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# Algebra I MTH 818

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Algebra Lecture Notes for MTH 818 Fall 2000 Ulrich Meierfrankenfeld January 197 2001 Chapter 1 Preface These are the lecture notes for the class MTH 818 which I m currently teaching at Michigan State University The text book used for this class is Hungerford7 Algebra Hun Much of the contends follows Hungerford s book but the proofs given here often diverge from Hungerford s CHAPTER 1 PREFACE Contents O CO 4 Preface 3 Group Theory 7 21 Latin Squares 7 22 Semigroups7 monoids and groups 9 23 The projective plane of order 2 14 24 Subgroups7 cosets and counting 15 25 Normal subgroups and the isomorphism theorem 17 26 Cyclic groups 22 27 Simplicity of the alternating groups 23 28 Direct products and direct sums 28 29 Free products 33 210 Group Actions 41 211 Sylow p subgroup 47 Rings 53 31 Rings 53 32 ldeals and homomorphisms 60 33 Factorizations in commutative rings 66 34 Localization 74 35 Polynomials rings7 power series and free rings 80 36 Factorizations in polynomial rings 85 Modules 93 41 Modules and Homomorphism 93 42 Exact Sequences 97 43 Projective and injective modules 101 44 The Functor Hom 109 45 Tensor products 113 46 Free modules and torsion modules 120 47 Modules over PlD s 124 48 Composition series 127 5 0 CD 03 49 Matrices Fields 5 1 Extensions Z0rn7s Lemma Categories C ON TEN TS 133 Chapter 2 Group Theory 21 Latin Squares llsl De nition 211 binaryop Let G be a set and j G x G a G a map a j is called a binary operation on G We write ab for gtab G gt is called a pre group b e E G is called an identity element if ea ae a for all a E G c We say that G gt is a Latin square if for all ab in G there exist unique elements 71 in C so that am b and ya b d The multiplication table of G gt is the matrix abaeabea e The order of G gt is the cardinality lGl of G We remark that G7 gt is a latin square if and only if each a E G appears exactly once in each row and in each column of the multiplication table If there is no confusion about the binary operation in mind7 we will just write G for G7 gt and call G a pre group De nition 212 Let G and H be pre groups and 04 G gt H a map a 04 is called a pre group homomorphism if aab aaab for all ab E G 04 is ca e an isomorp 1sm i 04 is a omomorjn ism an t ere EIZStS a omomorjn ism b 39 ll 11 i h 39 39 h h39 d h 39 h h39 B Hgt G with a idH and a idG c 04 is an automorphism if G H and 04 is an isomorphism 8 CHAPTER 2 GROUP THEORY Let G be a pre group The opposite pre group GOP is de ned by GOP G has a set and 9 3901 h h9 Let G and H be pre groups An pre group anti homomomhisma G a H is a pre group homomorphism Oz G a HOP So aab abaa Lemma 213 basicbinary a Let G be a pTe group Then G has at most one identity b Let Oz G gt H be a pTe group homomomhism Then 04 is a isomomhism if and only ifa is a bijection Proof a Let e and e be identities Then b Clearly any isomorphism is a bijection Conversely7 assume 04 is a bijection and let 3 be its inverse map We need to show that B is an homomorphism For this let 071 E H Then as 04 is a homomorphism a aab a aa b ab a ab Since 04 is one to one or by applying 3 we get BMW So 3 is an homomorphism D Below we list up to isomorphism all Latin square of order at most 5 which have an identity element e It is fairly straightforward to obtain this list although the case lGl 5 is rather tedious We leave the details to the reader7 but indicate a case division which leads to the various Latin squares Order 12 and 3 Order 4 Here we get two non isomorphic Latin squares One for the case that a2 7 e for some a E G and one for the case that a2 e for all a E G e a b c e a b c e e a b c e e a b c 1a a b c e 20 a e c b b b c e a b b c e a c c e a b c c b a e 22 SEMIGRO UPS MONOIDS AND GROUPS Order 5 This time we get lots of cases Case 1 There exists e 7 a 7 b with a2 e 2 Case 2 There exists e 7 a with a2 7 e7 am2 e and 02002 Case 3 There exists e 7 a With a2 7 e7 am2 e and aza Case 4 There exists e 7 a with a2 7 e7 020 e and aa22 e This Latin square is anti isomorphic but not isomorphic to the one in case 2 Anti isomorphic means that is there exists bijection 04 with aab abaa Case 5 There exists e 7 a With a2 7 e7 020 e and aa22 7 e This Latin square is isomorphic and anti isomorphic to the one in case 3 Case 6 There exists e 7 a with a2 7 e7 020 act2 7 e Case 7 There exists e 7 a with a2 7 e a22 Case 8 There exists e 7 a with a22 7 e and e 7 020 7 act2 7 e In this case put 0 aaz Then 02 7 e and either 002 e or 020 e Moreover 0202 7 e respectively 0022 7 e and the latin square is isomorphic to the one in Case 3 e a b c d e a b c d e a b c d e a b c d e e a b c d e e a b c d e e a b c d e e a b c d a a e c d b a a b e d c a a b e d c a a b c d e 1 b b d e a c 2 b b c d a e 3 b b c d e a 4 b b e d a c c c b d e a c c d a e b c c d a b e c c d a e b d d c a b e d d e c b a d d e c a b d d c e b a e a b c d e a b c d e a b c d e a b c d e e a b c d e e a b c d e e a b c d e e a b c d a a b c d e a a b c d e a a b c d e a a b c d e 5 b b e d a c 6 b b c d e a 7 b b d e a c 8 b b d a e c c c d e b a c c d e a b c c e d b a c c e d m y d d c a e b d d e a b c d d c a e b d d c e y m 22 Semigroups monoids and groups 5mg De nition 221 Let G be a pTe group a The binary operation on G is called associative if abc abc for all abc E G If this is the case we call G a semigroup b C is a monoid if it is a semigroup and has an identity any ayb 10 CHAPTER 2 GROUP THEORY c Suppose that G is a monoid Then a E G is called invertible if there exists a 1 E G with 1 Such an a is called an inverse of a d A group is a monoid in which every element is invertible e G is called abelian or commutative if ab ba for all ab E G Examples Let Z denote the positive integers and N the non negative integers Then ZJF is a semigroup N is a monoid and Z is a group Z and R are monoids Let I R Then 1 is a group The integers modulo n under addition is another example We denote this group by ZnZ All the examples so far have been abelian Note that in a group a lb is the unique solution of am b and ba 1 is the unique solution of ya I So every group is a Latin square with identity The converse is not true lndeed of the Latin squares listed in section 21 all the once or order less than ve are groups But of Latin squares of order ve only the one labeled 6 is a group Let K be a eld and V a vector space over V Let EndMV the set of all K linear maps from V to V Then EndMV is a monoid under compositions Let GLMV be the set of K linear bijection from V to V Then GLMV is a group under composition called the general linear group of V It is easy to verify that GLMV is not abelian unless V has dimension 0 or 1 Let I be a set Then the set SymI of all bijection from I be I is a group under composition called the symmetric group on I If I 1 n we also write Symn for SymI Symn is called the symmetric group of degree n SymI is not abelian as long as I has at least three elements Above we obtained various examples of groups by starting with a monoid and then considered only the invertible elements This works in general Lemma 222 basicmonoid Let G be a monoid a Suppose that a bc E G a is a left inverse ofb and c is right inverse of I Then a c and a is an inverse b An element in G has an inverse if an only if it has a left inverse and a right inverse c Each element in G has at most one inverse d Ifz and y are invertible then x71 and my are invertible Namely m is an inverse of x71 and yilzil is an inverse of my 22 SEMIGRO UPS MONOIDS AND GROUPS 11 e Let 1 be the set of invertible elements in G then 1 is a group Proof a a ae abc abc ec c b and c follow immediately from a d Clearly m is an inverse of m l Also y lw l w y 1 1xy y 1 1y y 1ey y ly 6 Similarly myy 1m 1 e and so y lm 1 is indeed an inverse for my e By d we can restrict the binary operation G x G a G to get a binary operation 1 x 1 a 1 Clearly this is associative Also e E 1 so 1 is a monoid By d m 1 E 1 for all m E 1 and so 1 is a group D 1 Every group G is isomorphic to its opposite group GOP Indeed the map m a m is an anti automorphism of G and an isomorphism G a GDP The associative law says that abc abc for all a b c is a semigroup Hence also abcd abcd ab Cd abcd abcd for all abcd in G That is for building products of four elements in a given order it does not matter how we place the parenthesis We will show that this is true for products of arbitrary length The tough part is to de ne what we really mean with a product of 11 an where ai E G for some pre group G We do this by induction on n For n 1 a1 is the only product of a1 For n 2 2 z is a product of a1 an if and only if z my where m is a product of 11 am and y is a product of am1an for some 1 lt m lt n The only product of 11012 is alag The products of a1a2a3 are a1a2a3 and a1a2a3 Associativity now just says that every 3 tuple as a unique product For later use if G has an identity we de ne e to be the only product of the empty tuple For the proof of next theorem we also de ne the standard product of a1 an For n 1 this is al while for n 2 2 it is man where m is the standard product of a1 an1 Theorem 223 General Associativity Law Let G be a semigroup Then any non empty tuple of elements has a unique product Proof By induction on the length n of the tuple For n 1 or 2 there is nothing to proof So suppose n 2 3 We will show that any product 2 of 11 Man is the standard product By de nition of 2 z my where m is product of a1 am and y is a product of am1 an Suppose rst that m n 7 1 By induction m is the standard product of a1 an71 Also 2 man and so by de nition 2 is the standard product 12 CHAPTER 2 GROUP THEORY Suppose next that m lt n 7 1 Again by induction y is the standard product of am an and so y 3a where s is the standard product of am an1 Hence 2 my Asa zsan As ms is a product of a1 an71 we are done by the m n 7 1 case B One of the most common ways to de ne a group is as the group of automorphism of some object For example above we used sets and vector spaces to de ne the symmetric groups and the general linear group If the object is a pre group G we get a group which we denote by AutG So AutG is the set of all automorphisms of the pre group G The binary operation on AutG is the composition We will determine the automorphism for the Latin squares in 21 As the identity element is unique it is xed by any automorphism It follows that the Latin square of order 1 or 2 have no non trivial automorphism any structure as the trivial automorphism which sends every element to itself The Latin square of order three has one non trivial automorphism It sends 5H5 agtb baa Consider the rst Latin square of order 4 It has two elements with m 25 namely a and 0 So again we have a unique non trivial automorphism 5H5 agtc bgtb caa Consider the second Latin square of order 4 Here is an easy way to describe the multipli cation ez mzz e and my 2 if z yz 1 130 It follows that any permutation of 5 a b c which xes e is an automorphism Hence the group of automorphism is iso morphic to Sym3 Consider the Latin square of order 5 labeled The multiplication table was uniquely determine by any pair z 7 y of non trivial elements with z2 yz 5 But 2 e for all z So every 5 7 z 7 y 7 5 there exists a unique automorphism with Lam bay Thus the group of automorphisms has order 12 The reader might convince herself that also the set of bijection which are automorphisms or anti automorphisms form a group In this case it has order 24 That is any bijection xing e is an automorphism or anti automorphism Consider the Latins square of order ve labeled This multiplication table is uniquely determine by any element with z2 7 e zzz e and x e a b and d have this property and we get two non trivial automorphism egteagtb bad CH6 dgtaand5gt5 gtd baa C C dab 22 SEMIGRO UPS MONOIDS AND GROUPS 13 That is any permutation xing e and c and cyclicly permuting a b d is an automorphism Consider the Latins square of order ve labeled This time only a itself has the de ning property It follows that no non triVial automorphism exists But it has an anti isomorphism xing a b and d and interchanging a and c The Latin square 4 and 5 had been anti isomorphic to 2 and So consider 6 All non triVial elements have the de ning property So there are 4 automorphisms They x 5 and cyclicly permute a b c 1 Finally consider the Latin square Here a c d have the de ning property So there are 3 automorphism They xe e and b and cyclicly permuted a cd Here all bijections xing a and b are automorphism or anti automorphism It might be interesting to look back and consider the isomorphism types of the groups we found as automorphism of Latin squares ZnZ for n 1 2 3 4 Sym 3 and a group of order 12 We will later see that Sym4 has a unique subgroup of order 12 called Alt4 So the group of order 12 must be isomorphic to Alt4 Another class of objects one can use are graphs We de ne a graph to be a tuple F 7 where P is a set and 7 7 7 is an anti re exive symmetric relation on P The elements are called vertices If a and b are vertices with a 7 b we say that a and b are adjacent An edge is a pair of adjacent vertices An automorphism of the graph P is an bijection Oz 6 SymF such that a 7 b if and only if aa 7 041 In other words a bijection which maps edges to edges AutP is the set of all automorphisms of P under composition As an example let F4 be a square 4 3 1 2 The square has the following automorphisms rotations by 0910180 and 270 degrees and re ections on each of the four dotted lines So AutP4 has order 8 To describe AutP4 as a subset of Sym4 we introduce the cycle notation for elements of SymI for a nite set I We say that 7139 E SymI is a cycle of length if the exists a1am E I such that 7ra1 a2 7ra2 a3 7ram1 am 7ram a1 and 7rj j for all otherj E I Such a cycle will be denoted by a1a2a3 am The set 11 am is called the support of the cycle Two cycles are called disjoint if their supports are disjoint 14 CHAPTER 2 GROUP THEORY It is clear that every permutations can be uniquely written as a product of disjoint cycle k k k 7139 jaguaim againnfm alazamk One should notice here that disjoint cycles commute and so the order of multiplication is irrelevant Often we will not list the cycles of length 1 So 13526 is the permutation which sends 1 to 3 3 to 55 to 1 2 to 6 6 to 2 and xes 4 and any number larger than 6 With this notation we can explicitly list the elements of AutF4 The four rotations e 1234 1324 1432 And the four re ections 1423 13 1234 24 23 The projective plane of order 2 In this section we will look at the automorphism group of the projective plane of order two To de ne a projective plane consider a 3 tuple E P R where P and E are non empty sets and R Q P x R The elements of are called points the elements of E are called lines and we say a point P and a line 1 are incident if P l E R E is called a projective plane if it has they following two properties PP1 Any two distinct points are incident with a unique common line PP2 Any two distinct lines are incident with unique common point We say that a projective plane has order two if every point is incident with exactly three lines and every line is incident with exactly three points Before studying the automorphism group will need to establish some facts about projective planes of order two Let P be any points Then any other point lies on exactly one of the three lines through P Each of whose three lines has 2 points besides P and so we have 1 3 2 7 points By symmetry we also have seven lines A sets of points is called collinear if they points in the set are incident with a common line Now let A B C be any three points which are not collinear We will show that the whole projective plane can be uniquely described in terms of the tuple A B C Given two distinct points P and Q let PQ be the line incident to P and Q Also let P Q be the unique point on PQ distinct from P and Q Since two distinct lines have exactly on point in common A B CA BA C B C are pairwise distinct Also the two lines AB C and BA C have a point in common and this point is not one of the six points we already found Hence they intersect in A B C B A C C A B the seventh and last point Also AB AC BC AB C BA C CA B are six pairwise distinct lines Now A BA C must intersect BC But B does not lie on A BA C since otherwise A BA C A BB AB Similar C is not on A BA C So 24 SUBGROUPS COSETS AND COUNTHVG 15 B C lies on A BA C and B C A B A C So the seventh line is incident with A B A C and B C So we completely determined the projective plane A An automorphism of E is a pair 04 6 SymP x Sym such that for each point P and each line 1 P and l are incident if and only if aP and 31 are incident Clearly B is uniquely determined by 04 namely ifl PQ then 31 aPaQ and after identifying a line with the set of incident points an automorphism of the plane is just a permutation of the points which sends lines to lines Let Aut be the set of automorphisms ofE Then Aut is a group under composition Let A B 639 be three non collinear points lts clear from the above that there exists a unique automorphism of E with A gt A B gt B C gt 3 llow A can be any one of the seven points Bis any of the six points different from A and C is any of the four points not incident to AB Thus lAut l764168 We nish this section with a look at the operation l we have introduced on the points Let G e U P Here e is an arbitrary element not in P De ne a binary operation on G as follows e g g e P P e and for distinct points P and Q P Q is as above It is easy to check that G is a group Also the points correspond to the subgroup of order 2 in G and the lines to the subgroups of order 4 In particular there is an obvious isomorphism between Aut and AutG 24 Subgroups cosets and counting De nition 241 defsubgroup Let G be a group and H Q G We say that H is a subgroup ofG and write H g G if 16 CHAPTER 2 GROUP THEORY a e E H b H is closed under multiplication that is for all a b E H ab 6 H c H is closed under inuerses that is for all a E H a 1 E H Note that any subgroup of G is itself a group where the binary operation is given by restricting the one on G We leave it as an exercise to the reader to verify that a subset H of G is a subgroup if and only if H is not empty and for all ab E H ab 1 6 H The following lemma is of crucial importance to the theory of groups Lemma 242 leftcosets Let H be a subgroup of G The relation N on G x G de ne by a N b if and only if there exists h E H with b ah is an equivalence relation Proof Let a E G then a ae and so N is re exive Let a N b and pick h E H with b ah Then bh l aha 1 ahh 1 ae a Since H is a subgroup of G h 1 E H and so b N a Hence N is symmetric Suppose next that a N b and b N c Then b ah and c bk with k h E H We compute c bk ahk ahk Since H is a subgroup hk E H and so a N c and N is transitive D The reader might have noticed that the re exivity corresponds to e E H the symmetry to the closure under inverses and the transitivity to the closure under multiplication lndeed N can be de ned for any subset of G and its a equivalence relation if an only if the subset is a subgroup The equivalence classes of N are called the left cosets of H The coset containing a E G is denoted by aH and so aH ah l h E The set of cosets of H is denoted by GH and so GH gH l g E G Observe that the map H a gH l h a gh is a bijection and so lng for all 9H 6 GH Theorem 243 Lagrange Let H be a subgroup of G Then lGl In particular if G is nite the order ofH diuides the order of G Proof Just observe that is the numbers of cosets each coset contains elements and G is the disjoint union of the cosets B Let 73G be the power set of G that is the set of subsets For H K C G put HKhklheHkeK 25 NORMAL SUBGROUPS AND THE ISOMORPHISM THEOREM 17 This binary operation is associative and e is an identity element So 73G is a monoid lf K is a subgroup then HK is a union of cosets of K namely HK UhEH hK We write HKK for the sets of cosets of K in HK In general if J Q G is a union of cosets of H JH denotes the sets of all those cosets The same argument as in the proof of Lagrange s Theorem shows lJl Lemma 244 Let H and Kbe subgroups of G a The map HH O K gt HKKhH K gt hK is a well de ned bijection b lHKl Proof a Since H KK K hK hH KK hH KK and so is indepen dent of the choice of h E hH The map is clearly onto Finally if hK jK for some hj E H then h le K h lj E K and so h lj E H K and hH K jH K Thus the map is one to one b By a we have So lHl lKHl lH Kl Thus b holds D 25 Normal subgroups and the isomorphism theorem Just as we have de ned left cosets one can de ne right cosets for a subgroup H of G The right cosets have the form Hg hg l h E In general a left coset of H is not a right coset De nition 251 defnormal Let G be a group a For ab E G put b Lbo f1 and for Q G put I alail i l i E I The map i G gt G b gt b is called conjugation be a b A subgroup H of G is called a normal subgroup ofG and we write H S G if HH9forallg G Lemma 252 basicnormal Let N g G Then the following are equiualent a N 31 G b gN Ng for allg E G C Every left coset is a right coset 18 CHAPTER 2 GROUP THEORY d Euery left coset is contained in a right coset e N9 Q N for allg E G Proof Suppose a holds Then gNg 1 N for all g C G Multiplying with g from the right we get gN Ng Suppose b holds Then the left cosets gN equals the right coset Ng so c holds Clearly c implies d Suppose that d holds Let g C G Then gN Q Nh for some h C G Since 9 C gN we conclude g C Nh and so Ng Nh7 as the right cosets partition G Thus gN Q Ng Multiplying with 9 1 from the right we get gNg 1 Q N Thus e holds Finally suppose that e holds Let g C G Then also 9 1 C N and applying e to 9 we obtain g lNg C N Multiplying with g from the left and 9 1 from the right we obtain N C gNg 1 N9 Since also N9 C N7 N N9 and so a holds D 1 We will now start to establish a connection between normal subgroups and homomor phism Lemma 253 basichom Let o5 G a H be a group homomorphism a MS 6 b W171 gta 139 gt may gtltagt ltggt c d If A g G then gtA H e If B H then 1B G f Let ker gt g E G l e Then ker gt is a normal subgroup of G g j is one to one if and only if ker gt e h UN 1 G and gt 239s onto S H i IfM 51 H 1M 51 G Proof Straightforward D For H Q G de ne H 1 If1 l h CH Lemma 254 basiCGN Let N 31 G a The product of any two cosets ofN is again a coset of N Namely aN bN abN b For each 9 E G gN 1 is a coset of N namely gN 1 gilN 25 NORMAL SUBGROUPS AND THE ISOMORPHISM THEOREM 19 c GN is a group d The map G gt GN7 g gt gN is an onto homomorphism with kernel N Proof a aNbN aNbN abNN abNN abN b 4 LN 1 Nof1 ailN c By a multiplication is binary operation on GN Clearly N is an identity element and by a g lN is an inverse for gN Also multiplication of subsets is associative and so c holds d By a the map is a homomorphism Clearly it is onto 9 is mapped to eaN N if and only if gN N and so if and only if g E N Thus N is the kernel of the map E Lemma 255 The Isomorphism Theorem IT Let 64 G a H be a homomorphism of groups The map 64 Gker64 a 64H7 gker64 a 649 is a well de ned isomorphism Proof Let a E ker64 and g E G Then a9n a9an a96 Mg and so 64 is well de ned By de nition 64 is onto Let gker64 E ker64 Then 649 e and so 9 E ker64 and gker64 ker64 eakem Hence by 253g7 64 is one to and so an isomorphism From the two preceding lemmas we see that the normal subgroup of G are exactly the kernels of homomorphism Also every homomorphism G a H can be factorized as 64 o64n397 where 7139 G a G ker64 is the canonical homomorphism from 2547 64 is the isomorphism from 255 and p 64H a H7 h a h is the inclusion map Note here that 7139 is onto7 64 an isomorphism and p is one to one Lemma 256 capg Let G be a group and Chi E I afamz39ly of subgroups Then mid Gi is a subgroup If all of the Ci are normal in C so is mid Ci Proof Left as an exercise D De nition 257 Let G be a group and J Q G a The subgroup J of G generated by J is de ned by 20 CHAPTER 2 GROUP THEORY b The normal subgroup JG of G generated by J is de ned by ltJGgt n H JQHSIG lf Jhi E I is a family of subsets we also write Ji ll 6 I for normal if J9 J for all g E G idJ J Q G is called Lemma 258 basicgen Let I be a subset of G a Let Oz G gt H be a group homomorphism Then altIgt ltaIgt b Letg E G Then I9 ltIgt9 c d IfI is normal in G so is ltIgt consists of all products of elements in I U I71 e IO I9 l g E G and consists of all products of elements in UQEGU U I71 Proof a Let A I and B ltaIgt As aA is a subgroup ofH and contains aI we have B aA Also a 1B is a subgroup of