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# Algebra I MATH 200

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This 74 page Class Notes was uploaded by Sienna Bradtke III on Monday September 7, 2015. The Class Notes belongs to MATH 200 at University of California - Santa Cruz taught by Robert Boltje in Fall. Since its upload, it has received 18 views. For similar materials see /class/182209/math-200-university-of-california-santa-cruz in Mathematics (M) at University of California - Santa Cruz.

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Algebra I Math 200 Groups and Rings Robert Boltje UCSC Fall 2008 Chapter I Groups 1 Semigroups and Monoids 11 De nition Let S be a set a A binary operation on S is a map I S x S a S Usually bzy is abbreviated by my z y z y z o y z o y z y etc b Let x y H z y be a binary operation on S i is called associative if z y z z gtk y gtk z for all myz E S ii is called commutative if z y y z for all z y E S iii An element e E S is called a left resp right identity if e z z resp z e z for all z E S It is called an identity element if it is a left and right identity c S together with a binary operation is called a semigmup if is associative A semigroup M is called a monoid if it has an identity element 12 Examples a Addition resp multiplication on No 0 1 2 is a binary operation which is associative and commutative The element 0 resp 1 is an identity element Hence N0 and N0 are commutative monoids N 1 2 together with addition is a commutative semi group but not a monoid N is a commutative monoid b Let X be a set and denote by X the set of its subsets its power set Then PX U and PX O are commutative monoids with respec tive identities Q and X c z y m y2 de nes a binary operation on Q which is commu tative but not associative Verify d Let X be a set Then composition f g gt gt fog is a binary operation on the set FX X of all functions X a X FXX o is a monoid with the identity map idX X a X z gt gt x as identity element In general it is not commutative Verify 13 Remark Sometimes a binary operation is given by a table ofthe form For instance the binary operation 77and77 on the set true false can be depicted as true false true false false false false 14 Remark Let S 9 be a semigroup and let m1 mn E S One de nes m1 gtk 2 zn 1 m2 zn Note that this element equals the element that one obtains by any other choice of setting the parentheses This can be proved by induction on n We omit the proof 15 Proposition Let S be a set with a binary operation Ife E S is a left identity and f E S is a right identity then e f In particular there exists at most one identity element in S Proof Since e is a left identity we have e f 1 But since 1 is a right identity we also have e f e Thus e e f f 16 Remark ldentity elements are usually denoted by 1 resp 0 if the binary operation is denoted by 0 o resp 17 De nition Let M 9 be a monoid and let x E M An element y E M is called a left resp right inverse of x if y z 1 resp z y 1 If y is a left and right inverse of m then y is called an inverse of m If x has an inverse we call m an invertible element of M 18 Proposition Let M be a monoid and let x E M Ify E M is a left inverse ofm and z E M is a right inverse of m then y 2 In particular every element ofM has at most one inverse Proof Wehaveyy1yzzyzz1zz l 19 Remark If x is an invertible element in a monoid then we denote its unique inverse by m 1 resp ix if the binary operation is denoted by 9 0 o resp 110 Example Let M a b labeEZ 397 0 e 7 39 3 M is a non commutative monoid under matrix multiplication The element 3 has an inverse if and only if 10 6 i1 In this case7 one has 71 a b 7 a iabc 0 c 7 0 c 39 Verify 111 Proposition Let M7 9 be a monoid and let 71 6 M a Ifz is invertible then also z 1 is invertible with m 1 1 m b If x and y are invertible then also z y is invertible with inverse 1 gtk 71 0 The identity element 1 is invertible with 1 1 1 Proof a Since z m l 1 z l z the element x is a left and right inverse of m 1 b We have myy 1m 1 myy 1z 1 myy 1z m 1 m m z 1 17 and similarly7 we have y l z 1m y 1 This implies that y 1 m 1 is a left and right inverse of z y c This follows from the equation 1 1 1 l 112 De nition In a semigroup S we set x z z n factors for any x E S and n E N If S is a monoid we also de ne 0 1 for all z E S If additionally z is invertible7 we de ne m m m 1 n factors for any n E N 113 Remark For an element x in a monoid we have the usual rules mm gtk m xmn and mm for all m7 n E N If x is invertible7 these rules hold for all m7 n E Z This can be proved by distinguishing the cases that m7 n are positive7 negative or equal to 0 2 Groups From now on through the rest of this chapter we will usually write abstract binary operations in the form x y gt gt my 21 De nition A group G is a monoid in which every element is invertible A group is called abelian if it is commutative The order of G is the number of elements of G It is denoted by 22 Remark If G is a semigroup with a left resp right identity e in which every element has a left resp right inverse with respect to e then G is a group cf Homework 23 Examples a Z Q R C are abelian groups but NO is not b QO RO CO are abelian groups but ZO and Q are not c and 0 are groups of order 1 A group of