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# Class Note for MATH 6302 at UH

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Modern Algebra Math 63026303 Klaus H Kaiser Chapter 1 Algebraic Systems 11 Operations Algebraic Systems An operation f on the set A is a map f z A 7gt A Where n is a natural number n 2 0 The number n is called the arity of nl sz7gtA andfisjustamaponA7 n 2 f z A2 7gt A and fis a binary operation on Al The case n 0 deserves special attention For any set S one has that AS ala S 7gt A In particular7 for n 0717 i i 7 n 7 l7 A alaz07l7ui7n717Aaa07iu7an1layeA and this is the set of all n 7 tuples a07i i i 7 an11 of elements in A An n7ary operation assigns to any n7tuple a07 i i i 7 an11 of elements in A an element fa07 i i i 7 an11 as operation value N0W7 A0 ala 0 7gt A A nullary operation assigns to the empty map an element a E A i A nullary operation is therefore called a constant For A 0 one has that A0 and A D for n gt 0 That is7 only operations of positive arity existi An algebraic system consists of 1 a set A7 2 a family of n 7 ary operations on Al We use the notation A A7 faker and A 710161quot is called the similarity type of Al Examples 1 Z Z7f17f27f3 Where f11Z2 7gt Z7 n7m gt gt nm7 f21Z 7gt Z7 1 gt gt 717 f3 7 Z7 0 gt gt 0 The binary operation fl is the addition7 the unary operation f2 is the additive inverse and f3 is the zero We use for binary operations most of the time the symmetric notation zfy instead of fz7 2 Let S be any set MapS is the algebra of all maps of S into itself with composition 0 as a binary operation and ids as a nullary operation Thus we get the algebraic structure Maids Me I 5 A S 07 ids 001571 WW 3 Similarly as before we de ne Bij590l9015 E s o ids The familiar algebras of integers real numbers and complex numbers Z Z70 71 R R70 71 C 07770 3971 are algebras of the same type A 2 l 0 2 0 They are similar Generalizations of the concept of an algebraic system are partial algebras The operations are not everywhere de nedl On the set of real numbers 1 R0 7gt R z gt gt 11 can be added as a partial unary operationl in nitary algebras These are algebras with operations of in nitary arityl Taking the limit of an in nite sequence may be considered as a partial in nitary operation the operation is de ned only for convergent sequences The projections pie A 7gt A aim61 gt gt aio are in nitary operations if I is in nite multivalued algebras The operation values are subsets of A ie f z A 7gt 73A fa1l l l an Q 73A One has that cardfa1 l l i an 1 iff f is an operation and cardfa1 l l l an S 1 iff f is a partial operationl relational algebras A relational system is a set A together with a family f teT of n 7 ary operations and a family R5565 of ms 7 ary relations R5 where one has that each R5 Q Ami An n 7 ary operation f corresponds uniquely to an n l 7 ary relation Rf Rf a1l l l anan1lan1 fa1 l l i an graphf An Example of a relational systems is Z ZHn oy 3971 S Any algebraic system may be considered as a relational system where S 0 Also nitary partial algebras are relational systemsl Note that the operations of the algebraic system as well as the relations of a relational system are nitary but that the number of operations of an algebra A is in general in nite Example Let V be a vector space over the eld of real numbers Then multiplication of a vector v by a real number a may be perceived as a unary operation and we have for each a E R the unary operation 1 gt gt alvl As an algebraic system such a vector space looks like V 7 70 10461 12 Homomorphisms of Algebras Subalgebras and Direct Prod ucts Let A and B be algebras of the same type A A A7 fzteT7 B B7 909 Where the arity of ft is equal to the arity of g t 6 T A map go A A B of the underlying set A of A to the underlying set E of B is called a homomorphism if for every t E T one has that fza077777an71 9z80a077 7 7780an71 This means for a binary operation eigi multiplication Which is denoted in both algebras by 77 z ao 39 a1 80a0 3980a1 For a unary operation eg a multiplicative inverse 1 this reads as ail a 1 For a nullary operation say at a and b b this is a b Let for a homomorphism go A A B Ev be the equivalence that is induced by go that is al N a m0dE4p and only al a An equivalence relation E on an algebra A is a congruence relation if a1 N ai7 a2 N a277777 am N ah yiEZdS ftlta177777am N ftltai77 7 7 7am Proposition 11 Let go A A B be a homomorphism of algebras Then the equivalence for go Ego a1a2lgoa1 a2 is a congruence relation on A PROOFi Assume al N a3 a N a 2i ant N agti This is al a liugoam agt Therefore 900301177 7 7 am 9490011 7 7 7 70077 97soa 177 7 7 790012 90030137 7 7 7 77 This is fta1uiam tha1Hiati D Proposition 12 Let E be a congruence relation on the algebra A Then there is exactly one algebraic structure on the set AE of equivalence classes for E such that the canonical projection qE A A AE becomes a homomorphism PROOFi We rst show uniquenessi lf qE a A a ME is homomorphic then M017 7 7 7 70m ltla1l77 7 7 7 lamb 130112011 7 774Eam 4Efza177777am7 Therefore if C1 a1i i Cm ant then necessarily filt0177 7 7 7 Cm lfzlta177 7 7 7an l Assume now for the congruence E on A that C1iH Cm a1iu amD ftgah i ant then because E is a congruence the choice of a1 6 C1 i i am 6 Cm does not matter ft is properly de ned by means of representatives The map qE A A AE is homomorphic 4Efza177777am fm4Ea1777774Eam is true by the de nition of D Proposition 13 Let go A A B be a homomorphism between algebras with associated congruence E Ev a N a a a Then there is a unique injective homomorphism AE A B such that o qE so A E AE L B A L B PREDOFl We only have to show that is homomorphicl But ltftlt01wvcm fz 1 m LIEAMLH A A A 7 800501 A want 94900107 A A A 790am 9z 0 4EgtWa17 A A A 7 Sb 0 4EWan 92 Cl A A CmDA D A bijective homomorphism is called an isomorphisml Let go A A B be an isomorphism between algebras Then soil B A A is homomorphic 90 1921717AAAybm fz 1b17AAA7 1bniff fzsf1bl7A790 1bn 92517AAAybniff 92soso 1b1AA79090 1bn 92517AAAJM Corollary 14 Homomorphism Theorem Let go A A B be a surjective homomorphism between algebras Then B is isomorphic to a factor algebra of A Note7 that the composition of homomorphisms is a homomorphism and that the identity on an algebra is a homomorphisml Thus7 the class of algebras of similarity type A is a category With the algebras as objects dots and the homomorphisms as morphisms arrows lt is very easy to see that for every algebra A7 the intersection of congruence relations is a congru ence Recall that an ordered set P7 S is a complete lattice if every subset S of P has a largest lower bound or a smallest upper bound Thus we have Proposition 15 For every algebra A the congruence relations on A form a complete lattice ConAl A aala E A is the smallest congruence on A and AA 2 Al A X A is the largest congruence on A and one has that AA X A 1A iel7 the oneelement algebra of type A Let A be an algebra and let C be a subset of Al C is said to be closed if a17 an E C implies that fza1 am 6 C ln particular7 a closed subset contains all constants Proposition 16 Let C be a closed subset of the algebra A Then there is exactly one algebraic structure g on C such that the inclusion icAZClt gtA is homomorphic PROOF We rst prove uniqueness igta1 am ftia1iam fta1 am gta1am That is 9 leCm On the other hand for any closed subset C of A we may de ne gta1 am fta1 am a1ant E C and i C lt gt A is homomorphic B Any injective homomorphism is called an embedding A closed subset becomes an algebra such that the inclusion is an embedding Any closed subset C with the operations ft restricted to C is a subalgebra It is easy to see that for any algebra A the intersection of closed subsets is closed Proposition 17 For every algebra A the subalgebras ofA form a complete lattice SubA A is the largest element of SubA and the smallest element of A is Z if there are no constants otherwise it is the subalgebra C0 that is generated by all constants C0 m C m C where M0 CLO ftolto E T and mo 0 CESubA 03MO CeSubA Let M0 be any subset of the algebra A Then de ne CAM0 0 CESubA CQMO Then CA M0 is the smallest subalgebra of A that contains M0 For any set M and ntary operation ft of A de ne f M ala E A a fta1 am a1 am 6 M M07 M1 U HMS UM07 M2 U ftltMiLtUM1 yields BET BET M0 M1 M2 Then let M U My We notice 0 M is closed Let a1 7am E M Then there is some k such that al am 6 Mk This implies that fta1 am 6 Mk1 E M o M contains M0 0 Let C be any subalgebra containing M0 Then C contains M We have C 2 M0 and assume that C 2 Mk Let a E M1644 Then a E Mk or one has that a fta1ant with a1 am 6 Mk In both cases we get that a E C Hence7 M Proposition 18 C CA de nes an algebraic closure for subsets M of the algebra A That is o C is extensive ie CM 2 M o C is monotone ie If M1 Q M2 then CM1g CMg o C is idempotent ie CCM Moreover a E CM if and only if there is a nite subset F of M such that a E CF CM U FgM CF 2 mm PROOF It is quite obvious that C satis es the properties of a closure operator We only have to show the algebraicity of C We need to show that U Cm is closed Let a17 am 6 U CF We then have that FgM F finite a1 6 CF17am E CFm implies that a17 am 6 CF1 U H U Fm CF where7 of course7 F is nite But then ftltahwamgtecltFgtg U cltFgt Now7 U CF is a closed subset that contains M This is clear because for any element a E M one has that a E Ca Q U CF and therefore M Q U CF and therefore CM Q U CF The converse inclusion is of course obvious D Proposition 19 Let Alig Ai7 fitteTl61 be a family of similar algebraic systems Let AHAZ a a1gt LJAZ7 ai 6 Ai i6 i6 be the cartesian product of the carrier sets Ai of the algebras Ai Then there is exactly one algebraic structure f teT on the set A such that all projections pi A A Ai a gt gt ozi7 are homomorphic PROOF We rst have to show uniqueness Let 11 7am E A and let a 1610117 701m Then one has that W Ma pifza17 e e 7 ffltpelta17 a pliant fflta1i7ami and that is I a aiiei fia1i7amii61 This yields7 I fz017am fla1i wamltigtgtgti61 On the other hand7 if we de ne ft on the cartesian product A by this last formula7 then Piltftlta1w 7am fla1i7 7am 139 fl0ial7piam shows that the projections pi are homomorphismsi D Note that we have in particular ct cal61 for nullary operations C For any algebra A7 and any set S A5 is called the direct power of Al The plane R2 is a typical example of the second power of the vector space R Proposition 110 Let goi B A Ai i E I be an initial family of homomorphisms Then there is exactly one homomorphism go B A A H Ai such that pi a go 301 i E It i6 PROOFi One de nes 0 11 I B A A7 805 90ibiEIA D i6 De nition 1 Gt Birkhoff A class 73 of algebras is called primitive if it is closed 0 under the formation of direct products 0 under taking homomorphic images 0 under taking subalgebrasi Groups and rings are examples of primitive classesi However7 elds are not closed under direct productsi De nition 2 At It Mal cev A class Q of algebras is quasiprimitive if it is closed under 0 isomorphic copies 0 direct productsi Cancellation semigroups are an example of a quasiprimitive class that is not primitivei Remarks on Ordered Sets and Lattices A relational system 07 S is called an ordered set if the relation S is refleacive7 ii transitive and if iii antisymmetry holdsi That is i zSzholds for alleOi ii lfzSyandySzthenzSzi iii lfzSyandySzthenzyi For I f y but I S y one writes I lt y Also7 I S y means the same as y 2 I An ordered set is totally ordered if one also has trichotomy iv Eitherxgyorygxorxyi Prominent examples for ordered sets are the natural numbers with divisibility N l and the power set 735 Q for the set 5 The set of real numbers with their ordinary ordering is the prototype of a totally ordered set Very often ordered sets are called partially ordered and totally ordered sets are called linearly ordered A relation that is re exive and transitive is called a quasiordering The integers with division are an example nlm and mln only yields n imi If S is a subset of the ordered set 0 then u is called an upper bound for S if u 2 s holds for all s 6 5 An upper bound that actually belongs to S is called the maximum of 5 A maximum if it exists is of course by antisymmetry unique If u is an upper bound for S and if v 2 u then v is an upper bound for 5 That is the upper bounds for a subset S for the ordered set 0 form an upper end of 0 Lower bounds and minima are de ned similarlyi By 77default every element of O is an upper as well a lower bound of the empty set If the set of upper bounds for S has a minimum then this minimum is called the supremum of S or the least upper bound of S supS minulu is an upper bound for 5 Similarly the of S or largest lower bound is the maximum of all lower bonds for 5 An ordered set is called a complete lattice if every subset S has a supremum as well an in mumi By the very de nition the in mum of the empty set must be if it exists the maximum of 0 Similarly the supremum of the empty set must be if it exists the minimum of 0 Only for the empty set the in mum may be greater than the supremumi An ordered set is bounded if it has a maximum as well a minimumi The following proposition is an easy but useful fact The proof is an easy exerciser Proposition 111 Assume that every subset S of the ordered set 0 has an Then every subset S of O has a supremum An ordered set 0 is called a complete lattice if every subset S has an in mum and then as well a supremumi The power set 735 of a set S is the prototype of a complete latticei Here the in mum of a subset iieia collection C of subsets of S is the intersection of C and the supremum is the union of C A collection C of subsets of S is called a closure system of S if the intersection of every subcollection S of C belongs to Cl As a corollary to the last proposition we state Proposition 112 Let T be a closure system of subsets of the set S Then T Q is a complete lattice An ordered set is a lattice if every nite nonempty subset has an in mum as well a supremumi The natural numbers with divisibility form an example of a complete lattice The in mum of a set of natural numbers is the greatest common divisor and the supremum is the lowest common multiplei The number 0 is the maximum of N l and l is the minimum of N An ordered set is called well ordered if every subset has a minimumi The natural numbers