G and contains I Thus A a 1B and so aA B Hence B aA b Apply a to the homomorphism z a my c Follows from d Let H be the subset of G consists of all products of elements in I U I l7 that is all elements of the form a1a2 an7 with n 2 0 and ai E I U I 1 for all 1 g t g n Then Clearly H is contained in any subgroup of G containing I Thus H Q But H is also a subgroup containing I and so I g H e By b I9 l g E G is normal subgroup of G It is also contained in every normal subgroup containing I and we get IO I9 l g E G The second statement follows from D c For a7 b E G put a7 b abailb 1 and for A7 B Q G de ne A7B 0119 l a 6 A71 6 B a7 1 is called the commutator of a and b Note that a7 1 e if and only if ab ba Also a b bab l ab where we used the abbreviation b b 1 17 Finally observe a7bli1 babila 1 b7a Hence A7 B B7A for any AB Q G 25 NORMAL SUBGROUPS AND THE ISOMORPHISM THEOREM 21 Lemma 259 comnormal Let G be a group a LetN G Then N S G if and only if N7G N b Let AB 51 G Then A B A o B C Let AB S G with A B 5 Then A7B e and ab ba for alla E Ab E B Proof n9 Nltgtn9n 1 eNltgt mg EN Thus a holds b follows from a7 and 6 follows from aamp D For H Q G de ne N0H g E H l H9 H N0H is called the normalizer of H in G If K Q G we say that K normalizes H provided that K Q N0H Lemma 2510 normalize Let G be a group a Let A7 B be subgroups of G Then AB is a subgroup of G if and only if AB BA b Let H Q G Then N0H is a subgroup of G c If KH G and K NaH then ltKHgt KH d Let Kin E I be a family of subsets if G If each Ki normalizes H so does lie I Proof a If AB is a subgroup of G7 then AB AB 1 B lA l BA Conversely suppose that AB BA The above equation shows that AB is closed under inverses Also 5 ee 6 AB and ABAB ABAB AABB A2B2 AB So AB is closed under multiplication b Readily veri ed e Let k E K Then kH Hk and so HK KH So by a HK is a subgroup of G d Follows directly from D 22 CHAPTER 2 GROUP THEORY 26 Cyclic groups A group is cyclic if G for some z E G In this section we will determine all cyclic groups up to isomorphism and investigate their subgroups and homomorphisms Lemma 261 subgroupsZ a Let H be a subgroup of Z7 Then H nZ for some n E N b Let n7 m E N Then nZ mZ if and only ifm diuides n Proof a If H 07 then H OZ So we may assume that H 7 Since H is a subgroup7 m E H implies 7m E H So H contains some positive integer Let n be the smallest such Let m E H and write m Tn 37 T7 8 E Z with 0 g s lt n We claim that Tn E H Tn E H if and only if T n E H So we may assume T gt 0 But then Tnnnn hf x Titimes and as n 6 H7 Tn E H So also 8 miTn E H Since 0 g s lt n7 the minimal choice ofn implies s 0 Thus m Tn E nZ and H nZ b nZ mZ if and only if n E mZ So if and only if m divides n D Lemma 262 freel Let G be a gToup and g E G Then 1 Z a G7 n a g is the unique homommjnhism from Z7 to G which sends l to 9 Proof More or less obvious E if lt ForT ZUoode neT T T 00 if T 00 This de nition is motivated by the fact that lZnZl n Lemma 263 clfcyclic Let G be a cyclic gToup and put n WV a The map ZnZHG7 mnZHzm is a well de ned isommphism b Let H g G and put m lGHl Then m diuides n and H Proof By 262 the map 1 Z a Gm a 9 quot is a homomorphismAs G z j is onto By 261 ker gt tZ for some non negative integer t By the isomorphism theorem the map ZtZHGmtZgtmm 27 SHVIPLICITY OF THE ALTERNATIN G GROUPS 23 is a well de ned isomorphism Hence ZtZ E G Hence t lZtZl lGl n and a is proved b By 261 gt 1H 3 for some 3 E N Since nZ 3 261 implies that s divides n As 1 is onto7 gt5Z H and so H lt gtsgt Moreover GH ZnZsZnZ ZSZ Thus 3 m andb is proved D Lemma 264 Let G be a cyclic Let Hbe any group and y E H Put n lGl and m Then there exists a homomomhism G gt H with m gt y if and only ifm divides n Proof Exercise D 27 Simplicity of the alternating groups In this section we will investigate the normal subgroups of symmetric group Symn7 n a positive integer We start be de ning a particular normal subgroup called the alternating group Altn Let R be the eld of real numbers and GLYLOR the group of invertible linear transformation of the vector space R De ne Oz Symn a GLWOR by oz7rrl rn 7 7 77117 77 7r71n It is easy to check that 04 is a homomorphism Now de ne sgn det 004 As the determinant is a homomorphism from GLWOR to R sgn is a homomorphism form Symn to R Also if z Lj E Symn is a 2 cycle7 we see that sgnz 71 So ifm 11012 am 6 Symn is a m cycle then sgnz sgn a17a2a2a3 am1am 71m 1 So we conclude that 1 if x has an even number of even cycles sgn 71 if x has an odd number of even cycles De ne Altn ker sgn Then Altn is a normal subgroup of Symn7 Altn consists of all permutation which have an even number of even cycles and SymnAltn sgnSymn 17 71 Z2Z ln particular7 X when 24 CHAPTER 2 GROUP THEORY Before continuing to investigate the normal subgroup of Symn we wish to de ne con jugacy classes in an arbitrary group G We say that two elements m y in G are conjugate if y 9 gmg 1 for some 9 E G It is an easy exercise to verify that this is an equivalence relation The equivalence classes are called the conjugacy classes of G The conjugacy class containing z is m0 mg l g E G Proposition 271 normalcc A subgroup of G is normal if and only if it is the union of conjugacy classes of G Proof Let N g G The following are clearly equivalent N51 G N9 Nforallg G ngeNforallnEN796G nGQNforallnEN NUnENnG N is a union of conjugacy classes D To apply this to Symn we need to determine its conjugacy classes For x E Symn de ne the cycle type of z to be sequence 7710121 where mi is the numbers of cycles of z of length 2 So the cycle type keeps track of the lengths of the cycles of z counting multiplicities Proposition 272 Two elements in Symn are conjugate if and only if they have the same cycle type Proof We will show that the conjugate of the cycle 9 a17 7am by the element 7139 E Symn is the cycle n39a1n39a27 77ram Let 1g k g n If k 7 7raj for some j then 7r 1k is xed by 9 So 7rg7r 1k 7r7r 1k k If k 7raj then W9W71Wa 9WD 17 Suppose now that g has cycle type Then 9 H H gij 13921 l j mi where gig a1lj7 ailj is a cycle of length 2 Moreover for each 1 g l g n there exists uniquely determined lj and k with l aklj ln particular7 9quotH H 9 13921 l j mi 27 SHVIPLICITY OF THE ALTERNATIN G GROUPS 25 where 97 7ra1ij 7raiij is a cycle of length 2 So also g7r has cycle type Conversely suppose also 71 has cycle type so h H H 71 13921 l j mi where hij 61ij is a cycle of length 2 De ne 7r 6 Symn by WWW 5W3 But then g7r h and g and h are conjugate D Lets now investigate the normal subgroups of Sym3 We start by listing the conjugacy classes 5 1 element 123 132 2 elements 12 13 23 3 elements Let 5 34 N 31 Sym3 If N contains the 2 cycles then N 2 4 Since N divides Sym3 6 we get N 6 and N Sym3 If N does not contain the 2 cycles we get N 5 123 132 Alt3 So Alt3 is the only proper normal subgroup of Sym3 Lets move on to Sym4 The conjugacy classes are e 1 element 123 132 124 142 134 143 234 2431234 1324 1423 3 elements 12 13 14 23 24 34 6 elements 1234 1243 1324 1342 1423 1432 8 elements Let N be a proper normal subgroup of Sym4 then N divides 24 Sym4 Thus N 234 68 or 12 So N contains 1 2 3 57 or 11 non trivial elements As Ne is a union ofconjugacy classes N712 3 So N71 E 3 5 7 11 In particular N71 is odd But Sym 3 has a unique non trivial conjugacy class of odd length namely the double 2 cycles So K Q N where K 5 1234 1324 1423 Then E 0 38 All remaining conjugacy classes have length 6 or 8 and we conclude that N K or 8 In the second case N consist of K and the 3 cycles and so N Alt4 Note also that 1234 1324 1423 and so K is indeed a normal subgroup of Sym4 Thus the proper normal subgroups of Sym4 are Alt4 and K Just for fun lets as determine the quotient group Sym4K No non trivial element of K xes 7 4 So Sym3 O K e and Sym3K 64 s 3K7 77 Wm Sym3 K 1 24 Sym4 26 CHAPTER 2 GROUP THEORY Thus Sym3K Sym4 And Symlt4gtK Symlt3gtKK Symlt3gtltSymlt3gt m K Symlt3gte Symlt3gt So the quotient of Sym4 by K is isomorphic to Sym3 Counting arguments as above can be used to determine the normal subgroups in all the Symn s but we prefer to take a different approach Lemma 273 3cycles N Altn Proof a By induction on n If n g 3 all non triVial elements in Altn are 3 cycles So we may assume n 2 4 Let H be the subgroup of Symn generated by all the 3 cycles By induction Altn 71 g H Let g E Altn lf gn n n E Altn 71 g H So suppose gn 7 n and let a lt n with a 7 Let h be the 3 cycle gnna Then hgn hgn n Hence by E Altn 71 g H and so also 9 h 1hg E H b Let h abc be a 3 cycle in Symn Since n 2 5 there exist 1 g d lt e g n distinct from a b and 0 Note that a bc a bd e b cd e and so the subgroup generated by the double 2 cycles contains all the 3 cycles Hence b follows from a c Let h abc by a 3 cycle in N and 9 any 3 cycle in Symn By a it suffices to prove that g E N Since all 3 cycles are conjugate in Symn there exists if E Symn with ht g lft E Altn we get 9 ht E N N as N is normal in Altn So suppose that 25 Altn Then tab E Altn Note that h 1 c b a b a c and so h 1 gtb b a 20117 a b c h Thus h71tab h71abt ht 9 As the left hand side is in N we get 9 E N d A similar argument as in c shows that b implies d D Theorem 274 altnsimple Let n 2 5 Then Altn has no proper normal subgroup 27 SHVIPLICITY OF THE ALTERNATIN G GROUPS 27 Proof If n gt 5 we assume by induction that the theorem is true for n 7 1 Let N be a non trivial normal subgroup of Altn Case 1 N contains an element g 344 5 with i for some 1 g i g 71 Let Hh Altn l i ThenHAltn71g EH NandsoH Nisa non trivial normal subgroup We claim the H O N contains a 3 cycle or a double 2 cycle Indeed if n 5 then 71 7 1 4 and the claim holds as every non trivial element in Alt4 is either a 3 cycle or a double 2 cycle It is also true for n gt 5 since then by induction H O N H By the claim and 273cd we conclude N Altn Case 2 N contains a element 9 with a cycle of length at least 3 Let a 30 be a cycle of g of length at least 3 Let 1 g d g n be distinct from ab and 0 Put h 901 Then h has the cycle dba Also as N is normal in Altn h EN So also by EN We compute hga 711 a and hgb 710 344 hd I So by 344 e and by xes 77of So by case 1 N Altn Case 3 N contains an element 9 with at least two 2 cycles Such a g has the form abcdt where t is a product of cycles disjoint from a 313 1 Let h 9W Then h bcadt Thus ghil abcdtt 1bcad acbd As h and gh l are in N Case 1 or 273d shows that N Altn Now let e 344 g E N As 71 2 4 9 must ful ll one of the three above cases and so N Altn D A group without proper normal subgroups is called simple So the previous theorem can be rephrased as For all n 2 5 Altn is simple Proposition 275 normalsym Let N lt1 Symn Then either M eAltn or Symn or n 4 and N e 1234 1324 1423 Proof The case n g 4 was dealt with above So suppose n 2 5 Then N O Altn is a normal subgroup of Altn and so by 274 N O Altn Altn or e In the rst case Altn N Symn Since lSymnAltnl 2 we conclude N Altn or N Symn In the second case we get lNl lNN Altnl lNAltNAltNl lSymnAltnl 2 Suppose that lNl 2 and let e 344 n E N As 712 e n has a 2 cycle ab Let a 344 c 344 b with 1 g c g n The Mal has cycle be and so n 344 Mal A contradiction to N 571 and N 1 Symn D 28 CHAPTER 2 GROUP THEORY Lemma 276 absim The abelian simple groups are exactly cyclic groups of prime order Proof Let A be an abelian simple group and e 7 a E A Then a 31 A and so A a is cyclic Hence A ZmZ for some m 2 0 If m O7 2 is a normal subgroup Hence m gt 0 If m is not a prime we can pick a divisor 1 lt k lt m But then kZmZ is a proper normal subgroup 28 Direct products and direct sums ldpd Let G1 and G2 be groups Then G G1 x G2 is a group where the binary operation is given by 0117 a2b17b2 011191701252 Consider the projection maps 7r1G a G1 a17a2 a a1 and 7r2 G a G2 a17a2 a a2 Note that each 9 E G is uniquely determined by its images under m and 7r2 Indeed we have 9 71197 7r2g We exploit this fact in the following abstract de nition De nition 281 Let Cl7i E I be afamily of groups The direct product of the Cl7i E I is a group G together with a family of homomorphism m G gt Chi E I such that WheneuerH is a group and 04 H gt Hhi E I is family of homomorphism then there exists a unique homomorphism oz H gt G such that the diagram G H at commutes for alli E I We will now show that the direct product in the above sense always exists As a set let G HE G the set theoretic product of the G23 That is G consists of all function f I a Uid G with E G for all i E I The binary operation is de ned by fhi It is easy to check that G is a group De ne mHerfa i39eI Obviously m is a homomorphism Let H a group and 04 H a G a family of homomor phism De ne oz H a G by 28 DIRECT PRODUCTS AND DIRECT SUMS 29 ahi Olilthgtf0r all i E I But this is equivalent to and so to ma 04 In other words7 04 is the unique map which makes the above diagram commutative lt trivial to verify that 04 is a homomorphism and so mi E I meets the de nition of the direct product Lets go back to the above example G G1 x G2 This time we consider the inclusion maps p13G1 a G791 a ghe and p2 G2 a G792 a e792 Note also that p1G1p2G2 e Given a group H A family of commuting homomorphism is a family of homomorphism 041GlH Hi E I such that aigiaj9j a g adgi foralliy j Igi EC andgj EGJ39 De nition 282 A direct sum of the family of groups Chi E I is a group G together with a family of commuting homomorphism pi Gi gt Gi E I such that Wheneuer H is a group and 04139 Gi gt Hi E I is a family of commuting homomor phism then there exists a unique homomorphism Oi G gt H so that the diagram G a H i commutes for alli E I Before we proceed and show that directs sums exists we prove a couple of lemmas7 which are useful in the construction Lemma 283 directsumgen Let pi G a Gi E I be a direct sum of Chi E I Then a Each phi E I is one to one M G ltpiltaigt HED 30 CHAPTER 2 GROUP THEORY Proof a For Lj E I de ne ifz aijGiHG7ga ifii Then for each j 6 I7 0417 l i E I is a family of commuting homomorphisms So there exists 04739 C G H Gj With 041739 Oszi As 04 idGi is one to one7 also pl39 is to one b Let H ltpiGi l i E I De ne 5 Gi a H79 a pig Hence there exists map 3 G a H with pi 51 3 de nes a map G a G79 a 39 Then pi pi for all i E I Now also idG is a map from G a G with idapi pi so the uniqueness part in the de nition of the direct sum implies B idG Thus for all g 6 G7 9 idGG 39 6 H D Lemma 284 alphaunique Let 043 G a H be group homomoTphz39sm Let Chi E I be a family of subgroups of G Suppose that a 04 02 lai for alliEI b GltGili I Thenoz Proof Let D g E G l 049 39 Clearly D is a subgroup of G By a G g D for alliEIThusbybG DandsoDG D It follows from the two preceding lemmas that the uniqueness statement in the de nition of the direct sum can be equivalently replaced by the condition G ltpiGi l i E I We will construct the direct sum as a subgroup of the direct product Hid G We will View the elements of the direct products as tuples g gm61 with 91 6 Ci for all i E I De ne the support suppg of g by supplt9gt i E I l 9139 3quot 6 63G g E HG l suppg is nite id tel NOW suppe 0 supplt9gt supptcfl and supp9h Q supp9 U supph SO 961 Gi is a subgroup of Hid G De ne pi Gi a Bid C by Pi9j 5 m g ifji 28 DIRECT PRODUCTS AND DIRECT SUMS 31 Then for all i 7 j gi if k i pi9ipj9jk 9739 if k j pj9jpigik e ifi 7g k and so phi E I is a family of commuting homomorphism Note that 3 the subgroup of Hid Gl generated by the piGil E I Let H be a group and al Gl H Hi E I a family of commuting homomorphism De ne oz 1616 H H by id Ci is exactly would Hatlt9 id The product on the right hand side does not seem to make sense at rst But all but nitely many of the 91 s and so also of aigi s are triVial So the product is effectively a nite product Secondly since the OliltGigt commute with each other7 the order in which the product is taken does not matter To verify that 04 is homomorphism we compute 0490 h Haigi Haihi Haigiaihi Hai9ihi 04971 id id ieI id Let g 6 Ci Then pigj e for all j 7 1 and ozpg Otig Thus 04 041 The uniqueness of oz follows from G and 284 B We say that G is the internal direct sum of Chi E I provided that each Ci is a subgroup of G and that G7 together with the inclusion map from Gl to G7 is a direct sum of Chi E I In particular this means that for each family og Gl H H of commuting homomorphism there exists a unique homomorphism oz G a H with 04 02 041 Proposition 285 internal sum Let G be a group and Chi E I a family of subgroups The following are equivalent 1 G is the internal direct sum of the Chi E I 2 Each of following three statement hold a Each Gi is normal in G b GltGi liEI c For each i Gi Cg E I e 3 Gi7 Cg e for alli E G and eachg E G can be uniquely written as Hid 9139 there 91 6 Ci and all but nitely many 91 s are triuial 32 CHAPTER 2 GROUP THEORY Proof 1 gt 2 Suppose 1 holds By 283b7 2b holds As the inclusion maps commute7 Gi7 Gj e for all i 7 j In particular Gj N0Gi for all j E I So 2a follows from 2b Let 0417 and al be as in the proof of 283a Then 1g g for all g E G In particular7 G O kerai e For all j 7 i Gj kerai so 2c holds lt2 lt3 Suppose 2 holds H Gi Ci 1 i j E I By assumption each Gj is normal in C so both C and H are normal in G Moreover by assumption G O H e and so Gi7Hi Gi Q Hi 6 Thus Gl7 Gj e for all i j E I By assumption G is generated by the G23 and so 9 E G can be written as Hid 91 Suppose that Hg H 61 61 for some 9139 L E G Then aigil H 619739 jel As the left side lies in Gi and the right side in H we conclude that Ll9171 e and so ai 91 Thus 3 holds 3 5 1 Suppose that 3 holds Let 04139 Gi a HJ I be a family of commuting homomor phism De ne Oz G a H by 049 Hid Ozigi where g H619 and 9 E G Hence 1 holds D The direct sums are slightly more natural in the category of abelian groups7 namely every family of homomorphisms is automatically a family of commuting homomorphisms In particular we see that the direct sum of a family of abelian groups is the coproduct in the category of abelian groups The direct sum can also be used to de ne the free abelian group FAI on a set I Namely put FAI EB z ieI ldentify i with pi1 Then we see that every element in FAI can be uniquely written as iEI where m E Z and almost all m s are 0 Note here that in an abelian group we write na for a FAI as the following universal property 2 9 FREE PRODUCTS 33 Theorem 286 freeabelian Let A be an abelian group and ahi E I a family of ele ments in A Then there exists a unique homommphism aFAI HA withiaai for alli GI Proof This follows from 262 and the de nition of the direct sum 04 is given by 042 2 mai i6 i6 D Direct products and direct can also be de ned for semigroups and monoids Direct sums cause a slight problem for semigroups since then the condition 9 e for almost all i makes no sense Maybe the easiest way to go around this problem is to embed each semigroup Gl into a monoid G Gi U ei where el is an identity in Gil Then de ne Gi a Cl US61 i39eI i39eI The free abelian monoid on a set I is EB N id and removing the identity from this we get the free abelian semigroup 29 Free products Let Chi E I be a family of groups In this section we will construct a group called the free product of the Cl7i E I and denoted by Hid G On an intuitive level this group is the largest group which contains the Gi s and is generated by them On a formal level it is the coproduct of the Gi s in the category of groups See theorem 295 below To simplify notation we will assume without loss of generality Hypothesis 291 Chi E I is a family of groups ii 5G2 ea for all ij E I Call this common identity element 5 iii Gi Gj e for alliy j GI Let X Uid G For x E X with z 7 5 let im 6 I be de ned by z E Gian Note that by iii of our Hypothesis im is well de ned Let n E N A word of length n in X is a tuple 34 CHAPTER 2 GROUP THEORY 17 2 with mm 6 X for all 1 g m g n W denotes the set of all words The empty tuple n 0 is denoted by we De ne a binary operation 7 7 on W by x17z27 7mn gtk yhyl 7gm 1 7zng1yn Clearly 7 is associative As we is an identity7 W is a monoid ldentify each x E X with the word Call ab E X comparable if there exists l E I with ab E G So either a e7 or b e7 or i la lb We would like that ab a b for such a7 b7 but this is nonsense7 since ab is a word of length 1 and a b a7 1 is of length 2 To x this problem we will introduce an equivalence relation on W Let v w E W Write v lt w if one of the following holds 77 lt 1 There exists my 6 W and comparable ab E Xwith w maby andv zaby lt 2 Thereexists 71 6 Wsothat wzey andwzy Note that v lt w is only possible if lv lw 7 17 where lv denotes the length of the word 12 Also since e is comparable with any x 6 X7 condition lt 2 implies condition lt 17 with one exception though If x and y are both the empty word we we see from lt 2 that we lt e As lt is not symmetric we de ne v 7 w if v lt w or w lt 1 Finally to achieve transitivity de ne v N w if UUe7U17U277vn17vnw for some 1 E W Here we allow 71 07 so u N v for all words 1 The minimal such 71 is called the distance of v and w Lemma 292 basicsim N is an equivalence relation and N 1 ab N a gtk b for all comparable ab in X N2 Iftvw E W withUNw thentgtkUNtgtkw N3 Iftvw E W withUNw thenvgtkthgtkt Proof By lt 1 with z y we7 ab lt ab for all comparable pairs a7b So N 1 holds To prove N 2 let u N w By induction on the distance of v and w we may assume that U 7 w7 say 1 lt w So w maby and v maby for appropriate m7y7a and b Thustw tzaby andtv twaby Hence 25 v lt tw and in particular 25 v N tw N 3 is proved with a similar argument D For w E W let w be the equivalence class of 7 N 7 containing w Let W be the set of all such equivalence classes De ne a binary operation on W by 17 w m We need to verify that this is well de ned So let 121 N 122 and w1 N wg Then the previous lemma 292 121 w1 N 122 w1 and v2 w1 N 122 102 So by transitivity of N7 121 122 N w1 wg7 as required 29 FREE PRODUCTS 35 Lemma 293 tildewgroup a W is a group b Leti E I Then map pi Gi gt W a gt is a homomorphism e W We 12 e Igt Proof As W is a monoid sois W By N 17 pi is a homomorphism of monoids Also pie we is the identity of W and it follows that a is invertible for all a 6 Ci that is for all a E X But every element in W is product of elements in X Thus c holds and all elements in W are invertible Hence also a is proved D We have found the group we were looking for Next we will proceed to nd a canonical representative for each of the equivalence classes Call a word w ml reduced if zk 7 e for all 1 g k g n and zk and zk are not comparable for all 1 g k lt n So w is not reduced if and only if there exists 1 E W with u lt w Also each 5 7 a 6 Ci is reduced We will show that every equivalence class contains a unique reduced word For this we consider one further relation on W De ne u ltlt w if uueltu1ltugun1ltunw for some 1 E W Again allow for n 0 and so u ltlt 12 Also note that u ltlt w implies u N w but not vice versa Lemma 294 reducedwords a For each w E W there exists a unique reduced word wT E W with wT ltlt w wT is called the reduction of w b u N w if an only UT wT c Each w contains a unique reduced word namely wT Proof a Choose a word 2 of minimal length with respect to 2 ltlt w If u lt z for some 1 we conclude u ltlt w and 12 2 7 1 lt 2397 a contradiction to the minimal choice of 2 Hence no such 1 exists and z is reduced This establishes the existence part To show uniqueness let 21 and 22 be reduced with 21 ltlt w If 21 w then w is reduced and 22 ltlt w implies 22 w 21 So we may assume that 21 7 w 7 22 By de nition of 7 ltlt 7 there exists ul 6 W with 21 ltlt ul lt w Suppose that vi we for somei E 17 2 Then w e and 21 we 22 Suppose next that U1 7 we 7 122 Then ul lt w ful lls lt 1 and there exist mi yi and comparable ai bi E X with vimiaibiyi wmiaibiyi 36 CHAPTER 2 GROUP THEORY Let Z be the length of 21 Without loss ll lg We will consider various cases If ll lg we get 121 122 As xi ltlt 121 v2 and 121 lt w we can apply induction and conclude 1 m2 Suppose next that lg 11 1 Then 2 ml a1 1 a2 and y1 b2 yg So 121 ml a1a2 b2 yg v2 1 a1 agbg yg lf a2 5 then 121 1 a1 b2 yg v2 and as above 1 m2 So suppose a2 7 5 Then ahag b1 and 2 all lie in a common G In particular alag and 2 are comparable The same holds for al and agbg Put u ml alagbg yg It follows that u lt 121 and u lt 122 Then zl ltlt 12 u ltlt 12 and by induction 1 u m2 Finally suppose that ll 1 lt lg Then 2 1 a1 b1 d and y1 e a2 b2 yg for some maybe empty words 5 and 1 Hence U1x1a1a2dea2b2y2 U21a1a2deazbzy2 Put u 1 a1a2 1 e agbg yg Then again u lt 12114 lt 122 and 1 u 2 This completes the proof of a Let v w be words If UT w7 then u N UT w N w and so u N w Conversely suppose that v N w By induction