order 1 is called a triuidl group d For any set X the set SymX f X a X l f is bijective is a group under composition It is called the symmetric group on X lts elements are called permutations ofX lf le n then lSymXl 71 We just write Symn instead of Sym1 2 n and call Symn the symmetric group of degree n We use the following notation for 7139 E Symn 1 2 n 12 3 7rlt7r1 7r2 7rgtv 597Wlt2 1 3gt Sym3 e If G1 G2 Gn are groups then also their direct product G1 x G2 x x Gn is a group under 9017 7ny17 Mm 901917 796mm f For every n E N the sets GLAQ GLAR GLMC of invertible matrices form groups under multiplication 24 De nition Let G and H be groups A map f G a H is called a homomorphism if fzy fzfy for all zy E G The set of all ho momorphisms from G to H is denoted by HomG A homomorphism f G HH is called a a monomorphism if f is injective b an epimoijohism if f is surjective c an isomoijnhism if f is bijective d an endomomhism if G H e an automoijohism if G H and f is bijective AAAA 25 Remark Let f G H H be a homomorphism between groups G and H Then lo 1H and fz 1 fz 1 for all z E G Moreover if also 9 H H K is a homomorphism between H and a group K then 9 o f G H K is a homomorphism If f G H H is an isomorphism then also the inverse function f l H H G is The automorphisms f G H G form again a group under composition called the automoijohism group of G and denoted by AutG 26 Examples a For each n E N the map Z H Z k H nk is a monomorphism b R H R4 x H em is an isomorphism c Let G be a group and let 9 E G Then 09 G H G x H gmg is an automorphism of G with inverse 0971 One calls 09 the inner automoijohism induced by g or conjugation by 9 Note that cg 0 ch cg for 971 6 G Thus G H AutG g H 09 is a group homomorphism d For each n E N the sign map 1 sgn Symn H i17 7 H H 1ltiltiltn 7 2 is a homomorphism see homework lf sgn7r 1 resp sgn7r 71 then we call 7139 an euen resp odd permutation e For every n E N the determinant map det GLAR H R 0 is an epimorphism 27 De nition Two groups G and H are called isomoiphio if there exists an isomorphism f G H H In this case we write G E H 28 Remark a The relation E is isomorphic to is an equivalence rela tion ie for groups G HK we have i G E G ii If G E H then H E G iii lfGEHandHEK then GEK b lsomorphic groups G and H behave identically in all respects In fact every statement about G can be translated into a statement about H by an isomorphism and vice versa In other words G and H are basically the same group one arises from the other by renaming the elements but keeping the multiplication de nition 29 De nition Let G be a group A non empty subset H of G is called a subgroup if for any zy E H also my 1 6 H We write H g G if H is a subgroup of G A subgroup H of G is called a proper subgroup if H 7 G In this case we write H lt G 210 Remark a Let H be a subgroup of a group G Since H 7 0 there exists some element y E H Then also 10 yy 1 6 H Moreover for every y E H also y l 1 y 1 E H Therefore with zy E H we also have my 1 E H and then my my 1 1 E H That shows that H is again a group with the original binary operation restricted to H b Conversely it is easy to see that if H Q G such that the restriction of the multiplication on G de nes a group structure on H then H is a subgroup of G 211 Examples a For each group G one has 10 g G and G g G We usually just write 1 instead of 1C and call it the triuz39al subgroup of G b The intersection of any collection of subgroups ofa group G is again a subgroup Warning In general the union of subgroups is not a subgroup 0 27 Q lt R lt 37 d For any non empty subsets X1X2Xn of a semigroup G we de ne X1X2Xn 3 12n l 1 X1n EXn In general this is not a subgroup even if X1 Xn are For subgroups HK g G one has HK g G ltgt KH HK e If X is a non empty subset of a group G its normalizer is de ned as NGX 9 6 G l ng 1 X Note that ng 1 X ltgt 09X X ltgt gX Xg One always has N0X g G Moreover the oentralz39zer of X is de ned as 00X g e G 193594 m for all m e X Note that g E 00X ltgt cg is the identity on X ltgt gm mg for all z E X It is easy to check that CaX NaX is again a subgroup If X consists only of one element we also write 00z instead of 00H f The subgroup ZG CaG g E G 1 gm mg for all z E G is called the center of G It is an abelian subgroup g If f G a H is a group homomorphism and if U g G and V g H then fU g H and f 1V g E G 1 y 6 V g G In particular the image of f imf fG is a subgroup of H and the kernel of f kerf f 11H are subgroups of H and G respectively Note 1 is injective if and only if kerf 1 The kernel of sgn Symn a i1 is called the alternating group of degree n and is denoted by Altn The kernel of det GLMR a R 0 is called the special linear group of degree n over R and is denoted by SLMR 212 Theorem The subgroups ofZ are the subsets of the form nZ nk l kEZ fornENO Proof For every n E Z the map Z a Z k gt gt kn is a group homomorphism cf Example 26a with image nZ By Example 211g it is a subgroup of Z Conversely assume that H lt Z If H 0 then H OZ and we are done So assume that H 7 Then H contains a non zero integer and with it its inverse So H contains a positive integer Let n be the smallest positive integer contained in H We will show that H nZ First since n E H also nnnnn E H Also in7n in E H Thus nZ g H On the other hand