are well ordered by g It is an axiom of set theory that every set can be well ordered This statement is equivalent to the Axiom of Choice The axiom of choice is also equivalent to Zorn s Lemma Assume that every totally ordered subset C of the ordered set 0 S has an upper bound in 0 Then 0 has a maximal element An element m of O is maximal if s 2 m implies that s mi Finally we remark that if R is an ordering the dual relation aR b iff bRa is also an ordering Instead of saying that a certain map go A A B is an order reversing isomorphism from A to B we say that it is an order isomorphism from A to B3 Note that go is an order isomorphism if it is bijective and go as well as go 1 are order preserving Chapter 2 Basic Facts about groups rings and elds Modules and vector spaces 21 Groups A groupoid is an algebraic system With just one binary operation A A groupoid becomes a semigmup if the operation is associative ie one has for all z y 2 E A IyzIy2 If A is a groupoid then 6 E A is called a unit for 77 if one has for all z E A A groupoid can have at most one unit Assume that e and e are units We then have 6 e 6 because 6 is a leftunit and e e 6 because 6 is a ghtunit Hence 6 equot A semigroup With unit is called a monoid S e Where we consider the unit 6 as a nullary operation Let S e be a monoidi Then I E S is called invertible if there is an 1 E S such that An element 1 of a monoid can have at most one inversei Assume that z is a leftinverse and that z is a ghtinverse of 1 ie 1 z e and z z er But then 1zez11 zzz ez z The inverse of an element 1 in a monoid if it exists is denoted as I ll A group is a monoid Where every element has an inverse Therefore the algebraic system G G 1e is a group if 0 the binary operation is associative 1y2Iyz o The constant e is a unit for 1 o The unary operation associates for every I E G the inverse o A group is abeliah if the binary operation is commutative z y y z A commutative group is also called a module The binary operation then is denoted as and called addition The unit is called zero and 7 is the operation that takes the additive inverse Examples 1 Z Z 70 is a module 2 R R 0 1 1 is a group ie the multiplicative group of the reals 3 N N 0 and N N l are monoids Theorem 21 The class of all semigroups is a primitive class of algebras The same is true for the class of all mohoids the class of all groups and the class of abeliah groups PROOF Let C be a closed subset of the semigroup S Then C C is obviously a semigroup Let go C A A be a surjective homomorphism from a semigroup to a groupoid Let abc E A Then a z b y c z and a b C 901 909 02 901 y 902 901 y w 901 2 I 9092 90190y902 a b e Let Ai i61 be a system of groupoids All operations are denoted by Let 15 6 H 14139 Recall i6 that a I A U Ai ai 6 14139 We also Write a aii61 and consider the function a as an i6 I tupleWhere the ith coordinate is in Ai a aiieI amer az39 01139 pita The operation on H Ai was de ned by i6 MW 395 Pz39a 39Pi 7 236 1 39BW 04139 39 i7 i6 11 1395 Mi ame that is aihd 39 bii61 ai 39bii 61 10 Thus in case of associativity for all Ai az heI 39 bike 39 6mg ai 39bii61 CiieI az 39bi 39Ci39i39eI ai bi39cii61 0161 bi39cii61 aii61 biieI 6mg Therefore semigroups form a primitive class If Si ei are all monoids then e elm61 is a nullary operation of H S and ame 6mg M a Luke shows that e is a rightunit and a similar argument shows that e is a leftuniti If all Ci 1 ei are groups then aiieI 39 ailheI az a1iei eii61 6 shows that G H G is a group i6 We got that monoids and groups are closed under direct products Closure under subalgebras is also easy to see Notice that a closed subset C must contain the unit e and in a group the inverse of an element 6 6 C We are going to show that monoids are closed under homomorphic imagesi Let g0 S e A A e be a surjective homomorphism from the semigroup S onto the groupoid Al We already know that A is a semigroupi We need to show that e is the unit of Al But e e and a e z g0e z e z a shows that e is a leftunit for every a E A and similarly e is a rightunit for Al Thus the class of monoids is primitivei If g0 G 1 e A A f e is a surjective homomorphism from a group G onto a similar algebra A then we already know that A is a monoidi We wish to show that f is the inverse operation of Al But if a g then a fa 909 fsoy 909 909 1 9099 1 906 6 shows that A is a group Thus the class of all groups is primitivei All proofs were very similar and showed that equations are preserved under taking subalgebras homomorphic images and direct productsi Thus for example abelian groups are a primitive class D Proposition 22 Let S S e be a monoid Then the set of all z E S which have an inverse is a closed subset and therefore a group PROOFi Let InvS is invertible in S We then have 0 e6 InvS eee o lfze InvS andyE InvS thenzz z zeforauniquez ESandyy y ye for a unique y 6 5 But then omitting the operation symbol my y z zyy z zez 11 e Thus y z is a rightinverse of my and a similar argument shows that it is a leftinverse Hence I y E Inv o If I has inverse 1 then 1 has inverse 1 Thus InvS is a group D 11 Examples 1 Let S be any set Then MapS go S A S 0 ids is a monoid and InvMapS BijS is the group of invertible maps on Si 2 Let S 17 i i n Then BijS Sn is called the symmetric group of degree ui Let go G A G be a homomorphism between groups Then 8011 8012 i 8012718011 em 15 1271951 em 15 13111 E 807150 Nw Therefore7 11 N 12 m0dgo iff Iglzl E N4p N4p N is called the kernel of go and has the properties of a normal subgroup of G o N is a subgroup of Gr This is obvious because N is the counter image of the subgroup eg of Gquot o N is normal That is lszNandyEGthenyilzyEN This is also obviousi If I is mapped to e then y lzy is mapped to ei Let H be any subgroup of Gr Then 11 NL 12 iff Iglzl E H de nes an equivalence relation on the set Gr It is called the leftequivalence modulo Hi 0 We have that z N I because 1 11 e 6 Hi 1 olszythenyNz ThatzNymeansy IEHandyNzisz lyEHi Butz ly y lz 1 6 Hi 0 lfz N y and y N 2 the z N 2 We have y lz E H and 2 1y E H thus z lyy 1z 2 11 6 Hi Similarly7 11 NR 12 iff zlzgl E H de nes an equivalence on G it is called the rightequivalence modulo Hi Let T1 and T2 be subsets of the groupoid Al Then T1 T2 ala 1112 for some 11 6 T1 and I2 6 T2 is called the complex product of T1 and T2 We put IT IT my 6 T We then have for the leftequivalence modulo H I 1H zh for some h E H 12 We have zH Q z y zh yields h I ly Which is y NL z I QIH yNLziffzilyEHiffzilyhforsome hiffyzhforsomehiffy zH The equivalence classes With respect to the leftequivalence modulo H are the leftcosets Similarly In Hz are the rightcosets modulo Hi H is a normal subgroup iff HIQIH holds for every 1 We have Hz Q 1H for every 1 iff for every h E H and z E G one has some h E H such that hr zh iff I lhz h E H iff I le E H iff H is normal Similarly H is normal iff zH Q Hz holds for every 1 Thus H is normal i the leftequivalence modulo H is equal to the rightequivalence modulo H We call it the congruence modulo H Let N be a normal subgroup Then the equivalence modulo N is a congruence relationi We need to show i If 11 N 12 m0dN and y1 N yg m0dN then zlyl N Igyg m0dN ii If 11 N 12 m0dN then 1171 N 1271 m0dN That is i If Iglzl E N and yglyl E N then zgygflzlyl 6 Ni But I2y2 lz1y1 y 1z 111y1 ygly1yfl13111y1 E N ii We need to show that if Iglzl E N then Iglflzfl 6 Ni But Iglzl E N implies that 12z111z1 6 Ni Thus 111271 E N but then also zlzglfl 1211 E N Assume that the leftequivalence NL modulo H is a congruence Then H is normal Assume that NL is a congruence Then one has that 11 NL 12 iff zlzgl NL e iff zlzgl E H iff 11 NR 12 Hence NL NR i That is NL is a congruence The next proposition summarizes our observations on groups and their homomorphismsi 13 Proposition 23 Let H be a subgroup of the group G Then NL 11 NL 12 i czglzl E H and NR 11 NR 12 i czlzgl E H are equivalence relations on G The equivalence classes are the cosets modulo H 1 zH 1 Hz The subgroup H is called normal i NL NR or equivalently zH Hz holds for every I this is the same as I le E H where actually equality follows In case of a normal subgroup N the equivalence modulo N is a congruence and for the factor algebra which is denoted as GN is as a homomorphic image of G also a group GN E G 1e N where the operations are de ned by NI Ny Nzy Nz 1 N171 The map pN G A GN z gt gt N1 is homomorphic Corollary 24 Homomorphism Theorem for Groups Let go G A G be a homomorphism between groups Then kergo is the equivalence de ned by the normal subgroup NW go 1eG and WW 900 GNu Proposition 25 Correspondence Theorem for Subgroups Let go G A G be a surjective homomorphism between groups Then the lattice SubG of all subgroups of G is order isomorphic to the interval LP kergo G E SubG of all subgroups that contain kergo39 907115ub900o 14970 H 90710 has inverse go le PROOFl For any subgroup C of G one has that go 1C is a subgroup of G and go 1C 2 kergo Also for any subgroup B of G one has that goB is a subgroup of G and the maps go and go are both monotone We have 0 go o go 1C C because go is surjectivel Note Q holds in general 0 go 1 a go B B if B 2 kergo Note 2 holds in general Let I E go lgoBl Then goz gob for some b E B but this is the same as gozb 1 el Hence zb l E kergo Q B ilel zb l b Where b E B Thus I b b E B D Proposition 26 Correspondence Theorem Let go G A G be a surjective homomorphism between groups Then the lattice N0rmG of all normal subgroups of G is order isomorphic to the interval LP kergo G E N0rmG of all normal subgroups that contain kergo39 8071 I N0rm80G o 1w 70 H 90710 has inverse go le l4 PROOF Let N be a normal subgroup of 1 N 4 G Then go 1N N 4 G Let I E N Then one has that z 6 NC Let y E G But then yzy l ysorsoy 1 6 N Let N 4 G Then N 4 G 1 Let u z Where I E N If v E G then by surjectiVity of go one has that v y for some y E G But then vuv l ysorsoy 1 yzy l 6 N This proves the claim D For the lattice of subgroups of Z we have that 0 Every subgroup A is of the form hZ hnln E Z for a unique k 2 0 k minnln E A o lZ 2 kZ iff llk Hence Proposition 27 The lattice of subgroups on is order isomorphic to N l SubZ A N lhZ gt gt k Corollary 28 For any positive number Is one has that the lattice of subgroups of Z which contain hZ ie the interval hZZ in SubZ is order isomorphic to the lattice of divisors of h lkzyzl g 1776 l PROOF We have by Proposition 27 lkZZl g 1776 l but llvklv l g llvklv Ml H k1 D Corollary 29 For any h gt 0 one has that the lattice of subgroups of Zk ZkZ is order isomorphic to the lattice 1 k of divisors l of h PROOF SubZk E kZZ E 1 k D Corollary 210 Let hZ and lZ be subgroups of Z Then a hZ lZ hZ lZ lcmhlZ b kZVlZ kZlZ gcdklZ In particular gcdlh l nok m0l for suitable n0m0 E Z The map 1H A Hz zh gt gt hr is a bijection from the leftcoset that contains 1 to the rightcoset that contains 1 Hence all cosets modulo H have the same cardinal namely the cardinal of H The cardinal of the set of cosets is called the index of H in G 15 Theorem 211 Lagrange s Theorem Let G be a nite group and let H be a subgroup ofG Then cardG cardH indezH In particular cardHlcardG and indezHlcardGl Corollary 212 Let G be a nite group If cardG p where p is a prime then G is generated by a single element That is G is cyclic Corollary 213 Let G be a nite group where cardG n and let I be any element of G Then if the cyclic group that is generated by z in G has m elements then Theorem 214 Main Theorem for Cyclic Groups Let G be a cyclic group ie G lt z gt for some I E G Then either G is in nite and in this case one has that G Z or in case that G is nite one has that G E Zn where n cardG PROOFl Z A G n gt gt I is a surjective homomorphisml Therefore ZnZ G n l V 53 B One has in every group the left translations A75 G A G y gt gt my the right translations pg G A G y gt gt yr the inner automorphisms 05 G A G y gt gt zyzill All Ag p75 and 05 are bijective the 05 are automorphisms Theorem 215 Cayley s Theorem For any group G the map A Z G G71e A BijGo 1idG I gt gt A75 is an injective homomorphism ie G is isomorphic to its group of left translations 22 Rings Domains and Fields An algebraic structure A A 70 l is a ring if a The additive reduct A 7 0 is an abelian group b The multiplicative reduct is a monoidl z Addition and multiplication are related by distributivity Dl zyzzyzz D2 yzzyzzz The following very basic facts hold in every ring The proofs are simple and are based on distributivityl O a 0 0 a 0 0 7a b 7a b a7b 7a b 7a 7b a b o 71aa717a 16 One also has the rules of working with the sum sign If one de nes n Zaia1a2 39an i1 then Zai 212139 alijuianij ZalijruiJrZanbj Z aibj i1 j1 j1 j1 j1 j1 ij1 A ring is commutative if the multiplication is commutative In a commutative ring one has the classical binomial theorem and a great deal of the theory of determinants still holds We will always assume that a ring has a unit 1 for multiplicationi Rings without unit have been called by Jacobson 2 77rngs77i Obviously he doesn7t want to deal with objects nobody can pronounce properly However Hungerford l develops rings without unit For us rings will always have a unit If a ring doesn7t have a unit one can in a canonical fashion add one Examples 1 Z Z E 01 is a commutative ring 2 It is easy to see that the congruence modulo kZ in Z is actually a ring congruence ie if z E z and y E y then my E z y hence de nes a multiplication on ZkZ and Z A ZkZ z gt gt is a homomorphismi It is easy to see that the homomorphic image of a ring is a ring 3 Let A be an abelian group Then EndA golgo A A A go is a homomorphism 70A a idA is a ring Here 0A is the zeromap on Al Of course is the addition of functions iiei 01 goga 01 a 02 a One needs the commutativity of addition in A in order to show that the sum of homomorphisms is homomorphic of course a is the composition In general a homomorphism of an algebra to itself is called an endomorphismi Because rings are de ned by equations we have as before Proposition 216 The class of all 7ings is a primitive class of algebras Let g0 A A A be a homomorphism between ringsi Then 1 NW 12 iff zl 9012 iff zl E 12 0 iff 11 E 12 6 LP 0 is a congruence relation on Al I LP is called the kernel of go and is an ideal of Al That is o I is closed under E and contains 0 This is obvious because so is a homomorphism between the additive module structures of the rings 17 o Ifz