on the distance of v and w we may assume that v 7 w and say u lt w Then UT ltlt w and so by a7 UT w Thus b holds Let U be any reduced element in d Then 1 N w and so by b v UT w Thus also c holds B Let WT be the set of reduced words By the previous lemma the map WT a W w a L is a bijection Unfortunately7 WT is not closed under multiplication that is the product of two reduced words usually is not reduced But it is not dif cult to gure out what the reduction of the product is lndeed let x 17 7x and y yl7 be reduced words Let s be maximal with y1 zn for all 1 g t lt 3 Then m ky N117quot397n797y57quot397ym39 lf mks and ys are not comparable7 this is the reduction of z y On the other hand if mks and ys are comparable we have y N 17 39 quot 7n75717nisysyys1H 79m and this is the reduction of z y We now de ne the free product Hid G of Chi E I to be the group W together with the family of homomorphism phi E I The free product has the following important property which we could have used to de ne the free product 2 9 FREE PRODUCTS 37 Theorem 295 COGI39 Let H be a group and 04 i gt a family of homomorphisms Then there exists a unique homomorphism 64 Hid Gi gt H so that the diagram 04 Hiel Gi H commutes for alli E I Proof We de ne a map 04 W a H by induction on length of the words Put aw0 e and if w y 91 with 91 6 Ci de ne aw ozyozig Clearly 04 is a homomorphism of monoids We claim that 041 aw whenever u N w Indeed it suf ces to show this for u lt w Then w zaby and u zaby with ab 6 Ci for some i Hence WU aaaabay aaiaaibay aaiabay aaabay MU By the claim we get a well de ned map 64 W a H771 a aw Also as 04 is a homo morphism7 64 is7 too 64 is unique7 as the restriction of 64 to piGi is determined by the commuting diagram and as the piGl generate W see 284 We remark that free products also exists in the categories of semigroups and of monoids lndeed7 everything we did for groups carries over word for word7 with one exception though In case of semigroups we do not include we in the sets of words and omit lt 2 in the de nition of u lt w Recall that the only role lt 2 played was to identify 5 with we A relation in X is an expression of the form w e where w is a word in X Let R be a set of words De ne HGiltw ew ER i39eI to be the group ll7N7 where N OilW WN is called the group generated by Chi E I with relation w e7 w E R Put 5139 3 Gt HHGiltW57w 673 gapilg v i39eI Let H be a group and al Gi a Hi E I a family of homomorphism We say that the family ful lls the relations w e7 w E R if aw e for all w E W Here 04 is as in the proof of2957 that is 04w17 7um 04i1w1 ainwn where wj E Gij 38 CHAPTER 2 GROUP THEORY Theorem 296 productrelation Let H be a group and 64 Gi a Hi E I a family of homomomhism which ful lls the relations w ew E R Then there exists a unique homomomhism 64 HGiltw ew E R gt H with 6igi gt Oliltgigt id for allgi 6 Chi E I Proof By 295 there exists a unique homomorphism 64 Hi6 Gi a H with 64pigi 64igi Since 641i E I ful lls the relations7 R ker64 and so N ker64 Thus the map 64 WN a HgN a 649 is well de ned and has all the desired properties The uniqueness of 64 follows from the uniqueness of 64 D We will now work out an easy example Let I 17 2 and suppose G1 and G2 are both of order 2 Let G1 a and G2 Then it is easy to list all the reduced words we ababab x V a n times bababa x V a n times abababa 39 ntimes bababab x K ntimes Put 2 Mb Then 2 1 b lssa l Ina So the above list now reads 20 z 2 z a and z msb The last words is equivalent to z baa z m lssa Let D Hi612Gl We conclude that D znz a l n E Z Let Z Then Z Z7 and Z has index two in G1 G2 In particular Z 31 D Also 2 aza aaba ba 2 1 Thus 2 1 2 and zna az ln particular7 if A Z then both Z and a normalize D and A 1 G Here is a property of D which will come in handy later on All elements in D Z are conjugate to a or 1 Indeed znaz znzna zzna and znbz 22W zznbaa zzn1a SO zzna is conjugate to a and zzn a is conjugate to b 29 FREE PRODUCTS 39 Fix it E Z Consider the relation 2 e Put N Then N 1 D and so H iiab egt DN ie12 Since ZD E ZnZ7 DN has order 271 DN is called the dihedral group of order 271 or the dihedral group of degree 71 Suppose now that D is any group generated by two elements of order two7 7 and I Then there exists a homomorphism Oz D a D sending a to 7 and b to I Let 2 713 and Z Since neither a nor 1 are in kera and all elements in D Z are conjugate to a or b7 kera Z Thus kera for some n E N and so D E Dkera DN So any group generated by two elements of order 2 is a dihedral group Next we will use the free products of groups to de ne the free group FI on a set I First for eachi E I pick a group Gi with i 6 Ci Gi E Z7 and Gi So Then let FI H Gi id and de ne 0IgtFI7 ldentify i and i Then every element in FI can by uniquely written as W1 4W2 Wm Z1 22 Zm where m E N ik 6 I7 ik 7 ik and 0 7 nk E Z Indeed these are exactly the reduced words Theorem 297 freegroup Let I be a set and H a group Fori E I let hi 6 H Then there exists a unique homomomhism Oz FI gt H with ai hi Proof De ne 04 Gi a H7 in a hf The theorem now follows from 295 B The concept of relations also caries over For R a set of words de ne FIltw ew E R FIN7 where N END FIltw ew E R is called the groups de ned by the generators i E I and relations w e7 w E R Often this group is also denoted by ltIlwew Rgt ForiEI7putiiN 40 CHAPTER 2 GROUP THEORY Let H be group7 and hi 6 H We say that hhi E I ful lls the relations w e provide that hml 2 hm 5 ii 12 im W1 W2 4n where w 72172277ZWZ Theorem 298 genrel Let H be a group and hhi E I a family of elements in H ful lling the relations w 57 w E R Then there exists a unique homomorphism o7ltIlwew RgtgtHwithigthi foralliEI 64 is onto ifand only ifHlthi lief Proof The rst statement follows directly from 296 The second one follows from 258 B Some examples The group 11 l a2 eb2 5 is the in nite dihedral group 071 l a2 2192 57 011 e is the dihedral group of degree n If u and w are words7 we also call u w a relations It just stands for uw i17 ilnm 1 7le Jim 1 e7 where ltabclabcabbacz a763 b765 5 3 5 is the triVial group lndeed7 c ab 020 c e Hence also a 02 eandb02e ltabla3b3ab2egt is isomorphic to Alt4 To see this let G be this group and put 2 ab Then 22 1 Put K lt2 2 Since both 2 and 2 have order two or 17 K is a dihedral group We compute 2 122 12 a2aba 2 aaba 1 ab a3ba 2a2ba71ab 1313 5 Thus 2 12 2quot 2 2 12 In particular 2 222 1 and so K is a quotient of the dihedral group of order Thus K ezz z Now 2amp2 2 13 25 z and so a E N0K Thus ltagtK is a subgroup of G It contains a and z ab and so also I a lz Thus G ltagtK As K has order diViding 4 and a has order diViding 37 G has order diViding 12 Thus to show that G is isomorphic to Alt4 it suf ces to show that Alt4 is a homomorphic image of G le we need to verify that Alt4 ful lls the relations 210 GROUP ACTIONS 41 For this let 1 123 and If 124 Then ab 1324 and so ab2 5 Thus there exists a homomorphism j G a Alt4 with gta f and gtb 13 As 1 and If generate Alt4 j is onto As lGl lAlt4l we conclude that j is an isomorphism Similarly to the free group on a set one can de ne the free semlgmup and the free monoid on a set I lf 1 1 these are Z1 and N respectively In the general case every element in the free semigroup free monoid can be uniquely written as 47114712 nm 21 22 2m where ik E I ik 7 ik1 and nk E Z1 Also m E ZJr in the semigroup case and m E N in the monoid case 210 Group Actions De nition 2101 defgroupaction An action of a group G on a set S is a function GXSgtSas acts such that GAl es s for all s E S GAQ abs abs for all a b E G s E S Note the similarity with the group multiplication In particular the binary operation of a group de nes an action of G on G called the action by left multiplication The function 15 a agtk8 sa is not an action unless G is abelian since abs sab asb bas For this reason we de ne the action of G on G by right multiplication as 13 a sa l There is a further action of G on itself namely the conjugation 15 a s asa l It might be interesting to notice that an action of G on a set S can also be thought of as an homomorphism j G a SymS lndeed given such a 1 de ne an action by 15 a gtas Since gte esym5 ids GAl holds And GA2 follows directly from we M o Mb Conversely given an action of G on S and a E G de ne gta S a S s a as GA2 translates into gtab gta o gtb and GA1 into gte ids We still need to verify that gta E SymS But this follows from ids we gtaa 1 Ma 0 gta 1 For X Q S de ne StabGX g E Glgssfor all 36X 42 CHAPTER 2 GROUP THEORY Then StabGS is exactly the kernel of j and in particular it is a normal subgroup By the isomorphism theorem7 G StabGS G g SymS lf StabGS e7 we say that the action is faithful This is the case if and only if j is one So in some sense the groups acting faithfully on a set S are just the subgroups ofSymS There are lots of actions of groups besides the actions on itself For example SymS acts on S The automorphism group of the square acts on the four corners of the square The automorphism group of a projective plane acts on the points and also on the lines of the plane The group of invertible matrices acts on the underlying vector space For the rest of the section we assume Hypothesis 2102 G is a group acting on a set S Lemma 2103 orbits De ne a relation N on S by s N t if and only t as for some a E G Then N is an equivalence relation on S Proof Since s es7 s N s and N is re exive lft as7 then ailt a 1as ailas es s Thus s N t implies t N s and N is symmetric Finally if s at and t br then s at abr abr Thus s N t and t N r implies s N r and N is re exive D The equivalence classes of N are called the orbits of G on S The set of orbits is denoted by SG The orbit of G containing s is Gs gs l g E G A subset T of S is called G inuariant if gt 6 T for all t E T and g E G In this case C also acts on T Observe that the orbits of G on S are the minimal G invariant subsets We say that G acts transitively on S if G has exactly one orbit on S That is for each st 6 S there exists 9 E G with t gs Equivalently G is transitive on S if S Gs for some or all s E S So the G orbits can be also described as subsets of S on Which G acts transitively Some special cases of orbits Let H g G The right cosets of H are the orbits for the action of H on G by left multiplication The left cosets are the orbits for H by the right multiplication Finally the conjugacy classes are the orbits for the action G on G by conjugation De nition 2104 Let G be acting on the sets S and T and 04 S gt T a function a 04 is called G equivariant if Mas 9M8 forallgEG andsES b 04 is called a G isomorphism if 04 is G equiuariant and an bijection 210 GROUP ACTIONS 43 If G acts on a set S it also acts on the power set 73S7 that is the set of all subsets Indeed for T Q S and g E G put gT gtlt E T If H g G then G acts on GH by a7bH a abH Clearly this is a transitive action It turns out that any transitive action of G is isomorphic to the action on the coset of a suitable subgroup Lemma 2105 transorbits Let s E S and put H StabGs a The map ozGHgtSaHgtas is well de ned G eqvivariant and one to one b 04 is an G isomorjnhism if and only if G acts transitively on S c Stabas H for all a E G d If G is transitive on S lSl lG Stabasl Proof a Let h E H Then ahs ahs as and 04 is well de ned Also aabH aabH abs abs aabH So 04 is G equivariant lf aaH abH we get as bs so a lbs s As H Stabas this implies a lb E H and so I E aH and bH aH Thus 04 is one to one and a is established b By a 04 is a G isomorphism if and only if 04 is onto But this is the case exactly then S Cs and so if and only if G is transitive on S C gas as ltgt ailgas s ltgt ailga E H ltgtg E aHa 1 H d follows directly from b D Lemma 2106 Suppose that G acts transitively on the sets S and T Let s E S andt E T Then S and T are G isomorjnhic if only if Stabas and Stabat are conjugate in G Proof Suppose rst that 04 S a T is a G isomorphism Since ags gas and 04 is one to one7 StabGs Stabaas Since G is transitive on T7 there exists h E G With gas t Thus StabGt StabGgas StabGas9 StabGs9 Conversely suppose that StabGs9 StabGt Then StabGgs StabGt and so by 2105b applied to S and to T S g G StabGgs G StabGt g T 44 CHAPTER 2 GROUP THEORY D A subset R C S is called a set of representatives for the orbits on S7 provided that R contains exactly one element from each G orbit In other words the map R a SG7 r a Gr is a bijection An element s E S is called a xed point of G if gs s for all g E G The xed points correspond to the orbits of length 1 the trivial orbits FixsG denotes the set of all xed points Note that FixSG R for any set of representatives R for SG and R FlXGR is a set of representatives for the non trivial G orbits Lemma 2107 orbiteq Let R C S be a set of representatives for SG 5 Z lG StabGrl mime Z lG Stabarl TER TERFiXSG Proof Since S is the disjoint unions of its orbits7 lSl ETER lGrl By 2105d7 lGrl lG StabGrl and the lemma is proved D For the case of conjugation the equation in the preceding lemma is called the class equation De ne the center ZG of a group by ZGgEGlghhgforallh G Note that g is a xed point for C under conjugation if and only if g gh hgh l for all h E G and so if only if gh by for all h E G That is FlXGG ZG Also setting Caa b E G l ba ba we have Gaa Stabaa With this notation we have Lemma 2108 Class Equation classeq LetR be a set of representatives for the con jugacy classes of G Then G Z lGCcrl lZGl Z lGCcrl rER reRZG These formulas become particular powerful if G is a nite p group7 that is lGl pk for some non negative integer k Lemma 2109 Smodp Letp be a prime and P a p group acting on a set S Then lSl E lFixSPl mod p Proof If s E S is not a xed point7 then Stabps g P and so lPl PStabp3l W E 0 mod p 210 GROUP ACTIONS 45 The lemma now follows from 2107 D For H g G de ne a E G l H Ha Note that is submonoid of G7 but not necessarily a subgroup7 as it might not be closed under inverses Lemma 21010 H xG LetH G a The xed points ofH acting by left multiplication on GH are b With respect to the action of G on the subgroups of G by conjugation StabGH N0H c If H is nite then Proof a Clearly gH H if and only ifg E H So StabGH H and StabGaH H Hence H xes aH if and only if H Ha7 that is if and only if a E Thus a holds b Obvious c As conjugation is an bijection7 ngl So for nite H7 H g H9 implies H H9 D Lemma 21011 CenterP LetP be a non triuialp group a ZP is non triuial b If H g P then H g NpH Proof a Consider rst the action by conjugation By 2109 OE Pl E lZPl mod p Thus lZPl 7 1 b Consider the action of H on PH By 21010 and 2109 0 2 lPHl 2 iNPltHgtHgtl media So leHHl 7s 1 a As a further example how actions an set can be used we give a second proof that Symn has normal subgroup of index two For this we rst establish the following lemma Lemma 21012 equivr ep Let A be a nite set and N a non triuial equivalence relation on A so that each equiualence class has size at most 2 Let Q R Q A l R contains exactly one element from each equiualence class ofw De ne the relation x on Q by R x S if and only if lRSl is euen Then z is an equiualence relation and has exactly two equiualence classes 46 CHAPTER 2 GROUP THEORY Proof For 1 E A let d be the equivalence class of N containing 1 and let A be the set of equivalence classes Let A B E Q and de ne AABX AlA X7 B X LetdeA Thend BifandonlyifA d7 B d So lABllAABland AzBltgtAAB is even In particular z is re exive and symmetric Let R S T E 9 Let X E Then X O R 7 X TexactlyifeitherX Ry X SX TorX RX S7 X T Thus ART ABS AST U AST ABS Hence lARTl lARsl lASTl 7 QlARs ASTl If R x S and S x T the right side of is an even number So also the left side is even and R z T SO x is an equivalence relation Let R E Q As N is not trivial there exist rt E A with r N t and r 7 t Exactly one of r and t is in R Say r E B Let T R U Then T E Q and lT Rl 1 Thus R and T are not related under Let S E 9 Then the left side of odd and so exactly one of lARsl and lASTl is even Hence S x R or S x T Thus x has exactly two equivalence classes and all the parts of the lemma are proved D Back to Symn Let A ij l 1 ij ni 7 j Then Symn acts on A De ne ij N kl iff kl ij or kl j De ne Q as in the previous lemma Clearly Sym acts on Q and also on Qz the set of equivalence classes of 7 z 77 Let R ij 1 g i lt j g ThenR E Q The 2 cycle 12 maptho RU2 Thus R and 1 2R are not related under z and so Symn acts non trivially on Qz which is a set of size 2 The kernel of the action is a normal subgroup of index two Lemma 21013 20ddlLet G be a nite group of order 2n with n odd Then G index a normal subgroup of index 2 Proof View G has a subgroup of SymG via the action of G on G by right multiplication Let t E G be an element of order 2 Since 25 has no x points on G t has n orbits of length 2 Thus if is the product of an odd number of 2 cycles and so sgnt 71 Hence ker sgn la is a normal subgroup of index 2 D The following lemma is an example how the actions on a subgroup can be used to identify the subgroup Lemma 21014 sym5 211 SYLOW P SUBGROUP 47 a Let H Sym6 with H 120 Then H g Sym5 b Let H Alt6 with H 60 Then H g Alt5 Proof Let G Sym6 in a and G Alt6 is case In both cases g 6 Let I GH Then G acts on I by left multiplication Let 1 G a SymI be the resulting homomorphism Since StabGH H ker gt H and so ker gt 7 Alt6 and ker gt 7 Sym6 Also ker gt is a normal subgroup of G and we conclude from 275 that ker gt 5 Thus 1 is one to one and so l gtGl lGl E 61 Since SymI Sym6 we conclude that gtG has index 1 or 2 in SymI Therefore gtG 1 G Thus gtG SymI in case a and gtG AltI in case Note that gtH xes an element i in I namely i H Thus gtH Stabsym1i Suppose that a holds Note that Stabsym1i E Sym5 lSym5l and j is one to one Thus gtH Stabsym1i and H E Sym5 Suppose that b holds The same argument as above shows gtH StabAMI and H g Alt5 D We remark that although any subgroup H of order 120 in Sym6 is isomorphic to Sym5 it does not have to be one of Sym5 s in Sym6 which x some k in 1 2 3 4 5 6 Indeed there does exist one which acts transitively on the six elements To see this consider the action of Sym5 on its six cyclic subgroups of order 5 On the other hand the above proof says that H xes a point i in the set I This seems to be contradictory but isn t The set I is a set with six elements on which Sym6 acts but it is not isomorphic to the set 1 23 4 5 6 Sym6 has two non isomorphic action on sets of size six Luckily this only happens for Sym6 and not for any other Symn but we will not prove this 211 Sylow p subgroup In this section G is a nite group and p a prime A p subgroup of G is a subgroup P g G which is a p group A Sylow p subgroup P of G is a maximal p subgroup of G That is P is a p subgroup and if P g Q for some p subgroup Q then P Q Let SylpG be the set of all Sylow p subgroups of G For the following it will be important to realize that G acts on SylpG by conjugation The following lemma implies that G acts on SylpG by conjugation Lemma 2111 Let P E SylpG and Oz 6 AutG Then aP E SylpG Proof Since 04 is an bijection lPl laPl and so aP is a p group Let aP Q Q g G a p group Then P a 1Q and the maximality of P implies P 04 1Q Thus aP Q and aP is indeed a maximal p subgroup of G D 48 CHAPTER 2 GROUP THEORY Theorem 2112 Cauchy Cauchy Let G is a nite group andp a prime diuiding the order of G Then G has an element of orderp Proof Let X be any cyclic group of order p Then X acts on GP by 95 a1 7 a1ka2k ap7a1 7 Consider the subset Sa1 7ap 6 GP l alagape Note that we can choose the rst p 7 1 coordinates freely and then the last one is uniquely determined So 5 lGlp l We claim that S is X invariant For this note that 71 a1a2 apa1 af1a1apa1 a2apa1 Thus alag ap e if and only if a2 apal e So X acts on S From 2109 we have 5 E lFiXSXl mod p Asp diVides lGl it diVides 5 and so also lFiX5Xl Hence there exists some a17 a27 ap E F1X5X distinct from e7 e7 7e But being in FiX5X just means a1 a2 ap Be ing in S implies all7 alag ap e Therefore a1 has order p D Theorem 2113 Sylow sylow Let G be a nite group p a prime and P E SylpG a All Sylow p subgroups are conjugate in G b Syuan GNGW 21 mod p c d Every p subgroup of G is contained in a Sylow p subgroup of G lPl is the largest power ofp diuiding Proof Let S PG P9 l g E G7 the set of Sylow p subgroups conjugate to P First we show 1 P has a unique xed point on S and on SylpG7 namely P itself lndeed7 suppose that P xes Q E SylpG Then P N0Q and PQ is a subgroup of G Now lPQl and so PQ is a p group Hence by maximality of P and Q 2 S E 1 mod p By 1 F1X P 1 and so 2 follows from 2109 211 SYLOW P SUBGROUP 49 3 Sy1pG 5 Let Q E SylpG Then E S E 1 mod p Hence Q has a xed point T E 5 By 2 applied to Q7 this xed point is Q So Q T E 5 Note that N0P is the stabilizer of P in G with respect to conjugation As G is transitive on S we conclude is S GNGP Thus 2 and 3 imply a and 4 p does not divides NGPP Suppose it does Then by Cauchy s theorem N0PP has a non trivial p subgroup QP Since Q QP P 7 Q is ap group with P 3 Q7 a contradiction to the maximality of P By b and 4 p divides neither GNGP nor NGPP Since G GNGP NGPP P c holds G is nite so any p subgroup of G lies in a maximal p subgroup7 proving D As an application of Sylow s theorem we will investigate groups of order 12715730 and 120 We start with a couple of general observation Lemma 2114 easysylow Let G be a nite group p a prime and P E SylpG Also let N the kernel of the action of G on SylpG a SylpG diuides and equals 1 mod p b P 1 PN In particular P is the unique Sylow p subgroup of PN c d N P E SylpN andN P S G In particularP N if and only ifP S G VVVV PNN E SylpGN Moreouer the map Sy1pG H Sy1pGN Q A is a bijection Proof a Follows directly from 2113b b Just note that N N0P c Since P 1 inPN7 NO P 1 N Also NN P NPP and so p does not divide NN O P So N O P is Sylow p subgroup of N As N O P 1 N it is the only Sylow p subgroup of N Thus N O P 1 G d Egg The latter number is not divisible by p and so PNN is a Sylow p subgroup of GNN Since every Sylow p subgroup of GN is of the form PNN9N PgNN the map is onto Suppose that PNN QNN Then Q PN and so by b Q P Thus the map is also one to one D 50 CHAPTER 2 GROUP THEORY Lemma 21 15 order30 a Let G be a group of order 12 Then either G has unique Sylow 3 subgroup or G Alt4 b Let G be group of order 15 Then G E Z3Z X Z5Z c Let G be a group of order 30 Then G has a unique Sylow 3 subgroup and a unique Sylow 5 subgroup Proof a By 2114a the number of Sylow 3 subgroups divides L32 and is 1 mod 3 Thus lSy13Gl 1 or 4 In the rst case we are done In the second case let N be the kernel of the action on Sy13G By 2114 GN still has 4 Sylow 3 subgroups Thus lGNl 2 42 12 lGl N e and G is isomorphic to a subgroup of order 12 in Sym4 Such a subgroup is normal and so G E Alt4 by 275 b The numbers of Sylow 5 subgroups is 1 mod 5 and diVides 5 Thus G has a unique Sylow 5 subgroup S5 Also the number of Sylow 3 subgroups is 1 mod 3 and diVides 5 Thus G has a unique Sylow 3 subgroup S3 Then S3 S5 1 lS3S5l 15 and so G S3S5 As both S3 and S5 are normal in G S3S5 S3 S5 e and so G g 53 x 55 g 131 x 151 c By 21013 any group which as order twice an odd number has a normal subgroup of index two Hence G has a normal subgroup of order 15 This normal subgroup contains all the Sylow 3 and Sylow 5 subgroups of G and so c follows from D Lemma 2116 order120 Let G be a group of order 120 Then one of the following holds a G has a unique Sylow 5 subgroup b G E Sym5 c lZGl 2 and GZG E Alt5 Proof Let P Sy15G and put I Sy15G If in 1 a holds So suppose that 1 gt 1 Then by2114a l E 1 mod 5 and 1 diVides lGPl 24 The numbers larger than 1 and less or equal to 24 which are 1 mod 5 are 1 6 11 16 and 21 Of these only 6 diVides 24 Thus 1 6 Let 1 G a SymI be the homomorphism given by the action of G on I Put N ker gt and H gtG The H is subgroup of SymI E Sym6 and H E GN By 2114d GN and so also H has exactly 6 Sylow 5 subgroups In particular the order of H is a multiple of 30 By 2115c 344 30 Suppose next that 120 Note that N 1 and so G E H in this case Now H SymI E Sym6 Thus 21014a implies G E H E Sym5 211 SYLOW P SUBGROUP 51 Suppose next that 60 If H f AMI7 then H O AMI is a group of order 30 with six Sylow 5 sulogroups7 a contradiction to 2115 Thus H AMI E Alt6 So by 21014107 H Alt5 Since N 2 and N 31 G7 N ZG Also gtZG is a abelian normal subgroup of H E A1t5 and so gtZG 5 Hence N ZG and GZG GN g H g Alt5 CHAPTER 2 GROUP THEORY Chapter 3 Rings 31 Rings De nition 311 A ring is a tuple R7 such that a 137 is an abelian group b R7 is a semigroup c For each 7 E R both left and right 39r quot I by r are L 39 39 of 137 In other words a ring is a set R together