take an arbitrary element h of H and write it as h qnr with q E Z and r E 01n7 1 Then we have r h 7 qn E H which implies r 0 by the minimality of This shows that h qn E nZ So H g nZ l 213 De nition Let G be a group and let X Q G be a subset a The subgroup generated by X is de ned to be the intersection of all subgroups of G which contain X It is denoted by It is immediate that X is the smallest subgroup of G containing X ie it contains X and is contained in every other subgroup that contains X If X 0 we have ltXgt 1 If X we also write instead of b If X G then we call X a generating set or a set of generators of G If G is generated by a single element then G is called cyclic 214 Proposition Let G be a group and let X Q G Then one has ltXgtxi1mzk lkENO 1zk EX 61Ek i1 Here we de ne 1 1 ifk 0 Proof Every subgroup of G that contains X also has to contain the right hand side Therefore the intersection of all such subgroups has to contain the right hand side Thus X contains the right hand side On the other hand it is easy to verify that the right hand side is a subgroup of G that contains X Therefore it also contains l 215 Examples a Let G be a group and let zy E G The element my zym ly l is called the commutator of z and y One has my z Thus m y 1 if and only if my gm ie z and y commute The subgroup of G generated by all the commutators zy m y E G is called the commutator subgroup or the deriued subgroup of G and it is denoted by G or C1 G Note that mgfl Therefore Gl17y1l 39lk7ykllkENmx177k7y177yk G b The elements 7 1 2 3 4 7 1 2 3 4 3 2 1 4 3 y 3 4 1 2 generate a subgroup V4 of Sym4 which is called the Klein 4 group One checks easily that z2 1 y2 1 and A7 7 71234 27myiyzi 4 3 21 This shows that V4 1 m y z and we obtain the following multiplication table 216 De nition Let G be a group and let H g G For zy E G we de ne mm y if m ly E H This de nes an equivalence relation on G verify The equivalence class containing z E G is equal to zH verify and is called the left coset of H containing z The set of equivalence classes is denoted by GH The number is called the index of H in G and is denoted by G H 217 Remark Let G be a group and let H g G In a similar way one de nes the relation AH on G by z my if my 1 6 H This is again an equivalence relation The equivalence class of z E G is equal to Hz the right coset of H containing m The set of right cosets is denoted by HG We will mostly work with left cosets 218 Example Fix n 6 N0 and k E Z Then the set k 712 is a left and right coset of 712 in Z For example 2SZH8H32712 For this particular choice G Z and H 712 we also write x E y mod 71 instead of szw y and call x congruent to y modulo n The coset k 712 is called the congruence class of k modulo n One has ZnZ0nZ1nZn71nZ and ZnZn 219 Proposition Let G be a group and let H g G a For each 9 E G the map H H gH h H gh is a bijection In particular any two left cosets of H have the same cardinality namely lH b For each 9 E G the map H H Hg h H hg is a bijection In particular any two right cosets of H have the same cardinality namely lH c The map GH H HG gH H Hg 1 is well de ned and bijective In particular Proof a It is easy to verify that gH H H x H g lm is is an inverse b One veri es easily that Hg H H x H zg l is an inverse c In order to show that the map is well de ned assume that 9192 6 G such that 91H 92H We need to show that then Hg1 Hggl But we have 91H 92H ltgt gflgg E H ltgt Hg1 Hggl Finally the map HGHGH HgHg lH is an inverse l 220 Corollary Lagrange 173671813 Let H be a subgroup of a group G Then lGl GiHllHl With the usual rules for the quantity 00 KG is a nite group ie lGl lt 00 then and G H are divisors oflGl Proof G is the disjoint union of the left cosets of H There are C H such cosets and each one has elements by Proposition 219a l 221 Examples a The subgroups V4 and Alt4 of Sym4 have order 4 and 12 which are divisors of 24 in accordance with Lagrange s Theorem By Lagrange Sym4 cannot have a subgroup of order 10 We will see later Alt4 does not have a subgroup of order 6 although 6 divides 12 10 Let G be a nite group Whose order is a prime p Then by Lagrange 1 and G are the only subgroups of G Moreover G is cyclic generated by any element x 7 1 Note that H is a subgroup of G with 1 lt Thus H G 3 Normal Subgroups and Factor Groups 31 Theorem Let G be a group and let N be a subgroup of G Then the following are equivalent i gNg 1 Q N for allg E G ii gNg 1 N for all g E G iii gN Ng for all g E G iv GN isagroup under giN ggN H 91N92N where glNggN is de ned as the product of the subsets glN and ggN of G as in Exam ple 2lld v There exists a group H and a group homomorphism f G a H such that kerf N Proof iii Let g E G Then applied to 9 1 yields g lNg Q N Applying 09 then gives N gg lNgg 1 Q gNg l Together with for 9 we obtain ii for 9 ii a m For each EGwe have N N 1 N 9 9 9 9 9 9 iiiiv For any 9192 6 G we have 91N92N 9192NN 9192N 3180 so that glNggN gt gt glNggN is a binary operation on GN Obviously it is associative Moreover by 31a N 1N is an identity element and for any 9 E G g lN is an inverse of 9N ivv Set H GN and de ne f G a GN g gt gt gN This is a homomorphism In fact for 971 6 G we have fgfh gNhN which must be again a left coset by iv But it also contains gh Therefore it must be ghN