E I and y E A then one has that LI yr 6 I and RI my 6 I Indeed if I 0 then Iy Isoy 0 yWW A nonempty subset of A that is closed under addition is a leftideal if LI holds and it is a rightideal if RI holds a subset that is a leftideal as well a rightideal is an ideal Let I be an ideal of A Then 11 N 12 m0d1iff 11 7 12 E I de nes an equivalence on A This equivalence is compatible with the ring multiplication and therefore a congruence That is If 11 N yl and 12 N yg then 1112 N ylyg We have 11 7 yl E I and 12 7 yg E I But then 1112 7 ylzg E I and ylzg 7y1y2 E I because I is a right as well a leftideal But then 1112 7 y112 ylzg 7 ylyg 1112 7 ylyg E I This is 1112 N ylyg On the other hand let I be a submodule of A Then I is an ideal if the equivalence modulo I is compatible with the ring multiplication Indeed we have that z E I iff 1 N1 0 But then yr N 0 and my N 0 Proposition 217 Let I be an ideal of the ring A Then the equivalence modulo I is a ring congruence The factor algebra AI is as a homomorphic image of A a ring AI Izlz e A70I1I1 where 1I1y 1Iy7 IH 1I7 11 39 19 1Iy and p1 A 7gt AIz 7 1 1 is homomorphic Corollary 218 Homomorphism Theorem for Rings Let go A 7gt A be a homomorphism between rings Then kergo is the equivalence de ned by the ideal I4p go 10 and A E AIW Examples 1 hZ is an ideal in Z and therefore ZhZ Zk is a ring the ring of integers modulo k 2 We de ne for any ring with multiplicative unit a 77 scalar multiplication 01 0 n 1l nl 1 where n is a positive natural number This de nes ml for every natural number Furthermore we de ne for any n E N 7nl 7nl Hence hl is de ned for every h E Z One easily veri es the following rules kl1 k1l1 7k1 7191 01 0 k l1 b1 11 121A 1A This tells us that HHAZ7gtAh gt gt hl is a homomorphism We have that kerH hZ for some number k 2 0 The natural number k is called the characteristic of A It follows from this de nition that the characteristic of a ring is either 0 or the smallest positive natural number k for which hl 0 if there is such a h 18 Let A be any ringi Any subring of A must contain 0 and l but then also ml for every k E Z Obviously is the smallest subring of Al It is also called the prime 7ing of Al Hence ZkZ Zk E prime ring of A where k chaTA Any ring A may be considered as an extension of Zk where k is the characteristic of Al In this sense we may identify the unit of a ring with 1 or 1 modulo 6 if k 2 0 A ring is called a domain if it is without zero divisors That is DO If I f 0 and y f 0 then zy 0 or equivalently zy 0 implies that z 0 or y 0 Examples 1 Z is a domain Because of this example domains are sometimes called integral domains 2 The zeroring U is a domain It is sometimes excluded 3 Zn n gt 0 is a domain iff it is a primer Indeed if n is not a prime then it TLan where n1 and n are proper divisors of n But then n1 n2 0 shows that Zn is not a domain On the other hand if p is a prime then Zp is a domain Indeed if n1 n2 0 then plnlngi This means that plnl or plngi hence n1 0 or n2 4 Z X Z is not a domain 01 10 00 5 A subring of a domain is a domain But the class of domains is not closed under direct products and not closed under homomorphic imagesi Equivalent to D0 is the restricted cancellation law D07 lfzyzyand17 0thenyyi lfyzyz andza Othenyy Assume DO I f 0 and my zyi We then have 1y 7 y 0 and therefore by DO y 7 y 0 ie y y The other half of DO7 follows similarlyi Assume DO7 and my 0 Then if I f 0 one has that my IO and therefore y 0 Similarly if y f 0 then my 0y yields 1 0 A ring A is called a division 7ing if DR holds the nonzero elements form a group under multipli cationi That is o l f 0 o lfz7 0andy7 0thenzy7 0 o If I f 0 then there is some y f 0 such that my yr 1 One has that DR is equivalent to the apparently weaker condition DR7 o l f 0 19 0 If x f 0 then there is some x such that xx 1 If x f 0 and y f 0 then y x 1 yields xy 0 If x f 0 then xx 1 yields x f 0 Therefore x x 1 and x xx x xxx x Thus x x 1 For a domain one has that it is either the zeroring ie 1 0 or the elements different from zero form a monoid under multiplication For a division ring this monoid is a group Proposition 219 A nite domain D y U is a division ring PROOF Let x f 0 Then A75 DO A DOy gt gt xy is by DO7 injective Because D is nite A75 is bijective Therefore xy 1 for some y This is DR7 D Proposition 220 A ring A is a division ring if and only ifA has exactly two right ideals U and A PROOF Let I be a rightideal of the division ring D Then either I U or there is an x f 0 in I But then xx e E I and one has for any y E D that 1y y E I This is I D For any ring A and x E A one has the right ideal that is generated by x xT x A E A Assume now that A has exactly two rightideals Then A f U and therefore 1 f 0 Let x E Ax 0 Then xT A yields some y such that xy 1 This is DRl D Corollary 221 Let go D A A be a homomorphism from the division ring D into A Then either U or E D A commutative division ring is called a eld Q R and C are elds Zn is a eld if and only if n is a prime number One can show that any nite division ring is a eld Let I be an ideal of the commutative ring A I is said to be a prime ideal iff the complement A 7 I is multiplicatively closed That is if x I and y I then xy I Equivalently if xy 6 I then x E I or y E I The following proposition is only a reformulation of the de nition in terms of factor algebras Proposition 222 Let A be a ring and let I be an ideal Then AI is a domain if and only I is a prime ideal 1n Z every ideal I is principal ie I hZ h for a unique It 2 0 and one has that h is prime iff h is prime An ideal I of the commutative ring A is called maximal if and only if I f A and there is no ideal J strictly between I and A That is I is maximal in the ordered set of all ideals different from A Proposition 223 Let A be a commutative ring and let I be an ideal Then I is maximal and only if AI is a division ring PROOF Let M be maximal Then if x M one has that the ideal that is generated by M and x is the whole ring It is easy to see that the smallest ideal that contains the ideals I and J is just the sum I J jli E Ij E J The ideal which is generated by x is the principal ideal xA Hence we have that MxmxylmEMyEAA and therefore in xy 1 for certain m E M and y E A For the factor ring AM this means that for every class f 0 M one has some such that Thus AM is a eld If I is not maximal then there is some ideal J such that I C J C A This yields a homomorphism AI A AJ xlj gt gt xlj whose kernel is J E J which is a proper ideal that is different from the zero ideal M and the whole ring A Hence according to Corollary 221 AI is not a eld E 20 Proposition 224 Krull s Theorem Let A be any ring di erent from the zero ring and let I be any ideal di erent from A Then I is contained in some maximal ideal ofA PROOF Let S JlJ 2 1 AJ E IdealA Then S is an ordered set Any chain in S has an upper bound because the union of a chain of proper ideals is a proper ideal Here we use that the ring has a unit Hence by Zorn s lemma 8 has a maximal element ie a maximal ideal M that contains 1 D 23 Modules over a Ring Vector Spaces Let A be a ring An Amodule M is a structure of type M M 7 707 10490 such that i M MHr 70 is a module ii The familiar vector space axioms hold aab aaab a a aa a a a al a La a We denote elements of the ring the 77scalars by lower case Greek letters and the elements of the module M by lower case Latin letters Each 1 E A leads to a unary operation a gt gt 0404 We do not make a notational distinction between the zero of A and the 77zero vector that is the he zero of M While not always required we will assume from now on that A is commutative Here are some elementary algebraic facts on Amodules 0 0a 0 0 10 0 o 71a 7a 70 04 7aa Examples 1 Every module is a module over Z where na has been de ned previously see page 18 Let A be a ring Then MA A 7 0 is an Amodule by means of am a z 2 In particular the rings Zn are modules over Z as well as over Zn For example 235 2 3L Hence it may happen that 0104 0 even if A is without zerodivisors 3 If A is a division ring then aa0iffa00ra0 The proof uses all Amodule axioms Assume 0404 0 and a f 0 Then 071 a 1 But then a 1aa 0710 0 La a 4 Let S be any set and let A be a ring Then the module AS is an Amodule according to the pointwise de ned scalar multiplication eg of is the map 3 gt gt Ozf8 For A R and S 0 n 71 n we get the familiar vector space R For 5 0 1 A1 is the ring A perceived as a module over itself More generally we have similarly as before 21 Proposition 225 For any ring A the class of Amodules is a primitive class of algebras A homomorphism go between Amodules M and N is called a linear map it is very easy to see that one only has to stipulate the following two conditions 0 go is additive ie goa b goa gob 0 go is homogeneous ie goaa agoa Note that multiplication by a ie the map a gt gt ab is homogeneous only if the ring A is commutative A subset C of an Amodule is already closed if 0 C is nonempty o lfaandbare in Cthen abis in C o lfaisinCthenaaisinCaEA Let BteT be a family of Asubmodules of M Then one has in the complete lattice of Asubmodules of the Amodule M ABE Bt and VBaabtlmbtk weal tkeTJh 631213 tET tET tET tET Let go M 7gt N be linear Then al N a m0dgo iff goa1 goa2 iff goa1 7 a2 0 iff a1 7 a2 6 kergo algoa 0 NW We already know that NW is a module But it is also a Asubmodule That is NW is closed under multiplication by scalars Assume that a 6 NW Then goa 0 and goaa agoa 0 This is an 6 NW Let N be an Asubmodule of M Then al N a iff a1 7 a2 6 N is a congruence relation Again we only have to show that this equivalence is compatible with the unary operations a But this is trivial On the other hand let N be a submodule of the Amodule M Then the induced equivalence modN is a congruence if and only if N is an Asubmodule Proposition 226 Let N be an Asubmodule of the Amodule M Then al N a a1 7 a2 6 N is an Amodule congruence The factor algebra AN is as a homomorphic image ofM an Amodule Note that aN a N an de nes the multiplication by scalars a E A The canonical map pN a gt gt N a is a linear map from M to MN with kernel N Corollary 227 Homomorphism Theorem for Amodules Let go M1 7gt M2 be a linear map between Amodules Then NW algoa 0 is an Asubmodule ofM and one has that MNw E A subset S of the Amodule M is generating if CS M That is every element of M is a nite linear combination of elements in S A subset S is linearly independent if only trivial linear combinations are zero That is If F s1sk are k elements of S then 1181 eggszc 0 only if 11 0 04k 0 A subset B that is generating and linearly independent is called a basis for M We recall from elementary linear algebra the following Theorem 228 Let V be a vector space with a nite generating set Then V has a basis and all bases have the same number of elements This number is called the dimension of V 22 The existence of a basis is quite trivial Any minimal generating set is a basis The uniqueness of the dimension is based on the fact that m linear combinations of n vectors are linearly dependent if one has that m gt n This is the same as saying that a linear system of m homogeneous linear equations in n unknowns has a nontrivial solution But this is an immediate consequence of the row echelon form The details of all of this are very easy to prove Vector spaces that are not nitely generated have a basis by Zorn s lemma A maximal linearly independent set is a basis That any two bases are equivalent follows from simple cardinal arithmetic De nition 3 An A module M is called the free Amodule freely generated by B if for every A module N and every map go B A N one has a unique linear map so M A N that extends go One writes M It is easy to see that M is uniquely determined by the set E up to a linear isomorphism that leaves B xed Theorem 229 The Amodule M is free freely generated by the set E and only B is a basis for M PROOF Assume that M has a basis B Because two homomorphisms that agree on a generating set must be equal uniqueness of g2 is obvious Now every element m of M is a unique linear combination of elements in B Then in an obvious way a map can be de ned and it is linear The details of this linear extension principle are easy Next we show that FB exists Let M Aw S A Afs 0 for almost all s This is clearly an A submodule of the direct power AS and the maps eb form a basis where eb b l and eb 0 if b b The map b A eb de nes a bijection from B to the basis eblb E B of M Thus we may identify the set of unit vectors ebb E B with B Because free modules are uniquely determined up to isomorphisms any free module over B has B as a basis D The representation of a linear map T R gt gt R by a matrix A is an important application of this theorem An algebraic structure M M 707 10493 lt gt is an algebra over A if 0 M 70 10493 is an A module o The map ltgt M X M A M mm gt gt lt ab gt is bilinear That is i ltaa bgtltabgtlta bgt ltabb gtltabgtltabgt ii ltaabgtlt aab gt altabgt A commutative ring is an algebra over itself A familiar example of an algebra over the reals is the cross product on R3 Also the eld C of complex numbers is an algebra over the reals where the underlying module is the plane R2 For any eld F the n X n matrices form an algebra gnF with respect to the ordinary matrix multiplication 7 o 77 Unlike the cross product of R3 this algebra is associative However with the auxiliary operation the commutator 5 T gt gt S o T 7 T o S it becomes the prototype of a Lie Algebra ie the Jacobian identity holds lav lbv Cll lbv lcv all lcv lavbll 0 Any nite dimensional Lie algebra is isomorphic to a subalgebra of a Lie algebra of matrices This is a fairly deep theorem for which no elementary proofs are known 23 The free module over B becomes an algebra if a multiplication has been de ned on the basis That is given a multiplication for any pair of basis elements 12 bj lt 12bj gt 2 as 1263 one has a unique bilinear extension Here we assume that the elements of B are indexed by a set 1 Of course for almost all I the 01ng are zero Again the proof is rather straight forward The familiar construction of the complex numbers is an example of this construction principle We remark that commutativity as well as associativity are carried over from the base to the whole algebra 24 Factorial Monoids and Domains In the ring Z of integers any element n gt 0 is a unique product