7 We can thus extend f to an automorphism of RR Now y f f1x7 and f1 6 P with inverse f 11 D 127 Show that for a semisimple module M over any ring R7 the following conditions are equivalent 1 M is nitely generated 2 M is noetherian 3 M is artinian 4 M is a nite direct sum of simple modules 13 Structure of Semisimple Rings Theorem 1 Any ideal I of MAR is of the form MnA for some ideal A 31 R 2 If R is simple7 then MAR is simple Proof Part a obviously implies part Given an ideal A 31 B it is clear that MnA is an ideal of Conversely let I 31 R and let A an calj E I It is easy to check that A 31 R Claim I First note that if Tab 6 MAR then EjrabEkl rjkEZl Now given Tab 6 I we have E1jTabEk1 TjkEll E I SO Tjk E A and I Q Let Tlj E Given 1 3 ab S 71 there exists M mkl E I with rub mu Then for such an M we get EalMElb rabEab E I Then rub ZrabEab E I ie MnA Q I Conclude MnA I D Theorem Let D be a division ring R Then 1 R is simple and left semisimple 2 R has a unique up to iso left simple module V R acts faithfully on V with RR 2 71V 3 EndRV E D as rings Proof H Simplicity follows from the previous theorem Semisimplicity follows from P Let V D the n dimensional right D vector space Note that R acts faithfully on V by left multiplication R g EndVD and V is a simple left R module Now RR A1 69 EBA where A denotes the set of matrices with a nonzero 24th column and zeros elsewhere Have that A V as R modules hence RR g nV a direct sum of simple submodules and R is left semisimple From the above decomposition of RR we have that all composition factors of R are isomorphic to the simple R module V Any simple left submodule of R is a composition factor of B so isomorphic to V 9 Claim The homomorphism A D a EndRV Ad right multiplication by d on V is an isomorphism This map is injective since D acts faithfully on V D Let f E EndRV Then 10 0tf dgtk gtkt for some d E D Then 11 a1 0 0 1 a1 0 0 d ald la l la 0 bl lbl la 0 bl lbl la dl So f acts as right scalar multipliation by d Conclude that A is an isomorphism D Lemma Given left semisimple rings R 1 S 239 S r R 1 1L R is a left semisimple ring Proof Viewing each R as an ideal in R a minimal left ideal of R is also a minimal left ideal of R Since each R is a direct sum of minimal left ideals we can write R as a direct sum of minimal left ideals Then R is semisimple D Lemma Schur7s Lemma Given a ring R and a simple left R module V EndRV is a division ring Proof Let 0 31 f E EndRV Now fV 31 0 is a nonzero R submodule of V7 and kerf 31 V is a proper R submodule of V By the simplicity of V7 we then have fV V and kerf 07 ie7 f is invertible 1 Theorem WedderburniArtin Let R be a left sernisirnple ring Then R E Mn1 D1 gtlt gtlt Mm D where the D are division rings7 m 6 N r E N is determined uniquely by R7 and m7 Di 1 S 2 S r is unique up to permutation ln particular7 R has exactly r pairwise nonisomorphic left simple R modules Proof A ring R of the above form is left sernisirnple by previous results Conversely7 let R be a left sernisirnple ring Write RR n1V1 EB EB nTVT for pairwise nonisomorphic left simple R modules Vl minimal left ideals Then V1L7 7 W is a complete set of left simple R modules by the JordaniHo39lder theorem By Schur7s Lemma D EndRVZ is a division ring7 and EndniVZ g by Exercise 1120 We have HomVi7 0 for 2 79739 since Vth are simple and V 5 for 27 j Then R E EndRR E MmD1 gtlt gtlt MmD Applying the JordaniHo39lder theorem to the decomposition RR 2 n1V1 B B727147 r and m7 1 S 2 S r is uniquely determined up to ordering7 and D EndVZ is determined by Vi so the uniqueness claim follows E Corollary A ring R is left sernisirnple if and only if it is right sernisirnple Proof This follows since a nite direct product of matrix rings over skew elds is simulta neously left sernisirnple and right sernisirnple Remark Let k be a eld7 R a nite dimensional k algebra Then R satis es the ascending and descending chain conditions on ideals because an ideal is a k vector subspace of bounded dimension If R is simple7 then R E If V denotes the unique left simple module of R7 then D EndRV is a k algebra If R g Mn1 D1 gtlt gtlt MmDT7 D division k algebras7 and if additionally P k ie7 k is algebraically closed7 then D g k for all 1 S 2 S r B is a nite dimensional k algebra7 hence all elements of D are algebraic over k P i D g Lemma Let R be a ring7 A a minimal left ideal of R7 BA the sum of all left ideals of R isomorphic to A isotypical component of R 1 BA is an ideal in R 2 If AA39 are nonisomorphic minimal left ideals of R7 then BABA Proof 1 Let I E A be a left ideal of R7 I Q BA Then Vr E R7 the map I a Ir7 2 a 2r is an R module homomorphism By the simplicity of I7 Ir 0 or Ir 2 I E A Either way7 I r Q BA 2 It suf ces to show AA lf AA 31 07 then Hr E A such that Ar 31 Consider the homomorphisms f A a Ar 7 g Ar gt A By the simplicity of AA and since Ar 31 07 fg are isomorphisms Then A 2 A7 Conclude AA D Assume R is left semisimple Write RR A1 69 69 A 69 where A1 AT is a complete set of pairwise nonisomorphic minimal left ideals7 and let Bl 8142 31 R Then RB1EB BT Proposition Let R be a left semisimple ring with notation as above Then each B1 is a simple left artinian ring Proof That each B1 is a ring and R 2 B1 gtlt gtlt B follows from Exercise 117 Each B1 is a quotient of the left artinian ring R7 hence is also left artinian Let 0 31 I 31 Bi Then I 31 R7 so I contains some minimal left ideal A7 isomorphic to some Ak Since I Q Bl k 239 A is a direct summand of B because R is semisimple Then by Exercise 1177 A R6 for some idempotent e E A Let A g Al be a minimal left ideal of R7 and let p A a A be an R module isomorphism Then A ltpA Ae Altpe Q I because I is an ideal7 and consequently Bl Q I Conclude Bl I D Theorem Let R be a simple ring Then the following are equivalent 1 R is left artinian 2 R is left semisimple 3 R has a minimal left ideal 4 R MnD for some n E N and some division ring D Proof 2 and 4 are equivalent by the WedderburniArtin Theorem 1 i 3 is immediate 3 i 2 Let A be a minimal left ideal of R7 and consider BA 31 Then BA R by the simplicity of R7 and RR BA g A 69 69 A7 A simple left R module Then R is left semisimple 2 i 1 A left semisimple ring has a composition series7 and we already proved that a left module has a composition series if and only if it is left artinian D Theorem Double Centralizer Property Let R be a simple ring7 0 31 A a left ideal of R7 D EndRA acting on A from the right Then the natural map L R a EndAD7 r gt gt LT left multiplication7 is an isomorphism Proof Let E EndAD L R a E is injective because R is simple Claim 1 For all h E E7 r 6 A7 h Lr Lhr E E lndeed7 given a 6 A7 h Lra hra hra since Ru 6 D Lhra7 ie7 h Lr Lhr Claim 2 LR is a left ideal of E AB is an ideal in R7 so AR R by the simplicity of R Then LR LAR LALR Now E LR E LALR g LALR by Claim 1 LR So LR is a left ideal of E The result now follows7 because 1 E LR LR E D 15 Example Nonartinian Simple Ring Given a chain of simple rings R1 Q R2 Q R3 Q sharing the same identity7 their union R RZ is also simple Let D be a division ring7 RZ M2 Then RZ Q R via the embedding M De ne el 6 RZ as the E11 matrix unit in RZ M2239D7 so under the embeddings RZ gt Rj j gt 239 61 will be a sum of diagonal matrix units Then elm1 6H1 but 61 Rel1 lf 2 E Rj then considering the images of 6136141 in Rj 26i1 will have smaller column rank than 6139 Then R60 2 R61 2 is an in nite strictly descending chain of left ideals in R7 and R is not artinian De ne fl 6 RZ by fl I1 761 Il the identity matrix in R Then fifZH fi fi1 Rfl7 and Rfo Q RfL Q ng Q is an in nite strictly increasing chain in R7 so R is not noetherian De nition Let k be a ring and let 6 E Derk An ideal A 31 K is called a 6 ideal if 6A Q A Call k 6 simple if it has no proper 6 ideals Theorem Let k e a Q algebra with derivation 6 Then the differential polynomial ring R Mob is simple if and only if k is 6 simple and 6 is not an inner derivation ie7 is not a commutator Proof i If k has a proper 6 ideal A7 let I llx 1139 E A Q R It is clear that I AR7 while llxi Edi Zach ll 6 I7 so I a proper ideal of R and R is not simple If 6 is an inner derivation7 R Md 6 g Mt is not simple7 because It lt1 R is a proper ideal It is equivalent to show that if k is 6 simple but R is not simple7 then 6 is an inner derivation Suppose k is 6 simple but R has a proper nonzero ideal 0 31 I lt1 B Let n mindegf 0 31 f E I7 and let A an E k anxn lower terms 6 I U lf andn 6 I7 then xanz 7anz x6anx E Ii an 6 A7 so A is a nonzero 6 ideal of k i A k by the 6 simplicity of k Then there exists 9 E I of degree n ofthe form 9 x dx 1 for some d E k Claim For b E k dnb bx m5bz 1 and if g x dx 1 as above7 by 7 gb bd 7 db 7 7n6bz 1 We have bd 7 db 7 715bz 1L by 7 gb E I By the minimality of 717 by 7 gb 0 So bd 7 db 7 n6b 0 If n 07 then 9 1 and I R7 in contradiction to the choice of I Then since Q Q k we can write 6b n 1bd 7 db7 ie7 6 7dn7 is an inner derivation Corollary Let k be a simple Q algebra Then R Md 6 is a nonartinian simple ring for any non inner derivation 6 Proof R is simple by the previous theorem7 and is nonartinian since Rd 2 Rzz Q Rzg Q is an in nite strictly descending chain in R 1 Corollary Let k0 be a simple Q algebra7 n 2 1 Then the Weyl Algebra Ank0 is nonar tinian simple Proof Since Ank0 A1An1k07 it suf ces to prove the case n 1 Recall A1k0 g k0y6 6 1 R A1k0 is nonartinian because Pa 2 Rzz Q is an in ntie strictly decreasing chain of ideals in R 6 is not an inner derivation of mm because 6y 1 and an inner derivation would map central elements to zero Let 0 31 A 31 mm be a 5 ideal7 and let f ay E A be a nonzero polynomial of minimal degree Then 6f nayn l E A Have a 31 0 by choice of f7 but 6f 0 by minimality of n Conclude n 0 and f a E A k0 Now 0 31 A kg 31 k0 By the simplicity of k0 A k0 k0 so 1 E A and A rob Conclude that mm is 6 simple D Exercises for 3 131 Show that if R is semisimple7 so is Proof Let J 31 Then J MnI for some uniquely determined ideal I 31 R Say R I EBI for another ideal I 31 R Then MnI 1 MAR7 and MAR MnI EB MnI Conclude MAR is semisimple D 133 Let R be a semisimple ring H Show that any ideal I 31 R is a sum of simple components of R E0 Show that any quotient ring of R is semisimple 9 Show that a simple artinian ring S is isomorphic to a simple component of R if and only if there exists a surjective ring homomorphism from R onto S Proof 1 Have R B1 69 69 B for some simple ideals B1BT Moreoever7 there exist central idempotents 61 6 E R such that 61 e 17 eiej 0 for 239 31 j and B Roi Let I lt1 R Given a 1161 ae 6 I7 have 16 16 E I So I 691 B I Now B simple implies B I 0 or Bi Conclude I is a direct sum of some subcollection of the simple components B1 7B of R 2 Let I 31 R and consider RI By 17 RI E B I 07 a direct sum of simple left ideals Then RI is semisimple 3 Suppose S E B Let I 691 Bi Then R a RI E B E S is a surjective ring homomorphism Conversely7 suppose there exists a surjective ring homomorphism f R a S for some simple artinian ring S Then S RI for some I 31 R Now S g a direct sum of some subcollection of the B1 7 ET But S simple implies that only one of the B appears in the direct sum 1 134 Show that the center of a simple ring is a eld7 and the center of a semisimple ring is a nite direct product of elds Proof Let R be a simple ring7 0 31 7 E Z0quot Then 0 31 7 31 R i r R i r 6 Pi So ZR is a commutative subring of R in which every nonzero element is invertible7 ie7 ZR is a eld Now if R is semisimple7 R 2 B1 gtlt gtlt B for some simple subrings B1 7 B7 Then ZR ZB1 gtlt gtlt ZBT7 a direct product of elds D 135 Let M be a nitely generated left R module7 E EndRM Show that if R is semisimple resp simple artinian7 then so is E Proof Suppose M is semisimple Let 1 S 2 S m be a