and we obtain fgfh ghN fgh Finally we have kerf N In fact f1 N must be the identity element of GN and f 1N N vi For each 9 E G and each n E N one has fgng 1 f9fnf9 1 f9 391f9 1 1 which shows that 9719 1 E kerf N Thus gNg 1 Q N for all g E G l 32 De nition If the conditions i7v in Theorem 31 are satis ed we call N a normal subgroup of G and write N 1 G We write Nlt1 G if N is a proper normal subgroup of G If N 1 G then iv and iii in the previous theorem imply that the set GN of left cosets is again a group under the binary operation 91N792N H 91N92N 9192NN 9192N It is called the factor group of G with respect to N7 or for short G modulo N The homomorphism 1 G a GN7 g gt gt gN7 constructed in the proof of Theorem 31 is called the canonical epimo rphlsm or natural eplmomhlsm 33 Examples a We always have 1 1 G and G 1 G If G and 1 are the only normal subgroups of G and if G 7 17 we call G simple By Lagrange s Theorem7 groups of prime order are always simple If G is not simple7 there exists 1 lt N lt1 G and we think of G as being built from the two groups N and GN This is often depicted as o G GN or GN o N NgNi i o 1 We may think of GN as an approximation to G the multiplication in GN determines the multiplication in G up to an error term in N In a way7 the simple groups are the atoms of arbitrary groups The determination of all nite simple groups was one of the largest projects in mathemat ics About 507100 mathematicians were involve and the results cover about 10000 pagers scattered in journals The project was more or less completed in 1980 b lfG is agroup and H g ZG7 then H 1 G In particular7 ZG 1 G In an abelian group G7 every subgroup is normal since G ZG The center of G is even more special For every 1 E AutG one has fZG ZG verifyl A subgroup N g G with fN N for all f E AutG is called characteristic in G In this case we write N 1 G Note that N hi G implies that N 1 G 09 E AutG for all g E G ar C ar c Let G be a group and let G g H g G7 where G denotes the commutator subgroup of G7 cf Example 215a Then H 1 G and GH is abelian ln fact7 for any 9 E G and h E H one has ghg l ghg llrlh hlh e G H g H and for any m y E G one has mHyH xyH zyly lw llH ywH yHH 13 Here the second equality holds since y lw l E H In particular with H G we obtain that G is normal in G and that GG is abelian Conversely if N is a normal subgroup of G with abelian factor group GN then G g N g G In fact let zy E G Then one has 907le zyz ly lN mNyN 1Ny 1NlxNnyl N which implies that m y E N Thus we have G g N The above two considerations show that G is the smallest with respect to inclusion normal subgroup of G with abelian factor group This factor group GG is called the commutator factor group of G and it is denoted by Gab d lfH g G with G H 2 then Hlt1 G In fact for g E H we have gHHHg and forg E GHwe havegHGHHg sincethere are only two left cosets and two right cosets and one of them is H e For every subgroup H of G one has H 1 N0H g G Moreover NaH G if and only ifH 1 G f For each subset X of a group G one has 00X 1 N0X see homework In particular setting X G we obtain ZG 1 G g For each n E N one has Altn kersgn 1 Symn h For each n E N one has SLMR kerdet 1 GLMR i Let G Sym3 and let me 3 2H 3 2M 3 2 Then H G since 12312312371123 lt132gtlt213gtlt132gt lt321gt H39 34 Theorem Fundamental Theorem of Homomorphisms Let f G a H be a group homomorphism and let N 1 G with N g kerf then there exists a unique homomorphism F GN a H such that To 1 f where 1 G a GN g gt gt gN denotes the natural epimorphism Moreover kerf kerfN and imf imf In particular choosing N kerf one obtains an isomorphism Gkerf a imf Proof a Existence Let 011 6 G with aN bN Then a lb E N and fb faa 1b fafa 1b fa since N g kerf Therefore the function GN H H aN H at is well de ned It is a homomorphism since faNbN fabN ab fafb faNfbN for all 011 6 G Moreover for all a E G we have Va aN fa Thus F o 1 f b Uniqueness lf also f GN H H satis es f0 1 1 then aN f0 1a f0 1a aN for all a e G Thus f f c For all a E G we have aNekerG ltgt 7aN1ltgt fa1ltgt a6kerf Therefore ker LN E GN l a E kerf kerfN Finally imf faN l a E G fa la 6 G imf El 35 Example For n 2 2 the sign homomorphism sgn Symn H i1 is surjective with kernel Altn By the Fundamental Theorem of Homo morphisms we obtain an isomorphism SymnAltn i1 Therefore Symn Altn 2 and lAltnl nl2 Note that the groups Z and ZnZ for n E N are cyclic generated by 1 and 1 712 The next theorem shows that up to isomorphism there are no other cyclic groups 36 Theorem Classi cation of cyclic groups Let G be a cyclic group generated by g a IfG is in nite then G E Z G gk l k E Z and gi 97 if and only ifi j for all ij E Z b IfG is nite oforder n then G ZnZ G 1992 gn l and gi gj ifand only ifi gj mod n for all ij E Z Proof We consider the function 1 Z H G k H gk It is a homomorphism since 9kgl 9k By Proposition 214 we have C gk l k E Z which implies that f is an epimorphism By Theorem 212 we have kerf n2 for some n 6 N0 By Theorem 34 we obtain an isomorphism ZnZ H G k 712 H gk This implies that G is in nite if and only if n 0 Now all the assertions follow from considering the isomorphism l 37 Theorem Fermat 160171665 Let G be a nite group let 9 E G and let k E Z Then gk 1 if and only if divides k In particular lGl 1 g 4 Proof