of primes Such a factorization theorem holds for many other rings and concerns mainly the multiplicative structure of the ring Thus we are rst studying factoiial monoids Let M be a commutative cancellation monoid That is ab ac implies that b c We then de ne all iff there is some c E M such that ac b and say that a divides b or that b is a multiple of a The relation 77 is o re exive ala o transitive If all and if blc then alc That is the relation l is a quasiordering For any quasiordering on a set A the relation a and b are associated is an equivalence where a and b are associated if they divide each other a N 1 iff all and bla On the set A N of equivalence classes an ordering is de ned by MW 16 alb This is called the contraction of a quasi ordering What are in our case the equivalence classes If a N b then ac b and Id a Hence acc bc a This is cc l where l is the unit of our monoid This is c and c are inverse to each other Now let U InvMl be the group of invertible elements of M We have shown that a N 12 implies that b E aU On the other hand if b E aU then b an and bu l a show that a N 12 Therefore aaU aEM is the partition of M into equivalence classes of associated elements The equivalence relation of being associated is actually a congruence relation on M We need to show that if a N a and b N b then ab N a b We have that an a In b but then abuv a b 24 Thus M A MN Ma gt gt aU is a monoid homomorphismi Of course U is as the class of 1 the unit of Mr M is also a cancellation monoid 1f ab ac then ab ac iiei abu aci But then bu c ie 12 If all where a is not a unit and where l l a then a is called a proper factor of 12 A unit cannot have a proper factor If alu then av u and a uv 1 is a unit Let a E M and let u be a unit Then uu 1a a shows that u divides a If a is a proper factor of b and ac b then c is a proper factor of 12 Indeed assume that b divides c iiei 12d cl But then acd c iiei ad 1 and a is a unit and therefore not a proper factor of 12 1f ac b where a is not a proper factori Then one factor is associated to bb and the other one is a unit An element 4 E M is called irreducible if 0 q is not a unit 0 4 does not have any proper factorsi That is if q ab then either a is a unit or a is associated to q 1n Z 0 1 exactly the prime numbers are irreducible The group of units is U 171 and n N m iff n imi De nition 4 A factorization a ql i i i qs is an essentially unique factorization of a into irreducible elements qi if for every other such factorization a q H i 4 one has that t s and qi N qi for a suitable permutationi gt gt ii De nition 5 Let M be a commutative cancellation monoidi Then M is called factorial if every non unit of M has an essentially unique factorization into irreducible elementsi An integral domain D is factorial if its monoid D 0 1 is factoriali Let M be a factorial monoidi Then either 0 a E U ie a N 1 and a is the empty product of irreducible elements We de ne in this case la 0i 0 a is the product of s many irreducible elements where s 2 1 and s la is uniquely de ned The natural number la is called the length of a Let a bc where b is a proper divisor of ai Then also c is a proper divisori Now the factorizations of a and b can be combined to a factorization of ab We see that la lb lc and where 1 S lblc lt lai Hence the length of a proper divisor b of a is less than the length of ai De nition 6 A binary relation R is said to satisfy the minimal condition if there is no in nite chain alRagRag i i H The following Lemma is obvious Lemma 230 Let M 1 be factorial Then the relation aRb that says quotI has a as proper divisorquot satis es the minimal condition For this we say that M 1 satis es the divisor chain condition De nition 7 An element p of M is prime if 25 o p is not a unit o If plab then pla or plbi A prime p is always irreducible Assume that ab pi Thus alp and My But p 1 ab also shows that plabi Because p is prime we have pla or plbi Hence a N p and b is a unit or vice versa Thus p does not have proper divisors ie p is irreducible Lemma 231 Let M 1 be factorial Then every irreducible element is prime This is Euclid s condition PROOFi Assume that qlab iiei qc ab where q is irreducible If a is a unit then qca 1 b and we have shown that qlbi Thus we may assume that a and b are not units and therefore a q l a i i w and b q l i i i 11 Thus abq 1Hiqq1Hiq i If C was a unit then 4 c lql i H q shows that q is not irreducible Hence c ql qri Thus 4 ql qr q l i H q q l L and because of uniqueness of the factorization we have thatqwq orqwq i D Theorem 232 Let M M 1 be a commutative cancellation monoid Then M is factorial and only if M M 1 satis es the divisor chain condition M M 1 satis es Euclid s condition PROOFi Assume that M satis es the divisor chain condition We are going to show that every element a in M is a product of irreducibles Let a 6 Mi If a is a unit or irreducible we are done Otherwise we are going to show that a has a proper divisor ql that is irreduciblei We assume that a is not a unit and not irreducible Hence a has a proper factor all If a1 is not irreducible then a1 has a proper factor a etc Because of the divisor chain condition the chain aRalRag i H must terminate with some aj that is an irreducible element 41 We now have a ql 121 lf 1 is not irreducible 121 has an irreducible divisor L12 and a qlbl qlqg 122 The chain ble2 i i i must terminate for some at some point s and we get a ql i i i qsi Because of Euclidls condition the factorization of a into irreducibles is essentially uniquei Assume a ql qs q l 4 Assume by induction that every element that has some factorization with less than s many factors has an essentially unique factorization The case s 1 is obvious The element a then is irreducible and cannot have more than one factori That is a q is the only factorizationi For the general case we have 41qu i i i q and according to Euclidls condition one has qllq for some i i But 49 is irreducible Thus ql N L19 iiei q ql ul for some unit all We may rearrange the second factorization and get a ql i H qs ql ul q 4 But then we may cancel on both sides 41 and we are done by induction D In every factorial monoid the contraction of the divisibility relation is a lattice orderingi That is modulo equivalence every two elements a and b have a greatest common divisor gicidi a b and a 26 lowest common multiple llclmla bl Let 73 be a complete set of representatives for the set of irreducible elements of Ml Then for a and b in M one has a N Hpeao 17673 b N Hp b17 17673 H pminea17gteb17 1767j llclmlab N Hpmweal7gtebp 1767j glcldla b 2 Of course one has for every a that for almost all p eap 0 Let D be an integral domainl D is called a principal ideal domain if every ideal I is generated by some element a ie I a blalb E D The generator a of I is unique up to equivalencel By the very de nition D 0 l is a commutative cancellation monoidl We extend the divisibility relation alb iff ac b for some c so that it includes zerol But then lla and alO that is 0 is the maximum of the ordered set DN Now w wgw shows us that IDle 9 IdeaK17g Of course IdealD Q is a complete latticel Hence we have Proposition 233 Let D be a principal ideal domain Then every nonempty subset ofD has a gcd and a lcm PROOFl To be more explicit let S Q Dl We claim that the generator d for S is the glcldlSl lndeed because of d 5 one has that dls for every 8 E 5quot On the other hand assume that els for every 8 6 5 Then e 2 S This is eldl D Lemma 234 Let D be a principal ideal domain Then d glcldl a b is the only divisor of a and b that is of the form d cla 621 ie the only divisor that belongs to the ideal ab PROOFl For any commutative ring A one has that ab cla Cgblci E A is the ideal that is generated by a and b Thus if D is a principal ideal domain then one has for the generator d of this ideal d c1a Cgbl Assume now d cla 6211 is a divisor of a and b Then i d 2 a b d because dla and dlbl ii d E d because d 6 ab Thus d d ilel d glcldlabl D 27 Theorem 235 Let D f U be a principal ideal domain Then D 7 01 is factorial ie every element di erent from zero that is not a unit has an essentially unique factorization into irreducible elements PROOFl We show that D satis es the ascending chain condition for ldealsl That is any ascending chain I1 Q I2 Q of ideals becomes stationary after some n ilel In In1 l l H To show this one observes that I UneN In is an ideal and therefore I a for some a 6 I But then a E In for some nENandIInHn But this implies the divisor chain conditionl That is if a1 a2 l l l is any sequence of elements in D such that ai1lai then an an1 l Next we show that irreducible elements are prime For the domain of integers this is usually referred to as Euclid s lemma Assume plab where p is irreducible one says primel Then pla or plbl Assume that p X 12 As an irreducible element p has only trivial divisorsl Thus the glcldl of p and b is 1 By the previous lemma 1 cp dbl But then a acp adbl Now plab by assumption and placpl Thus we have that plal D De nition 8 An integral domain E is called Euclidean if there exists some map go E 7gt N such that if ab E E where I f 0 then there is some 4 E E such that a In or a 124 r where r lt b Examples 1 Z is Euclidean with n 2 Let E be the set of Gaussian integers ie the complex numbers a of the form 2 mni where m and n are integers Then E is an integral domain We de ne go E 7gt N by m7ni m2 n2l Of course a ago l This formula holds for arbitrary complex numbers For any complex number a r1 7 rgi one can nd mn E Z such that lrl 7 ml S 12 and lrg 7 nl S 12 This yields 1 gtgtM7 4 we a 771 712 90T1 m T2 70239 ln W W W S Let now a m ni and m ni 0 Then choose some 7 E E such that a l 7 7 lt 1 But then goa 1 7 7 a 7 76 lt proves our claiml Theorem 236 A Euclidean domain is a principal ideal domain PROOFl Let I be an ideal of the Euclidean domain El If I is the zeroideal then we are done Otherwise there is some I E I where I f 0 We choose some a E I for which a is minimall That is b 2 a for all I 6 I We claim that I a Indeed I 2 a because a E I and I is an ideall On the other hand if b E I then b is a multiple of a If not then we have b aq r where r lt a But r E I and r lt a is a contradiction D Corollary 237 Every Euclidean domain is factorial 28 We have Z E El However 5 is irreducible in Z but as 5 2 2 7 shows 5 is not irreducible in El ln Z the gicidi of two integers can be obtained without going through the prime factorization rst This Euclidean Algorithm is based on the observation that if a 124 T one has that a b bTi Assume that a gt b gt 0 Then in a 124 T b qlT 7 1 T qul 7 2 etc we have that a gt b gt T gt 7 1 gt 7 2 gt Hi ab b T TT1 i i i and eventually we must have Tk1 0 where Tk gt 0 But then ab m0 shows that Tk is the gicidi of a and 12 For a commutative ring A we have that I p is a prime ideal if ab 6 I iff pla or plb that is p is a prime element The ideal I q is maximal amongst all principal ideals iff a 2 4 implies that a N l or a N 4 This is alq implies that a N l or a N 4 Hence 4 is maximal amongst principal ideals iff q is irreducible Hence we have Proposition 238 Let D be a pTincipal ideal domain Then the following statements aTe equivalent a is a pTime element a is a piime ideal a is a maximal ideal iv a is iTTeducible 25 Polynomial Rings ln elementary courses on algebra polynomials are de ned as formal expressions a0 alzi i i anz and one learns that the sum and product of polynomials is a polynomial where these operations are de ned in an obvious way Furthermore every polynomial stands for a certain function fa is what one gets by substituting a for 1 But what are polynomials and what is I They should be elements of a ring But what ring These questions were answered in the thirties by Hans Hasse who gave a rigorous de nition of the somewhat mysterious concept of a polynomial and the tTanscendental element 1 Even today there are still textbooks that take the concept of a polynomial ring for granted without going through its somewhat lengthy 77Hasse construction h We will construct polynomials as certain universal objects built upon the commutative ground ring A and a xed element 1 The universal character of the polynomial ring means that every substitution of I by an element a leads to a homomorphism from the polynomial ring to Al De nition 9 A pointed ring is an ordered pair Aa consisting of o a commutative ring A 0 an element a 6 Al De nition 10 A momhism g0 Aa A Bb between pointed Tings is o a homomorphism g0 A A B where o a 12 29 Theorem 239 For every commutative ring A there is a pointed ring A7 I together with a homo morphism 1 z A A A such that there is for every homomorphism go A A B7 12 from A into a pointed ring B712 a unique morphism 51 ACE A 1371 between pointed rings where 2 o 1 go The pointed ring ACE is uniquely determined up to isomor phisms between pointed rings PROOFl We rst show Existence of the 77polynomial ring77 over Al We de ne AlIl Plpi N A Apn 0 for almost all 11 That is7 Az is the free Amodule with base lwhenin7 elni 0 wheni7 nl ilel7 ei 5 where 5 is the Kronecker symboll We know already that any multiplication on the base makes AM to an algebra over A by linear extensionl We de ne a multiplication on the base by 6i 3961 em Then Az7 707 alaeA7 e0 is an associative and commutative algebra with e0 e as unitl Let p E Then p 226A ailei where the asterisk at the sum sign indicates that almost all coef cients are zero Now7 en e1en e1 If we de ne e1 as I then we have for every p in AM a unique representation as a 77polynomial n s E N aizi E aizi where n is the last coefficient different from zero ie i1 n is called the degree of p and an is called the leading eoe cient of p The empty sum is de ned as zero and p 0 does not have a degree We now de ne 1AgtAz7 a gt gt are Then 1 is an injective ring homomorphisml We have 1a b a ble ale be 1a 11 and 1ab able are ble 1a 117 11 lie e and injectiVity of 1 is obvious7 e is linearly independent We are going to show that 1 z A A Az7 z is universal for all go A A A7 b In order to see this7 we de ne the map 2 by Z ai Ii gt gt Z 0ai bi 30 and it is easy to see that this map has the desired properties7 ie7 it is a ring homomorphism for which aa ae a b0 a and l The multiplicativity of so needs the commutativity of A and of B The proof of Uniqueness up to isomorphism of the universal object in the