complete list of pairwise noniso morphic simple left R submodules of M the list is nite because M is nitely generated as an R module and is a direct sum of simple submodules Let Bl denote the direct sum of all submodules of M isomorphic to Mi so M 69pr Now HomBiB 0 for 2 31 j for if 0 31 f E HomBlBj and 2 31 j then le 31 0 for some simple submodule N 2 Bi By the simplicity of N we have kerle 07 so M g N fN Q B ie7 Bj contains a simple submodule isomorphic to Bi 2 31 j a contradiction to the de nition of the Bi Conclude E EndRM End 6921Bi g EndBZ Write Bl niMZ for some m 6 N Then EndBZ E EndniMZ E MmEndRMZ But Di EndRMZ is a diVision ring by Schur7s Lemma Conclude E E is semisimple Suppose R is simple artinian Then R is semisimple7 so R has a unique up to iso morphism left simple module V Any simple submodule of M must be isomorphic to V Conclude that m 1 and M B1 2 n1M1 E an Then E MmD1 is simple ar tinian D 137 Let R be a simple ring which is nite dimensional over its center k k is a eld by Exercise 134 Let M be a nitely generated left R module and let E EndRM Show Proof Since R is nite dimensional over its center k it is simple artinian7 hence semisimple Let V denote the unique up to isomorphism simple left R module7 and let D EndRV a nite dimensional k diVision algebra by Schur7s Lemma Then R E MnD for some n E N and V E D So dimkR n2 dimk D Let M be a nitely generated left R module Then RM is semisimple7 and since R has a unique up to isomorphism simple left R module7 M g mV for some 772 E N Then dimk M m dimk V mn dimk D Now E EndRM g EndmV MmEndV Then dimk E m2 dimk D Conclude dimk M2 m2n2dimk D2 n2 dimk D m2 dimk D dimk R dimk D 1311 Let R be an nZ dimensional algebra over a eld k Show R E as k algebras if and only if R is simple and has an element whose minimal polynomial over k has the form xialzian for some a1an E k Proof Suppose R as k algebras Then R is simple because k hence is simple Let A E be the matrix with 1s on the rst superdiagonal and zeros elsewhere Then the isomorphic image of A in R has minimal polynomial over k equal to x x 7 0 18 Suppose R is simple and 3 r E R such that 7 has minimal polynomial over k equal to H1x 7 oi al 6 k Since R has the structure of a nite dimensional k vector space7 R is simple artinian Then R E MmD for some division k algebra D and for some m E N Let 6739 1r7ai Then R61 31 B because 61 is a left zero divisor7 hence not invertible in B By Exercise 1267 Re Reg1 if and only if 671 uej for some u 6 Pi But if this were the case7 we7d have 0 u 0 u 1110 7 11 Hi lr 7 oi a contradiction to the minimality of 1110 7 11 Then 0 Ben C Reykl C C R61 C R is a strictly increasing chain of left ideals in B This chain can be re ned to a composition series for R i R must have at least 71 simple left modules i m 2 71 But 712 dimkR m2 dimkD and m 2 n i m n and dimk D 17 ie7 D k Then R E 1 1312 For a subset S in a ring R7 de ne Annls a E R aS 07 and Anns a E R Sa 0 Let R be a semisimple ring7 I a left ideal in R7 J a right ideal in R Show Ann AnnTI I and Ann AnnlJ J Proof Write R I 69 I for some complimentary left ideal I of R Now I R6 for some idempotent e E I Let f 1 7 e Certainly fR Q AnnTI If a E AnnTI7 then a a7ea 17ea fa7 so a E fR Conclude fR AnnTI Now I Re Q AnnlAnnTI If a E AnnlAnnTI7 then a a 7 of a17 f 16 so a E R6 I Conclude Ann AnnTI I The proof of Ann AnnlJ J is argued similarly D 1319 True of false If I is a minimal left ideal in a ring R7 then MnI is a minimal left ideal in Proof False Let I C R be a minimal left ideal Then MnI is certainly a left ideal of Let A C MnI denote the set of matrices with entries equal to zero outside of the 24th column Then for 1 S 239 S 717 Al is a nonzero left ideal in MAR7 and MnI 691 Ai so MnI is not a minimal left ideal in D 1324 A subset S of a ring R is said to be m39l resp nilpotent if every 5 E S is nilpotent resp if S 0 for some m7 where S denotes the set of all products 51 5m with s E S H Let R MAD where D is a division ring Let S Q R be a nonempty nil set which is closed under multiplication Show that S P Let R be any semisimple ring Show that any nonempty nil set S Q R closed under multiplication is nilpotent Chapter 2 Jacobson Radical Theory 24 The Jacobson Radical De nition The Jacobson radical of a ring R is de ned as the intersection of all maximal left ideals of R7 and is denoted by radR or JR radR JR C R M maximal left ideal of R Remark The Jacobson radical is invariant under automorphisms of R If M is maximal left ideal of R and f E AutR7 then fM is a maximal left ideal of R Example Have radZ If D is a division ring7 then radD Lemma For y E R7 the following are equivalent 1 y E radR 2 l 7 zy is left invertible Vm E R 3 yM 0 for any simple left R module M y E AnnM 417zy2 E RVzz E R Proof l i 2 Let y E radR7 z E R Then xy 6 rad R7 since if y is an element of every maximal left ideal of R7 then so is xy Now if xy is an element of every maximal left ideal of R7 then 1 7 zy is not an element of any maximal left ideal of R else the left ideal would contain the identity7 which implies that l7y is not an element of any proper left ideal of B because every proper left ideal is contained in some maximal left ideal Then Rl 7zy R and l 7 zy must be left invertible 2 i 3 Suppose l 7 zy is left invertible for all z E R but that yM 31 0 for some simple left R module M Then 3m 6 M such that ym 31 0 By the simplicity of M7 M Rm Then 3 z E R such that m7 ie7 l 7 xym 0 By the left invertibility of l 7 zy this implies m 07 a contradiction to the initial choice of m Conclude yM 0 for all simple left R modules M 3 i 1 Suppose yM 0 for all simple left R modules M Given a maximal left ideal I C R7 RI is a simple left R module Then yRI 0 i y E I Then y E radR 4 i 2 Put 2 1 2 i 4 If 1 7 is left invertible7 then 31 E R such that 111 7 zmy 1 Then 1 1 ozxy is left invertible by 2 i o 6 Pi since it is both left and right invertible As the right inverse of o 1 7 22y must then also be the left inverse of 1 so 1 7 22y 6 Pi Then by Exercise 1167 17 6 Pi D Corollary Let R be a ring Then 1 radR AnnM RM simple left R module ln particular7 radR is an ideal 2 radR is the largest leftrighttwo sided ideal A of R such that 1 A Q Pi Moreover7 the left and right radicals of R coincide Proof 1 This is condition 3 of the above lemma 2 Let A be a left ideal of R such that 1 A Q Pi and let y E A Then 7xy 6 A for all z E R and 1 7 zy 6 Pi Then y E radR by condition 2 of the above lemma Conclude A Q rad R7 and radR is the largest left hence two sided7 as well ideal A satisfying 1 A Q Pi That the left and right radicals of R coincide follows from this symmetric characterization of rad R D Example Let R be a semisimple ring By the Wedderburn7Artin Theorem7 we have R Hz with simple modules RR 691 Have AnnVZ Hj MnjD Conclude radR Proposition For any ideal I of R such taht I Q rad R7 radRI rad R I In particular7 rad R rad R Proposition Simple R modules are in bijective correspondence with simple R radR modules Proof This follows by property 3 of the above lemma D De nition A ring R is called Jacobson semisimple if radR 0 Example Z is Jacobson semisimple but not semisimple All semisimple rings are necessarily Jacobson semisimple by the previous example A simple ring is Jacobson semisimple Example Let p E Z be prime The localization of Z at p is Zp E Q 7110 1 m nn nn 7 pmm n n E Z077 1 7 p is invertible in Z07 with inverse since p l n and p l n p l 7171 7 pmm 7 so pZp Q rad Zp Conversely7 let E Z07 and suppose pl m Then Haj E Z such that ambp 1 Then 1 7 an bp has no inverse in Z07 Then rad Z07 Conclude rad Z07 pZp Given m De nition A subset S of a ring R is nil resp nilpotent if every 5 E S is nilpotent resp Sm 0 for some m E N Proposition Any nilpotent left ideal A of any ring R is contained in rad R Proof Given a simple left R module RM7 AM is an R submodule of M lf AM 31 07 AM M In this case7 AkM MVk 21 But Ak 0 for k gt 07 implying that M 07 a contradiction to the choice of M Conclude AM 07 so A Q AnnM RM simple rad R 1 Proposition Any nil left ideal A of R is contained in rad R Proof Let A be a nil left ideal of R7 and let y E A Then zy E A for all z E R and zy is nilpotent Now 1 7 zy 6 P with inverse well de ned by nilpotentcy of Then y E rad R Conclude A Q rad R D Theorem Let R be a left artinian ring Then radR is the largest nilpotent left right or two sided ideal N of R Proof By the above proposition7 any nilpotent left ideal of R is contained in radR7 so it suf ces to show that radR is nilpotent Let J radR Then J 2 J2 2 J3 2 Since R is artinian7 the chain stabilizes7 say at I Jk Suppose I 31 Let f left ideals A Q R IA 31 Then f 31 Q in particular7 J E 7 so there exists a minimal element A0 6 f Now 3a 6 A0 such that Ia 31 0 Since I2 I7 we have Iza Ia 31 then Ia E f Ia Q A0 Ia A0 by minimality of A0 Now 3239 E I such that m 1 Then 17 a 0 But 239 E J i 172396 P a a 07 a contradiction to the choice of a Conclude I 07 ie7 J radR is nilpotent D Corollary Let A be a left right or two sided ideal of a left artinian ring Then A is nil if and only if A is nilpotent Proof Any nilpotent ideal is automatically a nil Conversely7 if A is a left nil ideal of the left artinian ring R7 then A is contained in radR by the above proposition But radR is nilpotent by the above theorem7 so A Q radR is nilpotent as well D Example Recall that in a commutative ring R7 the nilradical NilR consists of all nilpotent elements in R and is the largest nil ideal in B Let R Zan7 n 2 1 Then Nil Zp Z 19 WWW Theorem For any ring R7 the following are equivalent 1 R is semisimple 2 R is Jacobson semisimple and left artinian 3 R is Jacobson semisimple and has the DCC on principal left ideals Proof 1 1 i 2 As previously noted7 a semisimple ring is Jacobson semisimple By Exercise 1337 any ideal I 31 R is a sum of simple components of R7 of which there are only nitely many7 say r Then any descending chain of ideals of R must stabilize after at most r steps Conclude R is artinian 2 2 i 3 Trivial 3 3 i 1 Since R has the DCC on principal left ideals7 every left ideal of R contains a minimal left ideal Let B 31 0 radR be a minimal left ideal of R Then there exists a maximal left ideal M of R such that B Q M else B Q rad R7 the intersection of all maximal left ideals Now RR B 69 M because B M R and B M 0 by the minimality of B and the fact that B Q M In particular7 B is a direct summand of R Fix a minimal left ideal B1 of R Then R B1 EBAl for some left ideal A1 of R7 and by Exercise 1177 A1 R61 for some idempotent 61 6 A1 Choose a minimal left ideal B2 of R in A1 Then A1 BZEBAZ for some left ideal A2 of A1 and A2 R62 for some idempotent 62 6 A2 Repeat this process to obtain a descending chain of principal left ideals A1 2 A2 2 This chain must stabilize7 since by assumption R has the DCC on principal left ideals Say the chain stabilizes at Ak Then R B1 69 EB Bk is a direct sum of minimal left ideals Conclude R is semisimple D Example Let k be a eld7 and let R be a nitely generated commutative k algebra Then R E klz17znI for some ideal I 31 klz17znl De ne the radical of I by f kx1xnfTEIforsomeTEN By Hilbert7s Nullstellensatz7 M lt1 k1n maximal 7I Q It follows that radR Nil R7 NilR E I7 and R is Jacobson semisimple if and only if R is reduced ie7 it has no nonzero nilpotent elements Example Let k be a division ring7 R C the set of upper triangular matrices Then the set J C R of strictly upper triangular matrices is an ideal of R Have Jn 0 so Q rad R And RJ E Dnk the set of diagonal marices But Dnk is semisimple Then radRJ 0 i radRJ 0 i J radR Now by the fact that the simple R modules are the simple RradR modules7 conclude that R has n nonisomorphic simple modules Ml g k The upper triangular matrix akl acts on Ml as scalar