Since G is nite the order of 9 is nite By Theorem 36b we have 9k190 ltgt 9k90 ltgt kEO mod ltgt l9ldividesk l 38 De nition Let G be a group and let 9 E G One calls the order of9 and denotes it by 09 If G is nite then 09 divides G by Lagrange and 09 minn E N l 9 1 39 Theorem 15 lsomorphism Theorem Let G be agroup and let N H g G be subgroups such that H g N0N this is satis ed ifN 1 G Then HNNHltG NgHN H Nng and HH NHHNN hH NHhN is an isomorphism Proof For all h E H and n E N we have hn hnh 1h 6 NH and nh hh 1nh E HN since H g NaN Thus HN NH By Exam ples 211d HN is a subgroup of G Moreover for n E N and h E H we have nhNnh 1 nhNf17f1 nN7f1 N since h E NaN Thus N 1 NH The composition of the inclusion H g HN and the natural epi morphism HN H HNN is a homomorphism f H H HNN h H hN It is surjective since hnN hN fh for all h E H and n E N lts kernel is H O N By the Fundamental Theorem of Homomorphisms we obtain an isomorphism HH N H HNN hN N H hN 310 Theorem Correspondence Theorem and 2nd lsomorphism Theorem Let G be a group let N 1 G and let 1 G H GN denote the canonical epimorphism The function LpHlNltHltGHXlXltGN HHHN1H is a bijection with inverse 1 X H V 1X For subgroups H H1 and H2 ofG which contain N one has H1ltH2 ltgt HlNgHZN and Hgo ltgt HNglGN Moreover ifN g H 1 G then E GH 16 Proof Since images and preimages of subgroups are again subgroups see Examples 211g applied to 1 the maps 1 and 1 take values in the stated sets and obviously respect inclusions With regard to the map 1 note that N kerz is contained in V 1X for every subgroup X of GN For every N g H g G we have V 1VH H since N g H And for every X g GN we have 1V 1X X since 1 is surjective Thus g5 and 1 are inverse bijections Moreover for N g H g G h E H and g E G we have ghg l e H ltgt ghg lN e HN ltgt gNhNg 1N e HN This shows that H is normal in G if and only if HN is normal in GN Finally for N g H 1 G the composition 1 G a of the two canonical epimorphisms G a GN and GN a is an epimorphism with kernel H Now the Fundamental Theorem of Homomor phisms induces an isomrophism GH a 311 Proposition Every subgroup and factor group of a cyclic group is cylic Proof If G is a cyclic group generated by g E G and if N 1 G then GN is generated by gN To prove that subgroups of G are again cyclic we may assume that G Z or G ZnZ for n E N using Theorem 36 But subgroups of Z are of the form kZ k E Z and k2 is generated by k Moreover by the correspondence theorem subgroups of ZnZ are of the form kZnZ with 712 g kZ lt Z But by our above argument with 2 also every factor group of k2 is cyclic l 4 Normal and Subnormal Series 41 De nition A subnormal series of a group G is a nite sequence GG0EG1EG2EEGi1 If G is normal in G for every i 1 1 then we call the sequence a normal series The groups CplCi i 1l are called the factors and the number 1 is called the length of the subnormal series A subnormal series is called a composition series if each of its factors is simple 42 Examples a Not every group has a composition series For exam ple 2 does not have one In fact every non trivial subgroup of 2 is again isomorphic to 2 and therefore not simple Every nite group has a composition series 10 One composition series of 262 is 262D 2262 D 6262 lts factors are isomorphic to 222 and 232 Another composition series of 262 is 262 D 3262 D 6262 lts factors are isomorphic to 232 and 222 c Sym3 D Alt3 D 1 is a composition series of Sym3 with factors isomorphic to 222 and 232 d Sym4DAlt4DV4Dlt1 23 4gtD1 is a composition series of Sym4 with factors isomorphic to 222 232 222 222 43 Theorem Jordan Holder Let G be a group and let GG0gtG1DG2gtDG11 and GH0DH1DH2DDHm1 be two composition series of G Then 1 m and there exists a permutation a E Syml such that CplG HU1HU for alli 1 1 Proof lnduction on n minlm If n 0 then G 1 l m 0 and there is nothing to show If n 1 then G is simple 1 m 1 and all assertions are obviously true We assume from now on that n 2 2 and that all assertions of the theorem hold for n 7 1 Since GG1 is simple and since GG1G1H0G1 EGlHlGl EEG1HmG11 438 18 is a subnormal series of GG1 note that GlHi 1 GlHi1 for i 1 m we have G1H13971G1 GG1 and GlHiGl GlGl for a uniquei E 1 m This implies Hi G1 G1Hi71 G and H141 y G1 43b The group homomorphism u H1 7 GG1h gt gt hGl has image Hi1G1G1 GGl Thus 1 is surjective Moreover since Hi G1 we otain Hi keru and there exists a surjective group homomorphism 11 Hi1H 7 GGl Since H 7 1H is simple and ker1 7 Hi1H since 11 is surjective we obtain that ker1 H171 G Hi Thus 11 is an isomorphism Hi71Hl39 g and Hi1 G1 Note that G1gtG2gtgtGl1 43d is a composition series of G1 of length l 7 1 Consider also the following subnormal series of G1 G1 Ho G1 EH1 G1 EHii1 G1 HiDHi1DDHm1 43e Note that Hj G1 1 Hj71 G1 forj 1 quotJ71 and that Hi71 G1 Hi by equation 43a We claim that H741 G1Hj G1 g Hj71Hj for 1j71 43f If we can show this claim then the sequence 43e with 77 Hi omitted is a composition series of G1 of length m 7 1 with factors isomorphic to HOH17 7Hi72Hi717HiHi17 7Hm71Hm Comparing this with the composition series 43d of G1 of length l 7 1 and using the induction hypothesis together with the isomorphism in 43c immediately yields the desired result So it suffices to show 43f To this end consider the group homomorphism 1 