category of pointed rings over A is a by now familiar exercise D For every xed a E A we have a homomorphism Ea Az A A which is de ned by A i Ami Au A g Au We have that Eap pa This is true because Eaz a and Ea ac c Eace Now7 Az g A7 a 6 A7 yields according to the de nition of the direct product with projections qa a homomorphism Am A AA such that E E Am AAALAQW AA Ep is a map from A to A which has at a the value 1aEp Eap pa lfp Eaizi then Epa Eaiai pa Theorem 240 The evaluation map EAz AAA pZalzi gt gt 5 a HZalai aEA is a homomorphism between the ring of polynomials and the ring of polynomial functions Because the map a a gt gt ae7 is an injective homomorphism we may identify a with ae7 and consider A as a subring of With this identi cation7 we have that ap 1a p me p alep ap The degree of a polynomial p7 degp7 has been de ned earlier We then have the degree rules degp 1 degp t1 maxdegP7 deg4 S S degltPgt deglt4gt where equality in the product formula holds if the leading coefficient of one of the factors is not a zero divisor If we de ne the degree of the zero polynomial to be 700 then these formulas are universally true We conclude from the product formula Proposition 241 For an integral domain D the polynomial ring DM is an integral domain Theorem 242 Division Algorithm Let p and d be polynomials in AM and assume that the leading coe cient of d is an invertible element in A Then there exist unique polynomials q and r such that p q d r deg lt degd 31 PROOFl We rst show Uniqueness of the quotient q and TemaindeT Tl lndeed f q1dT1 q2dTg yields 0 L11 7 qgd T1 7 T2 This is 41 7 qgd 7 2 7 T1 Now degT1 lt degd and degT2 lt degdl But degql 7 q2d 2 degd or 7 00 Because d is not a zero divisor the latter is only the case if 41 7 42 0 ie if ql 42 But then also 7 1 T2 The part of Existence is just the familiar long division First if degp lt degd then p 0d p is the desired decomposition Thus we may assume that degd S degp pa0alzluanzn db0blzlubmzm wheremgn The proof now goes by induction over it degpl If n 0 then Tn 0 and p a0 d 120 where 120 has an inverse and a0 a0 123120 0 Assume that n gt 0 Then a0alzulanzn 7 b0blzulbmzm anb1zn m p7 anb1znimdp1 where degp1 lt degpl By induction hypothesis p1 q1d 7 1 where degT1 lt degd or 7 1 0 Hence 0 41d T1 an b In m d an b1 In m and T1 q d T1 B Let F be a eld Then every element different from zero has an invertible leading coefficient Thus for every p E and d f 0 one has some 4 E such that p q d or degp7 q d degT lt degdl This shows that the degree map deg 0 7gt N makes to a Euclidean ring Theorem 243 FoT any eld F the polynomial Ting is a Euclidean domain One has that p is a unit if and only degp 0 and only if p E 7 0 Corollary 244 is a pTincipal ideal domain Corollary 245 is factoTial As a corollary of Theorem 24 we note Proposition 246 Let p 6 AM a E A Then p q I 7 a T wheTe T E A and pa T Corollary 247 The element a E A is a Toot ofp ie p01 0 and only if I 7 a l p Corollary 248 Let D be an integTal domain Then a polynomial p E Dz has at most n Toots in D PROOFl We prove this by induction on n degpl If n 0 then p a f 0 and p has no rootl Let n degp and let a be a root of p Then p I aP1 and 0 if and only if a b or 1311 0 Here we use that D is a domain Now degp1 n 7 l proves the claiml D 32 Corollary 249 For an in nite integral domain D the map E Dz 7 DD is injective Proposition 250 Let p E Az and let a E A Then the set l l I 7 a p has a maximum m and ifp I 7 am 4 then L a 0 Conversely each relation p I 7 al r where Na 0 implies that l m The natural number m is called the multiplicity ofa for p PROOFl Assume that p I 7 al 4 Then degp n l degql Thus l S n and a maximum m must exist If p I 7 am 4 then Lja 0 would yield 1 7 alq and therefore I 7 am1lp which is a contradiction to the choice of nm Assume now that p 17alr where Na 0 Thenm 2 l and 17alr7 17am l q p7p 0 Hence because I 7 al is not a zero divisor we have that r I 7 am l 4 Now m gt Z would yield ra 0 a contradiction Thus m l D De nition 11 The derivative of the polynomial p a0a1 1 i i lan I is de ned as the polynomial 6p a1 i i i nan zn ll One then has the familiar rules e gl 6p q 6p 64 and 6p q 510 4 10 64 Proposition 251 For any polynomial p of the ring AM one has that a E A is a multiple root if and only if it is a root ofp and of 6p PROOFl Assume that a is a root of p We then have p I 7 am q where L a 0 Assume that m gt 1 ie a is a multiple rootl Then 6p mz 7 am 1q z 7 am6q yields 510 a 0 If m 1 ie a is a simple root then 6p a a 0 D 33 Chapter 3 Modules over a Principal Ideal Domain 31 Free Modules Let M be a module over a principal ideal domain D and let I 6 Mi Then MIDHMdgt gtdiz is a homomorphism between Dimodulesi Clearly im7 dizld E D lt z gt span ofx is a Disubmodule of M and ker dldiz 0 is an ideal of D A generating element of ker is called the period perz of 1 Thus Dltperltzgtgt 2lt z gt Example Let A be an abelian group and let a 6 A Then A is a Z7 module and lt a gt2 Zpera Now pera minn l n gt 0nia 0 or pera 0 Thus ltagt Znor ltagtZ A cyclic group lt a gt is either nite and isomorphic to Zn or isomorphic to Z More generally A cyclic module lt z gt over a principal ideal domain D is isomorphic to a factor module Ddgc where dag perz 34 If I E M and if dz 0 then perzld Let X be a generating set of M ie M lt X gt Then one has dz 0 for all z E M i da 0 for all z E X i d 6 1001 for all z E X i d E mexperz i d E lcm experz Thus perM d 1 dz 0 for all x E M lcm experz Notice that If perz 1 then D1 Elt z gt thus 0 Elt z gt which is z 0 On the other hand one has that per0 1 Thus perz 1 iff z 0 1f perz 0 then DO 2lt z gt which is D 2lt z gt On the other hand assume D 2lt z gt and let go 7lt z gt be any isomorphism Assume that 1 gt gt am Then perz gt gt perz az shows perz 0 thus perz 0 Hence perz 0 iff lt z gt is free Recall that a module over the ring A is free freely generated by the subset B if every map go from B into any A7module mbe has a unique linear extension Such a module is isomorphic to A03 Modules over elds are always free and all bases have the same cardinal Theorem 31 Let M be a free module over the piincipal ideal domain D Then all bases B of M have the same cardinal which is called the dimension ofM over D PROOF Let p E D be a prime element Then D p is a eld and 11M is a D submodule of M Thus we can form the factor module MpM Let d p E Dp and z 11M 6 MpM Then d pM dz pM is a proper de nition That is if d1 7 d2 6 p and 11 7 12 6 11M then dlal 7 d212 6 11M Thus Mp becomes a module over the eld Dp Let X be a base of M Then X l I E X is a base of MpM Of course X is generating for MpM With respect to linear independence assume that Exexwgc z 0 ie Ed a py p EEGX ewz Exexpewz Then one has that dag p egg holds for all z E X Hence dag 0 for all z E X In particular I gt gt has to be injective or cardX cardX where cardX dimDp D Theorem 32 Let M be afree module over the principal ideal domain D and let N be a D7submodule of M Then N is also free and dimD N S dimD PROOF Let 11 In be a base of M and let NT N ltzlzT gtr1n Clearly Nn N We show successively that each NT is free and that dimNT S 7 For n 1 we have that N1 N lt 11 gt dal l d E D dzl E N Now d l dal Elt 11 gt N is an ideal ofD thus a principal ideal 11 d1 and we have N1 N lt 11 gtlt dlal gt 35 The case d1 0 is clear N1 U and dimN1 0 lt L Assume that d1 0 Note if dAd1Azl 0 one has that perzlldd1A But perzl 0 thus dd1 0 which yields d 0 Hence perd1Azl 0 and therefore N1 lt dlAzl gt is free and dimN1 l S L Now assume that NT is free and that dimNT S rA Similarly as before we get an ideal IT1 of D r1 NT1 N lt 11 A A A CL T L 7 1 gt eiAzi l ei E D N N i1 where 1 d l 3e1e 1A11 A A A ETUT I dAIr1 E Nr1 is an ideal of D and therefore IT1 dr1A lf dr1 0 then NT1 NT and NT1 is free and dimNT1 S r lt r 1A Otherwise we have some y E NT1 where y elAzl A A A eTzT dr1AzT1 where dr1 0 Let yl A A A yk be a base of NT where k S rA Then our claim is that y1A AA yky is a base of NT1 We rst show thaty1A AA yky generates NT1A Let I E NT1A Then I flAI1A A AfTAITdAzT1 where d E IT1 thus d fAdT1A We get fAy fern A A A ferAIr fdr1rr1 Therefore fdr1AzT1 fy 7 felAzl 7 A AA 7 feTAzT and I f1A11A A AfrAIrfdr1AIr1 f1A11AA AfrAIrfy f 1A11AAAiferAIr 91A11A A AyrAIrfAy Hence I7 E NT1 Q N but also z7fAy Elt zlAAAzT gt A This is z7fAy 6 NT lt y1AAAyk gt and nally I Elt y1AAAyky gt A We need to show that 31 A A A yk y is a linearly independent set Assume that flAy1A A AkaykfAy 0 Then f1Ay1 A A A kayk 7fAy 7fA 61ml A A A ETAIT dr1Azr1 eltxlxtgt 6lt11ET1gt Thus 7fAdT1 0 From this it follows that f 0 and therefore f1Ay1 A AA 0 which is f1AAAfk0A D Corollary 33 For modules over principal ideal domains submodules of nitely generated modules are nitely generated PROOFA Let M lt zlA AA 1k gt be nitely generated by k elements and let N be a submodule of MA Then let Dk be the free D7module which is freely generated by the k unit vectors eiA We then have a unique surjective homomorphism go which extends ei gt gt MA The counter image L go 1N of M is a submodule of Dk and nitely generated by l7many elements7 3 17 A A i yz where l S kA Of course L N and N lt yl A A A goyl gt D For a D7module M TorDM I l perz 0 is a submodule of MA The module M is called a torsion module in case that TorM M and M is called torsion free in case that TorM DA A nite module is always a torsion module The module Z1 is an example of an in nite torsion module 36 A product of torsion free modules is torsion free The sum of torsion modules is a torsion mod ule For a domain D the one dimensional Dimodule D is torsion free For a domain D any free Dimodule is torsion free Corollary 34 A nitely generated torsion free module over a principal ideal domain is free PROOFi Let Y y1 l l l yn be a set of generators for M where M is torsion free We may assume that Y y1luyk is a maximal linearly independent subset of Y and the submodule N of M which is generated by Y is of course free However one cannot conclude that Y is a base for Mi For example 23 is a linearly dependent set that generates Z but neither 2 or 3 is a base We can only conclude that M is isomorphic to a submodule of the free module Ni In order to show this we notice that for i gt Is one has some di 0 such that dlsyl i i erk yk di yi 0 Then let n d H d i39k1 Clearly dlyi Elt y1luyk gt for i k l l Then goMgtNmgt gtdlm is linear and it is injective because M is torsion freei D Corollary 35 A nitely generated module over a principal ideal domain is free if and only it is torsion free 32 The Decomposition into a Free and Torsion Part A Dimodule A is the direct sum of nitely many submodules A if every element a of A is a unique sum of elements ai E Ail This is the same as saying that A is the coproduct of its submodules Ail We also note that for nitely many modules product and coproduct are the same We use the 77El sign to indicate that a sum is direct If C A B then the sum is direct if and only if A N B 0 If C AGBBthen CAEBl De nition 12 Let A be a ring A module P is called projective if for any homomorphism go P A M and any surjective homomorphism 1 M a M there is a homomorphism p P A M such that rteJae 39U 37 Proposition 36 A free module over a ring A is projective PROOFl With the notation of the previous de nition let X be a base of P Where P is free Then let mac 6 w 1gozl There is a unique homomorphism p P A M I gt gt magi Clearly goz and therefore w o p 90 D Proposition 37 Let 1 M a P be a surjective homomorphism from the module M onto the pro jective module Pl Then kerw is a direct summand That is M keril 69 P where of course P g Pl PROOFl We put M P and let 45 idl Then there is some p such that w o p id and we set m m WNW WNW Then 0 For any 721 E kerw one has that m for some ml Therefore But then 0 thus m 0 This is kerW 7 P M U and M kerw E P E P Where P B Let M be a module over a principal ideal domain Di Then the factor MTorM is torsion freel Assume that dlz TorM TorMl Then dz 6 TorM for some d f 0 But this is perdlz 0 and therefore eldlz 0 for some e f 0 Thus perz 0 ie I E TorMl Theorem 38 Let M be a nitely generated module over a principal ideal domain D Then M is the direct sum of its torsion submodule and of a submodule which is free M TorM 69 N where N is free The decomposition is unique in the following sense If M M 69 N where M is a torsion module and N is torsion free then M TorM and N g N PROOFl The module MTorM is torsion free and as a homomorphic image of a nitely generated module also nitely generated Thus MTorM is free We apply the last proposition to q M a MTorM kerq TorM and get M TorM 69 N Where MT0rM 2 Ni Let M M 69 N be a similar decomposition and assume that m E TorMl We have m m n and dim 0 for some d f 0 gives dlm 0dln 0 Because N is torsion free we get that n 0 Hence m m E M and this is TorM Q Ml But M Q TorM holds by assumption Thus TorM M and N MM Ni D 38 33 The Primary Decomposition Theorem Let M be a module over the principal ideal domain D and let d E D Then multiplication by d hdMgtMzgt gtdz is linear and W kerad x 1 dz o x l perrld is a Disubmodule of M Lemma 39 Let d1 d2 1 and d d1 d2 Then Md Md1 69 Md2 PROOF We have 1 e1 d1 e2 d2 and therefore 96 z eldla egdga for every I E M Let now I E Md ie dz 0 Then one has that dlegdga 0 and dgeldla 0 ie egdga E Md1e1d1z E Md2 and of course Md1Md2 are submodules of M This is Md Md1 Md2 If I E Md1 Md2 then I 0 by B Let now M be a nitely generated torsion module over the principal ideal domain D If M lt 11zn gt then it is an easy exercise to show that perM lcmperzi d f 0 Then if d p 1023 one has that M Mp 1 69 69 Mp2 De nition 13 Let p be a prime of the principal ideal domain D A pimodule is a Dimodule where the period of every element is a power of p Clearly for every module M over the principal ideal domain D and every prime p E D one has that TPM I l Perm 107T 2 0 is the largest pisubmodule of M Lemma 310 Let M be a Dimodule of peTiod pr Then M is a p7 module and M contains an element 10 such that peTzo peTM IT PROOF We have perz l perM for every I E M and therefore perz p5 where s S T Thus M is a pimodule Let now 10 be an element of M with perzo p50 where so is maximal Then one has that p5 perzlperzo p50 for every 1 Thus psoa 0 for every I ie perMlp5 This is prlpsO or T S 80 Therefore T 8 because we have already shown that s S T D Lemma 311 Let M be a Dimodule of peTiod d p 