with two binary operations R x R a R7 a7b a ab and R x R HRab Hal such that abcabc sothat for all a7b706R R2 There exists 0 ER with 0aa a0 for all a ER R3 For each a E B there exists 7a E R with a 7a 0 7a a R4 abbafor all ab ER R5 abc abc for all a7 bc E R R6 ab c ab ac for all ab7 c E R R7 abcabac for all a7b7c6R Let R be a ring The identity element of 137 is denoted by OR or 0 If R7 has an identity element we will denote it by 13 or 1 In this case we say that R is a ring with identity R is commutative if R7 is Some examples Z7 7 7 ZnZ7 7 7 E JndvV7 7 0 There is a unique ring with one element 53 54 CHAPTER 3 RINGS 0 00 There are two rings of order two 0 1 HOD OHH 0 0 0 0 1 0 0 0 1 0 71 Here n 6 01 for n 0 we have a ring with zero multiplication that is ab 0 for all 011 6 R For n 1 this is Z2Z There are the following rings of order 3 l 0 1 71 0 1 71 0 0 1 71 0 0 0 0 1 1 71 0 1 0 n in 71 71 0 1 71 0 in 71 Indeed if we de ne n 1 1 then 71 1 71 1 in Here n E 0171 For n 0 this is a ring with zero multiplication For n 1 this is ZBZ If n 71 this is isomorphic to the n 1 case under the bijection 1 lt gt 71 At the end of this section we will generalize these argument to nd all rings whose additive group is cyclic Let A be an abelian group and EndA the set of endomorphisms of A that is the homomorphisms from A to A De ne 04 a aa 3a and 04 o a a a Then EndA o is a ring Direct products and direct sums of rings are rings lndeed let BlJ E I be a family of groups For fg E Hid Ri de ne 1 9 and fg by f gi gi and With this de nition both Hid R and 26113 are rings In the following lemma we collect a few elementary properties of rings Lemma 312 elementaryring LetR be a ring 0aa00foralla R ia a7b 7ab for all 11 6 R 7a7b ab for all a b E R na anb 71ab for all 11 6 R n E Z 221 102221 bi 21 221 aibj Proof a e hold since right and left multiplication are homomorphisms For example any homomorphism sends 0 to 0 So a holds We leave the details to the reader 31 RINGS 55 Let R be a ring and G be semigroup The semigroup m9 RG of C over R is de ned as follows As an abelian group RG 960 De ne Q s9 259 where 259 Z rhsl hleGXGlhlg Note that since the elements in 960 B have nite support all these sums are actual nite sums lts straightforward to check that RG really is a ring Here is an alternative description of the multiplication which makes the nature of this ring a little bit more trans parent For T E R and g E G write rg for the element in 63960 which has r in the g th coordinate and 0 everywhere else So rg pgr in the notation of section 28 Then TygeG 2960 T99 and Z rgg 2 shh Z rgshgh 960 heG geGheG Also note that rg sh implies r s and is r 7 07 g h lf RG has an identity7 then R has an identity lndeed r Z rgg is an identity in RG Let a Z We will show that a is an identity in B Let s E R and g E G Then sh rsh 2rgsgh Summing up the coef cients we see that s sumgeargb as Similarly sa s and so a is an identity in R If R has an identity 17 we identify 9 with lg Here is an example of a semigroup G without an identity so that for any ring R with an identity RG has an identity As a set C 1 b7 De ne the multiplication by m ifxy 34 i i ifzy y Then zygtzltzygtz 7 4 z otherw1se Hence the binary operation is associative and G is a semigroup Put r a b 7 i E RG We claim that r is an identity We compute ar m aa ab 7 ai a i 7i a7 brrbbabb7biibiibandirriiaib7iiiiiii As RG ful lls both distributive laws this implies that r is an identity in RG If R and G are commutative7 RG is too The converse is not quite true Suppose RG is commutative7 then Kg1 WNW 870W WWW WWW 56 CHAPTER 3 RINGS So if rs 7 0 for some r s E R we get gh hg and G is commutative But ifrs 0 for all r s E R then also my 0 for all 71 6 RG So RG is commutative7 regardless whether G is or not Here is an example for a semigroup ring Let G N Then RN is isomorphic the polynomial ring over R Indeed the map 00 Za l H allies i0 is clearly an isomorphism Put Eff R 07 the nonzero elements De nition 313 Let R be a ring a A left resp right zero divisor is an element a 6 Eli such that there exists b 6 Eli with ab 0 resp ba 0 A zero divisor is an element which is both a left and a right zero divisor b An nonzero element is called leftright invertible if it is leftright invertible in Elf7 An invertible element is also called a unit c A non zero commutative ring with identity and no zero divisors is called an integral domain d A non zero ring with identity all of whose nonzero elements are invertible is called a division ring A eld is a commutative division ring Note that a ring with identity is zero if and only if1 0 So in c and d the condition that R is non zero can be replaced by 1 7 0 We denote the sets of units in R with Pi Note that Ff7 is group see 222c A ring has no left zero divisors if and only the left cancellation law zayagtzy holds in R R is a eld7 Z is an integral domain For which n E Z is ZnZ an integral domain It is if and only7 nlklgtnlkornll so if and only if n is a prime The following lemma implies that ZpZ is a eld for all primes 19 Lemma 314 niteid eld All nite integral domains are elds 31 RINGS 57 Proof Let R be a nite integral domain and a E Ff As R is an integral domain multiplication by a is a one to one map from Ff a Ff As R is nite this map is onto Thus ab 1 for some I E B So all non zero elements are invertible D De nition 315 ringhom Let R and S be rings A ring homomorphism is a map 1 R gt S 50 j R gt S and j R gt S are homomomhisms of semigmups Note that j R a S is an homomorphism if and only if gtr s gtr gts and ltrsgt ltrgt gtltsgt For a ring R we de ne the opposite ring Rep by Rep 0P R and a 9quot b I a If R and S are rings then a map 1 R a S is called an anti homomomhism if j R a S0p is ring homomorphism So gta b gta gtb and gtab gtb gta Let EndR be the set of ring homomorphism Then EndR is monoid under compo sition But as the sum of two ring homomorphisms usually is not a ring homomorphism EndR has no natural structure as a ring The map Z a ZnZ m a m 71 is a ring homomorphism For T E R let RT R a Rs a M and ET R a Rs a rs By de nition of a ring Ru and ET are homomorphisms of R But left and right multiplication usually is not a ring homomorphism The map E R a EndRr a E is a homomorphism but the map R R a EndR r a R is an anti homomorphism Note that if R has an identity then both R and E are one to one Theorem 316 homgr Let Oz R gt S be a ring homomomhism G a semigmup and G gt S a semigroup homomomhism such that ozr g BgarVr E Rg E G Then v RlGl H 5 X T99 H ZQWWQCI 960 is a ring homomomhism Proof VltZ T99 Z 399 VltZltTg l Sg9 Z 04769 l Sg 9 960 960 960 960 ZWW a89 9 Z a7 g 9 t Z QWWQ NZ T99 Z 899 960 960 960 960 960 and W 727939 Z Skk Z Z Tgskgk Z Z a7 gsk 9k 960 keK 960 keG 960 keG 58 CHAPTER 3 RINGS Z Z a7 ga8k 9 k Z ammo 2 ammo X T99 Z skis 960 keG 960 keG 960 keG Let A be an abelian group De ne j Z a EndA by gtnr nr Then o is a ring homomorphism Since ker gt is an additive subgroup of Z7 ker gt nZ for some n E N n is called the exponent of A and denote by expA If n 7 07 n is the smallest positive number such that not 0 for all a E A that is nA 0 And n 0 if mA 7 0 for all m E Z Let R be a ring The characteristic charR of R is the exponent of 137 Suppose R has an identity The map p Z a R7 n a n1 is a homomorphism of rings We claim that kerp ker gt This can be veri ed directly or by observing that j E o 0 Lemma 317 Suppose R is a non zero Ting with identity and no zero divisors Then charR is 0 or a prime Proof Let n charR and suppose n 7 0 If n 1 then r 1r 0 for all r E B So n gt 1 Suppose n is not a prime7 then n st with st E Z So 31t1 st1 n1 0 As R has no zero divisors we conclude that 31 0 or t1 0 In each case the case the get a minimality of n B Let r E R If R has an identity we de ne r0 1 If B does not have an identity we will use the convention r05 s for all s E R Lemma 318 Binomial Theorem Let R be ring a17a2 an E R and m E Z a 71 Z aha 047 2 1 17171 m 2 myquot E aiming orquot m17m277mn 11 mi wl21miml Proof a follows form 312e and induction on m For b notice that ail H047 ainlagn2aglquot where mk l ij So b D follows from a and a simple counting argument Lemma 319 easygcd Let n7m7k E Z a If gcdmk 1 or gcdnm 1 then gcdfk 1 for some 1 E Z with f E n mod m 31 RINGS 59 b There exists 1 E Z so that gcdf7 k 1 and fn E gcdnm mod m Proof a Suppose rst that gcdm7 k 1 Then 17 n lm 1 sh for some integers l7 3 Thus 1 n 1 lm 1 sh Put 1 n lm7 then gcdn lm7 k 1 Suppose next that geoln7 m 1 Write k klkg where gcdh1m 1 and all primes dividing k2 also divide m By the rst part there exists l E Z with gcdn lm7 k1 1 Now any prime dividing k1 divides m and as gcdn7 m 17 does not divide m Hence it also does not divide m 1 lm Thus gcdn lm7 k gcdn lm7 k1 1 b Let 1 gcdn7 m Replacing n be and m by we may assume that d 1 Then nn E 1 mod m for some n E Z Since gcdn7 m 1 we can apply a to nim and k So there exists 1 with gcdf7 k 1 and f E n mod m Then also fn E 1 mod m D Lemma 3110 Let R be a ring with 137 cyclic Then R is isomorphic to exactly one of the following rings 1 Z with regular addition but zero multiplication 2 nZan7 1 where m E Nn E ZJr and n divides m Proof Let m E N so that 137 E ZmZ and let a be generator for 137 So a a na for some n E Z Then for all kl E Z7 ha la klna and so the multiplication is uniquely determine by n Note that EaEa na EnEa So replacing a be Ea we may assume that n E N Also if m gt 0 we may choose 0 lt n g m Suppose rst that n 0 Then by our choice m 0 as well So 137 Z7 and rs 0 for all 733 GR Suppose next that n gt 0 Then the map nZan gt R nh an a ha is an isomorphism If m 07 these rings are non isomorphic for different n lndeed R2 nR and so lRRzl n Therefore it is determined by the isomorphism type R For m gt 07 various choices of n can lead to isomorphic rings Namely the isomorphism type only depends on d gcdnm To see this we apply 319 to obtain 1 E Z with gcdfm 1 and fn E 1 mod m Then 1 ef 1 sm for some 53 E Z and so 1 mZ is invertible Hence also fa is a generator for 11 and fa fa fzna infa Ma Also R2 dB and lRRzl So at is determined by the isomorphism type of R D 60 CHAPTER 3 RINGS 32 Ideals and homomorphisms De nition 321 Let R be a ring a A subring ofR is a subset S Q B so that S is a subgroup of R7 and a subsemigroup of R7 b An left Tight ideal in R is a subring I ofR so that TI Q I Ir Q I for all 6 R C An ideal in R is a left ideal which is also a right ideal nZ is an ideal in Z Let V be a K Vector space Let W V be K subspace De ne AnnW oz 6 EndMV l aw 0 for all w E Then AnnW is an left ideal in EndMV Indeed it is clearly an additive subgroup and B o aw Baw 30 0 for all 3 E EndMV and 04 E AnnW We will see later that EndMV is a simple ring7 that is it has no proper ideals Lemma 322 basicring horn Let 1 R a S be a ring homomomhism a If T is a subring of R is a subring of S b If T is a subring ofS then 1T is a subring of R c ker gt an ideal in R d IfI is an leftright ideal in R and j is onto is a leftright ideal in S e If I is a left7ight ideal in S then 1I is an leftright ideal on R Proof Straight forward B Let oz R gt S be a ring homomorphism and B G a H a semigroup homomorphism Consider the ring homomorphism 39y RG a SH Z rgg a Emma 960 What is the image and the kernel of 39y Clearly 39yRG 04RBG Let I kera To compute ker39y note that zrgg Z a 2 TM 960 heH yawn and so ZTQQEkerV Z 73 e Iforallh G 960 te W 32 IDEALS AND HOM OM ORPHISMS 61 If B is a group homomorphism we can describe ker39y just in terms of I kera and N ker Indeed the 1h s h E BG are just the cosets of N and so Egg 6 kerg ltgt Zn 6 I for all T e GN 960 teT Let as consider the special case where R S 04 id and H e ldentify Re with R Via re lt gt r Then 39y is the map RlG gt 13ng gt ng The kernel of 39y is the ideal ma 2799 l ng 0 R G is called the augmentation ideal of RG For subsets AB of the ring R de ne A B a b l a E Ab E B Let A be the additive subgroup of R generated by A Also put AB ab l a E Ab E B More general de ne A1A2 An to be the additive subgroup generated by the products alag an ai 6 Ai If A is a left ideal then also AB is a left ideal If B is a right ideal then AB is a right ideal In particular if A is a left ideal an dB is a right ideal then AB is an ideal If A B are rightleft ideals so is A B and Am B Actually arbitrary intersection of leftrightideals are leftrightideals So for A Q R we de ne the ideal A generated by A to be A y l I is an ideal in R A g I The left ideal generated by A is just RA If R has an identity this is equal to RA Also A A RA AR RAB which simpli es to RAB if R has an identity Lemma 323 RmodI Let I be an ideal in the ring R a The binary operation RIxRIHRI aIbI HabI is well de ned b RI is a ring c The map 7rRgtRI rarl is a ring homomorphism with kernel I 62 CHAPTER 3 RINGS Proof a Let Lj E I Then a j ab 1 ib aj l ij As I is an ideal7 ibajij Gland so aibjab b and c follow from the corresponding results for groups and a Lemma 324 The Isomomorphism Theorem itr Let 1 R 7 S be a ring homo morphism Then the map 2 Rker gt 7 MR7 r ker gt 7 gtr is a well de ned isomorphism of rings By the lsomorphism Theorem for groups 2557 this is a well de ned isomorphism for the additive groups But clearly also ab da 3b and j is a ring isomorphism D We will see below that any ring R can be embedded into a ring S with an identity This embedding is somewhat unique Namely suppose that R g S and S has an identity Then for mm 6 Z and rs E R we have 711 rm1 s n m1 r s and 711 rm1 s nm1 mr 1 ms 1 rs So already Z1 R is a ring with 17 contains R and the addition and multiplication on Z1 R is uniquely determined But there is some degree of freedom Namely Z1 B does not have to be a direct sum Let 1 Z x R as abelian groups We make 1 into a ring by de ning n7r r7173 77771718 1 mr 1 rs Then 17 0 is an identity in The map 1 13 7 S7 71 r 7 n1r is a homomorphism with image Z1 B Let us investigate ker gt n r E ker gt iff r 7711 Let kZ be the inverse image of Z1 O R in Z Also put t k1 and D16 lk77lt l l E Z Then ker gt Dk Hence RDM Z1 R Now which choices of k E Z and t E R can really occur Note that as t 77117 tr kr rt This necessary condition on k and t turns out to be suf cient Let k E Z t E R is called an k element if tr rt kr for all r E R Note that a 1 element is an identity7 while a O element is an element with tR Rt 0 Also if a and b are k elements7 then a 7 b is a O element So if a k elements exists it unique modulo the zero elements Suppose now that t is a k element in R De ne Dm has above We claim that Bk Zk7 7t is an ideal in R For this we compute using rt kr n7r k 7t 71k kr 7 nt 7 rt 71k kr 7 nt 7 hr 71k 7nt nk 7t So D16 is a left ideal Similarly7 an is a right ideal Put Rm RDk Then Bk is a ring with identity7 contains R via the embedding r 7 07 r Dm and ful lls Z1 RkZ1Zt Note that if t is an k element and s an l element7 then 7t is an 7k element and t s is an k l element Therefore the sets of k E Z for which there exists a k element is a 32 IDEALS AND HOMOMORPHISMS 63 subgroup of Z and so of the form iZ for some i E N Let u be a i element Rm is in some sense the smallest ring with a identity which contains R Also if R has no O elements7 v and so Rm is uniquely determined For example if R nZ7 then i n v and Rm E Z lndeed B Z x nZ7 Dm jn7 ijn lje Z7 B Z10 63 Dim and the map Rn a Z7 j7r Dn Hj r is an isomorphism between Rim and Z Next we will show that R can be embedded into a ring with identity which has same characteristic as R Put it char R then 0 is an n element Also Dmo nZ x 0 and Rmo E ZnZ x R as abelian groups So Rmo has characteristic n On the other hand R R09 always has characteristic 0 De nition 325 dprime Let I be an ideal in the ring R with I 7 R a I is prime ideal if for all ideals A7 B in R ABQIltgt AQIorBQI b I is a maximal ideal if for each ideal A of R IQAQRltgtAIorAR Note that nZ mZ an and so kZ is a prime ideal in Z if and only if k is prime Also nZ Q mZ if an only if m diVides n So kZ is maximal if and only if k is a prime So for Z the maximal ideals are the same as the prime ideals This is not true in general Lemma 326 basicprime Let P 7 R be an ideal in the ring R a If for all a7 1 E R ab Pltgta Porb P then P is a prime ideal b If R is commutative the converse of a holds c If R is commutative then P is a prime ideal if and only RP has no zero divisors Proof a Let A and B are ideals in R with AB Q P We need to show that A Q P or B Q B So suppose A Q P and pick a E A P Since ab 6 P for all I E B we conclude b E P and B Q P b Suppose that P is prime ideal and a7 1 E R with ab 6 P Then for all n7 mZ and r s E R na ramb sa nmab ns mr rsab and so ab Q ab Q P As P is prime7 a Q P or 1 Q P Hence a E P or b E P c Just note that the condition in a is equivalent to saying that RP has no zero diVisors D 64 CHAPTER 3 RINGS Lemma 327 primeintegral Let R be a nonzero commutatiue ring with identity and P an ideal Then P is prime ideal if and only if RP is an integral domain Proof If P is a prime ideal or if RP is an integral domain we have that R 7 P So the lemma follows from 326c D Theorem 328 basicmaximal Let R be a ring with identity and I 7 R be an ideal Then I is contained in a maximal ideal In particular euery nonzero ring with identity has a maximal ideal Proof The second statement follows from the rst applied to the zero ideal To prove the rst we apply Zorn s lemma A1 For this let M be the set ofideals J of R with I Q J Q R Order M by inclusion and let C be a nonempty chain in M Let M UC Then M is an ideal and I Q M Also 1 is not contained in any member of C and so 1 M Hence M 7 R and M E M Thus every chain has an upper bound and so by Zorn s Lemma M has a maximal element M If M C A for some ideal A 7 R then I Q A7 A E M and so by maximality of M in 17 A M Thus M is a maximal ideal D Theorem 329 maximalprime Let M be a maximal ideal Then M is a prime ideal unless R has no identity R 7 R2 and R2 Q M Proof Suppose that M is not prime Then AB Q M for some ideals A and B with A f M and B f M The maximality of M implies R A M B M Thus R2 AMBM Q ABM Q M Has M 7 Mwe get R2 7 R In particular B does not have an identity since otherwise R R1 Q R2 D We remark that an ideal I 7 R is maximal if and only if RI is simple Lemma 3210 basicsimplerings a Let R be a diuision ring Then R has non proper left or right ideals In particular R is simple b Let R be a non zero commutatiue ring with identity The R is simple if and only ifR is a eld Proof a Let I be an nonzero left ideal in R and pick 0 7 i E I Then 1 i li 6 RI Q R andsoRR1QI b By a we only need to show that simple implies eld So suppose R is simple and 0 7 a E R Since R has an identity7 Ra is an non zero ideal As R is simple Ra R Thus ra 1 for some r As R is commutative7 ar 1 and so r has an inverse D lfI is an ideal we will sometimes write a E b mod I ifaI b l I7 that is ifaEb E I If R Z and I nZ then a E b mod nZ is the same as a E b mod 32 IDEALS AND HOMOMORPHISMS 65 Theorem 3211 Chinese Remainder Theorem CRT Let Algi E I be a family of ideals in the ring R a The map 9 R Ai a H RAl i39eI iEAi r Ai H r AiieI id is a well de ned monomorphism b Suppose that I is nite R R2 Ai and R Ai Aj for alli E I Then ba If m gt 1 then R Al WdAj bb 0 is an isomorphism bc Fori E I let bi E R be given Then there exists I E R with b E 1 mod Al for alli E I Moreover I is unique mod mid Ai Proof a The map r a r Amd is clearly a ring homomorphism with kernel mid A So a holds ba For 0 7 J Q I put AJ jd Aj We will show by induction on lJl that RAiAJ for all 0 7 J Q Indeed if lJl 1 this is part of the assumptions So suppose lJl gt 17 pick j E J and put K J Then by induction R Al AK and R Al Aj Note that as Aj and AK are ideals7 AjAK Q Aj AK AJ Thus R2 Ai AjAi AK Q Ai AjAK Q Ai A Hence R Ai R2 Ai AJ bb By a we just need to show that 9 is onto For lIl 17 this is obvious So suppose I 2 2 Let 95 mih39el E H RAi ieAi We need to show that z 01 for some I E B Let xi bl Al for some bl E B By ba7 we may choose bl E jei Aj So bl E Aj for all j 7 i Thus 0 b39 39 l 0 ifj 7 Put 1 bi Then 017 mi and so 01 m D i39eI 66 CHAPTER 3 RINGS be This is clearly equivalent to bb D The special case B Z is an elementary result from number theory which was know to Chinese mathematicians in the rst century AD To state this result we rst need to observe a couple of facts about ideals in Z Let nm be positive integers gcdn7 m denotes the greatest common divisor and lcmn7 m the least common multiple of n and m Then nZ mZ lcmn7 mZ and nZ mZ gcdn7 mZ In particular n and m are relatively prime if and only if nZ mZ Z So part bc of the Chinese Remainder Theorem translates into Corollary 3212 Let ml7 Hm be positive integers which are pairwise relatively prime Let b17 bn be integers Then there exists an integer b with bEbi modm foralllgign Moreover b is unique mod mlmg mn D 33 Factorizations in commutative rings De nition 331 Let R be a commutative rings and ab E R a We say that a divides b and write a l b if b Q a b We say that a and b are associate and write a N b if a b c We say that a is proper if 0 7 a 7 R Some remarks on this de nition First note that abltgtalbandbla Also alb is a symmetric and transitive relation a N b is an equivalence relation If R has an identity7 a l b if and only if b ra for some r E R Lemma 332 unitdivide Let R be a commutative ring with identity and u E R The following are equivalent 1 u is a unit Zull 33 FACTORIZATIONS IN COMMUTATIVE RINGS 67 3 ulrforallrER 4 R 5 a is not contained in any maximal ideal of R 6 rNurforallreR Proof Straightforward D In particular7 we see that if R has an identity7 a is proper if and only if a is neither 0 nor a unit Lemma 333 rorbits LetR be an integral domain Let a7 bu E Elf with b ua Then b N a if and only ifu is a unit Proof The 7 if 7 part follows from 3327 part 6 So suppose that b N a Then a vb for some v E R Thus 1b b va vvb uvb Since the cancellations law hold in integral domains we conclude that av 1 So a is a unit Recall that Pi denotes the multiplicative groups of units in R Ff acts on R by left multiplication The preVious lemma now says that in an integral domain the orbits of Pi are exactly the equivalence classes of N De nition 334 dpidlLetR be a ring a An ideal I is called a principal ideal if its generated by one element that is I r for some r E R b R is called a principal ideal ring if every ideal is a principal ideal c R is principal ideal domain PlD if R is an integral domain and a principal ideal ring De nition 335 Let R be a commutative ring with identity and c a proper element a c is called irreducible if for all a7 b E R cabgta orb is aunit b c is called a prime if for all a7 b E R plabgtpla orplb Lemma 336 primeirr Letp and c be nonzero elements in the integral domain R a The following are equivalent 68 CHAPTER 3 RINGS 1 p is a prime 2 p is a prime ideal 3 Rp is an integral domain b The following are equivalent 1 c is irreducible 2 For allaER alcgtaNuoraisaunit 3 c is maximal in the set of proper principal ideals c Euery prime element in R is irreducible d If R is principal ideal then the following are equivalent 1 p is a prime 2 p is irreducible 3 p is a maximal ideal 4 Rp is a eld e Euery associate of a prime is a prime and euery associate of an irreducible element is irreducible Proof a This follows from 326a and 327 b Suppose 1 holds and a l c Then c ab If a is not a unit7 then b is a unit and so a N c Hence 2 holds Suppose 2 holds and c Q a Then a l u If a N c7 a c and if a is a unit a R Thus 3 holds Suppose that 3 holds and c ab Then c Q a and so either a c or a R In the rst case a N c and so by 333 b us a unit In the second case a is a unit So 1 holds c Let p be a prime and p ab So p l a or p l b Without loss p l a Since also a l p we get p N a Thus 333 implies that b is unit 1 By c 1 implies 2 Suppose 2 holds By b3 says that p is a maximal proper principal ideal Since every ideal in a PlD is a principal ideal7 p is a maximal ideal So 3 holds By 3210 3 implies 4 Suppose 4 holds Then by 3927 p is a prime ideal So by a7 p is a prime e By a and b the properties 77prime77 and 77irreducible77 of an element only depends on the ideal generated by the element Thus e holds Lemma 337 uniquefactor Let R be an integral domain and a E R Suppose that a p1pn with each pi a prime in R 33 FACTORIZATIONS IN COMMUTATIVE RINGS 69 a IquRisaprime withqla thenquiforsomelgign b If a ql qm with each qi a prime Then n m and there exists 7r 6 Symn with qi39 N Pm a Put 1 p2pn Then a p11 and q l p11 Thus qlpl or q l b In the rst case as primes are irreducible q N p1 In the second case induction implies q N p for some j