Hj71 G1 7 Hj71Hj z gt gt mHj Since kerz H1 G1 Hj Hj G1 we obtain a monomorphism I73Hj71 G1Hj G1HHji1Hj Hj G1gt gtHj with the same image as 1 namely Hg1 G1HjHj By Dedekind s iden tity see homework we have Hj71 G1Hj HjAOGlHj But by 43b we have G1Hj 2 G1Hi1 G and obtain Hj71 G1Hj Hj71 G Hj71 Thus 17 is surjective the claim 43f is proved and the theorem holds l 44 De nition Let G be a group which has a composition series The length of a composition series of G is called the composition length of G It does not depend on the choice of a composition series The factors of a composition series of G are called the composition factors of G They are uniquely determined by G up to isomorphism and reordering 45 Examples a Sym3 and 262 are non isomorphic groups with the same composition factors namely 222 and 232 cf Examples 42 b The composition factors of Sym4 are 222 with multiplicity 3 and 232 with multiplicity 1 cf Examples 42 c If 2 g n E N has prime decomposition n p pit then 2n2 has composition factors 21912 with multiplicity el ZPTZ with mul tiplicity 5 46 De nition A group G is called soluable if it has a subnormal series with abelian factors 47 Examples a Every abelian group G is solvable since G E 1 is a subnormal series with abelian factors b The groups Sym 3 and Sym4 are solvable by the subnormal series given in Examples 42c and 48 Theorem Let G be a nite group Then G is solvable if and only if every composition factor of G is a cyclic group of prime order Proof s This is obvious since a cyclic group is abelian We choose a subnormal series of G with abelian factors After omit ting repetitions we can re ne it to a composition series of G The factors of this composition series are factor groups of subgroups of abelian groups and therefore again abelian lt suffices now to show that an abelian simple group S is cyclic of prime order Let 1 7 s E S be arbitrary Then the group generated by s is a non trivial and normal since S is abelian subgroup of S Since S is simple we obtain S ltsgt and S is cyclic and isomorphic to 2n2 for some n E N with n gt 1 Let p be a prime factor of n Then p2n2 is a proper normal subgroup of the simple group 2n2 This implies that p2n2 is the trivial group Thus p2 n2 and p n l 49 De nition Let G be a group The higher deriued subgroups or higher commutator subgroups G i 6 N0 of G are recursively de ned by Co G CO Ga Ga G1 Gi1 day 20 410 Proposition Let G be a group and i 6 No a Gltigt g G ch b IfH g G then Hltigt g Gltigt c IfN g G then ClNW GltigtNN Proof This is proved by induction on i see homework l 411 Proposition A group G is solvable if and only if there exists 3 6 N0 with 19 1 Proof HG 1then G 10 2 11 2 2 19 1isasubn0rmal series of G with abelian factors Thus by de nition G is solvable lt Assume that G is solvable and let G G0 2 G1 2 E G 1 be a subnormal series with abelian factors It suffices to show that G g Gt for i E 0 1 But this follows easy by induction on i In fact CO G G0 and if we have shown the statement for i E 0 l 7 1 then we can conclude by Proposition 410b that Ga Cil g G g Gt since GiGi is abelian cf Example 33c l 412 Proposition Subgroups and factor groups ofsolvable groups are solv able Proof Let G be solvable By Proposition 411 there exists 3 6 N0 with S If H g G then by Proposition 410b we obtain HltSgt g 19 1 Now Proposition 411 implies that H is solvable IfN g G then by Proposition 410c we obtain GNltSgt C9NN NN 1 Again Proposition 411 implies that GN is solvable l 413 Proposition Let G be a group and let N 1 G If GN and N are solvable then G is solvable Proof By Proposition 411 there exist r s 6 No such that ClNW 1 and N 1 By Proposition 410c we obtain CTlNN 1 and therefore Gltrgt g N Now Proposition 410b implies GltTSgt Gltrgtlt9gt g Nlt9gt 1 By Proposition 411 G is solvable l 21 414 Remark a Using representation theory7 Burnside showed 1911 that groups of order p qu 7 where 197 q are primes and 071 6 N0 are solvable b Feit and Thompson showed 1963 that groups of odd order are solV able The proof has 254 pages 22 5 Group Actions 51 De nition a A left action of a group G on a set X is a map 04 GX XgtX7 97m gt gtozgx 9x satisfying i 1zmforallz Xand ii 9 hm 9 for all g h e G and m e X A left G set is a set X together with a left action 04 of G on X Let X and Y be G sets and let 1 X a Y be a function We call 1 a momhism of G sets or G equiuariant if 9fm for all g E G and z E X We call the G sets X and Y isommphic notation X E Y7 if there exists a bijective G equivariant map f X a Y In this case7 f l Y a X is again G equivariant b Let X be a G set For x 6 X7 the set Gm g E G l 91 z is called the stabilizer of z in G It is easy to verify that Gm is a subgroup of G The element x is called a xed point if Gm G Moreover7 for an element x 6 X7 the set MG 91 l g E G Q X is called the orbit of z under G For m y E X one de nes the relation z N y if there exists 9 E G such that 91 y It follows immediately from the axioms and ii in a that this de nes an equivalence relation on X and that the equivalence class containing z is the orbit of z c Let X be a G set The kernel of the action 04 G x X a X of G on X is de ned as kera g E G l 91 z for all z E X It is a subgroup of G The action 04 and the G set X are called i faithful7 if kera 17 ii trivial7 