multiplication by ailam a De nition A ring R is semiprimary if radR is nilpotent and E R radR is semisimple Any left artinian ring R is semiprimary since the Jacobson radical of a left artinian ring is always nilpotent7 and because R R radR inherits the descending chain condition on ideals Theorem HopkiniLevitZki Let R be a semiprimary ring and M a left R module Then the following are equivalent 1 M is artinian 2 M is noetherian 3 M has a composition series In particular A a ring is left artinian if and only if it is left noetherian and semiprimary and B any nitely generated left module over a left artinian ring has a composition series Proof 3 i 12 This implication was already estbalished in 11 12 i 3 Assume that M is artinian resp noetherian Write J rad R P RJ R is semiprimary by assumption so P is semisimple and J 0 for some n Then M 3 JM 3 JZM D D J M 0 is a descending chain of submodules and JiMJiHM is artinian resp noetherian as an P module By the semisimplicity of P JiMJH lM is a direct sum of simple modules over P equivalently over R since J radR acts trivially Moreover because M is artinian resp noetherian this must be a nite direct sum It therefore follows that RM has a composition series A now follows from the established equivalencies and the above observation that all artinian rings are semiprimary A nitely generated left module over a left artinian ring is in particular noetherian hence has a composition series by the established equivalencies D Example Let k be a division ring a a non invertible endomorphism of k We established above that the skew polynomial ring R kxo is left noetherian but not artinian Rx D Rzz D Rzg D is an in nte descending chain Then R is not semiprimary Example Let R be a commutative Q algebra Let R Qz1 V j R is semipri mary but not noetherian Observe that radR 12 is nilpotent and R radR Q is semisimple but x1 Q 1 2 Q 123 Q is an in nite increasing chain of ideals LL LL Proposition Let 6 E R be an idempotent J rad R Then radeRe J 6R6 6J6 Note that 6R6 is a ring with identity 6 Proof Let r E radeRe y E R Then 6 7 eyer E eRe and 3b 6 6R6 such that e be 7 eyer be1 7 yer b1 7 yr since br E 6R6 i be b and er r Then 1 yrb1 7 yr 17 yr yrb17 yr 17 yr yre 17 yr yr 1 Conclude r E J so radeRe Q J 6R6 Let r E J eRe Then r ere E 6J6 so J eRe Q 6J6 Let r E 6J6 y E 6R6 Then r E 6J6 Q J so 3x 6 R such taht z17yr 1 Then exee 7 yr exe 7 yr ez17 yre 616 6 So r E radeRe so 6J6 Q radeRe Conclude radeRe J 6R6 6J6 D Corollary Let R be a ring Then rad MnR Mnrad R Proof Any ideal of MnR has the form MnA for some ideal A 31R Suppose rad MnR MnA for A 31 B Let 6 E11 6 Then eMnRe RE11 E R Now by the proposition A eMnAe e rad e radeMnRe rad R 1 Proposition Let R be a ring and suppose S UR U 0 is a division ring Then R is J semisimple Proof Have S radR 0 because radR contains no units of B Let y E rad R Then 1y E UR C S Theny 1y 71 E S radR So radR 0 and R is J semisimple D Example Let k be a division ring The following rings are J semisimple by the above proposition 1 R mm 6 I MR k 2 R kmz39 e 1 MR k 3 R kW 0 a an endomorphism or derivation of k7 UR k Proposition Let k be a eld7 and let R be a k algebra Let x E rad R Then z is algebraic over k if and only if z is nilpotent Proof Nilpotent elements are automatically algebraic Let x E radR and suppose x own 1 amxn n 0 for some 11 E k Then 0 x 1 1 am amzm But alz ammm E radR7 so 1 alz ame 6 Pi Conclude x 07 ie7 z is nilpotent D Corollary Let R be an algebraic k algebra Then radR is the largest nil ideal of R Proof By the above proposition7 radR is nil7 and any nil ideal is contained in rad R 1 Lemma Nakayama7s Lemma Let R be a ring7 and let J Q R be a left ideal Then the following are equivalent 1 J Q radR 2 For all nitely generated left R modules M7 JM M i M 3 For all left R modules M7 if RN Q RM is such that MN is nitely generated7 then N JM M i M N Proof 1 i 2 Suppose J Q radPt7 JM M7 but M 31 Take a minimal set of generators m1mk for M Since JM M7 we can nd r1rk E J Such that m1 rlml rkmk Then 17 r1m1 7quotng rkmk But 17 r1 6 Pi so m E ng Punk7 a contradiction to the minimality of this collection of generators Conclude M 2 i 3 Suppose JM M i M 0 for all nitely generated left R modules M Suppose RN Q RM is a submodule such that MN is nitely generated and N JM M Then JMN MN7 which implies by 2 that MN 07 ie7 that M N 3 i 2 Set N 2 i 1 Suppose JM M i M 0 for all nitely generated left R modules M Let M be a simple R module ln particular7 M is cyclic7 generated by any 0 31 m E M Now JM is an R submodule of M7 so by simplicity either JM M or JM lf JM M7 then M 0 by 27 a contradiction to the choice of M Conclude JM 07 and J Q rad R 1 Remark Let R1 Rk be rings Then rad R1 gtlt gtlt Bk radR1 gtlt gtlt rad Rk Exercises for 4 249 Let R be a J semisimple domain and let a be a nonzero central element of R Show that the intersection of all maximal left ideals not containing a is zero Proof Let N resp N denote the intersection of all maximal left ideals of R not containing a resp containing a If a 6 Pi then N radR Suppose 1 Pi Then there exists a maximal left ideal M of R with a E M Suppose N 31 Then 30 31 b E N Now ba ab 6 N M Note that radR N N Now ba 6 N 7 so ab ba 6 radPt7 a contradiction because radR 0 and ab 31 0 R is a domain Conclude N D 2410 Show that if f R 7 S is a surjective ring homomorphism7 then fradR Q rad S Give an example to show frad B may be smaller than rad S Proof Let r E radPt7 y fr Let x 6 S7 and say z ft fort E R Then 15 7 zy f1R 7 tr Since r E radPt7 13 7 tr is left invertible in R Then 31 E R such that o1R 7 tr 13 Say w fo Then 1015 7 xy fo1R 7 tr f1R1sie7 15 7 zy is left invertible in S Since z E S was arbitrary7 conclude fr y E rad S7 hence frad R Q rad S Let p be a prime7 n gt 17 and f Z 7 Zp Z the usual reduction mod p homomorphism Now radZ 07 but rad Zp Z equals the largest nilpotent ideal N in Zan This obviously contains pZp Z 31 D 2411 If an ideal 31 R is such that RI is J semisimple show that radR Q I Therefore7 radR is the smallest ideal I 31 R such that RI is J semisimple Proof If radRI 07 then 0 m M C RI M maximal left ideal M c R M maximal left idealI g M g I NowradRQ MCRMmaximal left idealIQMQI D 2412 Show that7 for any direct product of rings7 H Ri that rad R H rad 1 2416 A left R module is said to be cohop arz if any injective R endomorphism of M is an automorphism 1 Show that any artinian module M is cohop an 2 Show that the left regular module RR is cohop an if and only if every non right zero divisor in R is a unit In this case7 show that RR is also hop an Proof 1 Let RM be an artinian module7 and let f M 7 M be an injective R endormorphism Then M Q imf 2 im f2 2 is a descending chain of submodules Say the chain stabilizes at im fk Suppose k gt1 Let m 6 im fk 1im fk 31 0 Then fm 6 im fk im fkf l7 so 37716 imfk such that fm fm Now 0 fm 7fm fm7m but m 7 m 31 0 because m im fk7 a contradiction to the injectivity of f Conclude fM M and f is an automorphism 26 2 Suppose RR is cohop an Let a E R be a non right zero divisor in R De ne f R a R by fb ba Then f is an injective R endomorphism of RR i f is an automorphism Then 30 E R such that fc 1ieca 1 Now 0 aia aca 7a aci1a ac 1 Conclude a 6 Pi Suppose every non right zero divisor of R is a unit7 and let f R a R be an injective R endomorphism of RR Then f R l and f is injective f1 is a non right zero divisor f1 6 Pi Then f is an automorphism with inverse Rf171 R a R Suppose the above equivalent conditions hold By Exercise 11127 R is hop an gt R is Dedekind nite Let ab E R7 and suppose ab 1 Then ca 31 0 for all 0 31 0 E R Then a 6 Pi so 3d 6 R such that do 1 Now d dab dab b7 so ba 1 Then R is Dedekind nite7 hence hop an D 2418 The socle 500M of a left module M over a ring R is de ned to be the sum of all simple submodules of M Show that 500M Q m E M rad Rm 07 with equality if R radR is an artinian ring Proof The inclusion 500M Q m E M rad Rm 0 is clear from one of the equivalent characterizations of rad R Suppose P R radR is artinian Then P is semisimple Let N m E M rad Rm 0 Note that N is an R submodule of M because radR is an ideal in R Then N is a P module as well Now N BlNZ for some collection of simple left P modules Ni Each N1 is then also a simple left R module Then N Q 500M Conclude in this case that 500M N D 2421 For any ring R7 let GLAR denote the group of units in Show that for any ideal I Q rad R7 the natural map GLAR a GLnRI is surjective Proof Note that rad MAR MnradPt7 so if I Q radR is an ideal7 then MnI is an ideal contained in rad Also note that MARI It thus suf ces to prove the case n 1 by considering the new ring PU MAR and the new ideal I MnI Q rad PU Let x I E PuIf Then 3y I E PuIf such that xy I yr I 1 I Say xy 1z39z39 E I Q radR Then zy 6 Pi because 1 radR Q Pi ln particular7 z E R is right invertible7 xyzl 1 for some 21 E B By a similar consideration of yr 1j for some j E I Q rad R7 we get that z is left invertible in R7 and zzyx 1 for some 22 E R Now z is both left and right invertible i z 6 Pi Conclude z I E RI is the image under the natural projection map of z 6 Pi D 2422 Using the de nition of radR as the intersection of all maximal left ideals in R7 show radR is an ideal in R Proof Let y E rad R7 r E R7 and let M be a maximal left ideal of R If yr M7 then R MRyr by the maximality of M7 in which case 3 m 6 M7 5 E R such that 1 msyr Then m 1 7 syr 6 Pi because y E rad R a contradiction to the choice of m Conclude yr 6 M Since M was arbitrary7 conclude yr 6 rad R7 and radR is closed under right multiplication by elements of R Thus radR is an ideal 1 27 25 Jacobson Radical Under Change of Rings Lemma Let R be a commutative ring Then NilR H P P 31 R is prime Proof Q Ho 6 NilR then x 0 for some n Then inglR is a prime ideal x E P i z E P 2 Suppose z NilR Let S 1z2 so 0 S and S is multiplicatively closed Let f I 31 R I H S 0 Then T 31 Q E 7 so by Zorn7s Lemma there exists a maximal element P E f Suppose P is not prime Then there exist 1 P b P such that ab 6 P By the maximality of P there exist st E S such taht s E P a and t E P Then of E P 1 P Q P ie st 6 S P Q a contradiction Conclude P is prime Then z P P 31 R is prime D Theorem Let R be a commutative ring T flz39 E I a set of commuting independent variables Then rad RT NilRT NilR Proof Given a commutative ring S with ideal I Q NilS we have I NilS if and only if SI is reduced ie it has no nonzero nilpotent elements Now RTNilRT g R NilR Since RNilRT is reduced we conclude NilRT Nil RT Have NilRT Q rad RT by the above lemma because every maximal ideal of RT is in particular a prime ideal For the other containment it suf ces to consider the case T Let ft TO rlt mt E radRt Then 1 tft E I1Rt hence 1tft 6 I1 RP for all prime ideals P lt1 R Recall P lt1 R is prime if and only if RP is a domain and in this case I1RP I1 RP Now 1tft 1r0t rntn1 6 I1 i r E P V1 3 239 S 71 Since this holds for all prime ideal P lt1R we must have 7 6 Nil R V1 3 239 S n by the above lemma Conclude ft 6 Nil R T NilRT D Lemma Let R Q S be rings Assume that either 1 RR is a direct summand of RS as a left R module or 2 There is a group G of automorphisms of the ring S such that R is the subring of xed points R SG s E S 95 5V9 E G Then R rad S Q rad R Proof Let a E R radS Then for all z E R Q S 17 za 6 I1S Suppose RR is a direct summand of RS as a left R module Write RS RR 69 RT for some complimentary left R module T Then 35 r t such that 1 1 7 xas 1 7 xa 1 7 xat Note that 1 7 xa E R and 1 7 mat E T and 1 E R Conclude 1 17 xar Then 7 17 za 1 6 R and a E radR Suppose there exists a group of automorphisms G of S such that R is the subring of xed points Let s E S be the unique inverse of 1 7 za 6 R I1S Now 1 6 SG and 1 7 za 6 So so from the equation 1 7 