g 2 g n b By a7 ql N p for some j Without loss ql p1 As R has no zero divisor7 b qg qm and we are done by induction D De nition 338 A unique factorization domain UFD is an integral domain in which every proper element is a product of primes Let R be a UFD Then each irreducible element is divisible by a prime and so equal to that prime So primes and irreducibles are the same in UFD s Also 337 implies that the prime factorizations are unique up to associates Our next goals is to show that every PlD is a UFD For this we need a couple of preparatory lemmas Lemma 339 Chainideal LetI be chain of ideals in the ring R a UI is an ideal b If UI is nitely generated as an ideal then UI E I Proof a is obvious b Suppose that UI for some nite F Q UI For each 1 E F there exists If E I with f E If Since I is totally ordered7 the nite set If l f E F has a maximal element IthenIEI7 FQIandso U1FgIgUI Thus I UI and b is proved Lemma 3310 minchain Let R be an integral domain and I a set of principal ideals Then one of the following holds 1 HI 2 I has a minimal element 3 There exists an in nite strictly ascending series of principal ideals 70 CHAPTER 3 RINGS Proof Suppose that neither 1 nor 2 hold Then there exists an in nite descending series RalgRaZQRanRan1Qn and an element 0 7 a contained in each Ran Let a man with rn E R Since an 6 Ban an san for some 8 E R Thus Tnan a Tn1an1 Tn13an As R is an integral domain rn rn1s and so Rm Q Rm lf Rm Rm then Rm Rsrn1 So rn1 N srn1 and by 333 8 is a unit As an san we conclude that Ran Ran a contradiction Thus ergRrggangan1g is an in nite strictly ascending series of ideals D Lemma 3311 maxmin Let R be a ring in which euery ideal is nitely generated a Any nonempty set of ideals in R has a maximal member b Suppose in addition that R is an integral domain Then euery non empty set of principal ideals with nonzero intersection has a minimal member Proof a By 339 B has no strictly ascending series of ideals Thus a holds b follows from 339 and 3310 D Lemma 3312 PIDUFD Euery principal ideal domain is a unique factorization domain Proof Let S be the set of proper elements in R which can be written as a product of primes Let a be proper in B Let Ssla ss We claim that S is not empty lndeed let 5 be a maximal ideal with a C Then 8 is irreducible and so by 336 8 is a prime Hence 8 E S and s E 5 As a 6 HS ms is not empty So by 3311b S has a minimal member say 1 with b E S Since a E b a ub for some u E R Suppose that u is not a unit Then as seen above there exists a prime p diViding u Then pb diVides a and so a 6 pb But pl 6 S and 1 pb g b a contradiction to the minimal choice of D 33 FACTORIZATIONS IN COMMUTATIVE RINGS 71 De nition 3313 deuklidring An Euclidean ring is a commutatiue ring together with a function j Elf gt N such that For all ab E R with b 7 0 there exists q7 r E R with a qb r and either r 0 or qr lt qb An Euclidean domain is an Euclidean ring which is also an integral domain Some examples Z is an Euclidean domain with j the absolute value function Any eld is a Euclidean domain with o5 O The polynomial ring over any eld is a Euclidean ring with j the degree function Lemma 3314 ERPR Any Euclidean ring is a principal ring with identity An Eu clidean domain is a PID and a UFD Proof Let I be a nonzero ideal and let 0 7 a E I with gta minimal Let b E I Then by E13271 qa r for some qr E R with if r 7 07 gtr lt gta Suppose that r 7 0 Then r b 7 qa E I 7 a contradiction to the minimal choice of gta Thus I qa E a Hence I Ra a and R is a principal ring To show that R has an identity note that R Ra for some a In particular7 a ea for someeER NowforallrERrsaforsomesanderesaseasarsoeisan identity The last statement follows form 3312 B Let R be an euclidean ring and j Elf a N the corresponding function Let 0 7 a E a with Ma minimal From the proof of the previous lemma7 a a so a N 2 De ne gta Ma We claim that a gta gtab for all a b e Rt with ab 7g 0 b For all a7 1 E R with b 7 07 there exists qr E R with a ab 1 r and either r 0 or 0 lt W5 Let 1 ab Then at E a and so I E at Q a Thus Md gttilded So a holds By de nition of an euclidean domain7 there exists r and s with a qb and either t 0 or gtr lt gtb Since a mi and b vb with uu units we get a uuq uur If r 07 um 0 Suppose r 7 0 As u and u are units uur Thus gtW W gtltrgt lt w M So 10 holds D 72 CHAPTER 3 RINGS Next we introduce greatest common divisor in arbitrary commutative rings But the reader should be aware that often no greatest common divisor exist De nition 3315 dgcd Let X be a subset of the commutative ring R and d E R a We say at is a common divisor ofX and write at l X ifX Q d that is ifd l x for all d e X b We say that d is greatest common divisor ofX and write at N gcdX if le and eld for all e E R with elX c We say that X is relatively prime if all commons divisors ofX are units Note that if a greatest common divisor exists it is unique up to associates A common divisor exists if and only if X is contained in a principal ideal A greatest common divisor exists if and only if the intersection of all principal ideals containing X is a principal ideal Here we de ne the intersection of the empty set of ideals to be the ring itself The easiest case is then X itself is a principal ideal Then the greatest common divisors are just the generators of An element in X generates X if and only if it s a common divisor So if the ring has an identity7 X is a principal ideal if and only if X has a common divisor of the form ZmeX mm where as usually all but nitely many rm s are supposed to be 0 Note that from the above we have the following statement Every subset of R has a greatest common divisor if and only if any intersection of principal ideals is a principal ideal That is if and only if the set of principal ideals in R is closed under intersections ln particular7 greatest common divisors exists in PID s and can be expressed as a linear combination of the X Greatest common divisors still exists in UFD s7 but are no longer necessarily a linear combination of X lndeed let 73 be a set of representatives for the associate classes of primes For each 0 7 r E R7 m UT H pmp7quot 12673 for some mpr E N and a unit ur Let mp mpz Since mp mpz only nitely many of the hip are nonzero So we can de ne d H pmp 12673 A moments thought reveals that d is a greatest common divisor Here are a couple of concrete examples which might help to understand some of the concepts we developed above 33 FACTORIZATIONS IN COMMUTATIVE RINGS 73 First let R Zi the subring of C generated by i R is called the ring of Gau ian integers Note that R Z Zi We will rst show that R is an Euclidean ring lndeed put gta1 a2i a a3 Then gtzy gtz gty and gtz E Z So ERl holds Let zy E R with z 7 0 Put 2 g E C Then y 2m Also there exists 1 11 d2i E C with q z 7 d E R and g In particular gtd g g Put r y 7 gm then r zz 7 gm z 7 qz dm So gtr gtd gtz Hence also ER2 holds Let a be a prime in R and put P a Since gta Fm E P P Z 7 0 Also 1 P and so P O Z is a proper ideal in Z Since RP has no zero divisors Z PP ZP Z has no zero divisors Thus P O Z pZ for some prime integer p Let Q pR Then Q g P g R We will determine the zero divisors in RQ lndeed suppose that ab 6 Q but neither a nor 1 are in Q Then p2 divides gtab So we may assume that p divides gta Hence a 701 mod p lfp divides an it also divides a2 a contradiction to a Q Therefore we can divide by a2 mod p and conclude that the equation 2 71 has a solution in ZpZ Conversely if n2 E 7 mod p for some integers n we see that up to associates n i Q and n 7 i Q are the only zero divisors Suppose that no integer n with n2 E 71 mod p exists Then RQ is an integral domain and so a eld Hence Q P and a N p in this case Suppose that n is an integer with n2 E 7 mod p As P is a prime ideal and n 7i E Q g P one ofn ii is in P We conclude that a N 71 ii Next let R We will show that R has some irreducible elements which are not primes In particular R is neither UFD PlD or Euclidean Note that R Z ZxE For T E R de ne r1 r2 6 Z by r T1T2E De ne 77 r17r2xE and Nr r77 rf710rg Nr is called the norm of r We claim that r 7 77 is a ring automorphism of R Clearly it is an automorphism of R Let r s E R Then rs n T2V1081 32V10 7131 107232 7132 7232 V 10 It follows that 775 775 In particular N03 7377s 73775 r7335 NTNS and N R 7 Z is a multiplicative homomorphism Let r be a unit in R Since N1 1 we conclude that Nr is unit in Z and so Nr i1 Conversely if Nr i1 then 7 1 NE NO 6 R and r is a unit For example 3 10 is unit with inverse 73 10 As 10 is not rational Nr 7 0 for r E Pf We claim that all of 2 3 f 4 V10 and f are irreducible lndeed suppose that ab is one of those numbers and neither a nor 1 are units Then NaNb E 4 9 6 and so Na 6 i2 i3 and Na E 23 mod 5 But for any x E R we have Na E 1 014 mod 5 74 CHAPTER 3 RINGS So indeed 23 and f are primes Note that 2 3 6 eff Hence 2 divides ff but as f and f are irreducible 2 divides neither 1 nor So 2 is not a prime With the same argument none of 37 f and f are not primes We claim that every proper element in R is a product of irreducible Indeed let a be proper in R and suppose that a is not irreducible Then a be with neither b nor 0 units Then as Na NbNc both b and c have smaller norm as a So by induction on the norm7 both b and c can be factorized into irreducible Since R has irreducibles which are not primes7 we know that R can not be a PlD But let us verify directly that I 27f 2R fR is not a principal ideal First note that ff 76 6 2R Since also 21 6 2R we If 6 2R Since 4 does not divide Nf7 1 2B and so I does not contain a unit Suppose now that h is a generator for I Then h is not a unit and divides 1 So as f is irreducible7 h N f and I But every element in f has norm divisible by Nf 67 a contradiction to 2 E I and N2 4 34 Localization Let R be a commutative ring and Q 74 S Q Rln this section we will answer the following question Does there exists a commutative ring with identity R so that R is a subring of R and all elements in S are invertible in R 7 Clearly this is not possible if 0 E S or S contains zero divisors It turns out that this condition is also su icient Note that if all elements in S are invertible in R also all elements in the subsemigroup of R7 generated by S are invertible in R So we may assume that S is closed under multiplication De nition 341 dmult A multiplicative subset of the Ting R is a nonempty subset S with st 6 S for all st E S Lemma 342 fractions Let S be a multiplicative subset of the commutatiue ring R De ne the relation N ofR X S by T17 81 N T2 82 if tT182 7 T281 0 for some t E S Then N is an equiualence relation Proof N is clearly re exive and symmetric Suppose now that 7617 81 N r2 32 and T27 82 T37 83 Pick t1 and 252 in S With t1T182 7 T281 0 and 252T283 7 T382 0 Multiply the rst equation with t233 and the second one with t181 Then both equation contain the term t1t2T28183 but with opposite sign So adding the two resulting equations we see 0 t283t1T185 7 t181t2T382 t1t282ltT183 7 T381 34 LOCALIZATION 75 Thus T17 81 N T37 83 B Let SR and N as in the previous lemma Then S lR denotes the set of equivalence classes of N stands for the equivalence class containing r 3 Note that if 0 6 S7 N has exactly one equivalence class If R has no zero divisors and 0 S then 71 7 if and only if 7132 7231 Proposition 343 ringfrac Let S be a multiplicative subset of the commutatiue ring R and s E S a The binary operations 7 7 73 TS r r M 7 7 7 and s 3 ss 3 33 on SilR are well de ned b S 1F tpl77 is an ring A c i is an identity d The map gt5R7S 1R THE S is a Ting homommphism and independent from the choice of s e gtss is invertible Proof a By symmetry it suffices to check that the de nition of and does not depend on the choice of r s in Let 7 g so trsl 7 ms 0 for some t E S Then trs rssls 7 713 rslss trsl 7 T18SS 0 and so is well de ned Also tTT818 7 TlTSS trsl 7 nsrs 0 and so is well de ned b It is a routine exercise to check the various rules for a ring Maybe the least obvious one is the associativity of the addition 7712HT73 7 7182 7231 73 7 718283 728183 T38182 7 71 T283 7382 7 2T72T73 81 82 83 8182 l 83 818283 81 l 8283 81 82 83 We leave it to the reader to check the remaining rules i 2 i i c and d are obv10us For e note that gtss 9 has 9 as its inverse D The ring S lR is called the ring if fraction of R by S It has the following universal property 76 CHAPTER 3 RINGS Proposition 344 uringfrac Let R be a commutatiue ring So a non empty subset and S the multiplicatiue subset ofR generated by S Suppose that R is a commutatiue ring with identity and aR gt R is a ring isomorphism so that aso is a unit for all so 6 S0 Then there exists a unique homomorphism M S lR a R with Mr a am Moreouer Proof Note that as aS0 consists of units so does aS So once we verify that is well de ned the remaining assertion are readily veri ed So suppose that 71 Then tT182 7 7231 0 for some t E S Applying 04 we conclude atana82 a 2a81 0 As ata31 and asg are units we get 6v7 1a8171 a7 2a8271 Hence 04 is indeed well de ned B When is g 0 The zero element in S lR is Hence g 0 if and only if there exists t E S with 0 trs 7 Os trs for some t E S This is true if and only if tr 0 for some t E S Put R5r thr0forsomet S So 2 0 ifand only T E R5 What is the kernel of j gt57 gtr Hence r E ker gt if and only if rs E R5 and so if and only if r E R5 Thus ker gt R5 So 1 is one to one if and only if R5 O This in turn just means that S contains no zero divisors In this is the case we will identify R with it image in S lR Let R be the set of all non zero non zero divisors and assume R 7 Q We claim that R is a multiplicative set lndeed let st E R Suppose that rst 0 for some r E R Then as t E R rs 0 and as s E R r 0 so st 6 R The R lR is called the complete ring of fraction of R If R has no zero divisors then R R and the complete ring of fraction is a eld This eld is called the eld of fraction of R and denoted by lFR 34 LOCALIZATION 77 The standard example is R Z Then Q If K is a eld then lme Kz the eld of rational functions over K Slightly more general if R has no zero divisors then lFRM ll7Rz7 the eld of rational function over the eld of fractions of R We will now spend a little but of time to investigate the situation where S does contain some zero divisors De ne r W L H 7 s gt8 We claim that f is a well de ned isomorphism For this we prove the following lemma w S lR a gtSgt 1 gtltRgt Lemma 345 alp Leta R gt R be a homommphism of commutative Tings and S and S multiplicative subsets ofR and 13 respectively Suppose that aS Q S a aS is a multiplicative subset of R b as i 571R H SilR H 047 39 7 s 043 is a well de ned homommphism c Suppose that S aS Then kero l r 6 R73 6 SSr kera 7 0 and 04S 1R 04S 104R s Proof a Just note that asat ast for all s7t E S b Note that pg043 is invertible Hence 04 is nothing else as the homomorphism given by 344 applied to the homomorphism 15 o a R a SHR c Let g E ker 04 As seen above this means t ozr 0 for some t E S By assumption t at for some t E T Thus 2 0 if and only if tr 6 kera for some t E S That 04S 1R 04S 104R is obvious D Back to the map f By the previous lemma f is a well de ned homomorphism and onto Let g E ker f Then tr 6 ker gt for some and t E S As ker gt R5 ttr 0 for some t E S Hence r 6 R5 and g 0 Therefore f is one to one and so an isomorphism Note also that gtR E RRS Let R 13135 and S S RSVRS As f is an isomorphism we get S lR g 3 11 We have R3 0 So in some sense we can always reduce to the case where S has no zero divisors 78 CHAPTER 3 RINGS In the next lemma we study the ideals in S lR For A C R and T Q S put T lAlaeAteT Proposition 346 idealfrac Let S be a multiplicative subset of the commutative ring R a IfI is an ideal in R then SilI is an ideal in SilR b If J is an ideal in R then I gt 1J is an ideal in R with J SilI C The map I gt SilI is a surjection from the set of ideals in I to the set of ideals to S lR Proof Put 1 g a is readily veri ed b lnverse images of ideals are always ideals To establish the second statement in b letj J AsJisanideal Thus gtp E Jandr EI Soj S 1I Conversely7 if r E I and s E S then since gtp E J an dJ is an ideal 7 7 82 8 7 8 8 S 2 TgtE2J J So 10 holds 6 follows from a and D If R has an identity the previous proposition can be improved Proposition 347 idfraid Let R be a commutative ring with identity and S a multi plicative subset R a Suppose R has an identity Let I be an ideal in R Then g1511r e R l Sr o I 7g 0 b De ne an ideal I in R to be Sil closed if E I for all r E R with TS O I 7 Then T 1 a 5 1 is a bijection between the Sil closed ideals and the ideals in SAP The inverse map is given by 7 1 J a ll 34 LOCALIZATION 79 c I S 0 for all S 1 closed ideals with I 7 R d A prime ideal P in R is S 1 closed if an only ifP O S e r induces a bijection between the S 1 closed prime ideals in R and the prime ideals in S 1Pt Proof a Let r E R then the following are equivalent gtr 6 5 1 gtr for somei 6 I75 ES forsomei Is S trsii 0 for someiEIst E S tsr ti for some i E Ist E S erIforsomesES Sr O I 7 So a holds We write 1 for 15 and say 77closed77 for S 1 closed b follows from a and 346b7c c Suppose I is closed and s E S O I Then S lI contains the unit gts and so S lI S lR Thus I gt1S 1R and I R d Let P be a prime ideal in R Suppose that S O P Q and let r E R and s E S with rs E P Then by 326b7 r E P or s E P By assumption 8 P and so r E P Thus P is closed Conversely7 if P is closed7 c implies P O S 0 e Let P be a closed prime ideal We claim the S lR is a prime ideal in B First since 739 is an bijection7 S lP 7 R Suppose that E S lP Then also gtrr E S lP As P is closed7 rr 6 P As P is prime we may assume r E P But then i E S lP and so S lP is a prime Suppose next that I is closed and S lI is a prime lf rr 6 I7 then gtr gtr E S lI As S lI is prime we may assume that gtr E S lI As I is closed this implies r E I and so I is a prime ideal B Let R be a commutative ring with identity By 326 an ideal P in R is a prime ideal if and only if R P is a multiplicative subset of B Let P be the prime ideal The ring RP Rp1R is called the localization of R at the prime P For A Q R write AP for R p 1A Theorem 348 localprimes Let P be a prime ideal in the commutatiue ring with iden tity R a The map Q gt Qp is a bijection between the prime ideals ofR contained in P and the prime ideals in RP b Pp is the unique maximal ideal in RP r E R is a unit if and only if Pp 80 CHAPTER 3 RINGS Proof a Put S R P and let Q a prime ideal in R Then Q O S 0 if and only if Q C P Thus a follows from b Let I be a maximal ideal in RP Then by 329 I is prime ideal Thus by a I Qp for some Q Q P Thus I Q Pp and I Pp The statement about the units now follows from 332 Actually we could also have argued as follows all elements in RP Pp are of the form i and so invertible Hence by 332 Pp is the unique maximal ideal in R D De nition 349 A local ring is a commutative ring with identity which as a unique max imal ideal Using 332 we see Lemma 3410 Chlocall Let R be a commutative ring with identity The the following are equivalent a R is a local ring b All the non units are contained in an ideal M g R c All the non units form an ideal D We nish this section with some examples Let p be an prime integer Then Zp E Q l pf m Since 0 and p are the only prime ideals of Z contained in p7 O and 6 Q l pfm7p l n are the only prime ideal in Zp What are the ideals Every non zero ideal of Z is of the form t for some t E Z Write t apk with p f a Suppose t is closed As a E S Z p and apk E t we conclude that pk E Thus a 1 and t pk It is easy to see that pk is indeed closed So we conclude that the ideals in Zp are kaa g 6 lelmmk l n In particular Zp is a PlD We reader might have notice that in the above discussion Z can be replaced by any PlD R p by any prime in R and Q by lFR 35 Polynomials rings power series and free rings Let R be a ring and 147 a group written additively Since 147 is a subgroup of the multiplicative group of B it is convenient to the following 77exponential notation Denote a E A be m and de ne zazb z The the elements in BM can be uniquely written as f ZQEA fax where fa 6 Pt almost all fa 0 Also 35 POLYNOMIALS RINGS POWER SERIES AND FREE RINGS 81 f9 Z Z 121be Zlt Z fame 161417614 06A abc Let R be a ring and I a set Put ll 1613 the free abelian monoid on I The semigroup ring RBI is called the polynomial ring over R in the variables I and is denoted by RH We will use the above exponential notation Let i E I Recall that gti1 6 l is de ned as 1 ifz j 0 ifz 7g j 39 We write xi for Mia Let Oz 6 ll7 then 04 is a tuple 040161 al 6 N where almost all 04 are zero So 0 7 04239 3 i H 3 61 Pi1 Every element 1 6 RH now can be uniquely written as f 2 mg 046539 where fa E R and almost all fa are zero The element of RU are called polynomials Polynomials of the form rm with r 7 07 are called monomials As almost all 04 are zero rm rzzil mg for some lk E I The fax with fa 7 0 are called the monomials of 1 So every polynomial is the sum of its monomials Note that the map r a rm is monomorphism So we can and do identify r with rmo lfI 17 27 m we also write Rm17 mm for RU Then every element in Rm17 7mm can be uniquely written as 0 0 0 7L1 2 n E E E Tmrwnmxl m2 n10 n20 nm0 0 If I has a unique element l we write x for mi and RM for RU Then each 1 6 RM has the form f TO mm T22 mm where n is maximal with repect to Tn 7 O Tn is called the leading coe clent of f and n the degree of 1 Note that if I happens to have the structure of a semigroup7 the symbol RU has a double meaning7 the polynomial ring or the semigroup ring But this should not lead to any confusions 82 CHAPTER 3 RINGS If y ghi E I is a family of pairwise commuting elements is some semigroup and Oz 6 l we de ne y H 24 tel Note that as almost all 04 are zero and the y pairwise commute this is well de ned Also if we View the symbol x as the family zhi E I this is consistent with the z notation The polynomial ring has the following universal property Proposition 351 polev Let I R a S be a ring homomorphism and y gm61 a family of elements in S such that a For all r E R and alli E I WNW WP b for all ij E I yiya39 yjyi Then there exists a unique homomorphism byRIgtS with rmiHltIgtryiforallr Ri I Moreover ltIgty Ef e a Zoggy 046539 046539 Proof As l is the free abelian monoid on I b implies that there exists a unique homo morphism B l a S with 1a yi The existence of by now follows from 316 The uniqueness is obvious D The reader should notice that the assumption in the previous proposition are automat ically ful lled if S is commutative So each 1 6 RH gives rise to a function f4 SI a S with My Mi Example Suppose I 1 2 m R S is commutative and 04 id idR and 7 M rm f7 E Tn1gtvgtnm1 Then f1dylv 39 quot vyrl ZTn1nmy1 The reader should be careful not to confuse the polynomial f with the function fid Indeed the following example shows that fid can be the zero function without 1 being zero Let R ZpZ I 1 p a prime integerand fxm71z71x7p71 Then 1 is a polynomial of degree p in ZpZ But fidy 0 for all y E ZpZ 35 POLYNOMIALS RINGS POWER SERIES AND FREE RINGS 83 Lemma 352 RU Let R be a ring and I andJ disjoint sets Then there exists a unique isomorphism RIJ gt RI U J with rzi gt rm and rmj gt rmj forallrERi Ij J Proof Use 351 to show the existence of such a homomorphism and its inverse We leave the details to the reader B Let R be a ring and G a semigroup In the de nition of the semigroup ring RG we had to use the direct sum rather than the direct product since otherwise the de nition of the products of two elements would involve in nite sums But suppose G has the following property FP lab E G x G l ab gH is nite for all g E G Then we can de ne the power semigroup ring of C over R7 RGby RHGM H R7gt 960 T9960 39960 Z 510960 hkEGXGlhkg If G is a group then it ful lls FP if and only if G is nite So we do not get anything new But there are lots of in nite semigroups with For example G N is isomorphic to the ring of formal power series Other semigroups with FP are the free abelian monoids or semigroups over a set Let I be a set Then the power semigroup ring RHEB Nll i39eI is called the ring of formal power series over R in the variables I and is denoted by The elements of RHIH are called formal power series We use the same exponential notation as for the ring of polynomials Every formal power series can be uniquely written as a formal sum f Z fo a 046 Here fa E R But in contrast to the polynomials we do not require that almost all fa are zero If I 1 the formal power series have the form fanmnf0f1f2m2fnx n0 with fn E R Note that there does not