if kera G7 and ii transitive if X consists only of one orbit7 ie7 for any 71 6 X there exists 9 E G such that 9m y 52 Remark One can also de ne right actions of a group G on a set X in a similar way and use the notation 9 for z E X and g E G It is easy to verify that if one has a right action of G on X then gzx9 96G7mEX de nes a left action of G on X Conversely7 if one has a left action of G on X then x9 9x 96G7mEX de nes a right action of G on X 23 53 Examples a Let X be any set G SymX acts on X via H b Let G be a group H g G and set X GH Then GH is a G set under the action 04 G x GH H GH gaH H gaH Note that 04 is well de ned c Let G be a group and X G Then G acts on X by conjugation gz H cgz gzg 1 92 for g z E G The orbit of z E G is called the conjugacy class of m We call m y E G conjugate under G if there exists 9 E G such that 9m y The kernel of the conjugation action of G on G is ZG and Gm 00z Similarly G acts by conjugation on the set of subsets of G If Y is any subset of G and g E G then by 9Y we usually mean cgY ng l One has Gy NCO the normalizer of Y d Let G be a group and let X be the set of subgroups of G Then G acts on X by conjugation gH H 9H gHg l The orbit of H g G is called the conjugacy class of H The stabilizer of H is N0H We have H 1 G ltgt N0H G ltgt H is a xed point 54 Proposition Let X be a G set a For every 9 E G the map 7rg X H X x H 91 is bijective ie 7rg E SymX b The map p G H SymX g H 7rg is a group homomorphism c kerp is equal to the kernel of the action ofG on X Proof For g h E G andz E Xwe have 7rgo7rhz 9 hm 9mm wghz and 7r1z 1m z Thus 7rg o 717 g and 7r1 idX This implies 7rgo7r971 7r1 idX 7r1 wgilo n39g showing a Moreover 7rgo7rh wgh shows Finally an element 9 E G is in the kernel of the action of G on X if and only if 92 z for all z E X This happens if and only if 7rgz z for all z E X which in turn is equivalent to 7rg E kerp l 55 Remark a Conversely for a group G and a set X every group ho momorphism p G H SymX gives rise to an action of G on X via 990 p9 for g E G and z E X This construction is inverse to the construction in Proposition 54 b Let X be a G set and z E X Then Ggm gGmg 1 verifyl c Every orbit of X is a transitive G set in its own right and X is the disjoint union of its orbits 24 56 Theorem Cayley Every group G is isomorphic to a subgroup of a symmetric group IfG is nite oforder n then G is isomorphic to a subgroup of Symn Proof G acts on itself by left translation on X G via G x X H X g x H gm By Propositioin 54 we obtain a group homomorphism p G H SymX g H 7rg with 7rgz gm for g E G and z E X Since 7rg idX implies gm z for all z E X we obtain that kerp 1 so that p is injective and induces an isomorphism G E imp SymX Finally note that if 1 X1 H X2 is a bijection between two sets then SymX1 H SymX2 7139 H f 0 7r 0 f l is a group isomorphism l 57 Theorem Let G be a group and let X be a G set a For each x E X the map fiGGm lmlCL 9GmH97 is an isomorphism of G sets 10 Let R Q X be a set of representatives for the orbits ofX under G ie 9 contains precisely one element from each orbit Then one has the orbit equation le ZlG 3 Gl7 69 with the usual rules for oo In particular if X is a transitive G set then le G Gm for any element x E X Proof a The function f is well de ned since for g E G and h E Gm one has 9 91 By the de nition of MG the function f is surjective It is G equivariant since war26 fg192Gm lt9192gtz 9 gingham Finally 1 is injective lfghgg E Gsatisfy fgle fggGm then 91x 9 and gglglz z so that gz lgl E Gm and gle ggGm b The cardinality le is equal to ERR and by part a we have lGGml G Gm for every x ER l 58 Remark Part a of the previous theorem shows that every transitive G set X is isomorphic to GH with H Gm for any x E X This implies that GH E GgHg 1 as G sets for all g E G Conversely if H K g G and 25 if GH GK as G sets7 then H and K are conjugate cf Homework This completely classi es all G sets up to isomorphism Every G set is isomorphic to a disjoint union of G sets of the form GH7 where H runs through a set S of iepr for the nju a y classes of subgroups of G Two such 1 sets are isomorphic if and only if GH occurs with the same multiplicity in both G sets for all H E S 59 Example Assume that G acts on itself by conjugation Then Gm 00z and the orbit equation becomes lGl ZlG t CGl7 69 where R Q G is a set of representatives of the conjugacy classes of G Moreover7 for every x E G we have misa xed point ltgt GCam ltgt m ZG 510 Theorem Let p be a prime number and let G be a group of order pk with k E N Then ZG gt 1 Moreover G is solvable Proof The orbit equation from Example 59 yields 0 E lol 2p 00z E lzol modp m6 Therefore7 p divides Together with lZGl 2 1 this proves the theorem l 511 De nition Groups of order pk for a prime p and k 6 N0 are called p groups A subgroup H of an arbitrary group G is called a p subgmup if H is a p group An element of G which has order pl for some 1 6 N0 is called a p element 512 Theorem Sylow7 183271918 Letp be a prime and let G be a nite group of order n pam with a 6 N0 m E N plm A subgroup P g G of order p is called a Sylow p subgroup of G a For all I E 07177a the number napb of subgroups of G of order p17 satis es the congruence napb E1 mod p In particular G has a subgroup of order p17 for