xas 1 and by the uniqueness of 1 7 QTl we conclude s 6 SG B so a E rad R D Theorem For any k algebra R and any eld extension Kk we have R rad BK Q rad R If dimkR lt 00 or if Kk is algebraic then R rad BK radR In particular in this case radRK Q radRK lf K k n then radRKn Q radRK Proof Let 6 z39 E I be a k basis for K with 60 1 Then BK R 69 Eel01360 ie R is a direct summand of BK as a left R module Then R rad BK Q radR by the previous lemma Suppose dimkR lt 00 so dimK BK lt 00 as well New B is artinian radR is nilpotent rad RK is nilpotent Then rad RK Q rad BK and we have radR Q rad RK Q rad BK 1e radR Q R H rad BK Conclude in this case that radR R H rad BK Suppose K k n lt 00 Then the above direct sum decomposition of BK is nite and each 1 6 centralizes B Let M be a simple left RK module Then RM is a nitely generated left R module M BK a 21R1 6 a for some a E M Let J rad R Then JM is a simple left RK module since for each 1 ei 1 eJM J1 eM Q JM Since M 31 0 we have by Nakayama7s Lemma JM Q M But M is a simple RK module by assumption so we must have JM Conclude radR Q R rad BK Now apply a direct limit argument to prove the case where Kk is an arbitrary algebraic extension Suppose K k n and let 2 E radRKn Write z 217quot 6139 Let RV be a simple R module Then VK BLIV 6139 and VK as an R module has a composition series of length n K Then as an RK module any composition series of VK has length 3 71 Now zVK 0 for any element of rad BK reduced the composition length by one hence an element of radRKn must reduce the length to zero Then for all 1 E V we have 0 21 8 1 2 771 6139 Conclude 77V 0 and z E radRK Then rad BK g rad RK D Proposition Let Kk be a non algebraic eld extension For any k algebra R R rad BK is a nil ideal in R Proof Let a E R rad RK and let t E K be transcendental over k We may assume that K kt since by the previous theorem we have R rad BK Q R rad Bk a E R m rad RM Now liat E URK so liatftgt 1 for some ft b0b1tbmtm E Rt and gt co alt chrlthr1 E Comparing coef cients co b0 0 b 7 alt1 1 S 239 S m 1 with the convention bm1 0 Solving for the bss in terms of the cjs we have b aico ai lcl Ci In particular 0 bm1 am co acm cm and a is algebraic over k Then by a previous result x E radS is algebraic iff it is nilpotent a is nilpotent Conclude R rad BK is a nil ideal in R Remark lf radR is not nil then R rad BK Q rad R Lemma Let R be a k algebra lf Kk is a separable algebraic eld extension then R is Jacobson semisimple if and only if BK is Jacobson semisimple Proof By the above theorem rad R R rad BK so if BK is Jacobson semisimple radR R 0 0 and R is Jacobson semisimple Conversely suppose R is Jacobson semisimple Letz E radRK and write 2 21ri ai for some r E R a E K Let L ka1 an Then L is a nite eld extension of k and z E RL radRLK Q rad RL by the above theorem We may therefore assume that K k lt 00 Let E 2 K 2 k be the normal hull of K over k so Elk is a nite Galois extension By the previous theorem rad BK BK radRKE Q radRKE rad RE It therefore suf ces to show that if Klk is a nite Galois extension then radR 0 i rad BK Say G GalKlk Each 0 E G acts on BK by 1 o In particular we can identify GalKlk with a subgroup of AutRK Let 6 be a k basis for K and let 2 r ei E rad BK Let U E G Then 026j E rad BK because rad BK is invariant under AutRK Now 026j r 0eiej Consider the trace map tr 2060 o K a k The image ofthis map lies in k because it is invariant under G hence lies in F GalKlk This induces a map trR BK a R Then trzej 2060 026j E R rad BK Q radR 0 where the inclusion holds by the previous theorem Explicitly 0 trzej 2060 r oeej r Zaeadeie r 8 treej rtreej But the matrix treej is invertible so we must have r 0 Then 2 0 Since 2 E rad BK was arbitrary we conclude rad BK D Theorem lf Klk is a separable algebraic eld extension then rad BK rad RK Proof Note that the surjective homomorphism R k K a R rad R 8 K given by r 80 gt gt r rad R 8 c for r E R c E K establishes the isomorphism RKrad RK R rad RK R radR is Jacobson semisimple so by the above lemma R radRK g RKrad RK is Jacobson semisimple Then 0 rad BKrad RK rad BKrad RK because rad RK Q rad BK by the above theorem Conclude rad BK rad RK D Exercises for 5 252 Give an example of a ring R with radR 31 0 but radRt Proof Let S ley P lt1 S the principal ideal generated by x SP QM is a domain so P is a prime ideal of S Let R Sp the localization of S by P Then R is a local ring and has a unique maximal ideal M 31 So radR M 31 Since R is commutative rad th NilR But R Sp has no nonzero nilpotent elements because S has none Then radRt Nil Rt 256 For any ring R show that the Jacobson radical of the power series ring A RM is given by P a tft a E radRft E A 257 For any k algebra R and any nite eld extension Klk show radR is nilpotent if and only if rad BK is nilpotent Proof Finite eld extensions are algebraic so we have radR R rad BK In particular radRK Q radRK Say K k n Then we also have radRKYL Q rad RK Suppose radR is nilpotent Say rad R Then rad RKW Q radR 8 K Q radR 8 K 0 ie rad BK is nilpotent Suppose rad BK is nilpotent Say rad BK Then rad R k K Q rad BK 0 i radR 1Km 0 i rad R 0 ie radR is nilpotent D 30 26 Group Rings and the JSemisimplicity Problem Recall the notion of a group ring de ned in 11 De nition Let k be a eld A k representation of G is a group homomorphism b G a GLnk A left G module over k is an ordered pair V7p consisting of a vector space V and a group homomorphism p G a GLV A G module over k extends to a kG module Via the action lt2 M v Z agp9v 960 960 Conversely7 a kG module kV admits a G module over k Proposition Let G be an in nite group Then kG is not semisimple Proof Consider the ring homomorphism E kG a k the augmentation map satisfying elk idk7 and 69 10 for all g E G Let A ker 6 Suppose R kG is semisimple Then kG A 69 B for some left ideal B Q kG Now there exist idempotents ef E kG suchthatAReBRf7 1 efandef0 For anyUE G71706A In particular 17 Uf 6 Ref Then f Uf Write f 2960 fhh for some fh E k Let 039 E G The coef cient of 10 in f Zhea fhh is f1 but this is also the coef cient of 10 in f U lf Zhea fhaquotlh7 namely7 f0 Conclude that fh f1 for all h E G By de nition7 f is a nite sum Then we must have lGl lt 007 a contradiction to the choice of G Conclude kG is not semisimple D Theorem Maschke7s Theorem Let G be a nite group Then R kG is semisimple if and only if k is semisimple and lGl E Proof Suppose k is semisimple and lGl E Let W be an R submodule of the R module V By the semisimplicity of k there exists a k homomorphism f kV a kW such that le idW De ne f V a V by fv lGl l 2960 g 1fgv Since fV Q W and W is an R submodule7 we have imf Q W lf w 6 W7 then wGl lzg l gwlGl lzg 1gwlGl 1Zw w 960 960 960 ie7 W idW f is automatically a k homomorphism lf 1 E V and h 6 G7 we have w lGl l 29mm lGl lh 29h 1fghv marl ZT lfm h v 960 960 Tea so f is an R homomorphism Conclude that V W 69 ker Suppose kG is semisimple Let E kG a k denote the augmentation map Then k 6R is semisimple Write R Re EBRf for orthogonal idempotents e f E R satisfying 1 e f7 31 whereRe kere and Rf is a complimentary left ideal From before7 we have f 2960 09 for some constant c E k By the surjectivity of 6 we have 6Rf k so 3x 6 R such that 1 ef 6x6f 6C So 6 has a left inverse in k By a similar argument considering kere as a right ideal in R and nding orthogonal idempotents e f E R such that kere ER and R ER 69 fR7 6 has a right inverse in k Then 6 E D Exercises for 6 261 Let V be a kG module and H a subgroup of G of nite index 71 not divisible by char k Prove that if V is semisimple as a kH module7 then V is semisimple as a kG module Proof Let W be a kG submodule of V By assumption7 V splits as a direct sum of kH submodules7 so there exists a kH module homomorphism f V a W such that le idW Let 51 7 on be a complete set of right H coset representatives in G Note that for any 139 E G 51139 SnT is also a complete set of right H coset representatives in G De ne 9 V a V by 91 n 1151f5i1 9 is clearly k linear Note that 9V Q W because fV W and W is a kH submodule of V Also7 if 1 6 W7 then 91 714 21 51f5i1 714 21 5171511 17 ie7 9lW idW Let 139 E G 1 E V For each 1 S 139 S n7 write 517 Ill50M for some hi 6 H and some permutation O 6 Sn Then 91391 714 Zn 51f5i1391 11 N T7171 251771f517 11 T7171 Zlth15aiilfhisaiv 11 7714 Z sgih1hifsai1 11 7714 Z s1ifsai1 791 11 Conclude that 9 is a kG module homomorphism Then V WEBker 97 and V is a semisimple kG module D 263 Let k be a ring and G a nite group with 639 E 1 Let W Q V be left kG modules 1 If W is a direct summand of V as k modules7 show that W is a direct summand of V as kG modules 2 If V is projective as a k module7 show that V is projective as a kG module Proof 1 Suppose W is a direct summand of V as k modules Then there exists a k linear map f V a W such that le idW De ne g V a V by 91 lGl l 2060 U 1fav Then 9 is k linear7 gV Q W7 and gw w for all w E W Let 1 6 V7 739 E G Then 970 lGl 1U IgUTU 760 TlGl 1UT 1gUW 760 Gl jL Z UTlf039i 791 7 60 So 9 is a kG homomorphism Conclude V W 69 kerg 2 Suppose V is projective as a k module Let f A a B be a surjective kG module homomorphism7 and suppose g V a B is kG module homomorphism By assump tion there exists a k module homomorphism h V a A such that I o h 9 De ne h V a A by h1 lGl l 2060 flWm Check as above that h is a kG module homomorphism Let 1 E V Then NEW fltGIZUthgtgt 760 WW 2 f0 1hUv 760 W4 Z 0 10 0 WM 760 WP 20490771 760 WP Ego W 760 Conclude that f 0 9 hence V is projective as a kG module 1 267 Show that if k0 is any nite eld and G is any nite group7 then kOG rad kOG ko K is semisimple for any eld extension K 2 k0 For this Exercise7 we assume Wedderburn7s Theorem that nite division rings are commutative Proof We know that kOG rad kOG is semisimple7 so we can write kOG rad kOG 2 Ella for some m 6 N and some nite dimensional kO divison rings Di By Wedderburn7s Theorem7 each D is in fact a nite eld extension of k0 Now kOG rad kOG ko K g Hz ko K g ELI Mm DZ ko It now suf ces to show that each Di ko K is isomorphic to a direct product of elds7 for then kOG rad kOG ko K will be isomorphic to a direct product of matrix rings over elds7 hence will be semisimple Write D Di Recall that nite elds are perfect7 so the nite eld extension Dko is separable We can thus write D k0a for some 04 E D Let f 6 mm denote the minimal polynomial of 04 over k0 Then D E kOMf Over K7 we have a factorization f f1 fm for pairwise nonassociate irreducible polynomials f E Then D ko K kolzlf ko K g g KMfj7 a direct product of elds 268 Let k Q K be two elds and G a nite group Show rad KG rad kG k K Proof If chark 07 then kG and KG are semisimple by Maschke7s Theorem7 and the result follows trivially Suppose char k p gt 07 and let FF denote the nite eld of order p Since dime FPG lGl lt 007 we have FPG rad kG ralepG7 hence ralepG k k Q rade Note that FPG EFF krad FPG EFF k g FpGrad FPG EFF k which is semisimple by Exercise 267 Oonclude that rad kG rad FPG EFF k By a similar argument7 rad KG rad FPG EFF K Now rad kG k K rad no my k k K rad no my K rad KG D 269 Let k Q K be two elds and G a nite group Show that a kG module M is semisimple if and only if the KG module MK M k K is semisimple Proof Suppose MK is semisimple Let N Q M be a kG submodule Then there exists a KG module homomorphism f MK a NK such that leK ldNK Let 6 z39 I be a k basis for K Then MK k 6139 Now given m 6 M7 fm 1 61 for uniquely determined k linear maps gi M a N Let r E kG On the one hand7 frm 1 girm 6139 On the other hand7 frm 1 fltltr 1gtltm 1gtgtltr 1gtfltm 1gtltr 1gtgltmgt e 27mm m Oonclude that each 9 is a kG module homomorphism Assume that 610 1 E k Now leK ldNK i gio M a N is surjective and giOlN idN Then M N Bkergify Oonclude that M is semisimple Suppose M is semisimple