exist an analog for 351 for formal power series7 since the de nition of Pgf involves an in nite sum 84 CHAPTER 3 RINGS Lemma 353 invpower Let R be ring with identity and f E a f is a unit if and only if f0 is b IfR is commutative and f0 is irreducible then f is irreducible Proof a Note that fg0 fogo and 10 1 so if f is a unit so is 1 0 Suppose now that f0 is a unit We de ne g E by de ning its coe icients inductively as follows go 1 61 and for n gt O gt ital Zz omitsg Note that this just says 20 fn 39g 0 for all n gt 0 Hence fg 1 Similarly f has a left inverse h by222 g h is a left inverses 10 Suppose that f gh Then f0 goho So as f0 is irreducible one of go f0 is a unit Hence by a g or h is a unit D As an example we see that 1 7 z is a unit in lndeed 17x711mz2m3 Lemma 354 fdpow LetlD be a division ring a 90 f E llPllg ll lfo 0 b The elements of are exactly the non units of c Let I be a left ideal in Then I for some h E N d Every left ideal in is a right ideal and is a principal ideal ring e is the unique maximal ideal in f Ile is a eld is a PID and a local ring Proof a is obvious and b follows from 353 c Let k E N be minimal with zk E I Let f E I and let it be minimal with fn 7 0 Then 1 any for some 9 E with go 7 0 Hence 9 is unit and m 9 1 6 I So k g n and f zn kg k E Thus I de and f follow immediately form 36 FACTORIZATIONS IN POLYNOMIAL RINGS 85 36 Factorizations in polynomial rings Let R be a ring and I a set and f 6 RH We de ne the degree function deg RU a N U foo as follows a if f is a monomial rm then degf 2161 04 b if f 7 0 then degf is the maximum of the degrees of its monomials c if f 0 then degf foo Sometimes it will be convenient to talk about the degree dng f with respect to subset of J of I This is de ned as above7 only that WNWZ jeJ Alternatively dng f is the degree of f as a polynomial in R LI7 where R RI J A polynomial is called homogeneous if all its monomials have the same degree Let f E RX then f can be uniquely written as 00 f Z W i0 where hfi is zero or a homogenous polynomial of degree i Note here that almost all hfi are zero Let hf hf7 deg 1 Lemma 361 basdeg Let R be a Ting I a set and g 6 RH a degf g maxdeg f7degg with equality unless hg 7hf b Iff andg are homogeneous then fg is homogeneous Also either degfg degf deg9 or is 0 0 h1 9 hfh9 unless hfh9 039 d RU has no zero divisors if and only ifR has no zero divisors e deg fg deg f degg with equality if R has no zero divisors Proof ab and c are readily veri ed d If R has zero divisors7 then as R is embedded in RU RU has zero divisors Suppose next that R has no zero divisors Let g 6 RI We need to show that fg 7 0 By c we may assume that f and g are homogeneous 86 CHAPTER 3 RINGS Consider rst the case that lIl 1 Then 1 am 9 bzm and fg abz m Here ab E Pitt and so ab 7 0 Thus also fg 7 0 If I is nite7 RU and so by induction RU has no zero divisors For the general case just observe that f g E R e If R has no zero divisors7 d implies hfh deg deg1W deghfh9 degh J for some nite subset J of I g 7 0 Thus by b and c7 AA 1 deg19 deg degg Lemma 362 RP Let R be a ring P an ideal in R and I a set a LetPI 6 RH l fa E P for alla Ell Then PI is an ideal in RH and RlIlPlll g RPWl b If R has an identity PI P RU is the ideal in RU generated by P Proof a De ne j RU 7 PtPHILZadfaz 7 Zad a Pm By 351 1 is a ring homomorphism Clearly j is onto and ker gt PI so a holds b Let p E P then pm 6 P RU Thus PI P RI The other inclusion is obvious D Corollary 363 pstay Let R be a commutatiue ring with identity I a set and p E R Then p is a prime in R if and only ifp is a prime in RU Proof R is a prime if and only if RpR is an integral domain So by 361d if and only if RpRI is an integral domain So by 362 if and only if RIpRI is a prime ideal and so if and only ifp is a prime in RU D Theorem 364 Long Divison lngdiv Let R be a ring and g 6 Suppose that the leading coe cient ofg is a unit in R Then there exist uniquely determined qr E R with f qgr and degr lt degg Proof Let hf am and hg In If n lt m7 we conclude that q 0 and r f is the unique solution So suppose that m g n Then any solution necessarily has hf hqhg and so sq ab lmn m Now 1 qg 7 r if and only if f 7 abilmn mg q 7 abilmn mm r So uniqueness and existence follows by induction on deg f D 36 FACTORIZATIONS IN POLYNOMIAL RINGS 87 Let R be a ring and f E De ne the function fTR7gtRC7gtZfacD WEN The function fT is called the right eualuation of 1 Note here that as R is not necessarily commutative fac might differ from caf If R is commutative fT fid The map f 7 fT is an additive homomorphism but not necessarily a multiplicative homomorphism That is we might have fgTc 7 fTcgTc lndeed let 1 rm and g sz Then fg rsz2 fgTc T862 and fTcgTc TCSC Lemma 365 fgfg Let R be a ring g 6 RM and c E R If gTcc 6976 then fQWC 107039703 Proof As f 7 fT is a additive homomorphism we may assume that f 7 for some T6R7mEN Thus f9 Z mama WEN and so f9 c 2 WW WEN M gacacm rgTC m rcmgWC fTcgTc WEN Corollary 366 xrnc Let R be a ring with identity 0 E R and f E a Then there exists a unique q 6 RM with f q C 1 70 b fTC 0 if and only if f qm 7 c for some q E Proof a By364 f q m 7 c r with degr lt degz 7 c 1 Thus r E B By 365 1 70 WW0 C t T T Hence r fTc The uniqueness follows from 364 In follows from a D 88 CHAPTER 3 RINGS Corollary 367 xrncp Let R be an commutatiue ring with identity and c E R a 7 c g R b m 7 c is a prime if and only R is an integral domain Proof a Consider the ring homomorphism idc RM 7 R7 f 7 fc see 351 Clearly idc is onto By 366b keridc m 7 c so 10 follows from the lsomorphism Theorem for rings 10 Note that z 7 c is a prime if and only if 7 c has non zero divisors Thus 10 follows from a D Corollary 368 Let ll7 be a eld Then is an Euclidean domain In particular is a PID and a UFD The units in are precisely the nonzero elements in ll Proof Just note that by 364 is a Euclidean domain B Let R be a subring of the commutative ring S Write R 7 S for the inclusion map from R to S Let I be a set7 f 6 RH and c 6 SI We say that c is a rootoff if 23503 039 Let R be any ring7 f 6 RM and c E R We say that c is a root of f if fTc 0 Note that for R commutative this agrees with previous de nition of a root for f in R Theorem 369 sroots Let D be an integral domain contained in the integral domain E Let 0 7 f E Let in E N be maximal so that there exists 17 cm 6 E with m Hm7ci l f i1 in Let c be any root of f in E Then 0 ci for some i In particular 1 has at most deg 1 distinct roots in E Proof Let f gHZlm 7 ci with g E By maximality of m7 z 7 cfg By z 7 c is a prime in and so m m 7 c l H m 7 ci i1 By3377m7cwm7ciforsomei Thusm7cz7ciandcci D We remark that the previus theorem can be false for non commuative divison rings For example the polynomial 2 1 0 has six roots in the division ring H of quaternions7 namely ii ij ik 36 FACTORIZATIONS IN POLYNOMIAL RINGS 89 Let R be a ring7 f 6 RM and c b a root of f in D Then by 369 We can write 1 has 1 gz 7 c with m E Z1 9 6 RM and so that c is not a root of g m is called the multiplicity of the root g If m 2 2 we say that c is a multiple root As a tool to detect multiple roots we introduce the formal derivative 1quot of a polynomial f E f Z naza l a6 Put 1401 f and inductively Null My for all k e N Lemma 3610 diru LetR be a ring g 6 RM and c E R Then a CW Cquot b 2 g f g C fgy M J s d1fff ffi W nfnilfl Proof a and b are obvious c By 10 we may assume that f ram and g sz are monomials We compute awwmwwmmme4 fg 109 mrzT Lil Thus c holds d follows from c and induction on n D 3m rmmnsmnil n mrsxmn71 Lemma 3611 Let R be a ring with identity 1 6 RM and c E R a root of a Suppose that f gm 7 c for some n E N and g E Then Wamaa b c is a multiple root of f if and only if fc 0 c Suppose that deg is neither zero nor a zero divisor in R Then the multiplicity of the root 0 is smallest number m E N with flmlc 7 0 Proof a We will show that for all 0 g i n7 there exists hi 6 RM with x f 7 we 7 W w e For i 0 this is true with he 0 So suppose its true for i Then using 3610 90 CHAPTER 3 RINGS 7g ltz7cgtwgltn7igtltz7cgtW1gth2ltz7cgtw1hiltn7i1W This is of the form jimmz 7 0 holds for i 1 For i n we conclude flnl nl9 hnz 7 a Thus a holds b Since 0 is a root7 f 9z 7 a for some 9 E So by a applied to n 17 f 0 90 Thus b holds c Let in the multiplicity of 0 has a root of 1 So 1 9z 7 0 for some 9 6 RM with 90 7 0 Let n lt m Then 1 9m7 0m m7 0 and a implies f l0 0 Suppose that flml0 0 Then by a7 ml90 0 As in degf we get deg fl90 0 Thus by assumption 90 07 a contradiction This Fm 0 7 0 and c holds D fli1l flily i l plus a left multiple of 70 l So the statements Consider the polynomial mp in ZpZM Then z pzp l 0 This shows that the condition on deg f in part c of the previous theorem is necessary Let D be an UFD7 I a set and f 6 DH We say that f is primitive if 1 is a greatest common divisor of the coe icents of 1 Lemma 3612 cont Let D be a UFD ll7 its eld of fractions and I a set Let f E Then there exists af7bf E D and F E so that a F is primitive in DU b af and bf are relatively prime a f a Moreover af7bf and F are unique up to associates in D Proof We will rst show the existence Let f Ewe fax with fa 6 ll Then fa 7 with rmsa E D Here we choose sa 1 if fa 0 Let s Hadsa Then sf 6 DH Let r gcdadsfa and F r lsf Then F E DU7 F is primitive and f EF Let e the a greates common divisor of r and s and put af g and bf Then ab and c hold To show uniqueness suppose that f f with ab 6 D relative prime and f 6 DH primitive Then ha bfaf Taking the gretaes common divisor of the coe icents on each side of this equation we see that baf and bfa are associate in D In particular7 a divides baf and as b is realtively prime to a7 a divides af By symmetry af divides a and so a aaf for some unit a in D Similarly I vbf for some unit v E D Thus vbfafF abfaff As D is an integral domain we conclude f a lvF B Let f be as in the previuos theorem The fraction 0f 7 is called the content of 1 Note that 0f 6 lF and f 0fF 36 FACTORIZATIONS IN POLYNOMIAL RINGS 91 Lemma 3613 cont Let D be a UFD lF its eld of fraction I a set and 1 79 6 lFI lFlIl f Suppose f is primitive Then f is a prime in DU if and only if it is a prime in Proof Note that f9 ofcgf9 So a7 b and c will follow once the show that the product of two primitive polynomials is primitive Suppose not Then there exist primitive 1 79 6 DH and a prime p in D dividing all the coefficients of f9 But then p l f9 in DU By 363 p is prime in DH and so p divides f or 9 in DU A contradiction as f and 9 are primitive d Suppose that f l 9 Then 9 fh for some h E By b 9 fh and so d holds e Suppose that f is irreducible in lFU and f 91 with 97 h 6 DM Then by a both 9 and h are primitive On the other hand since 1 is irreducible in EU one of 9 or h is a unit in FU and so in l It follows that one of 9 and h is a unit in D So 1 is also irreducible in DU Suppose that f is irreducible in DH and f 91 for some 971 6 Then 1 f N 9h and as f is irreducible in DU7 one of 97 h is a unit in D But then one of 9 and h is in lF and so a unit in f Suppose that f is prime in DH and that f l 971 in By d f f l 9h and as f is a prime in DU we may assume 1 l 9 As 9 divides 9 in FU 1 does too So 1 is a prime in Suppose that f is a prime in FU and f l 971 in DU for some 971 6 DH Then as f is a prime in FU we may assume that 1 l9 in But d f f l9 in DU As 9 divides 9 in DU7 1 does too So 1 is a prime in DU D Theorem 3614 DXUFD Let D be a UFD and I a set then DH is a UFD Proof Let f be in DU We need to show that f is the product of primes Now 1 E DJ for some nite f and by 363 a prime factorization in DJ is a prime factorization in DU So we may assume that J is nite and then by induction that 1 1 Note that f eff with f 6 DM primitive and of E D As D is a UFD7 of is a product of primes in D and by363 also a prodcut of primes in So we may assume 92 CHAPTER 3 RINGS that f is primitive Suppose that f gh with 97 h 6 DM with neither 9 nor h a unit As 1 is primitive7 g and h both have positive degree smaller than 1 So by induction on deg 1 both 9 and h are a product of primes So we may assume that f is irreducible Let lF lFD By 3613 1 is irreducible in As is Euclidean7 f is a prime in Hence by 3613 1 is a prime in D Chapter 4 Modules 41 Modules and Homomorphism In this section we introduce modules over a ring lt corresponds to the concept of group action in the theory of groups De nition 411 Let R be a ring A left R modules is an abelmn group M together with a function RXMHM7 Tmgtrm such that for all r s E R and 071 6 M Ma Na b ra Tb Mb 7 sa ra 3a Mc rsa rsa Given a ring R and an abelian group M M is an R modules if and only if there exists a ring homomorphism Q R a EndM Indeed if M is an R modules de ne Q R a EndM by Qr m rm By Mot7 Qr is indeed an homomorphism And Mb and Me imply that Q is a ring homomorphism Conversely7 given a ring homomorphism Q R a EndM7 de ne rm Qrm Then M is a R module Analogue to a module we can de ne a right R module via a function M x R a R ful lling the appropriate conditions It is then easy to verify that a right modules for R is nothing else as a left modules for Rep ln particular7 for commutative ring left and right modules are the same Every abelian group M is a Z m0dule via 71 m H mm Indeed this is the only way M becomes a Z module under the additional assumption that 1m m 93 94 CHAPTER 4 MODULES De nition 412 Let R be a ring with identity and M a R modules a M is a unitary R module provide that 1m m for all m E M b If R is a diuision Ting and M is unitary then M is called a uector space ouer R Examples If M is an abelian group then M is a module for EndM Via gtm gtm If R is a ring and I is an left ideal then I and RI are R modules by left multiplication If S is a subring of R then R is an S module Via left multiplication Direct sums and direct products of R modules are R modules In particular7 if Q is a set7 then R9 is a R module Via rfw rfw Or TSw rsw Let R is a ring and G a semigroup Suppose that G act on a set 9 Then R is an RG modules To see this we rst de ne an action of G on R9 by 9fw f9 1w Then extend this to RG by Z 799 Z 7 ny 960 960 De nition 413 Let V and W be R modules An R module homomoijohism from V to W is a function 0sz such that fa c fa and fra rfa for allac Rr ER We often will say that f V a W is R linear instead of f V a W is a R modules homomorphism Terms like R module monomorphism7 R module isomorphism7 kerf and so on are de ned in the usual way If V and W are R modules7 HomRV7 W denotes the set of R linear maps from V to W Since sums of R linear maps are R linear7 HomRV7 W is an ablian group EndRV denotes set of R linear endomorphsims of V Since compositions of R linear maps are R linear7 EndRV is a ring Note that V is also a module for EndRV Via to gtu De nition 414 Let R be a Ting and M a R module An R submodule A of M is an additiue subgroup A ofM such that Ta 6 A for all r E R a E A 41 MODULES AND HOMOMORPHISM 95 Note that submodules of modules are modules If the modules is unitary so is the submodule If V is a submodules of M7 then MV is a R module by rm V rm V It is easy to verify that this gives a well de ned module structure Also the map M H MV7 m H m V is R linear7 is onto and has kernel V If f V a W is R linear7 then kerf is a submodule of V and fV is a submodules of W Theorem 415 Isomorphism Theorem for Modules IMT Let R be a m9 and f V gt W and R lz39near map Then i vW H mm W a u is a well de ned R lz39near isomomhz39sm Proof By the isomorphism theorem for groups 2557 this is a well de ned isomorphism of additive groups We just need to check that it is R linear So let T and v E V Then JFWU W 13W W v THU W Let M an R module and S Q R and X C M SXsmls Sx X that is the additive subgroup of M generated by the sz 3 6 S7 z E X Note that for all X Q M7 RM is a submodule of M Let T Q R Recall that ST st l s 6 525 6 T Note that this agrees with the above de nition of SX7 when we view T is a subset of the R module R It is easy to verify that STXSTX stx l 3 65 Tz EX Here we wrote stm for stz 3tz De ne the submodule X of M generated by R as the intersection of all the R submodules of M containing X Note X M RM and if M is unitary X RM For X C M de ne the annihilator of R in X as AnnRX r e R TX 0 For S Q R de ne AnnMS m E M l Sm 96 CHAPTER 4 MODULES Lemma 416 bann Let R be Ting M a R module and X Q M Then a AnnRX is a left ideal in R b Let I be a right ideal in R Then AnnMI ls R submodule in M c Suppose that one of the following holds 1 R is commutative 2 All left ideals in R are also right ideals 3 AnnRX is a right ideal Then AnnRX AnnRX d Let m E M Then the map RAnnRm gt R7717 gt mm is a well de ned R lsomomhlsm a Let as E AnnRX7 t E R and z E X Then T sm rm sm 0 and tsx tsm So AnnRX is a left ideal in R b RAnnMI IRAnnMI Q IAnnMI Thus RAnnMI AnnMI and 10 holds c Note that 1 implies 2 and by a 2 implies 3 So in any case AnnRX is a right ideal in R Hence by In W AnnMAnnRX is an R submodule Since X Q W we get X W Thus AnnRX annihilates So AnnRX AnnRX The other inclusion is obvious d Consider the map f R gt M7 7 gt rm Clearly f is Zrlinear Also for r s E R ms mm rem ms So f is R linear Since AnnRm ker f7 d follows from the isomorphism theorem D Example Let K be a eld Let R be the ring of n x n matrices over K Then K is a module for R Via 221 mijkj Let e1 em 6 K be de ned 42 EXACT SEQUENCES 97 by e11 1 and em 0 for all 2 g i g n Then AnnRe1 consists of all matrices whose rst column is zero Note that e1 Rel K indeed if k E K and M is any matrix with k as it rst column7 then Mel k and so k E Rel Hence AnnRe1 0 and AnnRe1 7 AnnRe1 So the conclusion in part c of the preViuos lemma does not hold in general 42 Exact Sequences De nition 421 dexact A nite or in nite sequence of R linear maps fAZEgtA1LAf gt H1 is called exact if for all suitable j E Z 1111 fj ker fj1 We denote the zero R module with 0 Then for all R modules M there exists unique R linear maps7 0 a M and M a O The sequence 0 L A L B is and only if f is one to one A L B gt 0 is exact if and only if f is onto 0 L A L B L 0 is exact if and only if f is an isomorphism A sequence of the form 0 L A L B L C L 0 is called a short sequence If it is exact we have that f is one to one7 kerg lm f and g is onto Since 1 is one to one we have lmf E A and so kerg E A Since 9 is onto the isomorphisms theorem says Bkerg E C So the short exact sequence tells us that B has a submodule which isomorphic to A and whose quotient is isomorphic to C Given two exact sequences A homomomhism of exact sequences 4p A a B is a tuple of R linear maps h1 Al a 13 so that the diagram fiil fi fi1 fi2 gt Aiil gt gt i1 gt lhiil lhi1 923971 g fi1 9i2 L Bi1 L Bl L Bi1 L 98 CHAPTER 4 MODULES commutes idA A a A is de ned as idAi 4p is called as isomorphism if there exists i9 B a A with 194 idD and m9 idg It is an easy exercise to show that 4p is an isomorphism and if and only if each hi is Theorem 422 Short Five Lemma ve Given a homomorphism of short exact se quences f 9 O gtA gtB gtC gt0 la lquot l7 O gtAgtBLgtC gt0 Then a fa and 39y are one to one so is b fa and 39y are onto so is c If 04 and 39y are isomorphisms so is Proof a Let b E B with 31 0 Then also g b 0 and as the diagram commutes 39ygb 0 As 39y is one to one 91 0 As kerg lm f b at for some a E A Thus Bfa 0 and so f aa 0 As 1quot is one to one aa 0 As 04 is one to one a 0 So I fa 0 and B is one to one 10 Let b E B As 39y and g are onto so is 39yog So there exists I E B with g b 39ygb As the diagram commutes 39ygb g b Thus at b i b E kerg As kerg lm f and 04 is onto kerg lmf o 04 So at f aa for some a E A As the diagram commutes d Bfa So 5 7 35 d BUM Hence 3 31 fa and B is onto c follows directly from a and D Theorem 423 split Given a short exact sequence 0 gt A i B i C gt 0 Then the following three statements are equivalent a There exists a R linear map 39y C gt B with g o 39y idC b There exists a R linear map 7 B gt A with 7 o f idA c There exists 739 B gt A EB C so that f 9 O gtA gt B gtC gt0 H l7 H 0ALA CLCO is an isomorphism of short exact sequences 42 EXACT SEQUENCES 99 Proof a c Consider 0ALA CLCO H W H 0AL B LCMO Here f7 39y A 63 C a B7 10 a fa 39yc It is readily veri ed that this is a homomor phism The Short Five Lemma 422 implies that is an isomorphism b c This time consider OMAL B LCMO H W H o vALA CLgtC gtO c aamp b De ne 7rlo739and 7 1 2 Then 7 39Y P 7of7rloTof7rlop1idA 90v90T 10p27r10p2idc D An exact sequence which ful lls the three equivalent conditions in the previous theorem is called split To make the last two theorems a little more transparent we will restate them in an alternative way First note that any short exact sequence can be viewed as pair of R modules D g M lndeed7 given D g M we obtain a short exact sequence 0 gt D gt M gt M D gt 0 Here D gt M is the inclusion map and M a MD is the canonical epimorphism Con versely7 every short exact sequence is isomorphic to one of this kind f 9 O gtAgtBgt C gt0 if H if OelmfeBeBlmfeo Secondly de ne a homomorphism Q A g B a A g B to be a homomorphism Q B a B with QA g A Such a Q corresponds to the following homomorphism of short exact sequences 100 CHAPTER 4 MODULES 0AgtB gtBA gt0 14gt 14gt l bBA 0A gtB gtBA gt0 Here Q4 A 7 A a 7 ltQa and QBA BA 7 B A 1A 7 ltQ1 A Since ltQA A both of these maps are well de ned Lets translate the Five Lemma into this language Lemma 424 ve77 Let Q A g B 7 A g B be a homomorphism a If Q4 and QBA are one to one so is I b If Q4 and QBA are onto so is I c If Q4 and QBA are isomorphism so is Q Proof This follows from the ve lemma7 but we provide a second proof a As kengA O7 kerltQ A So kerltQ kerQ4 O b As QBA is onto7 B ltQB A As ltQA A we conclude B ltQB c Follows from a and D The three conditions on split exact sequences translate into Lemma 425 split77 Given a pair of R modules A g B The following three conditions are equivalent a There exists a homomorphism 39y BA 7gt B with 1 39y1 A for 0111 6 B b There exists a homomorphism 71 B 7gt A with 71a a for all a E A c There exists a R submodule K of B with B A K Proof Again this follows from 423 but we give a second proof a c Put K 39yBA Then clearly K A B Also if 39y1 A E A we get 1AAOBA Thus39y1A 0andK AO b c Put K ker71 The clearly K O A 0 Also if 1 E B Then 711 6 A and 711 7 711 711 7 711 0 Thus 1 711 1 7 711 6 A B Thus B A K c a a De ne 39yk A k for all k E K c a b De ne 71akafor allaEA7k K D Finally ifA is a R submodule of B we say that B splits over A ifthe equivalent conditions in the previous lemma hold 43 PROJECTIVE AND INJECTIVE MODULES 101 43 Projective and injective modules In this section all rings are assumed to have an identity and all R modules are assumed to by unitary Wewrite gtA Bif gtAHBis onto And AH Bif gtisonetoone De nition 431 Let P be a module ouer the ring R We say that P is projectiue provided that P A P A B B where both diagrams are commutatiue Let I be a set and R a ring Then the free module FRI on I over is the module 9161 R We will usually write FI for FRI We identify i with pi1 Hence FRI Bid Ri Lemma 432 universalfree LetM be an R module and ml7i E I a family of elements in M Then there exists a unique homomorphism oz FRI gt M with i gt mi 04 is giuen by 042161 mi 2161 rimi Proof Obvious Lemma 433 freeproj Any free module is projective Proof Given oz A B and B FRI a B Let i E I Since 04 is onto7 04ai for some ai E A By 432 there exists 39y FRI a A with ai Then MW Mai W So by the uniqueness assertion in 4327 1 o 39y B B Let A and B be R modules We say that A is a direct summand of B if A g B and BACforsomeC B Note that if A is a direct summand of B and B is direct summand of C then A is a direct summand of C Also if Al is a direct summand of Bi then Bid Al is a direct summand of i61 Bi39 Lemma 434 dsproj Any direct summand of a projectiue module is projective 102 CHAPTER 4 MODULES Proof Let P be projective and P P1 P2 for some sub modules P1 of P We need to show that P1 is projec tive Given oz A B and B P1 a B Since P is l l l l 7T1 7 prOjective there ex1sts 39y P a A With pl 04 o 7y 0 7r1 y A Put 39y am Then 5 040V040 YOP1507V10P15 Theorem 435 Chproj Let P be a module ouer the ring R Then the following are equivalent a P is projective b Euery shot exact sequence 0 gt A L B i P gt 0 splits c P is isomorphic to a direct summand of a free module Proof a b Since P is projective we have idp P So the exact sequence is split by 423a b a c Note that P is the quotient of some free module F But then by b and 423c7 P is isomorphic to a direct summand of F c a Follows from 433 and 434 Corollary 436 Direct sums of projectiue modules are projective Proof Follows from 