1025 Then one has that Mp7 T171 PROOF We have already shown that M Mp 1 69 H 69 Mp2 holds for any module of nite period d p 1 1025 Here nite means larger than 0 We have Mp 1 I l p ila 0 I l perzlp 1 Q Tpl For the other direction let I E Tpl M ie psa 0 for some 3 We have that z 11 1k and therefore pix pial piak 0 It follows that pial H piak 0 Now pix 0 yields perzglpi but perzg is a power p of p2 But this is only possible if T 0 Thus 12 0 etc and we get I 11 E Mp f1 D A pTimaTy module is a pimodule for some p E D 39 Theorem 312 A nitely generated torsion module over a principal ideal domain D is a nite direct sum of primary modules If perM p il 10 then M Tp1MTpkM where each T1 is the largest piisubmodule of M and one has that perTpl p272 As an example let A be a nitely generated abelian group We may consider A as a module over the domain Z of integers We then have A B 69 TorA where B E Zr and where TorA is the direct sum of primary groups Tpk a l pka 0 We are going to show that each such Tpk is a direct sum of cyclic groups ie groups that are isomorphic to Zpll 34 The Decomposition according to Elementary Divisors and Invariant Factors We are going to show that every Dimodule M is a direct sum of cyclic primary modules For example all nite abelian groups of order 24 are given by the list Z3 69 Z3 Z3 69 Z4 69 Z2 Z3 69 Z2 69 Z2 69 Z2 This is a listing according to the 77 elementary divisors 2 2 2 3 4 2 3 8 3 A different listing of the same groups is Z24 Z12 EB Z2 Z5 GBZ2 GBZ2 according to the 77invariant factors 24 12 2 6 2 2 This will be generalized every Dimodule is a direct sum of cyclic primary modules where the p7 are called elementary divisors Every Dimodule is also a direct sum Dq1Dq269 GBDqk where qquk1l lql is the list of invariant factors The proof needs some preparations in form of several technical lemmas Lemma 313 Assume that perz 0 Then one has that 39 7 1001 M amen PROOF Let d cperz Then d f 0 d l c and Perm E 7 d or 7 dperzz 7 0 which is perczlx Now assume for 0 We have d we perz and therefore dfa acfa perzfz 0 which is perzld Now d is a divisor of perz and therefore per d lf This is w percz d D Corollary 314 If cperz 1 then percz perz D 40 Corollary 315 If peTltIgt pT then 1 istT pr s ifs lt T W PROOF The rst case is obvious If s lt T then p5peTz 1051 p5 and T petunia 7 p D Lemma 316 Let M be a p7m0dule with peTM pT7 2 1 Let 11 E M such that peTzl peTM pr Then eveTy coset y of M lt 11 gt M contains an element y such that peTy peT PROOF We rst remark that for every factor module and every y E 3 one has that peT pery This is obvious cy 0 implies that c 6 Let now per p n S T Then pn lt 11 gt which is p y Elt 11 gt or pny aal p5 czl where we assume that p c 1 We need to consider two cases s 2 T We have that p y c 10511 0 and therefore pery p per and by the pervious remark PerlperyA Thus Pery WW 1071 s lt T We have perczl perzl pT by the previous lemma cpT Thus 7 7 pST7s m P i Pipermh 1051 105 perp y perp5 CM We have pery pk where h 2 n and therefore by the second corollary of the last lemma perp y pk Thus pr s pk which is h n T 7 s Now h S T gives n T 7 s S T ien S s The element y y 7 10 czl is the element we are looking for We have yEy Inod ltzlgt ieyE and p y pny 7 p5 czl 0 gives pery p per We trivially have perlpery and therefore pery per D De nition 14 y1 ym are independent if and only if lty1ym gtlt in gt EBEBltymgt iff a1y1amym O alpl 0amym 0 Lemma 317 Let M be ap7m0dule wheTe peTM pT7 2 l and let 11 E M such that peTzl pr Then if E W aTe independent in M lt 11 gt theTe aTe yl E E with peTyl peTE such that foT eveTy 2 Elt z gt one has that z y1 ym aTe independent in M PROOF We choose y1 ym according to the last lemma and assume a2 a1y1 amym 0 Then a2 army 0 We have a 0 because 2 Elt 11 gt But then a1 army 0 which is a 0 amm 0 by assumption This yields perE peryilai therefore aluyi 0 and nally dz 0 D Corollary 318 Assume that M lt 11 gtlt E gt SD 69 lt yim gt Then one has foT suitable yi E E thatMltzlgtlty1gtltymgt D 41 Let M be a module over D and let p be a prime in D Then MP I l perrlp is a vector space over the eld Dp with respect to d dz We need to show that the multiplication is well de ned if d E d mod p then d d a p and dz d z apz d a Lemma 319 Assume that the elements yi7i l7 7 ym 0f Mp are di erent from zero Then one has that y17 7 ym are independent in the Dimodule Mp i y17 7 ym are linearly independent in the vector space Mp over Dp PROOF We rst note that yi 0 is the same as peryi p Assume rst that y17 7ym are independent Then if d1y1 0 one has that dyyi dm m Odlnyi 0 and therefore peryildi This is pldi or 0 in the vector space D over Dp For the converse7 assume that y177ym are linerarly independent in Mp over Dp Then dryl dmym 0 yields d1 E 0p This is diyi 0 because yi E D Lemma 320 Let M be a nitely generated pimodule where perM pT7r 2 1 Let 11 E M where perzl pr Then ClimDp M03 gt CllmDp lt 11 gtp PROOF Mp is as a submodule of a nitely generated module nitely generated A generating set of M over D is a generating set of Mp over D Thus Mp has a nite basis over D M lt 11 gt is as a homomorphic image of M also nitely generated Let now E7 7 be a base of M lt 11 gt p over Dp Then E7 7 are independent over D We pick yi E E where peryi perE and some 2 Elt 11 gt where perz p7 eg7 2 pr lal such that 17 y17 7 ym are independent in M and different from zero All 17 yi belong to Mp and we get that z7y17 7 ym are linearly independent in Mp over Dp This is dimDp MUD 2 dimDpM lt 11 gtP 1 gt dimDpM lt 11 gtP D Theorem 321 Let M be a nitely generated pimodule over the principal ideal domain D Then M is a direct sum of cyclic pimodules Mltzlgtlt1mgt where perzi py 711 2 121m 21andperMpV1 PROOF We proceed by induction on dimDp dimDp Mp 0 is M 0 because otherwise M has an element of power p Let dimDp Mp n and pick 11 E M7perzl perM pm Then dimDp lt 11 gt p lt n By induction hypothesis we get Mltzlgtlt gtmltmgt 42 where perTg perM lt 11 gt pV27perTj phat2 2 2 pm 2 1i Clearly7 12 S 11i We already know that we can pick zj E Tj such that perzj perTj and Mltzlgtmltzmgt We have lt 1 gt2 Dperz and therefore one has for any nitely generated pimodule M M 2 Dpm ea i i i ea Mpm According to Theorem 31 every nitely generated torsion module is the direct sum of its pimodules T171 and therefore one has Theorem 322 Every nitely generated torsion module over the principal ideal domain D is isomor phic to a direct sum of primary cyclic modules M E DWl 69 A A A 69 more 69 A A A 69 IDpr 69 A A A 69 Dp2k kgtgt The p27 are called the elementary divisors of M Note that perM pi nil172 We can also group the cyclic modules Dp according to the invariant factors Let k mmaimiandputvij01fmilt m Z and de ne Cl DPlll EB DPgml 69 A A A GED102 41 10 03 mpg Cz DPim EB DPgml 69 A A A 69 DPZH7 12 pi pg 02 cm Dle EB Dszm 69 A A A 69 DPZWL 4m 1051quot 105 AA A 02 We clearly have that glemill A A A lq2lq174m 1 and the qi are called the invariant factors We are going to show that the submodules C1 are actually cyclici This is a consequence of the Theorem 323 Chinese Remainder Theorem Let dhdgpiqdm be pairwise relatively prime ele ments of the principal ideal domain D Then the system of linear congruences z 311 mod d1 1 E 12 mod d2 1 E rm mod dm has modulo d d1 d2 i H dm a unique solution I 43 PROOFi Assume that rand z are congruent module d1 4 i i dmi This is dllz 7 1 i i dmlz 7 1quot Because the di are pairwise relatively prime we can conclude that dlz 7 1 ie I E 1 mod d In order to show existence of such an I we de ne againam Then d1 and di are relatively primei Hence we have a1 d 121 d1 14 This is al di E 1 mod dli Of course di is zero module d2 4 i i dmi We similarly de ne d as the product of the di with d2 missing We then get an a such that a d E 1 mod d2 etc Then 111a1dizga2diuzmama is obviously a solution B For elements d1 4 i i dm of the principal ideal domain consider the homomorhism D 7gt Ddl gtlt gtlt Ddmz gt gt The kernel of this homomorphism is licimid1 i i 4 dm and according to the Chinese Remainder Theorem the map is surjective in case the di are pairwise relatively primei In this formulation the Chinese Remainder Theorem invites generalization to more general rings and even to universal algebra We also see now that C E Dqi and get Theorem 324 Every nitely generated torsion module over the principal ideal domain D is isomor phic to a direct sum of cyclic modules M gDqINaMDqk where the qi are the invariant factors 35 Uniqueness of the Invariant Factors and Elementary Divi sors We are going to show that the invariant factors determine a nitely generated D7module M up to isomorphism uniquelyi The spaces Mp over Dp are again a basic tooli Lemma 325 Let L a 0 Then for any prime p E D one has 0 otherwise dimp Dltqgtltpgt 7 1 if PROOFi Assume rst that p does not divide qt Then one has that p d 4 iff p d E 4 iff pd cq iff d cp qi This is d E 4 which is d q Now assume that plq that is L p b where I f 04 We have that blq but if we also had qlb then b N L and it would follow that p is a unit Thus 5 4 47 1445 4 4 This is b q E and b q f Now let d q E Then pd cq cpl and therefore d cl which is d q cib Thus I q is a generator U of over D and therefore dimDp p 14 D 44 Lemma 326 Let L ab 0 Then aDq Db as D7modules PROOF ad gt7 d b is the desired isomorphism The map is well de ned adl E ad2q is ad1 7 d2 CL and therefore d1 7 d2 cl7 ie7 d1 E d2b Linearity and surjectivity are obvious Assume d E b7 ie7 d cl Then ad CL 6 q is injectivity D Theorem 327 Let M be a nitely generated torsion module over the principal ideal domain D and let MClCkC 1Ci be two decompositions of M as direct sums of nonzero cyclic modules C and C3 respectively with periods qi and 4 and forming divisor chains Cl 2 Dqi7 and CE 2 ii le ul h and all lql Thenhl andqiwqgforilh PROOF Let p p1 be any prime that divides qk Then p divides 4161 ql Assume that z E Wehave zzlzkpz0pzlpzk and therefore pal pak 0 This is MP 0110 EB a EB Cm and therefore dimDpP k Hence there must be exactly h7rnany q for which pllq This is h S l and by symmetry we conclude h l We also see that pllqlc Hence qk plbk q plbjc and pllqi pllq say qi plbi q plbg ilh By the last lernrna7 le E Db1EBDbk EDbl BHGBDb where bkl lb1 and Now7 a prime p2 which divides bk divides b2 and vice versa7 and we get 01ng E Dc1Dck E Dc l EB where ckllc1 and czl lcl and where we have p1p2ci 4i and 10110262 12 lqup1p5 then qu E Del EB GED1671 E Del EB EB De c where eiqk qi equ q and 1N eklek71lle1 eyegill le l On the lefthand side for the sum decomposition of qu there are at most h 7 1 summands different from zero7 so also on the righthand side Thus e2 N 1 which is qk N 41 45 If we assume the theorem for Dimodules with decompositions into lt h cyclic modules then we get ei N e and qi N L for i 1 he 1 The case It l is trivial Dq E Dq as Dimodules yields Perm4 q perm4 4quot D The invariance of the elementary divisors is now an easy consequence We have that M is a direct sum of its maximal primary submodules and each of these primary modules is a direct sum of cyclic primary modules subject to the divisor chain condition 36 Finitely Generated Abelian Groups Let A be a nitely generated abelian group Then A E Zr 69 TorA where r rankA and TorA E Zml EB EB Zm where mllmhll lml ml gt1 The isomorphism type of A is uniquely determined by r and the invariant factors mi Notice ordTorA m1 U ml perA m1 We also have a unique decomposition according to the elementary divisors TOI A E Zpin G9 GB Zp1t1 GB 69 szm 69 GB ZPZklk The product of all elementary divisors pgij is the order of the torsion and its period is the product of the p for i lh Theorem 328 Kronecker A nite abelian group G of order m is cyclic and only if there are for every divisor d ofm at most d elements I for which 1 1 e PROOF While we have formulated the theorem for multiplicative notation we use for the proof additive notation eg we consider G as a module over Z Let G be cyclic ie G E Zm Let d be a divisor of in Then I l dz 0 is a subgroup H of GNow H is generated by the class and this group is isomorphic to Z01 On the other hand if G is not cyclic then one of the primary components say the one for the prime p1 is not cyclic ie Tpl E prm 69 prm 69 and we have in the rst component at least plimany elements of order p1 but also in the second component That is G contains for the divisor p1 of the order m at least 2p17many elements of order p1 D Corollary 329 Gauss The multiplicative group of a nite eld is cyclic PROOF This is now obvious rd l is the same as 1 1 7 l 0 and there are at most d solutions in F D Recall that if F is a nite eld of order m then m p where p is the characteristic of F 46 37 The Structure of a Linear map over a Finite Dimensional Vector Space Let F be a eld and let be the ring of polynomials over Fl Let A be an Fz7modulei Then VA A n oy Clear is a vector space over F and 73975 VA AVAJ gt gtziv is a linear transformation on VAi Thus we have a correspondence A gt gt VA 739 from the class of Fz7modules to pairs consisting of a vector space V over F and linear maps 739 on Vi On the other hand let V 739 be such a pair For a polynomial E de ne HIM fTv This makes V to an Fz7module A and it is easy to see that the correspondence A gt gt VAT is bijectivei Let 45 VA A VA be linear Assume that 45 is also Fmilineari Then one has that which is T U or ab 0 739 7quot o qb In particular if 45 is an isomorphism between Fmimodules then lt15 0 739 o 1571 7 shows that 739 and 7quot are similar Let W be a subspace of VA Then W is an Fz7submodule of A if and only if 1 1 6 W in case that v E W and this is that 71 6 W for any 1 E W or that W is Tii vllTill t Let C be a cyclic submodule of the Fz7module A C lt v gtFM l f E lt Tliv l 1 E N gt117 and one has FlIlPerv g C as Flrlimodules where perv is the polynomial of smallest degree such that 727411 0 or mm 0 in case that Ci We assume from now on that mm has a degree ie that it is different from zero and let degmv it Of course 1 0 is equivalent to n degmv 0 m1 is called the minimal polynomial of vi Of course we may always assume that mm is monic m1 a0 alz i i i an1 znn degm4i n71 Now is also a vector space over F and as such has a base 1 i i z where denotes the congruence class modulo my in the polynomial ring For C this means that v739vui739n 1v 47 is a base of lt v gtFM over F1 The map 739 restricted