Then M is a direct sum of simple kG modules7 M j6JMj for some index set J Now MK idM 61 69161 6376 M7 6139 a direct sum of simple kG modules7 so MK is semisimple as a kG module Let W Q MK be a KG submodule ln particular7 W is a kG submodule of MK Write W NZ 8 e for some kG submodules NZ Q M For any 23972quot 6 I7 we have NZ 8 6y 6i61Ni 86139 Q oi61W Q W Conclude NZ Nix for all 23972quot 6 I So W N ei NK for some kG submodule N g M Say M N ea N Then MK NK ea N K W 69 N K Oonclude that MK is a semisimple KG module D Chapter 3 Introduction to Representation Theory 37 Modules over FiniteDimensional Algebras In this section we assume that R is a nite dimensional k algebra R is left right artinian7 because ideals are k subspaces of bounded dimension Then E R radR is artinian and Jacobson semisimple7 hence semisimple By the Wedderburni Artin theory of semisimple modules7 we can write E B1 gtlt gtlt B where Bl for some m 6 N and some nite dimensional division k algebra Di lf Ml denotes the unique simple left Bi module7 then M17 7MT is a complete list of isomorphism classes of simple left F modules and FF 691 Moreover7 we have the following 0 Di EndBiMZ EndRMZ and B1 2 EndMZDi Double Centralizer Property 0 The natural map R a EndMZDi is surjective o dimk Ml m dimk Di since Ml g o dimk R dimk radR 2171 dimk Di 0 If k ie7 k is algebraically closed7 then Di k Lemma Burnside7s Lemma Let M be a nite dimensional right k vector space7 and let A be a k subalgebra of EndMk such that AM is simple Assume that EndAM k Then A EndMk Proof This follows from the above observations with R A and D k D Example When does EndRM k for simple M 1 Let k R R C an lR algpbra7 and RM RR C Then EndRM C 31 k 2 Let Klk be a eld extension7 K 31 k Then EndkK K 31 k 3Let i kk 727 k Bil kl Mikilkl so M is a left R module M is not simple I5 EndRM Z M2k 2 0V6ETek aIa kk is a proper R submodule of M Now De nition Let R be a k algebra7 Klk a eld extension De ne the scalar extension of R to K by RK R k K If M is a left R module7 de ne the scalar extension of M to K by MK M k K RK is a K algebra7 and R can be identi ed with R R k k Q BK MK is a left RK module Via the action 7 clm 02 rm 8 0102 for 01 6 K7 7 E R7 m E M Lemma Let R be a k algebra not necessarily nite dimensional7 and let MN be left R modules7 dimkM lt 00 Let Klk be a eld extension Then the natural map 6 HomRM7 NK a HomRKMK7 NK is a k Vector space isomorphism Proof Fix a k basis oiWEI of K Let f E HomRKMKNK Then fm 1 2161 11 E N k K for uniquely determined k linear maps gi M a N Now for r E R7 we have Zgi7 m 0139 Tm 1 ieI fr 1m 1 r 1fm 1 1139 13961 2 WWW 1139 id r 1 Conclude that gi E HomRM7 N Moreover7 only nitely many of the gi can be nonzero Fix a k basis 7711 7mn for M For each 1 S j S n7 let 7 C I consist of those indices 239 such that gimj 31 0 Then each 7 is a nite set7 because every element of NK is a nite sum of simple tensors Now I 7 is a nite set7 and gi E 0 for all 239 1 because the 771 span M Set 9 216191 8 al 6 HomRM7 NK this sum makes sense by the above comment Then 99 f7 since in particular they agree on simple tensors m 1 E MK Conclude that 6 is surjective Let f 2161 fl ai E HomRMNK7 and suppose 0f 0 Then for each m 6 M7 0 fm 1 2161 8a Conclude that 0 for all m E M and f ZieIfi ai0 D Theorem Let R be a nite dimensional k algebra7 and let M be a simple left R module7 dimk M lt 00 Then the following are equivalent 1 EndRM k 2 The natural map 7139 R a EndMk is surjective 3 For all eld extensions Klk MK is a simple RK module 4 For an algebraically closed eld E 2 k ME is a simple RE module Under these equivalent conditions7 say the R module M is absolutely simple or absolutely irreducible over k Proof 1 i 2 Let A im 7139 Then M is a simple left A module7 and by Burnside7s Lemma7 A EndMk7 ie7 7139 is surjective 2 i 3 By the surjectivity of 7T we may replace R by EndMk Now M g k for some n E N so R Then MK g K and RK Then MK is a simple RK algebra 3 i 4 Trivial 4 i 1 Suppose ME is a simple RE module for an algebraically closed eld E 2 k EHdltREME is a division algebra by Schur7s Lemma ln particular7 it is a nite dimensional division E algebra7 so EHdltREME E E But EHdltREME E EndRME by the above lemma Conclude EndRM k 1 De nition Let R be a nite dimensional k algebra A eld Klk is called a splitting eld for R and we say that R splits over K if every irreducible RK module is absolutely irreducible The algebraic closure E of k is always a splitting eld for R Proposition Let R be a k algebra7 Klk ai eld extension Then Klk is a splitting eld for R if and only if Klk is a splitting eld for R R rad R Proof We established in 25 that radRK Q radRK and RradRK RKradRK From this it follows that radEK g radRKrad RK7 hence EK radEK RK radRK Let M be an irreducible EK module Then M is an irreducible EK radEK module7 hence an irreducible RK radRK module by the above observation7 hence an irreducible RK module The reverse implications are all also valid Conclude that M is absolutely irreducible over k as an EK module if and only if it is absolutely irreducible over k as an RK module Thus Klk is a splitting eld for R if and only if Klk is a splitting eld for R D Proposition Let k be a eld7 R a nite dimensional k algebra7 a complete set of simple left R modules Then the following are equivalent 1 R splits over k 2 E R radR splits over k 3 E E Mmk gtlt gtlt Mmk for some 7216 N 4 dimk R dimkrad R 21dimk Mi Proof 1 ltgt 2 This is the previous proposition 2 ltgt 3 This is immediate from the de nition of a splitting eld and the observa tions made at the beginning of the section 3 ltgt 4 The implication 3 i 4 is immediate Suppose dimk R dimkrad R 2171 dimk Mi We know that F E MmD1 gtlt gtlt Mm D for some division k algebras Di and m 6 N satisfying dimk Ml m dimk Di It follows that Z n dimk Di Zdimk M02 anwimk Di 21 i1 21 EDI this equality to hold7 we must have dimk Di 17 1 S 2 S r Conclude Di k and REMmkgtltgtltMmk D Example 1 Let k Q R QM or QM2 1 Let M Q2 and let 2 act on M by H 0 1 95 71 0 Because 2 has no real eigenvectors7 M is an irreducible left R module We have 0 1 N 2 EndltRMgt QI Q 1 0 QMz 1 So M is not absolutely irreducible 2 Let K C BK CM or CM2 1 MK is reducible7 MK C211 1 C212 for eigenvectors 21122 of 2 Then EHdltRKMK E C 69C Proposition A eld extension Klk is a splitting eld for R if and only if K is a splitting eld for ft E Proof Write f f1 f as a product of irreducible polynomials in KM fl 32 f for 2 32 j Then BK WW I 42 2 WW I x XKltl r and radRK E x x E where the bar denotes passage to the quotient Note that RK rad BK E Ktf1 gtlt gtlt Kltlfr7 and is a eld extension of K of degree deg fi Now K is a splitting eld for R if and only if RK rad BK is isomorphic to a direct product of matrix algebras over K This is true if and only if deg fl 1 for each 1 S 2 S 7 But deg fl 1 for all 1 S 2 S r if and only if K is a splitting eld for f D Recall that a eld k is called perfect if every algebraic extension of k is separable Recall that all elds of characteristic zero are perfect7 and a eld k of characteristic p gt 0 is perfect if and only if k k Proposition Every k algebra R over a perfect eld has a splitting eld K with K k lt 00 Proof Since Klk is a splitting eld for R if and only if it is a splitting eld for Rrad R7 we may assume that R is semisimple Let E k be the algebraic closure of k Then E is separable Moreover7 by a theorem of 257 we have rad RE rad RE 0E Then RE is semisimple Now RE Hz for some r E N m 6 N7 by a previous observation that the algebraic closure of k is always a splitting eld for B Let denote the ab matrix unit of if 24th factor Say has preimage under the above isomorphism El r1 8 ega b i Let K k 4 11 1 3 ab 919 Then K k lt 00 and BK 2 1 1 MMK Conclude that K is a splitting eld for k D Proposition Let k be a eld7 Klk a eld extension7 and R a k algebra 1 If M1 and M2 are nonisomorphic simple left PL modules7 then M5 and MZK have no common composition factors 2 Every simple RK module V is a composition factor of MK for some simple left R module M Proof 1 Let 7T P R a R radR Bl denote the projection map7 the B the simple components of P We may assume that M1 is the unique simple module of B1 and M2 is the unique simple module of B2 From the direct product decomposition P Bi we obtain idempotents e E B such that B Roi Let x E R such that 7Tx 61 Then z 8 1 E BK acts as the identity on any composition factor V1 of MILK7 and acts as zero on any composition factor V2 of Then V1 5 V2 P Let 0 g 1 g 2 g Im RR be a composition series for R Then 0 g If g If g g If BK is a ltration of BK not necessarily still a composition series Let V be a simple left RK module Then V must be a composition factor of Mil1K for some 239 But IglIiK E Ii1IiK Set M ILAIi Then M is a simple left PL module7 and V is a composition factor of MK D Proposition Let k be a eld7 R a k algebra Let Klk and LlK be eld extensions7 with K a splitting eld for R Then L is a splitting eld for B More explicitly7 if V1L7 7Vm is a complete set of pairwise nonisomorphic simple left RK modules7 then VILL7 7VWLL is a complete set of pairwise nonisomorphic simple left RL modules Proof Since K is a splitting eld and by part 1 of the above proposition7 V1L7 V75 are pairwise nonisomorphic simple left RL modules By part 2 of the above proposition7 this is a complete list of simple left RL modules Now each ViL is absolutely irreducible because each eld extension FlL is in particular a eld extension of K7 so is a simple left BLF BKF module lf K is a splitting eld for R the number of simple left R modules is less than or equal to the number of simple left RK modules and if LlK is a eld extension then the number of simple left RL modules equals the number of simple left RK modules Lemma Let k be a commutative ring R For ab E R let ab ab 7 ba denote the additive commutator 1 R the additive subgroup generated by all such ab Then 1 R M E trM 0 Proof The inclusion 1 R Q M E trM 0 is clear For the reverse inclusion note that M E trM 0 is spanned by the collection of E147 Ejj Elj Eji and Eij Em Eijlv i 7 7 D From now on let k be a eld B a nite dimensional k algebra and TR rad RR R a k subspace of R Theorem Assume that R splits over k Then the number of simple left R modules equals dimk RTR Furthermore TR contains all nilpotent elements in R Proof Let P R rad B By the WedderburniArtin Theorem and since R splits over k P Mn1 gtlt gtlt Mmk for some n E N Then there are precisely r simple left R modules Let A Mm Now AiAj 0 for 239 31 j i PP g H1AAil Then PPP g H1AiAiA Hz k since AAA is isomorphic as a k algebra to the collection of scalar diagonal matrices in Ala Then dimkPPPl r Since 1 R maps onto PP under the projection map R a P we now have dimk RTR dimk PEP r If x E R is nilpotent then E E P is a sum of nilpotent matrices in H1A Then E6PPso radRRRTR 1 Corollary Let R be a k algebra which splits over K 2 k Let r be the number of simple left R modules and let r be the number ofsimple left RK modules Then r S r S dimk RTR Proof The rst inequality follows from a previous observation As for the second inequality we have by the previous theorem that r dimK RKTRK The inclusion 1 RV Q RKRK is clear while the inclusion rad RK Q rad RK follows from a theorem of 25 so TRK Q TRK Then r dimK RKTRK S dimK RKTRK dimk RTR D De nition Let R be a nite dimensional k algebra and let M be a left R module of nite dimension over k with left R action given by p R a EndM Call the map XM R a k given by z gt gt tr p the character associated with the left R module M Remark Let M1 M2 M be nite dimensional k vector spaces with left R actions p1p2p Suppose 0 a M1 L M 3 M2 a 0 is a short exact sequence of R modules Say dimk M1 711 and dimk M2 712 Choose an ordered k basis for M such that