435c Next we will dualize the concept of projective modules De nition 437 A module J for the ring R is called injective if J A 1414 N gt R B B where both diagrams are commutatiue D 43 PROJECTIVE AND INJECTIVE MODULES 103 Above we showed that free modules are projective and so every module is the quotient of a projective module To dualize this our rst goal is to nd a class of injective R modules so that every R modules is embedded into a member of the class We do this into step steps First we nd injective modules for R Z Then we use those to de ne injective modules for an arbitrary ring with identity To get started we prove the following lemma which makes it easier to verify that a given module is injective Lemma 438 ezinj Let J be a module over the ring R Then J is lnjectz39ue if and only if for all left ideals I in R J R J l R N gt R I I where both diagrams are commutative Proof GivenaBHAand BHJweneedto nd39yBHJwith 39yoz Without loss B A and 04 is the inclusion map 3 ya now just means 39y 3 3 That is we are trying to extend B to A We will use Zorn s lemma nd a maximal extension of 3 Indeed let M6DHJBDA6BB Order M by 61 g 62 if D1 g D2 and 62lD1 61 We claim that every chain 6139 Di a J l i E I in M has an upper bound Let D Uid Di and defone 6 D a J by 6d 6id if d E Di for some i E I It is easy to verify that 6 is well de ned 6 E M and 6 is an upper bound for delta D a J l i E I Hence by Zorn s lemma M has a maximal element 6 D a J The reader might have noticed that we did not use our assumptions on J yet Maximal extensions always exists Suppose that D 7 B and pick I E B D Consider the R linear map aDRgtA drgtdrb Let I be the projection of keer onto D Then as keer is a submodule of D 63 R I is a submodule of R that is a left ideal Moreover keer 7ib l i E I and I consists of all r E R with Tl E D Consider the map E I a Ji a 6ib By assumption 5 can be extended to a map 5 R a J with W 62 b 104 CHAPTER 4 MODULES De ne E D 63R a J7 176 a 6d 50 Then E is R linear Also 37ib7 76ib 76ib 6ib 0 Hence kerlu kerE and we obtain a R linear map E D Rkerlua J So by the lsomorphism Theorem we conclude that DRbgtJ7 drbgt6d r is a well de ned R linear map Clearly its contained in M7 a contradiction to the maximal choice of 6 Thus D B and J is injective The other direction of the lemma is obvious D Lemma 439 hrmr Let R be a ring and M an R module Then A HomRR7 M HM j a gt1 is a Z isomomhism Proof Clearly A is Zrlinear To show that A is an bijective we will nd an inverse Let in E M De ne Pm RHM7TgtTm The claim that Hm is R linear lndeed its Z linear and Pmsr srm 3Tm for all st E B So Hm E HomRR7 Also and for j E HomRRM7 PA gtT TAM T gt1 M MT So PA gt j and P is the inverse of A B Let R be an integral domain We say that the R module M is divisible if TM M for all r E Ff Note that every quotient of a divisible module is divisible Also direct sums and direct summand of divisible modules are divisible If R is divisible as an R modules if and only if R is a eld The eld of fraction7 lFR is divisible as an R module Lemma 4310 divinj Let R be an integral domain and M an R module a IfM is injectiue then M is divisible 43 PROJECTIVE AND INJECTIVE MODULES 105 b If R is a PID M is injective if and only of M is divisible Proof a Let 0 344 t E R and m E M Consider the map Rt gt M7 rt gt rm As I is an integral domain this is well de ned and R linear As M is injective this homomor phism can be extended to a homomorphism 39y R a M Then t39y1 39yt1 39yt m Thus m E tR and R tR so M is divisible b Suppose that M is divisible Let I be a ideal in R and B I a M a R linear map As R is a PlD7 I Rr for some t E B As M is divisible7 Mt tm for some m E M De ne 39y R gt M7 r gt rm Then 39y is R linear and 39yrt rtm 3rt We showed that the condition of 438 are ful lled So M is injective D Proposition 4311 exinj Let R be a integral domain a Every R module can be embedded into an divisible R modvle b If R is a PID then every R modvle can be embedded into a injective module Proof a Let M a R module Then M AB where A 161 Rfor some set I and B is a submodule of A Let D ileR Then D is divisible and B A g D Also DB is divisible and AB is a submodule of DB isomorphic to M b follows from a and 4310 D An abelian group A is called divisible if it is divisible as module Let R be a ring and A7 B and T be R modules Let 1 A a B be R linear Then the maps f HomRBT a HomRAT7 f a f o gt and lt13 HomRAT H HOmRBT7 f H l O f are Z linear Suppose that 11 B a C is R linear Then ism oi and wow io Lemma 4312 exian Let R be a ring M a R modvle D a right R modvle and A an abelian group 106 CHAPTER 4 MODULES a H0m25DA is an R module via r gtd gtdr b The map E EM7 A HomRMH0m5RA a HornMA7 EltIgtm ltIgtm1 is an Z isomorjohism C EM7 A depends naturally on M and A That is ca Let B A gt B be Z linear Then the following diagram is commutative HomRM7 HangAR A M HomM7 A 2 a HomRM Hom5R 13 M HangAM B That is an o lt1gt 5 o 3a for all lt1gt e HomRMHom5R A Cb Let n M gt N be R linear Then the following diagram is commutative HomRM7 HangAR A M HomM7 A T t lt lt HomRN7 HornR7 A M H0111 N7 A That is o 7 E Il o n for all 1 E HOInRN7 Horn R d If A is divisible Hom R7 A is an injective R module Proof a Let r s E R7 171 E HornD7 A and d e E D WW 5 gtd t 5W Md 5i MW M T gtd WW Thus M E HornD7 A W wgtgtltdgt lt gt wgtltdrgt gtltdrgt W w wgtltdgt T sgt gtltdgt gtltdltr 8 W gtltdsgt w s gtgtltdgt WWW gtdTS gtdi 8 WNW WNW S0 HornD7 A is indeed a R module 10 Clearly E is Zrlinear Suppose that EltIgt 0 Then ltIgtm1 0 for all m E M Let r E R Then 0 1gtrm1r 1gtm1 1gtm1r 10700 43 PROJECTIVE AND INJECTIVE MODULES 107 Thus Qmr 0 for all r So Qm 0 for all m and so Q 0 So E is on to one To show that E is onto7 let Oz 6 HomM7 A De ne Q E HomRMHom5RA by Qmr 04rm Clearly Qm is indeed in Hom5RA We need to verify that Q is R linear Let s E R Then Phismr 04rsm and 3Qmr Qmrs 04rsm So Qsm sQm and Q is R linear Also 3 1 m 1gtm1 641m Mm and so EQ Oz and E is onto ca V B O E 1gtm 5E 1gtm 3 I m1 and V V E O B 1gtm B m1 B 1gtm1 m1 cb Let 1 E HomRNHom5R7 Then and E O n1 m En1 m n1 m1 nm1 d Let I be a left ideal in R and B I a HornR7 A By 438 we need to show that 3 extends to 39y R a HomzjRA Let E EI7 A be given by Put 3 Then 3 I gt A is Zrlinear Since A isdiVisible7 it is injective as an Z4nodule So 3 extends a Z4inear map 7y R a A That is B 7y 0 p7 where p I a M is the inclusion map By b there exists an R linear 39y M a HornR7 A with E39y By cb EWOP Evopampop33 As E is one to one7 we conclude B 39y o p and so 39y is the wanted extension of B D Theorem 4313 eminj Let R be a ring Every R module can be embedded into an injectwe R module Proof Let M be a R module By 4311 M is a subgroup of some diVisible abelian group A Let p M a A be the inclusion map Then 5 HomRR7 M H HomRA7 j a po is a R momomorphism By 439 M E HomRR7 M and so M is isomorphic to an R submodule of HornR7 A By 4312 Hom5RA is injective D 108 CHAPTER 4 MODULES Lemma 4314 prodinj a Direct summdnds of injectiue modules are injectiue b Direct products of injectiue modules are injectiue Proof a Let J J1 ean with J injective Given oz B H A and B B a J1 As J is injective there exists 7y A a J with oaplo Put 39y 7r1 o 7y Then ySoz7rlo7yoa7rloploaoz b Suppose that Jhi E I is a family of injective modules Given oz B a A and B B a Hid Ji Since Ji is injective there exists y A a Ji with yioa7rio Put Y Wid Then wio39yoog39yioa7rio and so 39y o 04 3 Hence Hid Ji is injective D Theorem 4315 Chrinj Let M be an R module Then the following are equiualent a M is injectiue b If A is a R module with M g A then A splits ouer M Proof a b Since M is injective we obtain M l A idM MaA M Hence by 3927 A splits over M b a By 43137 M is a submodule of an injective module So by assumption7 M is a direct summand of this injective module Thus by 4314 M is injective D 44 THE FUNCTOR HOM 109 44 The Functor Horn If A B7 idALB denotes the inclusion map A a B7 a a a Lemma 441 homeX Let R be a ring Given a sequence A L B i C of R modules Then the following two statements are equivalent a A L B A C is exact and A splits over ker b For all R modules D HomRD A L HomRD B i HomRD C is exact Proof We rst compute kerg and lm Let B 6 HomRD7 B Then 9 o B 0 if and only if lm 3 kerg Thus kerg HomRD7 kerg Also lm f f o HomRDA f o a l a e HomRDA HomRDim f Suppose rst that a holds Then kerg lm f and A ker feaK for some R submodule K of A It follows that f lK K a lm f is an isomorphisms Let 1 E HomRD7 lm 1 Let 04 idKHAo lK 1 o gt Then 04 E HomRDA and f 004 1 So mf HomRDlm f HomRDkerg kerg Suppose next that 10 holds Let D kerg Then idkergng E kerg 1mg HomRDlmf thus kerg lm 1 Next choose D A Then f f o idA E kerg HomRDkerg Hence lmf kerg and so 0 HA 13 H CH0 is exact Finally choose D lm 1 Then idImeB E kerg and so idlmfaB f O V for some 39y E Homlm f A So by 4257 A splits over A D Here is the dual version of the previous lemma 110 CHAPTER 4 MODULES Lemma 442 homexl Let R be a ring Given a sequence A L B L C equivalent a A L B L C is exact and C splits over 1mg b For all R modules D HomRA D L HomRB D 9 HomRC D is exact Proof Dual to the one for 441 We leave the details to the reader The following three theorem are immediate consequences of the previous two Theorem 443 homex2 Let R be ring Then the following are equivalent a 0 L A L B L C is exact b For every R module D 0 L HomDA L HomD B L HomD C is exact D Theorem 444 homex3 Let R be ring Then the following are equivalent a A L B L C L 0 is exact b For every R module D HomRA D L HomRBA 9 HomRC A L 0 is exact D Theorem 445 homex4 Let R be a ring Given a sequence of R modules 0 gt A L B a C gt Then the following three statements are equivalent 44 THE FUNCTOR HOM 111 a OAALBLCAO is split exact b For all R modules D 0 HomRDA L HomRD B i HomRD C a 0 is exact C For all R modules D 0 E HomRA D L HomRB D 9 HomRC D E 0 is exact D Theorem 446 homdir Let A and Bht E I be R modules Then a HomR id B1314 E Hid HomRBiA b HomRltA H B e H HomRltA B c HomRltA 696 B e e HomRltA B Proof Pretty oloVious7 the details are left to the reader B Let R and S be rings An RS b239module is abelian group M so that M is a left R module7 a right S module such that rms rms forallrER736SandeM For example R is a RR modules if we View R as a left R and right R module by multiplication from the left and right7 respectively Lemma 447 bimo Let R and S be Tings Let j A gt A be R llnear and let B a RS blmodule Then a HomRA7 B is a right S module by MW 13 0 f HomRA7 B HHOIIlRA7 3 f A f O l is S lln ear 112 CHAPTER 4 MODULES c HomRB7 A is a left S module with action ofS given by 8fb fa d V j HomRBA H HOIIIRB7 147 f H l f is S linear Proof Straightforward B Let R be a ring and M a R module The dual of M is the module M HomRM7 B As R is an R7 R loimodule7 M is a right R module The elements of M are called linear functionals on M From 446 we have EB MN H M i39eI i39eI By Pi R7 but the reader should be aware that here R is a right R module that is the action is given by right multiplication We conclude FI g R1 and so if I is nite then FI is isomorphism to the free right module on I An R module M is called cyclic of M Rm for some m E M Lemma 448 MCy LetR be a ring and M Rm a cyclic R modules LetI AnnRm andJr RlIr0 a I is an Tight ideal in R b 739MgtI7 is an isomomhism of right R modules Proof a Let j 6 I7 r E R and i E I Then ijr ijr 0r 0 and so jr 6 I Thus a holds 10 Let a E AnnRm Then afm fam f0 0 and so fm E I So 739 is well de ned It is clearly Z linear and WNW Hm So 7fr TfT and 739 is right R linear 45 TENSOR PRODUCTS 113 Let j E J Then j 0 and so the map MaR m Hrj is well de ned and R linear TEj WW j1m1jj and Tf m WU WW rm and so 5Tf f and 739 is a bijection D If R is commutative7 left and right modules are the same So we might have that M E M as R modules In this case M is called self dual For example free modules of nite rang over a commutative ring are self dual Let R be a ring7 the double dual of a module M is M M De ne 19 M a M19mf fm It is readily veri ed that 19 is R linear If M FRU is free of nite rang we see that 19 is an isomorphism If M FRU is free of in nite rang7 then 19 is a monomorphism but usually not an isomorphism In general 19 does not need to be one to one For example if R Z n E Z and M XnZ7 then it is easy to see that M 0 Indeed let 1 E M and m E M Then nm 0 and so n gtm gtnm 0 Thus gtm 0 Since M 07 also M 0 Let us investigate ker19 in general Let m E M then 19m 0 if and only if gtm 0 for all j E M 45 Tensor products Let R be a commutative ring and A7 B7 C R modules A function f A x B a C is called R bz39lz39near if for each a in A and b in B the maps fa Ba 07y a fay and fb AHC7 a fzb are R linear For example the ring multiplication is R linear Also if M is any R module Then M X M gt Rfm gt Let R be any ring7 A a right and B a left R module Let C be any abelian group A map f A x B a C is called R loalanced7 if is Z bilinear and an 5 1775 for all a 6 A71 6 B7 6 R M x M a B7 m a fm is an example of a R balanced map 114 CHAPTER 4 MODULES De nition 451 dtensor Let A be a right and B a left module for the ring R A tensor product for A7 B is an R balanced map AXBgtA RBabgta b such that for all R balanced maps f A X B gt C there exists a unique Z linear fA BHCwithfabfa b Theorem 452 extens LetR be a ring A be a right and B a left R module Then there exits a tensor product for A7 B Proof Let A R B the abelian group with generators za7 b l a 6 A7 b E B and relations Ma b Mal b xa a7b7 a7a E Ab 6 B7 Ma b zab Ma b b 7 a 6 A7 b7b E B and marb zarba E Ab E Br E R Write a X b for Ma b and de ne AXBgtA Babgta b We leave it as any easy exercise to verify that this is indeed an tensor product B Let R be a ring Then X R XR R a R7 r7 5 a rs is a tensor product Indeed given any R balanced map7 f R x R a C De ne fRRHCerr1 As f is Z loilinear7 f is Z linear Also fr 8 Hrs fr8 1 ns So indeed X is a tensor product With the same argument we have Lemma 453 AtenR Let R be a ring A a right and B a left R module Then A RRgA and R RBgB With a little bit more work we will prove Lemma 454 tensy Let R be a ring J a right ideal in R and I a left ideal in R Then RJXRRIHRJIrJsI 4 TH 1J is a tensor product for RJ7RI 45 TENSOR PRODUCTS 115 Proof Note here that I J is neither a left nor a right ideal in R It is just an additive subgroup7 RI J is an abelian group but in general not a ring First we need to check that X is well de ned rjsiIJrsjsrijiIJrsIJ Note here that as J is a right ideal jsji E J and as I is a left ideal M E I Clearly X is R balanced Suppose now that f RJ x RI a C is R balanced De ne fRIJHCTIJHfrJ1I Again we rst need to verify that this is well de ned frijJ1IfrJiJ1IfrI1If1Ji1I frI1If1Ji1IfrI1If1IORIfrJ1I So f is well de ned and clearly Z linear rJ 8IfrsIJfrsJ1IfrJs1IfrjsI and X is indeed a tensor product D If R is PlD we conclude RRm R RRnERgcdn7mR ln particular7 if n and m are relative prime RRm X RRn 0 Let M be a nite dimensional vector space over the division ring Let x 6 M7 1 6 M R EndDMI AnnRz and J AnnRy Then M E RI and M RJ Thus M RM E RI J We leave it as an exercise to verify that RI J E D We conclude that M X M gt Dfm gt is a tensor product of MM Lemma 455 tenbi Let RST be rings 04 A a A RS linear and B B a B S T l239near a A 85 B is an FLT bimodule in such a way that Ta b Ta X bt forallrER7a A7b Bs S 116 CHAPTER 4 MODULES b There exists a unique Z ltnear map a A sHBgtA B witha baaa b for all a 6 A71 6 B Moreover 04 X B is RT lmear Proof a Let r E R7 and t E Y We claim that gtrt AX B HA 5 B7 11 ara bt is S balanced Indeed its clearly Zrbilinear and ras X bt ras X bt ra X 8bt ra X sbt So its S balanced Hence we obtain a map a Z4inear ltIgtr7t A 5B HA 5 B7 With a bgtra bs Let r r E R and t E T It is easy to verify that ltIgtr r ta o b r t r ta o b and ltIgtrr 1a X b ltIgtr1o gtr 1a X 1 Thus by the uniqueness assertion in the de nition of the tensor product7 ltIgtr r t ltIgtr t T t and ltIgtrr 1 T 1 o r 1 Thus A X B is a left R module by re ltIgtr7 1v SimilarlyA X B is a right T module by vt ltIgt1tv Also ra bt ra bs ra bt So rvt rvt for all 6 Rt 6 T712 6 A R B Thus a holds b The map A X B a 14 85 BC 11 gt aa X is S balanced So 04 X 3 exists That its R7 T linear is easily veri ed using arguments as in a Proposition 456 seten Let D be a right R module and A L B A C a 0 an exact sequence of R Then D RAinD RBi gD CHO is exact sequence of Z modules 454 TENSOR PRODUCTS 117 Proof As D x C is generated by the d X c and g is onto7 idD 89 is onto Also 0 9 O id3 fd a d 9fa 0 So lmidD X f kergidD X 9 Let E lmf and HlmidD idEnB ltd ed D75 6E D RB Note that H lmidD X 1 We will show that H F keridD X 9 Without loss C BH and g is the canonical epimorphism We claim that the map D x BEH D BHdbE Hd bH is well de ned and R balanced Indeed at X b e H d X b d e H d 81 H So it well de ned lts clearly R balanced Hence we obtain an onto ZAinear map D RBEH D BH7 with d bE Hd bH idD X g induces an isomorphism D BFgtD RBE7 with d bFHd bE The composition of these two maps give on onto map 7 D BFH D BHwith d bFH d bE As D B is generated by the d b we get T UF vE for all u E D 813 Since 70 0 we conclude that f E E for all f E F Thus F g E and E F D Lemma 457 tendir a Let Al7i E I be a family of right R modules and Bi73 E J a family of left R modules Then ieltxm jngZZmwj ieI jeJ b Let R be a ring and I J sets Then FI FJ gFU X J as a Z modules 118 CHAPTER 4 MODULES b Let R and S be rings with R g S Let I be a set and view S as an SR b239module Then S R FRI g F5ltIgt as S module Proof a is readily veri ed using the universal properties of direct sums and tensor products 10 Since R R R E R7 10 follows from a c As S R R E S7 c follows from a D Lemma 458 Let A be a right R module B a RS btmodule and C a left S module Then there exits Z lmear isomomhtsm A RB SC with a b cgta b c for alla E Ab 6 B76 6 C Proof Straightforward form the universal properties of tensor products D In future we will just write A R B 25 C for any of the two isomorphic tensor products in the previous lemma A similar lemma holds for more than three factors A R B 85 C can also be characterized through R7 S balanced maps from A x B x C a T7 where T is an abelian group We leave the details to the interested reader Proposition 459 tehoas Let A be a right R module B a B7 S b239module and C a right S module Then the map E C Hom5A 8 B7 C AH HOII1RA7HOIIISB7 CEfab fa b is a Z tsomomhtsm Proof Note that Efa B a Cb a at7 b is indeed S linear Also EU A a Hom5B7 C is R linear and E is Zrlinear It remains to show that E is a bijection We do this by de ning an inverse Let Oz A a Hom5B7 C be R linear We claim that the map A X B gt C a7b gt aab is R balanced Indeed it is Zrbilinear and altargtltbgt ltaltagtrgtltbgt altagtltrbgt So there exist 9a A X B gt C with aa X b aab 45 TENSOR PRODUCTS 119 for all a E A b E B It is readily veri ed that 9a is S linear So 9 HomRAHomSB C a Hom5A R B C We claim that 9 and E are inverses Eaab 9aa 5 aab 9Efa 5 Efab fa 5 and so Ef f D Here is a special case of the previous proposition Suppose R is a commutative ring and A and B R modules Applying 459 with C R We get that A X B HomRA B Suppose that R is commutative and A B are R modules we obtain a Z4inear map a A X B gt A X B with 39yoz X 30 X b aa b Indeed this follows from 04 X B A X B a R X R R If A and B are free of nite rang it is easy to see that this is an isomorphism If A and B are free a is still one to one but not necessarily onto Suppose that Ak RIk k E 1 2 is a cyclic R module Put Jk AnnRUk Then by A E Jk Also A1 X A2 E RIl 2 Now AnnRIl 2 AnnRUl AnnRIZ J1 J2 Thus A1 R A2 J1 J2 a from above with A A1 B A2 now reads 0 3 J1 R J2 H J1 J2 j17j2 j1j2 We will know give an example where a 0 but J1 X J2 7 0 Let S be a ring and M an S S bimodule De ne M gt0 S to be the ring with additive group MeBS and multiplication 77117 81 39 77127 82 771182 8177127 8182 It is easy to verify that M x S is a ring As an example we verify that the multiplication is associative 77117 8139m278239m37 83 W18281m27818239m3783 W1828381m2838282m37818283 A similar calculation shows that the right side is also equal to 771131 mg 32 mg 33 ldentify m 0 with m and 0 s with s Then Mgt0SMS31823132smsmmsmsandm1m20 for all 55132 E S and mm1m2 E m Also M is an ideal in M gt0 S and M gt0 SM E S Indeed the map M gt0 S a S m s a s is an onto ring homomorphism with kernel M 120 CHAPTER 4 MODULES Suppose now that S is commutative and M a faithful S module Put R M gt0 S Then R is commutative As M2 0 and Ann5M O AnnRM M Also M O M M M M We conclude RM M RM RM RM RM RM M and UM RMgtMm1m2gtm1m20 Suppose that M F5I is a free S module Then as an R module M g EBBM i39eI Thus M R M EBEBRM ieI jeI and soM RM7 O unlessly fl 46 Free modules and torsion modules In this section B is a ring with identity and all R modules are assumed to be unitary Let M be a R module and mli E I a family of elements in M By 432 there exists a R linear map a RHMwithiHmi ieI If 04 is an isomorphism m l i E I is called a basis for M It is more or less obvious that a subset B of M is a basis if and only if every element m in M can be uniquely written as m 2163 ml with Tb E R for all I E B As usually we assume here that almost all Tb are zero If M has a basis B then M is isomorphic to the free module on B For this reason we call M itself a free module We say that a subset L of M is linear independent if L is a basis for RL note that this is the case if and only if an0gtn0foralll L lEL ls a submodule of a free module free The answer is almost always no For example if R ZnZ with n E Z n not a prime and m is a proper divisor of m then mZnZ is a submodule of R which is not free A obvious necessary condition for all submodules of all free modules for a ring R to be free is that all submodules of R itself are free The next theorem shows that this condition is also sufficient Theorem 461 subfree Suppose that all left ideals in R are free as R modules Then every submodule of a free module is free 46 FREE MODULES AND TORSION MODULES 121 Proof Let M be a free module with basis B and A a R submodule in M According to the well ordering principal A6 we can choose a well ordering on B For b E B de ne Mg 2 6R5 563563 and Mb Mg Let Ab Mb O A and A M O A Then AbAZ AbAb M EAbMljM MbM gRbgR By assumption every submodule of R is free and so AbAZ is free Let Eb C Ab so that e Aie E E5 is a basis for AbAZ Let E UbeB Eb We claim that E is a basis for A Let m E M Then m 2176b rbb with Tb 6 R and almost all mb 0 So we can choose bm E B maximal with respect rbm 7 0 Clearly for e E Eb7 be b In general7 bm is minimal in B with m E Mbm Now suppose that ZBEE Tee 0 so that almost all7 but not all re 0 Let b be maximal with b be and re 7 0 for some e E E The e E E5 for all e with be e and e E A for all e with re 7 0 and be 7 b Thus OZreeA Z reeAZ 56E 86E But this contradicts the linear independents of the e AZ e E E5 Hence E is linear independent Suppose that A f RE and pick a E A RE with b be miniaml Then a 6 Ab Hence aAZ ZreeA 86E Put 1 ZBEEb Tee Then at 6 RE A7 a 7 d 6 AZ By minimality of b7 a 7 d 6 RE and soa dRERE S0 A RE7 A RE and E is a basis for A D If R is commutative with identity only PID have the property that every left ideal in R is free Indeed if at7 b E Ff then ba 7 ab 0 and so a and b are linear dependent Hence every ideal is cyclic as a module and so a principal ideal Suppose that ab 0 for some non zero at7 b E R Then bRa 0 and so Ra is not free7 a contradiction Thus R is also an integral domain and so a PlD De nition 462 dtorsion Let M be a unitary R module and m E M a m is called a torsion element if AnnRm 7 0 b M is called a torsion module if all elements are torsion elements c M is called torsion free ifO is the only torsion element 122 CHAPTER 4 MODULES d M has nite exponent of AnnRM 7 0 Note that m is not a torsion element if and only if m is linear independent Lemma 463 torsion Let M be a module for the integral domain R a The torsion elements form a submodule b If M is generated by nitely many torsion elements then M has nite exponent c is torsion free Proof a Let ab be torsion elements and pick r s E Eff with ra sb 0 As R is an integral domain rs 7 0 Also rsRa Rb Rsra Rrsb 0 Hence all elements in RaRb are torsion So the torsion elements indeed form a submodule b Suppose that M RA where A is a nite set of torsion elements For a E A pick ta 6 R with taa 0 Put t HaeAta Then t 7g 0 and tM 0 c Let x TM be a torsion element Pick 0 7 r E R with rm 6 Then rm is torsion and so srz 0 for some 0 7 s E R Hence srm 0 and as R is an integral domain sr 7 0 So z E TM and is torsion free