to the cyclic space C has with respect to the base v 7v111 Tn 1v the matrix 0 0 1 0 fag 1 0 0 7a1 TD 0 1 0 fag 0 0 1 7an1 and T v 7a01v 7 a11739v 7 111an711rn 1v Theorem 330 Let V be a nite dimensional vector space over the eld F and let 739 V 7gt V be a linear transformation on V Then V is a direct sum of cyclic invariant subspaces Let m1 be the minimal polynomial of 739 Then there is a list mzltzgt1mzilltzgt11111m1ltzgt7degltmzgt gt o of monic polynomials each one dividing the next such that V7 7 2 F1z1ltm1ltzgtgteamea Flu72111 The list of invariant factors determines the similarity class UfT uniquely One has dimV degm1 111 degml and 71111 minimal polynomial of 739 One can nd in V a base such that the matrix T UfT is in rational canonical form T1 1 1 1 T T2 1 1 1 11 The matrices TV TUV are as before and all entries outside these blocks are zero For 739 V 7gt V and matrix T for 739 one de nes detT 7 111E Cz E as the characteristic polynomual for T1 Here 1 0 0 0 1 0 E 0 0 1 is the unit n X n7matrix1 The polynomial Cz does not depend on the chosen base for V1 If T is the matrix for 739 in any other base then one has some nonsingular matrix S namely the matrix which encodes the coordinate transformation between the two bases such that T S o T o S 1 and detT 7 detSTS 1 7 detST 7 11ES 1 detT 7 If V V1 6911 1 EB Vk is a direct sum of 77invariant subspaces then Cz c1z 1 11 ck where is the characteristic polynomial of 739 restricted to Vi1 For invariant subspaces C that are cyclic ie C lt v gtF lt v739v111739 1v gtF 739 has for C a matrix like TU ie a matrix that has units on the lower diagonal and the the last column is given by coef cients 7ai of the minimal polynomial my of v1 It is a rather easy exercise to prove that the characteristic polynomial of such a matrix is irm 48 Theorem 331 HamiltonCayley Let c cr be the characteristic polynomial for 739 V 7gt V Then cim1ml where mll lml is the list of invariant factors for 739 Corollary 332 The minimal polynomial in ml for 739 divides the characteristic polynomial ie CT 0 Corollary 333 The characteristic and the minimal polynomial for 739 have the same prime factors galc Theorem 334 Let 739 V 7gt V be a linear transformation on the nite dimensional vector space Let in 1011 1 Pk 1 he the prime factorization of the minimal polynomial 72171 for 739 Then V 739 Tp1739 EB Tpk 739 where 11717 v l mvzlpizT for some r 21 v l piTT U 0 for some r 21 and each Tp1739 is a direct sum of cyclic piz7modules Tp1739 Fzpizy 1 EB EB Fzpiz z 1 2 H 2 Dill 21 where the list of elementary divisors gal1 determines the similarity type of 739 uniquely Of particular interest is the case where F is algebraically closed Then 101 is prime if and only if 101 z 7 A for some A E F An example is F C where C is the eld of complex numbers Let C be a cyclic invariant subspace ie C E 7 Ml If C has v as a cyclic generator then v 7v Tl l v is a base of C We claim that also v0 vv1 739 7 Avv2 739 7 A2v vl1 739 7 Al 1v is a base Clearly C is invariant under 739 7 A and therefore all vi belong to C Hence we only have to show that they are linearly independent Assume that p0v p1739 7 Av pl71739 7 Al 1v 0 If we multiply this relation by 739 7Al 1 then we get pg 7397 Al 1v 0 and this yields p0 0 because the degree of the minimal polynomial n74 I 7 Ml is l We get p1 H M71 0 similarly Now 7v0 v1 Ave and v2 739 7 AT 7 Av 739 7 Av1 Tv1 v2 Av1 and v3 739 7 AT 7 A2v 739 7 Av2 7v2 v3 Avg and v4 H Tvl2 vl1 Avl2 and 739 7 Avl1 739 7 Alv 0 Tvl1 Avl1 With respect to this basis the matrix of the map 739 on C looks like A 0 0 1 x 0 0 EM 0 0 x The elementary Jordan matrix EM has l7many A s on the main diagonal and l 7 1 units on the lower diagonal lt corresponds to the polynomial I 7 Ml 49 Corollary 335 Jordan Normal Form If the minimal polynomial of 739 splits in into linear factors I 7 Aii 17 i i 7 h ie z7A1 1 I7Ak k Then one can nd in V a base such that the matrix T UfT is in Jordan Normal form EA1V11 T EMWI 7W W1 2 2 Dill EAWW The primary component for I 7 Ai is the space T057 1 l T7AiT 0 for some r 2 0 vl T7Ai 0ni Di1i 1411 dimTTl and its minimal and characteristic polynomial for 739 are mdr I AVE CM I i A For A E F we de ne EA vl 739 7 Av 0 ker7 7 A as the eigenspace for Al Clearly7 EA Q TEi lf EA D then A is called an eigenvalue for 739 and TE is called a generalized eigenspacei We have that A 0 is an eigenvalue iff 739 is singular We have TEI EAI ifl 1 l 739 7 AiTv 0 for some r 21vl 739 7 Aiv 0 iff 1 1 iff z 7 Ai is the minimal polynomial of 739 restricted to T057 ifl T I has an eigenbase iff the matrix T1 of 739 restricted to T057 is an ni gtlt ni diagonal matrix With Ai on the diagonal Corollary 336 Assume that the minimal polynomial UfT splits into linear factors I 7 Ai Then the following statements are equivalent V has a base of eigenvectors for 739 739 admits a representation by a diagonal matrix The minimal polynomial UfT is without multiple roots ie mzz7A1iiiI7Aki iv The characteristic polynomial UfT is Cz z 7 A1 1 I 7 A116 7 where ni dimEii 50 Chapter 4 Sylow Theorems 41 Transformation Groups Let S be any set The group of bijective mappings on S is sometimes called the symmetric group SymS on 5 Any subgroup G of SymS is called a transformation group on the set S and we say that G acts on 5 Examples 1 Let S n 07 i i 7 n 7 l Symn is called the symmetry group of degree n 2 Let G be any group We already established the injective group homomorphism A G A SymGr gt gt A 7A y z y Every group G acts on its carrier set by means of lefttranslations and G 2 MG Q SymGl This7 of course7 is Cayleyls theoreml The group G acts on itself by left translations 3 Let G be any group We already established the group homomorphism a G HSymGz gt gta a y zyITI The kernel of the homomorphism a is centGl Each 075 is an inner automorphism of G and G acts on itself by conjugation We have that G centG AutiGl De nition 15 A representation of the group G by transformations on a set S is a homomorphism p G A SymS De nition 16 For a representation p G A SymS the equivalence under p is an equivalence on S and de ned by a Np 1 iff a pzb for some I E G The equivalence classes for Np are called the orbits of pi We obviously have that the orbit of a under p is the set orbpa pzla l I E G and it is also called the trace of a under pl For a E S the set stabp I l pza a is called the stabilizer Ufa under p It is a subgroup of Gr 51 Examples 1 We get for the representation A of a subgroup H of G that aNAbiffahbforsomehEHiffaEHb The orbits are the rightcosets of H in Gi The stabilizer is the trivial group e for every a 6 Ci 2 For conjugation we get that aNUbiffazbz 1 for some I E G and the orbits are the conjugacy classes 12 b zaz hz E Ga 6 G7 of Gi The conjugacy class of a is the singleton a iff a E centGi We have stabUaz Ga aazEGzaz 1az GzaazcentGa Where centGa is called the centralize of a in Gi It is the set of all z E G Which commute With a The conjugacy class of a is the singleton a iff a E centG iff centG a Gi This is a special case of the next theoremi Theorem 41 Let G stabpa be the set of leftcosets 0f stabpa in G Then one has that G stabpltagt a orb7a z stabpltagt e pltzgtltagt is Z well de ned bijective map PROOFi We have that 11 stabpa 12 stabpa iff Iglzl E stabpa iff iff pzg 1pzla a iff pzla pzg a This proves that the map is well de ned and injectivei For surjectivity7 let I E orb7a This is b Mr a for some I E Gi But then zstabpa gt gt Mr a 12 D Corollary 42 Counting Lemma One has for every a S S that cardorbp a cardG stabp a D Recall that for any subgroup H of G7 cardGH is called the index G In case of a nite group G of order n and subgroup H of order m one has that nm G Let G act on S by pi The orbits of p partition the set 5 Let A be a complete set of representatives for the partition into orbitsi Then S w orbpa and cardS Z cardorbpa Z cardG stabpa ISA ISA ISA We apply this equation to the action by conjugacy on a nite group and notice that the center is made up by the elements of the center Corollary 43 ClassEquation Let G be a nite group Then 0TdG 0TdcentG Z G centGa ISA where A is Z complete set of nonconjugate elements which 1T6 not in the center 52 As an immediate consequence of the class equation we get Theorem 44 Any nite group of prime power has a nontrivial center PROOFi Each term G centGaa E A is divisible by p and therefore ordcentG must be divisible by p B Any action p G A SymS induces also an action on 735 by means of w G A Symlt73ltsgtgt7z e A H mm pltzgtltagt l a e ALA g s It is easy to see that pz y A Mr that is7 p is a group homomorphismi For every subset A of S 0rbpA PIA l I E G consists of all subsets B of S that are similar to A with respect to or Of course7 cardB cardA for every B E orbw For every subset A Q S stabw A z E G l pzA A is the subgroup of all z E G for which Mr permutes Al Let now a G A SymG be the representation of G by inner automorphisms Note that for every subgroup H of G the image of H under a is also a subgroup of Go Therefore7 0 G A SymSubGz gt gt 0H gt gt sz 1 is a representation of Go One has that orb0H K l K sz 1 for some I E G is the system of all subgroups K of G which are conjugate to Hi One has that H is normal i orb0H H The stabilizer of H in G is stabw z E G l sz 1 H and is called the normalizer NH of Hi Note a N 2 H b H is normal in NH C Let K 2 H be any subgroup of G in which H is normal Then sz l H for all z E Ki But then K E The normalizer NH of H is the largest subgroup of G in which H is normal Of course7 NH G iff H is normal The Counting Lemma 4 now yields cardK l K conjugate to H cardGNH If G is nite then the number of subgroups K that are conjugate to H equals the index of the normalizer NH in G 7 in particular this number is a divisor of Go Theorem 45 Isomorphism Lemma Let H and K be subgroups of the group G Assume that H E Then one has 53 a HKhhhEHhEK isosubgroup ofG b K is normal in H K and H N K is normal in H c H KK g HH K PROOFi We have for any h E H that hKh 1 K ilel hh hhi Thus hhfl h lh l h lh E HKi Similarly hh h h hh HUM E HKi This proves a We have that hhhhh 1 hhhh lh l hhHh 1 E Ki This proves that K is normal in HKi Let l E H Ki Then m hlh 1 is in K because 1 E K and H is within the normalizer of K m E H because 1 6 Hi Thus m is in H N K and this proves Finally H A HK K h gt gt hK is surjective because hhK hK and clearly homomorphici We have hK K iff h E K ilel h E H N Ki This proves c by the homomorphism theoremi D A special case is where K is a normal subgroup N of Gr Then HNN 2 HH n N This is sometimes called the second isomorphism theoremi 42 Sylow Theorems Let G be a nite group of order n pl plinmi As a partial converse of Lagrangels theorem we have Theorem 46 If the prime power pk divides the order n of the nite group G then G contains a subgroup of order pk PROOFi We prove this by induction on the order n of the group Gr Of course n 1 and n 2 are trivial Assume that G has order n and that the theorem is true for all groups of order lt n We will use the class equation ordG ordcentG Z G centGa aEA Assume rst that p does not divide ordcentGi Because p divides ordG one has some a E A such that p does not divide G centGai But ordG ordcentGa G centGa and because pk divides ordG one has that pk divides ordcentGai Now ordcentGa lt n because a centGi Hence centGa contains a subgroup of order pk by induction hypothesis Now assume that p does divide ordcentGi As an abelian group centG is the direct sum of its primary components and clearly centG contains an element 6 of order p We can form the factor group G lt c gt which has order np and therefore pk 1 divides ordG lt c gti By induction hypothesis G lt c gt contains a subgroup H of order pk li Under the canonical projection 15 G a G lt c gt the counter image 1H is a subgroup H 2lt c gt and H lt c gt Hi We have that ordH ordlt c gt ordH lt c gt p pk 1 pk 54 Let pm be the largest power of p that divides n ordGl Then G contains a subgroup of order pml Any such subgroup is called a pSylow subgroup of Go They are obviously maximal amongst pisubgropupsilel subgroups whose order is a power of p In the abelian case there is only one p7Sylow subgroup namely the primary component Tp Theorem 47 Ludwig Sylow 18321910 Let G be a nite group Then i If H is a pisubgroup of G then H is contained in some pisylow subgroup ii All pisylow subgroups are conjugate to each other iii The numberl of pisylow subgroups is congruent to 1 modulo p and a divisor of ordG hence 1orplor2plpuandllnl PROOFl Let Mp be the set of all p7Sylow subgroups of G and let G act on MI by conjugation a G A SymMpaz P gt gt sz 1 Let P0 6 Mpl Then orbUP0 G1 NP0 where NP0 2 Po Hence lorbaPollIP0171lI 6 GH is not divisible by p Now let H be any pisubgroup of Go Also H acts on orbUP0 by conjugation O39H H A SymorbUP0aH h P gt gt hPh 1 and orbUP0 orbUH P0 L L orbUH P191 each of the orbUH Py divides ordH and therefore they are powers of pi But p does not divide the sum orbUP0 and therefore orbUH PV R for some 1 or equivalently hPVh 1 PV for all h 6 Hi This is H E NPl By the lsomorphism Lemma 45 K PV HPVP g HH PV The group on the righthand side is a group whose order is a power of p The same must hold for the lefthand side and we conclude that ordHP ordHH N R ordP Hence HPV has an order which is a power of pi But PV is a maximal pigroup and therefore HPV Pyl This is H Q R and proves We saw that for any p7Sylow group P0 and any pisubgroup H one has a p7Sylow group PV which is conjugate to P0 such that H E Pyl If we take now for H any p7Sylow group P then we see that P is contained in a Sylow group PV which is conjugate to Pol But then by maximality PV Pi Thus all p7Sylow groups are conjugatel This proves ii 55 Thus we have for any p7Sylow group P0 and any pisubgroup H orbUP0 Mp orbUH P0 L L orbUH P1671 and H Q R for some 1 and orbUH R for all such 1 If we take H P0 then P0 B only for 1 0 and therefore powers of p Of course lln because 1 is an index This proves in D Examples 1 No group of order 20 is simpler Any such group has exactly one Sylow subgroup of order 5 ie a nontriVial normal subgroup 2 No group of order 30 is simpler Here the argument is somewhat more complicated We have that n 2 3 5 The possible numbers of Sylow subgroups for p 2 are l 5 and 15 for p 3 l and 10 for p 5 l and 6 All these Sylow subgroups are of prime order thus cyclic