the rst 711 basis vectors span imf and the the last 712 basis vectors span kerg Then with respect to this basis each px z E R is represented by a block upper triangular matrix PM N my p225 Conclude trp tr p1 tr p2 and thus XM XM1 XM2 In particular the composition factors of a module M counted with multiplicities completely determine XM 40 Theorem Let k be a eld of characteristic zero Let M be a left R module of nite dimension over k Then XM determines the composition factors of M Explicitly if XM XN for left R modules M N of nite dimension over k then M N have the same composition factors counted with multiplicity Furthermore if M N are semisimple then XM XN i M g N Proof Write P 5 S the simple components of R and let 7T R a P denote the projection map Let M denote the simple left Si module so is a complete list of simple left R modules Suppose M has multiplicity m as a composition factor for M Then XM miXMi and M completely determines XM cf previous remark Take 17 E R such that Maj is the identity of 57 Then 17 acts as 67 on Mi Now XM mixM XMa7 ZimixMia7 mj dimk M So 771 XMajdimk M and XM completely determines the composition factors of M D Example Let k be a eld of prime characteristic p gt 0 R a k algebra and M an R module Then XPM pXM E 0 E M0 but pM and 0 have different composition factors Proposition Let MN be left R modules N absolutely irreducible Assume that either 1 dimkM dimk N or else 2 M is irreducible Then XM XN i M g N Proof Let 7T R a P denote the usual projection and write P ULL 5 S the simple components of B Let M be the unique simple left Si module so that 1 S 239 S r is a complete list of simple left R modules Say N M1 By an equivalent characterization of absolute irreducibility the natural map R a 1 EndNk is surjective Let a E R such that 7Ta E 51 has trace 1 Then XNa 1 Write XM 2171 miXMi Now if XM XN we have 1 XNa XMa 2171 miXMia mixMia m1 Then by assumptions 12 we must have m2m0henceMN 1 Corollary Let R be a nite dimensional k algebra and suppose R splits over k Then two simple R modules are isomorphic if and only if they have the same characters Proof This is immediate from the previous proposition D Exercises for 7 371 Let M N be nite dimensional modules over a nite dimensional k algebra R For any eld K 2 k show MK and NK have a common composition factor as RK modules if and only if M and N have a common composition factor as R modules Proof Let 0 C M1 CC Mm M and 0 C N1 CC Nn N be composition series for M N respectively Suppose as RK modules MK and NK have a common composition factor V Then for some 0 S 239 S m 7 1 and some 0 S j S n 7 1 V is a common composition factor of g and g Nj1NjK Note Mi1Mi and Nj1Nj are simple left R modules so by an above proposition we must have Mi11Ml g NHlNj ie M and N have a common composition factor 41 Suppose M and N have a common composition factor Without loss of generality7 we may assume M1 2 N1 Then M5 2 NIX7 and MK and NK share the composition factors of M5 2 N5 1 372 Let R be a nite dimensional k algebra which splits over k Show that for any eld K 2 k radRK radRK Proof Since dimkR lt 007 we have automatically rad RK Q rad BK Note that R radR is a semisimple7 hence a direct sum of simple left R modules Since R splits over k all simple R modules remain simple upon scalar extension to K Then Rrad RK RKrad RK is a semisimple PtK module7 so rad BKRKrad RK 0 i rad BK Q rad RK Conclude rad BK radRK D 374 Let R be a left artinian ring and C Q ZR a subring Show that Nil 0 0 rad R If R is a nite dimensional algebra over a sub eld k Q 0 show rad C 0 rad R Proof Since R is left artinian7 radR is a nil ideal in R7 so 0 radR Q Nil 0 Let c E NilC Then RC is a two sided ideal in R Say 0 0 Then for r E R7 7 rnc 0 Conclude that RC is a nil ideal7 hence RC Q rad R In particular7 c E 0 rad R Conclude NilO 0 radR Suppose R is nite dimensional over a sub eld k Q C in particular7 R is left artinian Then G is a nitely generated k algebra radC Nil 0 0 rad R 375 Let R be a nite dimensional k algebra which splits over k Show that any subalgebra C Q ZR also splits over k Proof Since R splits over k R radR E Hz for some 77477quot E N Then 0 radC CC radR E C radRradR is a subring of ZR radR E Hz k Now 0 radC is a commutative semisimple ring7 so Crad C E H1 K for some nite eld extensions Kjlk and some 5 S r Consider H1 Kj as a subalgebra of Hz k and let 6739 denote the identity element of K Let 7139139 Hz k a k denote the 2 th projection homomorphism Then mfg 31 0 for some 1 S 2 S r Now mfg mejej mej7riej mfg 1 Restricting 7139139 to K we have a k algebra homomorphism 7139139 KJ a k and conclude K g k Then 0 radC E H1 k splits over k D 378 Let R be a nite dimensional k algebra7 and let L 2 K 2 k be elds Assume that L is a splitting eld for R Show that K is a splitting eld for R if and only if for every simple left RL module M7 there exists a simple left RK module U such that UL g M Proof Suppose K is a splitting eld for R7 and let M be a simple left RL module Let W 1 S 2 S n be a complete set of pairwise nonisomorphic simple left RK modules Then ViL 1 S 2 S n is a complete set of pairwise nonisomorphic simple left RL modules7 and MgVJLforsome1gjgn Conversely7 suppose the given condition on simple left RL modules holds Let V be a simple left RK module lt suf ces to show that VF is a simple RF module for all eld extensions ln fact7 it suf ces to show that VL is a simple RL module7 for then if 42 0 31 V1 VF were a proper nonzero RF submodule of VF7 we7d have 0 31 VlL VFFL VLFL7 ie7 VlL is a proper nonzero submodule of the simple RFL module VLFL7 a contradiction Let W Q VL be a simple RL module By assumption7 there exists a simple left BK module U such that W UL Since VL and UL share a common composition factor namely W g UL7 we conclude that V g U since U7 V are both simple RK modules Hence VL g UL g W a simple RL module7 and the result follows E 379 If K 2 k is a splitting eld for a nite dimensional k algebra R7 does it follow that K is also a splitting eld for any quotient algebra R of R Proof Yes We may assume K k Let p R a E denote the projection map to the quotient algebra E and let V be a simple left E module Then R acts on V Via p and V becomes a simple left R module For any eld extension LlK7 VL remains simple as a left R module7 hence simple as a E module as well Conclude that V is absolutely irreducible7 so K is a splitting eld for E 38 Representations of Groups In this section we assume that G is a nite group7 unless noted otherwise Let k be a eld7 and let G be a nite group Recall that the group ring kG is semisimple if and only if char kl Restating previous results in the context of the group ring kG7 we have 0 kG rad kG MmD1 gtlt gtlt MmDT for some n E N and diVison algebras D 0 As left G modules7 kG rad kG 691 niMi where M g o lGl dimkrad kG 2171 dimk Di o If k is a splitting eld for G7 then there are dimk kGkG7 kG simple kG modules De nition Say that a eld k is a splitting eld for the group G if the group algebra kG splits over k Theorem Let k be a eld7 G a nite group Then there exists a nite eld extension Klk such that K is a splitting eld for G Proof If char k 07 then k is perfect and such a K exists by a result of 37 Suppose k is of prime characteristic p gt 0 Then E 2 k 2 PP FF is a perfect eld7 so by a result of 377 there exists a nite extension 1le such that G splits over k1 Let K denote the unique smallest eld extension of FF containing both k and k1 Then K k lt 007 and k1 Q K i K is a splitting eld for G D Example 1 Let G be an abelian group Then necessarily kG rad kG D1 gtlt gtlt D for some nite eld extensions Dilk If in addition G splits over k then D k for all 1 S 239 S r and the dimension of every irreducible kG module is one 43 2 Let G On ltggt the cyclic group of order n Then kG E 71 a Suppose k C or k C7 C a primitive n th root of unity Then 2 Hi Ci g Hi k b Suppose k Q Let ltIgtdx denote the d th cyclotomic polynomial Recall that ltIgtdx is a degree d irreducible polynomial over Q and 111 ltIgtdz x 7 1 Then QM96 1 g H Ql lq dWD g H ltCd dln dln 9 acts on QCd as multiplication by a primitive d th root of unity d c Suppose k R If n 2m 17 then Rival96 71 7 R x HRlzl ltz 7 CW 7 2 5 There is a unique one dimensional module with trivial G action7 and m two dimensional simple modules7 with g acting as multiplication by 739 ie7 as a rota tion by the angle 27Tjn If n 2m7 then Rival96 71 7 R x R x Rlzl lt95 7 cm 73 There are two one dimensional modules7 with g acting as 1 and 717 respectively7 and m 71 two dimensional simple modules7 with g acting as multiplication by 7 ie7 as a rotation by the angle 27Tjn 3 Let G Sn7 the symmetric group on 71 letters a Let k Q Sn acts on Q Qlt61 767 by permuting the basis elements 0 M1 Q the trivial module7 XM1 E 1 0 M2 QnML 2 221 am 21 11 0 We will prove later that M2 is an absolutely irreducible QSn module o One dimensional sign representation7 sgn Sn g GLAQ dig i17 where p Sn 7 GLAQ denotes the inclusion of Sn into GLAQ as monomial matrices Note that Xpa the number of one cycles in 039 From the decomposition Q M1 M2 it follows that XMZ Xp 7XM1 Xp 71 b Let n 37 k F2 F253 is not semisimple by Maschke7s Theorem 0 M1 F2 trivial module 0 M2 FgMl two dimensional module Check that M2 is an absolutely irreducible FgSg module by checking dimensions 6 ngl dimralegSg 12 22dimD 21 44 For some eld extension D1192 Conclude that D F2 dimrad F253 17 and M1L7 M2 is a complete list of simple ngSg modules7 F2 is a splitting eld for 3 Check that ralegSg F2 20653 a Let n 37 k F3 F353 is not semisimple by Maschke7s Theorem A O V 0 M1 F3 trivial module 0 sgn sign representation The normal Sylow 3 subgroup lt123gt acts trivially on an irreducible Sg module to be proved later Then the irreducible representations of Sg are the same as those of Sglt123gt E SQ From the equation 6 dimkrade 2171 dimk Di we infer that 2 has at most two irreducible representations Conclude that M1L7 sgn is a complete list of irreducible ngSg modules7 Sg splits over F3 and dimralegSg 4 Theorem Clifford Let H G and let V be a simple kG module Then kHV is a semisimple kH module Proof Let M be a simple kH submodule of V Then for all g E G gM is again a kH submodule of V7 th 99 1hgM Q gM by normality of H Moreover7 gM is a simple kH submodule7 for if M were a proper kH submodule of gM7 then g lM would be a proper submodule of M Now 2960 gM is a kG submodule of V7 hence V 2960 gM by the simplicity of V So V is a sum of simple kH submodules Conclude that V is semisimple as a kH module D Theorem Let k be a eld of prime characteristic p gt 07 and let H be a normal p subgroup of G Then 1 H acts trivially on every simple kG module7 and the simple kG modules correspond bijectively to the simple kGH modules If G itself is a l39o group7 then the only simple left kG module is k itself7 with trivial G action 2 kH has a unique simple module7 namely7 the trivial module k Proof 1 This follows from 27 since by Clifford7s Theorem every simple kG module is a direct sum of simple kH modules 2 Argue by way of induction on lHl7 the case 1 being trivial Assume that gt 1 and let M be a simple kH module Recall that H has a nontrivial center7 so let 1 31 h E Say lhl 1 Then in kH7 we have h 71 hp 71 171 07 so h 7 1 acts as a nilpotent endomorphism of M Then kerh 7 1 31 0 and M0 m E M hm m 31 M0 is a left kH submodule because h E Conclude that M0 M by the simplicity of M Now M is a simple kHlthgt module But lt lHl7 so induction applies and we conclude that M is the trivial module D De nition Given a nite group G de ne OpG to be the intersection of all Sylow p subgroups of G Lemma OpG is the largest normal p subgroup of G Proof OpG P P E SylpG 31 G7 since for each 9 6 G7 we have gOpGg 1 ng 1 P E SylpG P P E