Theorem 464 freetorsion Let M be an R module a Any linear independent subset of M lies in a maximal linear independent subset b Let L be a maximal linear independent subset ofM Then MRL is a torsion module Proof a Let E be a linear independent subsets of M Let E be the set of linear independent subsets of M containing E Order E by inclusion The union of a chain in E an upper bound for the chain So Zorn s Lemma A1 implies that M has a maximal linear independent subset L b Let W RL Suppose that MW is not torsion and pick in E M so that m W is not torsion Then in is not torsion and Rm O W 0 Hence L U m is linear independent a contradiction to the maximal choice of L We remark that if L is a maximal linear independent subset of M then RL does not have to be a maximal free submodule Indeed the following example shows that M does not have even to have to have a maximal free submodule Zorn s lemma does not apply has the union of a chain of free submodules might not be free Let R Z and M Q with Z acting by right multiplication As Q has no zero divisors M is torsion free In particular every non zero element a is linear independent We claim 46 FREE MODULES AND TORSION MODULES 123 a is a maximal linear independent subset Indeed ab E Q Then a and b g with n7 m7 qp E Z Then mpa 7 nqb nm 7 nm 0 and a and b are linear dependent We conclude that every free submodule of M is of the form Za7 a E Q Let t E Z with t gt 1 then Za g me and so M has no maximal free submodules Q as a Z module has another interesting property every nitely generated submodules is cyclic lndeed7 if A is generated by 1 g i g k put in lcmlgigkmi and Then mA E A and mA Z So mA and A are cyclic Since diVision rings have no non zero unitary torsion modules 464 has the following corollary Corollary 465 vectorspace Let V be a vector space over the division ring D Then V has a basis and every linear independent subset of V can be extended to a basis of V D Lemma 466 idsubfree Let M be a torsion free R module for the integral domain I Suppose that one of the following holds 1 M is nitely generated 2 If N is a submodule ofM so that MN is a torsion module then MN has nite exponent Then M is isomorphic to a submodule of a free module W Moreover W can be chosen as a submodule of M Proof Note that by 463 condition 1 implies condition 2 So we may assume that 2 holds By 464 there exists a free submodule W of V so that MW is torsion By there exists 0 344 r E R with rMW 0 Hence rM W Consider the map 04MgtlVmgtrm As R is commutative7 04 is a R linear As M is 04 is one to one Thus in E aM rM W D Theorem 467 torfree LetM be a nitely generated module for the PID R Then there exists a free submodule F g M with M F EBTM Proof By 4637 is torsion free7 so by 466 is isomorphic to a sub module of a free module Hence by 461 is free Finally by splits over D 124 CHAPTER 4 MODULES 47 Modules over PID S A non zero R module M is called simple if 0 and M are the only R submodules of M For example if I is a left ideal in R7 then RI is simple if and only if I is a maximal left ideal Theorem 471 primary Let R be a PID and p E R a prime Suppose that M is an R module with pkM 0 for some k E Z3 Then a If M Rm is cyclic M Rpl for some 0 g l g k Moreover every R submodule of Rm is of the form Rptm for some 0 g t g l b Let W be a R submodule in M Then W is a direct summand of M if and only if ptM W gptW for alltE N c M is a direct sum of cyclic submodules Proof Put P Recall that P0 is de ned as R Without loss k is minimal with respect to pkM 0 a Let J AnnRm Then M E RJ Let W be a an R submodule of M Then W Im for some ideal I of R with J g I Since R is a PlD7 I Ri for some i E I Since pkM 07 pk E J g I and soi lpk Thus I pt and J pl for some 3 g lk Since W Im Rptm all parts of a are proved b 7 gt 7 Suppose that M W 63 K for some R submodule K of M Then ptM ptW ptK and so ptM O W ptW 7 lt 7 Suppose that PtW O W PtW for all t E N We proceed by induction on k If k 0 we get M 0 and there is nothing to prove Note that Pk 1PM 0 and P PM m PW Pith n W m PW P th m PW P PW Thus by induction PW is a direct summand of PM Let PM D 63 PW for some submodule D ofPM Note that DOW D PM W D PWO Let M be the set of all submodules E in M with D g E and E O W 0 The D E M Order M by inclusion The union of a chain in M is in M So by Zorn s lemma Al M has a maximal member E Suppose M 7 E W and let m E M E Then pm 6 pM D pW So there exists w E W with pm pw 6 D g E So replacing mbymiwweassumethatpm 6E LettERwithtm EW lft P7thenas PM E W and P is a maximal7 Rm E W7 a contradiction So t rp for some rERandtmrpm E ThusRm EW Eand RmE W RmE EW WRm EWE W E WO a contradiction the maximal choice of E So MEW EQBW and W is adirect summand ofM c Let D be the set of sets of cyclic submodules of M We say that D E D is linear independent if 2 D 69D and we say that D is a direct summand if 2D is a direct 47 MODULES OVER PID S 125 summand of M Let M be the set of all linear independent direct summands in 7 Order M by inclusion Let C be a chain in M and D UC Clearly D E 7 Also D is linear independent We claim that ZDCLEJCZC Clearly the right hand side is contained in the left Let x E 2 D Then z 6 U1 U2 Un for some U E D Since D UC there exits C E D with U 6 0 Since 7 is a chain we may assume 01 Q 02 Q Q On hence Ui E On for all 1g l g n and so x U1U2mUn ZCn U20 06C This proves the claim From the claim ZDl PtM UZCW M U Zpt0ptZD 06C 06C Thus by b 2 D is a direct summand of M So D E D and D is an upper bound for C So by Zorn s lemma 7 has a maximal member D By de nition of D M 2 D 63 E for some R submodule E of D Suppose that E 7 0 Pick l minimal with plE 0 Then pl lE 7 0 and we can choose e E E with pl le 7 0 We claim that Re is a direct summand of E For this we want to apply b to Re and B Let 0 g t lt l By a every submodule of Re is of the form psRe In particular Re ptE psRe for some 3 But pl tRe ptE 0 so pl tpse 0 and s 2 t Thus Re ptE psRe ptRe So by b Re is a direct summand of E But then 2 D Re is a direct summand of M Also Re 2 D g E 2 D 0 and D U Re is linear independent But thus contradicts the maximal choice of D Hence E 0 and M 2 D D and c holds Theorem 472 decprim Let R be a PID and M a torsion module forR Let be the set of non zero prime ideals in M For P E 73 let Mp Uke AnnMPk Then MEBMP P673 Proof Let m E M and pick r E R with rm 0 Then there exists pairwise non associate primes p E R and postive integer ki 1 g k g n with k k rp11pwf 126 CHAPTER 4 MODULES Put a 117 p Then r pfiai Also gchl a 1 and so 1 Z sia for some 3 E R Put m siaim Then in ELI m and pfimi pfisiaim sip iaim sirm Thus mi 6 AnnM MW and so M 2 MP P673 Let P E P Put Mp Z MQ P7 Qe73 ln remains to show that My Mp O For this let k E Z and 0 7 m E Mpl Then am 0 for some a E R with a P Let P Then 1 ra spk for some rs E R Thus in ram 1 spkm spkm Hence pkm 7 0 and m Mp D Theorem 473 pidmod Let M be a nitely generated module for the PID R Then M is direct sum of nitely many cyclic modules Moreover each of the summand can be chosen be isomorphic to RPk for some prime ideal P and some k E Z3 Proof By 467 M F 63 TM where F is free So F is a direct sum of copies of R Note that R E RO and 0 is a prime ideal Also by 472 TM FEP Mp As M is nitely generated and M17 is a homomorphic image of M M17 is nitely generated So see 463 pkMp 0 for some k E N Thus by 471 M17 is the direct sum of cyclic modules and each of the cyclic modules is isomorphic to Rpl We leave it as an exercise to verify that all the direct sums involved are actually nite sums D Corollary 474 a Let A be a nitely generated abelian group Then A is the direct sum of cyclic groups b Let A be an elementary abelian p group for some prime p That is A is abelian and pA 0 Then A is the direct sum of copies of ZpZ Proof Note that an abelian group is nothing else as a module over Z So a follows from 473 and b from 471 b can also by proved by observing that A is also a module over the eld ZpZ and so has a basis D 146 CHAPTER 4 MODULES Appendix A Zorn s Lemma This chapter is devoted to prove Zorn s lemma To be able to do this we assume throughout this lecture notes that the axiom of choice holds The axiom of choice states that if Ahi E I is a nonempty family of nonempty sets then also Hid A is not empty That is there exists a function f I a Uid Ai with 6 Ai Naively this just means that we can pick an element from each of the sets A A partial ordered set is a set M together with a re exive anti symmetric and transitive relation 7 g 7 That is for all a b c E M a a g a reflexive b a g b and b g a gt a a anti symmetric c a g b and b g c gt a g c transitiue We say that a and b are comparable if a g b or b g a M g is called linear ordered if any two elements are comparable Let C be a subset of M then C is also a partially ordered set with respect to 7 g 7 C is called a chain if any two elements in C are comparable that is if C is linear ordered An upper bound in for C is an element M in M so that c g m for all c E C An least upper bound for C is an upper bound in so that m g d for all upper bounds d ofC A function f M H M is called increasing if a S fa for all a E M An element in E M is called a maximal element if a g m for all a E M comparable to M ie if a m for all a E M with M g a We are now able to state Zorn s Lemma Theorem A1 Zorn Zorn Let M be a nonempty partially ordered set in which euery chain has an upper bound Then M has a maximal element As the main steps toward a proof of Zorn s lemma we show Lemma A2 xedpoint Let M be a non empty partially ordered set in which euery non empty chain has a least upper bound Let f M gt M be an increasing function Then fm0 mg for some m0 6 M 147 148 APPENDIX A ZORN S LEMMA Proof To use that M is not empty pick a E M Replacing M by m E M l a g m we may assume that a g m for all m E M We aim is to nd a subset of M which is a chain and whose upper bound necessarily a xed point for 1 We will not be able to reach both these properties in one shot and we rst focus on the second part For this we de ne a subset A of M to be closed if L a a E A L b fb E A for all I E A L c If C is a non empty chain in A then its least upper bound is in A Since M is closed there do exists closed subsets Suppose that there exists an closed subset which is a chain By L a A is not empty Let me be its least upper bound By L c m0 6 A and by L b fm0 E A Thus fm0 m0 and as g is anti symmetric fm0 me So all what remains to do is to nd a closed chain There is an obVious candidate It is immediate from the three conditions of closed that the intersection closed sets is still closed So let A be the intersection of all the closed sets De ne e E A to be extreme if fb eforallb Awithblte Note that a extreme so the set E of extreme elements in A is not empty Here comes the main point of the proof Claim 1 Let 5 be extreme and b E A Then I g e or fe b In particular 5 and b are comparable To prove Claim 1 put A5b Alb eorfe b We need to show that As A As A is the unique minimal closed set we just need to show that As is closed Clearly a 6 A5 Let b 6 A5 If I lt 5 then as e is extreme fb e and so fb 6 A5 If e b then fe fb and again fb 6 A5 lf fe I Then fe b S b and fe fb by transitiVity So again fb 6 A5 Finally Let D be a non empty chain in A5 and m its least upper bound lf 1 g e for all 1 in D then 5 is an upper bound for D and so m g e and m 6 A5 So suppose that d f e for some 1 E D As 1 6 A5 fe d g m and again m 6 A5 We proved that As is closed Thus A5 A and the claim holds Claim 2 E is closed Indeed a E E Let 5 E E and b E A with b lt fe We need to show that fb fe By Claim 11 g e or fe bThe latter case is impossible by anti symmetry If I lt 5 then 149 since e is extreme fb 6 S He If e bthen fb fe fe So fe is extreme Finally let D be a non empty chain in E and in its least upper bound We need to show that m is extreme Let b E A with b lt m As in is a least upper bound of D b is not an upper bound and there exits e E D with e f b By claim 1 e and b are comparable and so b lt e As e is extreme fb e g m and so in is extreme So E is closed D As E is closed and E Q A A E Hence by Claim 2 any two elements in A are comparable S0 A is a chain As remarked above the least upper bound for A is a xed point of f D As an immediate consequence we get Corollary A3 weakzorn Let M be a non empty partially ordered set in which every non empty chain has a least upper bound Then M has a maximal element Proof Suppose not Then for each m E M there exists fm with m lt fm The axiom of choice is used here But then f is a strictly increasing function a contradiction to A2 D Lemma A4 maXChain Let M be any partial ordered set Order the set of chains in M by inclusion Then M has a maximal chain Proof Let calM be the set of chains in M The union of a chain in M is clearly a chain in M and is an least upper bound for the chain Thus A3 applied to M yields a maximal member of M That is a maximal chain in M D Proof of Zornls Lemma By A4 there exists a maximal chain 0 in M By assump tion C has an upper bound m Suppose that m g a for some a E M Then C U m l is a chain in M and the maximality of C implies l E C Thus l g m and m l Thus in is maximal element D As an application of Zorn s lemma we prove the well ordering principal A partially ordered set M is called well ordered if it is a chain and if every non empty subset I of M has a least element that is there exists an lower bound i of I with i E I We say that a set can be well ordered if the exists a relation 7 g 7 on M so that M 7 g 77 is well ordered Theorem A5 Wellordering principal wellorder Every setM can be well ordered Proof Let M be the set of well ordered sets 04 Ma 04 with Ma M As the empty set can be well ordered M is not empty For a B E M de ne 04 g B if lt1 Magi 150 APPENDIX A ZORN S LEMMA lt 2 g lMa xMa a lt3 ag bforalla eMabeM Ma It is easy to see that g is a partial ordering on M We would like to apply Zorn s lemma to obtain a member in M For this let A be a chain in M Put MT UaeA Ma and for ab 6 Mi de ne a g b if there exits oz 6 A with a7 1 6 Ma and a 04 1 Again it is readily veri ed that g is a well de ne partial ordering on Mar ls it well ordered Let I be any non empty subset of M and pick 04 E A so that I O Ma 7 0 Let m be the least element of I O Ma with respect to 04 We claim that m is also the least element of I with respect to 3 Indeed let i E I lfi E Mm then m u i by choice of m So also m g i lfi Mm pick 3 E A with i 6 Mg As A is a chain7 oz and B are comparable As i 6 Mg Ma we get 04 lt B and lt 3 implies m g i Again m g i and we conclude that Mr7 g is well ordered Clearly it is also an upper bound for A So by Zorn s lemma there exists a maximal element 04 E M Suppose that Ma 7 M and pick m E M Ma De ne the partially ordered set Mr7 g by MT Ma U m7 gilMaxMa a and i lt m for all i 6 Ma Then clearly Mr7 g is a well ordered set and 04 lt Mr7 g a contradiction to the maximality of 04 Thus Ma M and u is a well ordering on M D The well ordering principal allows to prove statement about the elements in an arbitrary set by induction This works as follows Suppose we like to show that a statement Pm is true for all elements m in a set M Endow M with a well ordering a suppose that we can show Pa is true for all a lt m gt Pm then the statement is to true for all m E M lndeed suppose not and put I E M l is false Then I has a least element m Put then Pa is true for all a lt i and so is true by the induction conclusion A well ordering can also be used to de ne objects by induction Lemma A6 de nd LetI a well ordered set andS a set Fora E I let I E I l i g a and Ia E I l i lt a Suppose that for each a E I fa is a set offunctions from I gt S Also suppose that if f Ia gt S is a function with 1 me 717 for all I E Ia then there exists f6 fa with f lIa Then there exists 1 I gt S with 1 me fa for all a E A Proof Let I be the set of all subsets J of I so a g b E J implies a E J Note that either J R or J Ia where a is the least element of R I Put MfiJfHSlJf617fl1a fa7Va J OrderM by f g 9 if Jf C Jg andg Jf Let C be a chain in M Put J Ufa Jf Clearly J 61 De ne f J a S by fj gj where g E C withj 6 J9 Then also 1 lg 9 W 151 and so 1 E M Thus 1 is an upper bound for M By Zorn s lemma7 M has a maximal member f If Jf R we are done So suppose Jf 7 R Then Jf a for some a E I Buy assumptions there exists f 6 1 with f ha 1 But then 39 E M and f lt f a contradiction to the maximal choice of f D 152 APPENDIX A ZORN S LEMMA Appendix B Categories In this chapter we give a brief introduction to categories De nition B1 A category is a class of objects C together with i for each pair A and B of objects a set HomA7 B7 an element f of HomAB is called a morphism from A to B and denoted by f A gt B ii for each triple A7 B7 C of objects a function HomBC x HomA7 B a HomAC397 for f A gt B and g B gt C we denote the image of 97f under this function by g o f g o f A a C is called the composite off and 9 so that the following rules hold I Associative If f A a B 9 B a C and h C a D are morphisms then h090fh090f H Identity For each object A there exists a morphism idA A gt A such that for all fAHBandgBHA foidAf andidAogg A morphism f A a B in the categoryC is called an equivalence ifthere exists 9 B a A with fogidB andgofidA 153 154 APPENDIX B CATEGORIES Two objects A and B are called equivalent if there exists an equivalence f A a B Note that associativity implies that the composite of two equivalences is again an equivalence Examples Let S be the class of all sets For A7 B 6 57 let HomA7 B be the set of all functions from A a B Also let the composites be de ned as usual Note that a morphism is an equivalence if an only if it is a bijection The class of all groups with morphisms the group homomorphisms forms category 9 Let C be a category with a single object A Let G HomA7 A The the composite G x G a G is a binary operation on G I and U now just mean that G is a monoid Conversely every monoid gives rise to a category with one object which we will denoted by CC An object in CC is equivalent to ea idA if and only if it has an has inverse Let G be a monoid For 071 6 G de ne Homab x l ma b If x a a b and y b a 0 Then yma yza yb c so yz a a 0 So composition can be de ned as multiplication The resulting category is denoted by CG The class of all partially ordered sets with morphisms the increasing functions is a category Let I be a partially ordered set Let 011 6 I If a g I de ne Homab Q If a g b then Homor7 b has a single element7 which we denote by 77a a 1 De ne the composite by bacoagtbagtc this is well de ned as partial orders are transitive Associativity is obvious and a a a is an identity for A We denote this category by C1 Let C be any category Let D be the class of all morphisms in C Given morphisms f A a B and g C a D in C de ne Homfg to be the sets of all pairs 043 with ozAHGand BHDsothatgooz of7 that isthediagram AALHB al l CgiaD commutes Let C be a category The opposite category C0p is de ned as follows The objects of C0p are the objects of C HomopA7 B HomB7 A for all objects A7 B 1 E HomquotpA7 B will be denoted by AgB m AeB 155 Op 1 O 9 9 O f The opposite category is often also called the dual or arrow reuersing category Note that two objects are equivalent in C if and only if they are equivalent in COP De nition B2 a An object I in a category is called universal or initial if for each object O ofC there exists a unique morphism I gt C b An object I in a category is called couniversal or terminal iffor each object O ofC there exists a unique morphism C gt I Note that I is initial in C if and only if its terminal in COP The initial and the terminal objects in the category of groups are the trivial groups Let I be a partially ordered set A object in CI is initial if an only if its a least element lts terminal if and only if its a greatest element Let G be a monoid and consider the category CG Since 9 e a g is the unique morphism form e to G7 e is a initial object e is a terminal object if and only if G is a group Theorem 13 uniuni Any two initial resp terminal objects in a category I are equiv alent Proof Let A and B be initial objects In particular7 there exists f A a B and g B gt A Then idA and g o f both are morphisms A a A So by the uniqueness claim in the de nition of an initial object7 idA g 0 f7 by symmetry idB f o 9 Let A and B be terminal objects Then A and B are initial objects in C0p and so equivalent in COP Hence also in C D De nition B4 Let C be a category and Al7i E I a family of objects in C A product for Ahi E I is an object P in C together with a family of morphisms m P gt Ai such that any object B and family of homomorphisms B gt Ahi E I there exists a unique morphism f B gt P so that m o f for alli E I That is the diagram commutes P B up Al commutes for alli E I Any two products of Chi E I are equivalent in C Indeed they are the terminal object in the following category 5 156 APPENDIX B CATEGORIES The objects in E are pairs B7 tinyi E I there B is an object and 1 B a Ahi E I is a family of morphism A morphism in E from B gti E I to D7 uh7i E I is a morphism f B a D with if iJiozf for all i EI A coproduct of a family of objects Chi E I in a category C is its product in COP So it is an initial object in the category 5 This spells out to De nition B5 Let C be a category and Al7i E I a family of objects in C A coproduct for Ahi E I is an object P in C together with a family of morphisms m Ai gt B such that for any object B and family of homomorphisms Ai gt Bi E I there exists a unique morphism f P gt B so that f o m for alli E I Bibliography Hun Thomas W Hungerford7 Algebra Graduate Text in Mathematics 737 Springer Verlag 1974 New York 157 Index FAI 32 A 61 N0H 21 RG 55 RH 81 RllGlL 83 R G 61 BOP 57 RP 79 S lR 75 ZG 44 g 75 lFR 76 SylpG 47 FixSG 44 p group 44 p subgroup 47 73G 16 arrow reversing 155 degree 141 abelian 10 action 41 algebraic 141 algebraic closure 144 algebraically closed 144 annihilator 95 anti homomorphism 8 anti homomorphism 57 anti symmetric 147 associate 66 Associative 153 associative 9 augmentation ideal 61 158 automorphism 7 basis 120 bilinear 113 bimodule 111 binary operation 7 category 153 center 44 chain 147 characteristic 58 characteristic polynomial 138 Chinese Remainder Theorem 65 common divisor 72 commutative 10 commuting homomorphism 29 comparable 147 complete ring of fraction 76 composite 153 composition series 127 coproduct 156 cosets 16 couniversal 155 cyclic 112 degree 81 derivative 89 direct product 28 direct sum 29 direct summand 101 divides 66 divisible 104 105 division ring 56 double dual 113 dual 112 155 INDEX elementary abelian 126 equivalence 153 equivalent 154 equivariant 42 Euclidean domain 71 Euclidean ring 71 exact 97 exponent 58 exponential notation 80 factor 127 eld 56 eld extension 141 eld of fraction 76 nite exponent 122 nite extension 141 xed point 44 formal power series 83 free abelian group 32 free abelian monoid 33 free abelian semigroup 33 free group 39 free module 101 free monoid 41 free product 33 free semigroup 41 Gau ian integers 73 generated 19 greatest common divisor 72 group 10 homogeneous 85 homomorphism 7 97 ideal 60 Identity 153 identity element 7 initial 155 injective 102 integral domain 56 invariant 42 inverse 10 invertible 10 159 irreducible 67 isomorphism 7 42 Jordan canonical form 138 jump 127 Lagrange 16 Latin square 7 leading coe cient 81 least upper bound 147 linear 94 linear functionals 112 linear independent 120 linear ordered 147 local ring 80 localization 79 maximal element 147 maximal ideal 63 minimal polynomial 138 141 module 93 monoid 9 monomials 81 morphism 153 multiple root 89 multiplication table 7 multiplicative subset 74 multiplicity 89 norm 73 normal 17 normalizer 21 normalizes 21 opposite 8 154 opposite ring 57 orbits 42 order 7 PlD 67 polynomial ring 81 polynomials 81 power semigroup ring 83 power set 16 43 160 INDEX pregroup 7 upper bound 147 prime 67 prime ideal 63 primitive 90 principal ideal 67 principal ideal domain 67 principal ideal ring 67 product 155 well ordered 149 word 33 zero divisor 56 re exive 147 relation 37 39 relatively prime 72 representatives 44 right evaluation 87 ring 53 ring homomorphism 57 root 88 self dual 113 semigroup 9 semigroup ring 55 sequence 97 series 127 similar 134 simple 124 simple ring 60 split 99 splits 142 submodule 94 subring 60 support 30 Sylow p subgroup 47 terminal 155 torsion element 121 torsion free 121 torsion module 121 transcendental 142 transitive 42 147 UFD 69 unique factorization domain 69 unitary 94 universal 155

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