and every element different from 6 is a generator Thus the intersection of of any two different Sylow groups is 6 Hence any simple group of order 30 must have at least 4 elements of order 2 20 10 2 elements of order 3 and 24 6 4 elements of order 5 Of course this is impossible Hence any group of order 30 has for some p 2 or 3 or 5 a normal Sylow subgroupi 56 Chapter 5 Some Universal Constructions 51 Peano Algebras Let V be a variety of similar algebras of type Al An algebra of V is called the free Valgebra freely generated by M if any map go from M into any algebra A in V admits an extension to a homomorphism go from into Al Because we assume that M generates FM the extension is unique A free algebra is uniquely determined up to an isomorphism over the generating set Mi Examples ll Let V be the variety of vector spaces over the eld E Because every vector space has a base they are all free 2 Let M be the variety of modules over the principal ideal domain Di For the set M DM is free freely generated by the Mmany unit vectors We are going to show that varieties admit for any set M free algebras Actually this holds even for quasivarieties of algebras Recall a quasivariety is a class of algebras that is abstract is closed under isomorphic copies and closed under subalgebras and direct products A quasivariety is non trivial if it contains algebras with more than one element The direct product of the empty family of algbras of a given type A is the one element algebra with 0 as carrier seti Before we prove the existence of free algebras for arbitrary quasivarieties we prove the existence of free algebras for the class VA of all algebras of type Al Theorem 51 For any set M and algebraic type A nii6 there is an algebra P PM A such that the generalized Peano axioms hold P1 fia0iuam1 M P21 fia07ian11 fjboi Hbn71 only ifi j and ay by P3 The set M generates P PROOFi We rst remark that we get the familiar Peano axioms for the natural numbers N for the case where the type contains only one unary operation called the successor and where M is a one element set Then Pl is n f 0 P2 is n 721 only if n m and P3 is inductioni For the proof we take a set F of operation symbols where I A F is a bijection We also assume that F and M are disjoint We de ne the algebra P as the algebra of terms Terms are de ned as the smallest set which is closed under the following conditions 57 a Each element m E M is a term b If t0tn1 are terms and if is an n 7 ary operation symbol ie ni n then t fit0 tn71 is a term In particular if ni 0 then is a term According to property b P is an algebra of type A and the Peano Axioms are easily veri ed D Theorem 52 The Peano algebra PM A is the free algebra freely generated by M for the variety of all algebras of type A PROOF The proof is straight forward on the complexity of terms D We can de ne the Peano Algebra also as the set of wellformed words Let W be the set of all strings of elements over the set M U F If w0 wn1 are words and if f is an niary operation symbol then w fiwo wn1 is a word This way W becomes an algebra of type A The subalgebra that is generated by the set M is the algebra of terms ie the Peano algebra over M We can single out the terms amongst all words by a simple algorithm De ne the valency function on the algebra of words to the integers by a ini l vm 1 b For a word w de ne v as the sum of valencies of its letters c E M U F One has the following theorem of R Hall Theorem 53 A word w C1C2 Cu is a term and only a vltwgt 1 b vsi gt 0 for every right segment si ci cn Moreover w is a product of remany terms if and only the rst condition is replaced by vw r The proof is an interesting exercise Theorem 54 Let Q be any nontrivial quasiprimitve class of algebras of type A Then for any set M there exists a free Qialgebra freely generated by M PROOF Let 5 be the set of all congruences E of P PM A Where the factor algebra A is a member of Q The family of projections QEIPHAEE68 leads to a homomorphism q P a A H A E E65 such that pE 0 q 4E and where p are the canonical projections from the direct product A to the factors AE We have that kerq E0 m E E65 58 seperates the points of Mr That is if a and b are different points of M then ab Eoi lndeed let A be any algebra in Q with at least two elements a and Iquot Then choose any map so from M into A such that a a and b bquot For the kernel E of the homomorphic extension of go we have that ab E and therefore a 12 Eoi We have that the image B imq of q is a subalgebra of A which is a direct product of algebras in Q Thus B is an algebra that belongs to Q Because B 2 PEO AEO we have that AEo is an algebra that belongs to Q We claim that AEo is the free Qialgebra freely generated by the set M E M of equivalence classes of elements in Mr lndeed let 6 be any map from M into an algebra C E Q Because an equivalence class for E0 cannot have more than one element of M in it we get a map g0 M A C for which m The homomorphic extension sf has a kernel E 2 E0 and therefore we have a unique homomorphism 6 AEO A C such that 6 o 4E0 sf Replacing the by m proves our claimi D We mention some important facts and leave the proofs as exercisesi 1 If N E M then the subalgebra of FM Q that is generated by N is FN Qi Of course this does not imply that a subalgebra of a free algebra is free 2 A subalgebra of a Peano Algebra is a Peano Algebra 3 Let M L be a partition of the set M into disjoint subsets NE of Mr Then LET it FNt Q A FM Q is a coproduct system in Qi The it are the homomorphic extensions of the inclusions NE lt gt M and are injectivei In particular the free algebra FM Q is the coproduct of free Qialgebras that are freely generated by one element sets 4 The free Qialgebra freely generated by the empty set is an initial object of Qi That is for every algebra A E Q there is exactly one map from FUD Q into Al This map is the homomorphic extension of the empty map from 0 into Al For the class of groups 6 is initial For the class of rings it is Z If the type does not contain constants then the empty set is an algebra in Q and it must be the initial algebra of Qi 5 A trivial quasiprimitive class contains at any rate the oneelement algebra as the empty product It contains the empty set if the type is without constantsi If the type is without constants then the empty set is the free Qialgebra otherwise it is the oneelement algebrai Theorem 55 A quasiprimitive class of algebras admits sums PROOFi Let Q be quasiprimitive and let A A fii61 be given We form M Um X A ta Henley LET 59 and extend the type A of Q by an Miindexed set of constants N 71ernzazaeM7nza 0 An algebra of type A is of the form A A7 CtataeM Where A is of type A and Ct 6 Al 90 1 A7 Cam A 137 dam is a homomorphism of A7algebras if g0 A A B is a A7 homomorphism and 906ta dam La 6 M A subalgebra of A is a subalgebra B of A that contains all Ct 6 Mi The direct product of algebras Bs E S is given by 13 B7 Cfta565ta6M Where B is the direct product of the algebras le We call an algebra B B7 CtataeM special for Ant 6 T if B 6 Q7 and got At A Ba gt gt Ct are Aihomomorpicl If Ci is a constant that belongs to the type A With values c 6 A then one must have in B 80402 CiB Ctcf7t E T and zfao7anri fiBCtaow76tan11 Ctff 10an117lE T Which is CiB Cmg and fiBCtaov76tan11 Ctffaoan117t E T That is7 an algebra B is special for the family At iff B E Q and these equations hold between the constants emu La 6 Mi It is therefore obvious that 8 B lB special for Ant 6 T is a quasiprimitive class of algebrasl It is primitive7 in case that Q is Now let A Fab 8 be the free 8 ialgebra freely generated by the empty set The algebra A exists7 even if 8 is trivial For A A7 CtataeM we have A E Q and it At A A7 a gt gt Ctat 6 T7 are homomorphic We claim that Ait AL A A is a sum system in Ql 60 Let ft A A B be a family of homomorphisms in Qi Then one has that 13 By Cm ftata6M E 3 Because A is free there is a unique homomorphism f z Aquot A BCta gt gt fta t a E M Which is fita fta for all t E T and a E A Thus foil fhteTi 52 Ultraproducts Let T be any set A collection 1 of subsets of T is called a lter if the following holds a lfAEfandBEfthenA BEfi b lfAEfandifBQAthenBEfi c T 6 f A lter 1 is proper if d 25 1 Let T be an in nite set Then the system of all co nite subsets S of T ie the sets for Which the complement T S is nite is a proper lter It is called the Fr chet lter on T For a topological space X and a point z E X the sets N Which are neighborhoods of 1 form a lter De nition 17 A subset K of the lter 1 is called a base for f if for every A E b one has some B E 6 such that A 2 B The open sets that contain 1 form a base of the neighborhood lter For the set N the sets Nk 2 16 form a base of the Fr chet lteri Let 5 be a system of subsets of Ti Then lt5gtVlVQSl 3kkENSi65 is a lter It is a proper lter if and only if 5 has the nite intersection property A lter 1 is principal if it is generated by one of its sets ie f lt S gt i For a nite set T every lter is principali More generally a lter that contains a nite set has to be principali De nition 18 A proper lter 11 is called an ultra lter if for every subset S of T one has that either 5 or its complement T 5 belongs to 11 Filters that have a singleton as a base are certainly ultra ltersi They are the trivial ultra ltersi Filters are partially ordered by inclusion Because the union of a chain of proper lters is a proper lter we have according to Zorn s Lemma that every lter 1 is contained in a maximal lter mi 61 Theorem 56 A lter is maximal if and only it is an ultra lter PROOFl Assume that u is an ultra lter and that f 2 u where f is proper Then for every 5 E 1 one has that S E 11 because otherwise one would have T S E u and therefore 5 N T S ll 6 Thus 11 is maximal Now assume that m is maximal and that 5 ml Then because m is maximal m U 5 cannot have the nite intersection property This is S F Z for some F 6 ml But this is F E T S and therefore T S 6 ml Hence m is an ultra lterl D Ultra lters are sometimes called prime lters Recall that an ideal P in a ring is prime if ab 6 P only if a E P or b 6 Pl Filters are the dual to ideals Thus a lter p should be called prime if A U B E p only if A E p or B 6 pl We then have that a lter is a prime lter if and only if it is an ultra lterl Indeed a prime lter p is an ultra lter because A U T A T E p so A E p or T A 6 pl Now assume for the ultra lter u that S 51 U 52 6 11 If we had 51 11 and 52 11 then TSl TSg T 51 U52 6 ul Theorem 57 Every proper lter 1 is the intersection of all maximal lters In which contain In particular for an in nite set the Frechet lter c is the intersection of all proper ultra lters PROOFl If the set S does not belong to the proper lter 1 then fU T S has the nite intersection property Thus there is a maximal lter which contains T S and f and therefore not 5quot Let u be a proper ultra lterl Because it cannot contain a nite subset of T it must contain L D De nition 19 Let Att E T be a family of sets and let A H A teT be their direct product Then for any lter 1 on T one has that aaz 3501 i itlazbz l is an equivalence relation on Al The set Af0 Canaan of equivalence classes is called the 1 7 reduced product of the sets All It is quite easy to see that Ef is an equivalence on Al Re exivity holds because T E f and symmetry is totally obviousl If 51 t l at 12 E f and if 52 t l b ct E 1 then Sitlazcz2ftlazbzandbzcz51 52 l and this shows transitivityl For f T the reduced product becomes the ordinary product If f lt t gt then the reduced product is equivalent to the factor A Assume that the A are algebras of a xed type A and let f be one of the operations de ned on every All Assume for simplicity that thea rity of f is two We then de ne on the reduced product an operation f by flal7 l l lfa75l This is a wellde ned operationl Assume that a E 06 E Bquot We then have that 51 t l at ot E f and that 52 t l But then on S 51 N 52 one has that a 0 and and therefore fat t fot tl This is fa E fa l We have on the reduced product the operations de ned in a way that a gt gt a is a homomorphisml Thus we have 62 Theorem 58 Primitive classes are closed under reduced products If the algebras At carry also a relational structure of type A ie on every algebra relations Rj of arity mj are de ned then one de nes on the reduced product a relation similarly as we have done for equality For example if R Rj is binary then Rja should hold if and only if t l Rat t holds on At 6 Again it is easy to see that this is a proper de nitioni We have already de ned the algebra of terms over a set Mi In order to de ne elementary propositions we take for M a countable set X of variables 11 12 l l H Atomic formulas are de ned inductively a t s is atomic for terms s and t b Let R Rj be an miary relational symbol and let t1 l l l tm be terms Then Rt1i l l tm is atomici Formulas then are de ned according to the following rules Of course atomic formulas are formulas And then C If a and B are formulas then a VB 1 AB W are formulas d If a is a formula then Vza 31a are formulas A formula without free variables is called a rst order sentence p A sentence is either true or false One now has the famous Theorem 59 Ultraproduct Theorem A sentence holds in an ultraproduct ALl if an only if it holds in 117 almost all At That a property holds for fialrnost all t E T means that the subset S of those t for which the property holds is an element of The proof of the ultraproduct theorem goes by induction on the complexity of formulas and is omitted As a typical example one might want to prove directly that an ultraproduct of elds is a eld If we take for T the set N of natural numbers then an ultrapower R of R is a totally ordered eld that contains a copy of R The map d1 a H a anneNl is an order embedding The eld R is nonarchiInediani We have that w nneNl gt Ml for every k E N and is therefore an in nite hypernatural numberi ltls multiplicative inverse is an in nitesimal ie a number that is smaller than any ordinary positive real number Any ultrapower of the reals can serve as a simple model for a calculus with in nitesimals 63 Bibliography 1 Hungerford7 Thomas W7 Algelmz7 Graduate Texts in Mathematics 74 Springer Verlag7 New York7 Heidelberg7 Berlin 2 Jacobson7 Nathan7 Basic Algelmz7 WiHi Freeman and Company7 San Francisco 64

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