SylpG OpG by the conjugacy of Sylow p subgroups of G And any normal p subgroup of G is certainly contained in OpG because every p subgroup of G is contained in a Sylow go subgroup7 and all Sylow p subgroups of G are conjugate 1 Corollary Let k be a eld of prime characteristic p gt 0 Then OpG is the subgroup of G acting trivally on every sirnple kG rnodule Equivalently7 OpG h E G h 7 1 E rad kG Proof Let H be the subgroup of G acting trivially on all sirnple left kG rnodules By the previous theorern7 OpG S H Let h E H The left regular rnodule kG is a nite dirnensional k vector space7 hence has a composition series From this we infer that h 7 1 acts kG as a nilpotent endornorphisrn Say h 7 1 0 for some n gt 0 But h 71 hp 7 1 in kG7 so we conclude that lhl is a p power7 and H is a p subgroup of G lt7s clear that H 31 G7 since if M is a simple left kG rnodule and g 6 G7 9M is a simple left kG rnodule and for m 6 M7 g lhgm g 1hgm g 1gm m7 ieg 1hg E H Conclude H S OpG 7 H OpG D Corollary Wallace Let k be a eld of prime characteristic p gt 0 Suppose G has a normal Sylow p subgroup H Then rad kG ZheH kGh 7 1 ln particular7 dirnk rad kG G H llH l i 1 Proof Let I ZheH kGh 71 I is clearly a left ideal of kG Since h71g 99 119 71 and H 31 G7 conclude that I is a two sided ideal of kG Now I Q rad kG by the previous corollary Note that the surjective k linear rnap kG 7 satisfying 9 gt7 gH7 V9 6 G factors through a map o kGI 7 kGH7 while the surjective k linear map 1 7 kGI satisfying gH gt7 9 I is well de ned7 since for all g 6 G7 h 6 H7 we have gh I gh 7 gh 7 1 I g I Conclude that these maps are isornorphisrns7 establishing kGI 2 kGH Now pl lGHl7 so is sernisirnple by Maschke7s Theorern Conclude rad kG I Finally7 from the isomorphism kGI E we obtain lGl 7 dirnk rad kG lGHl7 ie7 dirnkradelGl7lGHlGHlHl71 D Recall the augmentation map E kG 7 k mapping 9 gt7 1 for all g E G Corollary Let k be a eld of prime characteristic p gt 0 Let G be a go group7 and set J rad kG Then 1 Jker6 296049 71 2 1101 0 3 If G is generated by 91 9 then J is generated by 91 71gn 71 as a left ideal Proof 1 Apply the previous corollary with H G Then J 2960 kGh 7 1 Note that for g 6 G7 h 6 H7 we have gh71 gh717g71 Conclude that J 2960 kg71 It is now clear that kere Q J But dimkker E lGl 7 1 dimk J7 so we conclude J kere 2960 kg 71 kG is a nite dimensional k vector space7 hence has a composition series Because G is a j39o group7 the only simple left kG module is k with trivial G action cf previous theorem Conclude that G must have lGl composition factors Now because J acts trivially on all composition factors of G7 we must have JlGl F 9 This follows from part a7 the observation that if 99 6 G7 then 9971 99 7 1 g 7 17 and induction on the length of words in G Theorem Let k be a eld with chark l Suppose k is a splitting eld for G Then the number of simple left kG modules equals the number of conjugacy classes in G Proof Since charkl lGl7 kG is semisimple by Maschke7s Theorem Write kG E Mmk gtlt gtlt Mm for some m 6 N so kG has precisely r simple left kG modules up to isomor phism Under this decomposition of kG7 observe that ZkG E Hz k and dimk ZkG r Let C17 7C5 be a complete list of conjugacy classes of G For 1 S 239 S 5 let cl deci 9 Then 01 E ZkG The 01 are clearly linearly independent7 hence form a k basis for ZkG Then 5 dimk ZkG r E De nition Let G be a group7 p a prime Say 9 E G is p regular or ap element if ifpl Say that g E G is p singular or a p element if 191 for some j E N Say a conjugacy class C of G is p regular if any 9 E C is p regular By convention7 say that every element of G is 0 regular Lemma Let G be a nite group For each 9 6 G7 there exist unique gingpz E G such that 917 is j39o singular7 917 is j39o regular7 and g gpgp gngp Proof Say lgl pkn with pln Then rpk 5n 1 for some 735 E Z Let 9 g and let 9 gfpk Then 9 is go singular7 917 is go regular7 and g gpgp gpgp Suppose g hph hphp is another such decomposition of 9 Note that hp7hp each commute with 9 hence commute with each of gingpz Now gPth l nghpz gthj l is p singular7 and gzjlhp is p regular Conclude gphzj l 1 gzjlhp and 97 hp 917 hp D Recall that if R kG splits over the eld k then the number of simple left R modules is equal to dimk RTR7 where TR radR 1 R We showed that TR contains all nilpotent elements of R Lemma Let R be a ring of prime characteristic p gt 0 Then 1 For all a1an E R and r 21 a1anpf ai a f mod R7R 2 If s E RR7 then spf E 1 R Proof 1 We have a1 dram Z 1 aip where the sum is taken over all words of length pT in the symbols 11 an The cylic group OPT of order pT acts on these words by cyclic permutations and two words in the same OPT orbit are congruent modulo 1 B This is clear since llla2 now 12 alp 1 11 12 owl and distinct elements in a OPT orbit are obtained through a sequence of such cyclic permutations There are precisely n singleton OPT orbits namely those of 15 afz and all other orbits are of cardinality p7 Conclude that modulo 1 R our sum takes the form al39 gt Eai ma 107mai a mod mi 2 It suf ces to prove the case r 1 Write s Emil 7 biai for some aib E R Then 5p E Emibi b p mod 13 R E 2 011301 0017 mod 13 R E Z calbia p lbi 7 b l pilbiai E 21 ba 1b E 0 mod 1 R 1es RR D Lemma Let k be a eld G a nite group and set R kG Then 04 E 1 R if and only if the sum of its coef cients over each conjugacy class of G is zero Proof i Let 04 E 1 R It suf ces to consider the case 04 ab 7 ba for some a b E R Write a 2960 agg b ZheH hh Then 04 Ema 049691 7 by But ghhg are conjugate in G by g 1ghg so in this case it is clear that the sum of the coef cients of 04 over each conjugacy class of G is zero Let 9192 6 G be conjugate with say 92 h lglh Then 91 7 92 hh lgL 7 h lglh IL1491 E 1 R Now let C 91 gn be a conjugacy class of G and let Oz 2 19 lf 21 1 0 then by the previous observation we have 04 21 llg E 2 191 E 0 mod 1 R ie 04 E 1 R D Corollary Let R kG as above and let 1 239 E I be a complete set of conjugacy class representatives of the group G Then 1 1 R 239 E I is a k basis for RRRl Proof Let B 1 1 R 2396 I B is a spanning set for RRR by the previous observation that conjugate elements of G are equivalent modulo 1 R and it is linearly independent by the previous lemma Conclude that B is a basis for RRRl Lemma Let R kG as above k a eld of characteristic p 2 0 and let J Q I be such that aj j E J is a complete set of the p regular conjugacy class representatives of G Suppose that k is a splitting eld for G Then aj TR j E J is a k basis for RTR Proof lfp 0 then radR 0 and TR 1 R All conjugacy classes are 0 regular by convention so in this case J I and we are done by the previous corollary 48 Suppose p gt 0 Let g E G By a previous lemma7 g gpgp gpgp for uniquely determined gingpz 6 G7 with 917 a p regular element and 917 a p singular element Say lgPl pk Then k k k k k k k k k k 9 i 91 9 7 9f 917 i g 95 g i g g i 9 0 So 9 7 9px is a nilpotent element of R7 hence is an element of TR by a result of 37 So 9 E gp mod TR We already established that conjugate elements of G are equivalent modulo R713 Conclude that aj TR j E J is a spanning set for TR Suppose 276 ejaj E TR Say 276 ejaj r b for some 7 E radR7 b 6 R713 Set m lcmlajl j E J Necessarily p 1 m7 ie7 p E ZR so EN 6 N such that pN E 1 mod m Set q p Then a 17 for each j E J Choose 71 large enough so that W 0 possible since 7 E radR is nilpotent By a previous lemma bq 6 R713 Now q 07 26711719 739 E Z 6341 7 bq 739 E Z 6341 mod 1 R 739 Then 6739 07Vj E J by the previous lemma Conclude that aj TR j E J is a linearly independent subset of TR7 hence is a k basis for TR 1 Theorem Brauer Let k be a eld of characteristic p 2 0 Suppose k is a splitting eld for the group G Then the number of simple kG modules equals the number of p regular conjugacy classes of G Proof Let R kG as above The case char k 0 has already been established Suppose chark p gt 0 We proved in 37 that the number of simple left kG modules equals dimk RTR The result now follows from the previous lemma 1 We now develop further the character theory for G From now on7 assume that k is a splitting eld for G7 and that chark l For a kG module M7 let p kG a EndkM denote the given representation Recall that XM is de ned by XMg tr pg for all g E G We make the immediate observation that if g h 6 G7 then Xghg 1 twig194 trp9php9 1 trphpg 19 tr WWW Xh ie7 X is constant on conjugacy classes of G Let R kG as above7 and recall the notations from the beginning of this section By our assumption that charkl 1G1 and that k is a splitting eld for G7 we have Di k and m dimk M1 for all 1 S 239 S 7 Write Xi XMi Let 1 1 S 239 S r be the central idempotents in R giving the decomposition of B into its simple components7 R g R61 gtlt gtlt R67 Then 1 acts on M as 617 Remark Let C 1 S 2 S r be a complete list of the conjugacy classes of G and let 0 deci 9 Then 0 1 S 2 S r and 6 1 S 2 S r are two k bases for ZkG Remark The character of the left regular module kG is given by X759 2171 72 For g E G we have 90599 0G 5 Theorem With the notations as above 1 e lGl lni 2960 Xig 1g So charklniV1 S 2 S 7quot 2 09 1Cyl 2171 Xigniiei where Cg denotes the conjugacy class of g in G Proof 1 Let g E G Write e Zhea ahh for some a E k Now 7 aglGl 9055497152 anijilei 71009 1 j1 by the previous remark that X759 2171 nix and the fact that 6 acts on Mj as 6 2 Write cg 2171 bio for some b E k Now nglXj9 MW X1 1W 52711 21 by the previous remark that characters are constant on conjugacy classes of G and that 6 acts on M as 67 We make the following notational convention Given a character X of kG and a conjugacy class C of G de ne 9C xg for any 9 E C This makes sense by a previous observation Theorem Character Orthogonality With the notations as above 1 lGlil 2960 Xilt971Xj9 52739 2 Let CC be conjugacy classes of G Then 39 lGllCl CC X2CX2C0 C7 c Proof 1 This follows by applying Xj to the expression 6 lGl lni 2960 Xig 1g 2 Combining the results of the previous theorem7 we have 7 1 7 7 nglZXAg G HEMh 1 21 1 heG 39 71 i G 2 ZXMMUI h heG 21 C Le and the result follows E Fix a set of representatives aj 1 S 2 S r for the conjugacy classes of a group G We construct an r gtlt 7 matrix with 2739 entry equal to X11217 and call it the character table of G with respect to the splitting eld If we let a1 1G7 then the rst column of the character table will always list the dimensions of the irreducible representations of G Example 1 Let G Sg Let M1 denote the trivial representation7 M2 klt61engtk61 en as before7 and sgn the sign representation Recall that XM2g is one less than the number of one cycles in g P Let G ltggt G 7 the cylic group of order 72 Note that each gi 6 G7 1 S 2 S n 71 is its own conjugacy class Fix a primitive n th root of unity C E C and let pj G a C denote the representation mapping 9 gt gt 739 so pjgi 1739 Then p17 7pn is a complete list of irreducible CCGn representations De nition A function M G a k is called a class function if 22hgh 1 9Vg h E G Let FkG denote the collection of class funtions on G Then FkG is a k vector space of k dimension r r the number of conjugacy classes in G De ne a symmetric bilinear form on FkG by 1 1 lt22 Vgt KEG49 My Theorem Let Xi 1 S 2 S 7quot denote the collection of irreducible characters of G Then Xi 1 S 2 S r is an orthonormal basis for FkG with respect to the inner product lt Proof We certainly have Xi 1 S 2 S 7 Q FkG by the previous observation that class functions are constant on conjugacy classes of G Moreover7 we have that the Xi are or thonormal with respect to the inner product lt by the previous theorem7 hence form a linearly independent set Since 1 S 2 S 7 l r dimk F G7 we conclude that Xi 1 S 2 S r is an orthonormal basis for D 51 Note that FkG and ZkG are dual with respect to the pairing M702 gt gt 02 M E F G7 04 E ZkG Furthermore7 nlel 1 S 2 S r and 61 1 S 2 S r are dual bases with respect to this pairing Remark Over k C once can show that Xig 1 XigVg 6 G7 and the bilinear form lt can be replaced by the Hermitian inner product Q2 Vgt lGl l deapgyg Assume now that char k 0 In this case kG is semisimple7 and the character X of a left kG module M completely determines the composition factors of M Corollary Let X E FkG be the character of some kG module M Then ltX7Xgt 1 if and only if X is irreducible Proof Write X 2171 niXi Then 1 ltX7Xgt 2171 if and only if m 1 for some 1 S 2 S r and 772 0 for 2 31 j ie7 X Xi is irreducible D Corollary Two G modules M7 N are isomorphic if and only if XM XN Proof This is immediate D Remark There are several ways to produce new kG modules 1 Given a kG module V7 the dual space V HomkVk becomes a G module under the action z f2 fx 12Vf E Vx E Gm E V 2 Let GH be nite groups7 V a G module7 W an H module Then V 8 W becomes a G gtlt H module via the action 97 h 2 20 go 8 hw on simple tensors g 6 G7 h 6 H7 2 6 V7 w E W7 and extending linearly Lemma Let GH be nite groups7 V a G module7 W an H module The character of V 8 W7 denoted XV XW7 is given by XV Xwg7 h XVgXWh Proof This follows from the de nition of how G gtlt H acts on V 8 W D Theorem Let G7 H be nite groups 1 If V is an irreducible G module7 and W is an irreducible H module7 then V 8 W is an irreducible G gtlt H module 2 If 1 S 2 S r is a complete list of irreducible G modules7 and 1 Sj S s is a complete list of irreducible H modules7 then Wj 1 S 2 S r1 S j S s is a complete list of irreducible G gtlt H modules Proof 1 Observe that 1 ltXV XW7XV XWgt W Z XV XWltlt9J1gtAgt XV XW97h ghEGgtltH 1 L W Z XV9 1XWh 1XV9XWh ghEGgtltH ZXVlt91XV9gt Z XWh1XWhgt 960 heH ltXV7XVgtltXW7XWgt 1 52 Conclude that V 8 W is irreducible P Let B 1 S 239 S r1 S j S s Mimicking the previous calculation we have ltXV XijXVk XW1gt M571 so the elements of B are mutually nonisomorphic irreducible G gtlt H modules But there are precisely rs conjugacy classes in G gtlt H so we conclude that B is a complete list of irreducible G gtlt H modules D Corollary Let G be a group V an irreducible representation of G V1 a one dimensional representation of G Then V 8 V1 is an irreducible representation of G Proof Note that since V1 is a one dimensional representation we must haVe XV191 XV19717V9 E G Then 1 ltXv Xvisz XV1gt W Z XV Xv197h 1gtltv XV197 h gJLEG 1 W Z XV9 llXV1h 1gtltv9gtltv1h gJLEG G Z mg lmu gJLEG 1 71 ZXVW XV9 960 ltXV7XVgt 1 Conclude that V 8 V1 is irreducible D Example Let G S4 Let M1 denote the trivial module M2 klt61 6ngtk61 en as above X H1 12 123 1234 1234 M1 1 1 1 1 1 sgn 1 71 1 71 1 2 0 71 0 2 M2 3 1 0 71 71 M2 8 sgn 3 71 0 1 71 The entries of the unmarked row can be inferred from the character orthogonality relations and the fact that the squares of the rst column entries must sum to 24 184 Let G be a group H S G A kG module V may also be viewed as a kH module by restricting the representation p G a GLV to H We call this the restriction of V to H and denote it by ResgV ResH V Restriction induces a map resg FkG a FkH which simply restricts elements of FkG to kH 53 De nition Let H be a subgroup of the nite group G7 and let V be a kH module affording the representation p of H The kG module lndg V lndG V kG kH V is called the in duced module of V and the representation of G it affords is called the induced representation of p lf 7 is the character afforded by V7 then the character of the induced representation is called the induced character7 and is denoted by lndg b Proposition Retain the notations from above Then H Let k denote the trivial G module Then lnd1k E kG7 the left regular module E0 Denote by 1H the trivial kH module Then lndg 1H kGH7 the left coset repre sentation of G 9 Let L be a group7 H S G 3 L7 and let W be a kH module Then indg 111de W 2 inde W Theorem Let V be a kH module affording the matrix representation p H a GLnk7 and let 91 7gm be representatives for the distinct left cosets of H in G There exists a basis for the induced module W lndG kG kH V of dimension n G H over k such that W affords the representation ltIgt G a GLnGHk de ned for each 9 E G by the block matrix 9 My 19979 where 9171997 is an n gtlt n block appearing in the z j block position of g and where pgilggj is de ned to be the zero block whenever gilggj H Note that g is a block permutation matrix in the sense that there is exactly one nonzero block in each row and column Proof Note that kG is a free right kH module and kG glkH EB EB gmkH Then W kG kH V E 91 8 V EB B 9m 8 V Let 111 1 be a basis of V affording the matrix representation p H a GLnk Then 918017918027791 Un792 017792 77gm is an ordered basis for W lndg FixgEG1 j m Sayggjgihforsome1Sigmandh6H7andsay ph am 6 GLnk Then for 1 S k S 717 we have 99739 8 99739 8 9139 hvk Z atkltgi lit t1 So g maps the j th block of basis vectors 97 111 797 8 ml to the i th block of basis vectors gi 111 791 8 on and then has the matrix ph in that block Since h gflggj7 this is precisely the action described by the matrix ltIgtg 54 Corollary lf 7 is the character afforded by V7 then the induced character is given by 1 Indi m Zd W 19 60 where 7y 0 ify 6 H Proof Let g 6 0 By the previous theorem7 1nd w9ztrp9 1991 EM 199139 l 1 2 Z wh 19199ih i1 h6H 1 Emma 60 because gih ranges over 0 as Ah ranges over 17 7m gtlt H D Theorem Frobenius Reciprocity Let H be a subgroup of the nite group 0 Let 7 6 F H7 p 6 Then lt1nd w ltpgta lt11 R832 ltpgtH Proof By the previous corollary we have lt1Hdg 7lt gtc zlndgwgm l 960 1 1 E 2 E Z 719 971 960 60 1 7 L W Z 1 1 9 19 y60 lm 2 wyltpy 1 y6H 60 7 Z mom1 w Resg m y6H ie ltInd2 w ltPgtG lt22 Rest m D Example Let k C and let Dn lt7y yz 17y 1ygt denote the Dihedral group of order 271 Note that On lt1 Dn Suppose n is even We have the following one dirnensional characters of D Let E 627W De ne ph D a GLnC 0 S h S n by ehk 0 0 eihk MW 0 67m 7 Phyk hk 0 We have the following character table of Dn 71 even X LZ zik y KVz yZi1Vz39 701 1 1 1 1 1 7112 1 1 1 i1 i1 713 1 71 71 1 1 714 1 71 41 1 1 ph 2 241 6hk hk 0 0 where in the third column we let 1 S k 3 12L 7 17 and in the last row we let 1 S h S g 7 1 Exercises for 8 381 Give an example of a pair of nite groups G7 G such that for some eld k kG E kG as k algebras7 but G 5 G as groups Proof Let k C Let G G4 the cyclic group of order 4 and let G V4 the Klein 4 group Then kG 2 69 c 2 kG but G 30 G 382 Let k be a eld whose characteristic is prime to the order of the nite group G Show that the following statements are equivalent 1 Each irreducible kG rnodule has k dirnension 1 2 G is abelian7 and k is a splitting eld for G 385 For any eld k and for any normal subgroup H of a group G7 show that kH rad kG rad kH Proof Let gi 239 E I be a complete set of right coset representatives for H in G7 with gio 1 Let I I 2390 Then as a left kH rnodule7 kG kH 69 161 ngi ie7 kH is a direct surnrnand of kG as a kH rnodule Then kH rad kG Q rad kH Let V be a simple left kG rnodule By Clifford7s Theorern7 V is sernisirnple as a left kH rnodule Say V 697 M7 for simple left kH rnodules Mi Then rad kH V jrad kH M7 Then rad kH Q kH rad kG Conclude rad kH kH rad kG D 56 Exercise Let k C Dn ltzy x yz 17yx z 1ygt the dihedral group of order 2717 and E 627W Dn contains the cyclic group On as a subgroup Denote by Uh 0 S h S n the one dimensional irreducible representation of the cyclic group On de ned by Uhx eh De ne ph D a GLAC 0 S h S n by MW 6 PM 7 Phyk egk 670M Show that 1 ph is a representation of D 2 ph g lndC Uh 339 pk g pnih Below in 4 and 5 we assume that n is even ph is irreducible unless h 07712 r p0 1 2 and pnZ 3 4 where 71 are the four one dirnensional representations of Dn de ned in the lectures Cf For Dn with n odd7 formulate sirnilar statements as in 4757 and determine the character table CT Chapter 7 Local Rings Semilocal Rings and Idempotents 719 Local Rings Theorem The following are equivalent for a ring R 1 R has a unique maximal left ideal 2 R has a unique maximal right ideal 3 R radR is a division ring If R satis es any of these conditions7 call R a local ring Proof lt suf ces to prove 1 gt The equivalence 2 gt 3 will follow by symmetry 1 i 3 If R has a unique maximal left ideal M7 then radR M and RradR has precisely two left ideals 0 and R rad R Conclude that R radR is a division ring 3 i 1 Suppose R radR is a division ring RradR is has precisely two left ideals7 0 and R rad R7 so radR must be a maximal left ideal of R But radR is the intersection of all maximal left ideals in R7 so we conclude that radR is the only maximal left ideal in R D Properties of a Local Ring R 1 R has a unique maximal left ideal M radR RUR 2 R has no nontrivial idempotents Suppose e E R is an idempotent lf 6 6 Pi then 6 6667166711lf6 Rithen6EradRgt16 Rgt61760gt60 Example Let k be a eld7 V an n dimensional k vector space7 R the exterior algebra on V Then R is a local ring with unique maximal left ideal M radR 1V EB EB Lemma Fitting Decomposition Let R be a ring7 and let M be a left R module that admits a composition series Let f E EndRM Then M ker fT 69 im fT for some 7 gtgt 0 58 Proof Since RM admits a composition series it has the ACC and D00 on submodules Consider the chains kerf Q ker f2 Q and imf 2 im f2 2 Choose r E N large enough that both chains stabilize Then for all z E M f7x f27y for some y E M Now z f7y x 7 f7y 6 im f7 ker f Suppose f7x E ker f7 Then 0 f27z a z E ker fZT ker f i f7z 0 Conclude that the decomposition M ker f im f is direct ie M ker f 69 im f D Lemma Let R be a ring and let RM be an indecomposable left R module that admits a composition series Then EndRM is a local ring Proof Let I be a maximal left ideal of E EndRM Let a E EI Then E Ea I by the maximality of I Write 1 go f for some 9 E E f E I Since f E I f cannot be an isomorphism of M In particular f and hence also f is not surjective By Fitting7s Lemma 3 r E Nsuch that M ker fTEBim f7 Since im f 31 M we conclude ker f M and im f7 0 by the indecomposability of M Now go 17f and 1f f7 1ga 1 ie a is left invertible Conclude that I is the unqiue maximal left ideal of E and E EndRM is a local ring D Proposition Let R be a ring and M a left R module whose submodules satisfy either the ACC or the DCC Then there exists a decomposition of M into a nite direct sum of indecomposable submodules M M1 69 69 Mm Call such a decomposition of M a KrulliSchmidt decomposition of M In particular any nitely generated module over a left artinian ring has a KrulliSchmidt decomposition Proof Call a submodule N Q M good77 if it has a KrulliSchmidt decomposition and call a submodule bad77 if it does not Note that the zero submodule is good as is any indecomposable submodule If N N Q M are good and N N 0 then NN NEBN is good Suppose M is bad Then in particular M itself is not indecomposable and we can write M M1 69 M for some submodules M1 Mi 31 0 of M Now at least one of M1 Mi must be bad say M1 Then M1 M2 EBM for some submodules M2 Mg 31 0 of M1 Repeating this process we obtain in nite chains of submodules M2M12M22M32m and 0MgMieM gMieM eM gw in contradiction to the fact that the submodules of M satisfy either the ACC or the DCC Conclude that M is good ie M has a KrulliSchmidt decomposition D Theorem KrulliSchmidt Let R be a ring and let M be a left R module that admits a composition series Suppose M has two KrulliSchmidt decompositions MM1eeMmMeeM Then m n and after reordering M E for all 1 S 239 S m Proof We argue by way of induction on m the case m 1 being trivial Note that if M admits a composition series then so do each Mi Mil so EndRM EndRMJ are local rings by the previous lemma

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