every I O7 7a b Every p su bgroup of G is contained in some Sylow p su bgroup of G c Any two Sylow p subgroups of G are conjugate 26 Proof a The statement is cleary true for b 0 So x I E 1 2 a and set 9XQG1 lepb Then 111 G acts on Q by translation 9X gX Let R Q Q be a set of representatives of the G orbits of 9 Then the orbit equation yields 27 191 Z GGx 512a X67 Claim 1 For every X E R one has lGXl lpb Proof Since GX g E G l gX X we have GXX X and GX acts on X by translation But each orbit of this action has length lGXl since gm 9 implies g g for all z E X and gg E GX Thus le pl7 equals lGXl multiplied by the number of orbits Claim 2 If lGXl p17 then the G orbit of X E 9 contains a subgroup of G Proof Since lGXl pb GX acts transitively on X and we have X GXz for some z E X Thus m lX 4me is a subgroup of G in the G orbit of X Claim 3 If the G orbit of X E 9 contains a subgroup U of G then this orbit equals GU In particular G GX G U pa bm lGXl p17 and U is the only subgroup in this orbit Proof The orbit of X equals the orbit of U which is GU lf gU E GU is a subgroup of G then 1 E gU and gU U The remaining statements clearly hold Claim4 One has 11 E paibm napb mod paib1m Proof We use the orbit equation 512a Let X E R If the orbit of X contains a subgroup of G then by Claims 1 and 2 we have lGXl l p and consequently G GX E 0 mod pa b m If the orbit of X contains a subgroup of G then it contains exactly one subgroup and G GX pa bm by Claim 3 This proves Claim 4 Claim 5 One has napb E 1 mod p Proof Let H be an arbitrary group of order n Then by Claim 4 for H we have 7 771 7 E p l mmHpb E pi 3p l mwdpl mod p b 27 and this implies nHpb E napb mod p Using H 2712 and noting that nHpb 1 Homework we obtain napb E1 mod p b Assume that U is a p subgroup of G and that P is a Sylow p subgroup of G Consider the set P ng 1 l g E G and note that it is a transitive G set under conjugation The stabilizer of P is N0P and since P g N0P we obtain lPl G N0P l G P m Thus p does not divide With C also U acts on P by conjugation Let 5 denote a set of representatives for the U orbits of P By the orbit equation we have lPl EQESW UQ Since U is a p group and since m is not divisible by p there exists Q E S such that U UQ 1 or equivalently such that U N0Q This implies that UQ is a subgroup ofG and UQQ E UU Q by the rst lsomorphism Theorem With UU O Q also UQQ is a p group and therefore lUQl lUQQl lQl is a power ofp Thus UQ is ap subgroup of G containing the conjugate Q Note that Q is conjugate to P and has order pb By Lagrange we obtain UQ Q and U g Q This proves ii c If U in ii is a Sylow p subgroup it follows that U Q 513 Remark Let G be a nite group let p be a prime We will denote the set of Sylow p subgroups of G by SylpG By Sylow s Theorem SylpG is a transitive G set under conjugation If P E SylpG the orbit equation gives us lSylpGl G NaP Since P g NaP this implies that lSylpGl divides G P m in the notation of Sylow s Theorem Moreover P is normal in G if and only if P is the only Sylow p subgroup 514 Theorem Cauchy 178971857 Let G be a nite group and let p be a prime divisor of Then G has an element of order p Proof By Sylow s Theorem 512a G has a subgroup of order p This subgroup must be cyclic and every generator of it has order p l 515 Proposition Let p and q be primes and let G be a group of order pq or 19 Then G is solvable Proof From Theorem 510 one obtains easily that every p group is solvable see also Homework So we may assume that p 7 q a Assume that lGl pq and that p gt q By Sylow s Theorem we have nap E 1 mod p and nap l q this implies nap 1 Thus G has only 28 one Sylow p subgroup P and it is normal in G Therefore 1lt1 P lt1 G is a subnormal series and it has abelian factors b Assume that lGl pzq By Sylow s Theorem we have 710192 E 1 q and naq E 1pp2 lf 710192 1 then G has a normal Sylow p subgroup P and G is solvable by Proposition 413 Therefore we may assume that 710192 q By Sylow s Theorem q E 1 modp and p l q 7 1 so that p lt q Since q lpi 1 we obtain naq 7 p lf naq 1 then G has a normal Sylow q subgroup Q and GQ is a p group and therefore solvable Thus G is solvable Therefore we may assume that naq p2 Since the intersection of two distinct Sylow q subgroups of G is the trivial subgroup we can count the elements of order q of G by naqq71 p2q71 1G1 7192 Since no element of order q can be contained in a Sylow p subgroup of G we obtain that G can have only one Sylow p subgroup But this contradicts our assumption that 710192 q 29 6 Symmetric and Alternating Groups 61 De nition Let n E N and a E Symn a is called a k cycle k E N or a cycle of length k if there exist pairwise distinct element a1 ak E 1 n such that 0a1 a2 0a2 a3 0ak a1 and 0a a for all a E 1 n 11 ak In this case we write a 11 ak If k 2 then a is called a transposition Two cycles a1ak an b1bl are called disjoint if 01 ak m 51 bj 62 Remark a Elements of Symn are functions 1 n a 1 The binary operation on Symn is given by composition 0739 means 77first apply 739 then 0 Warning Some books or computer programs write functions from the right and have the opposite convention 0739 means 77first apply a then 739 b a1a2ak a2a3aka1 c 11012 ak 1 ak ak1 a1 d If a a1a2ak and 739 b1bl are disjoint cycles then 0739 T0quot e Every element a E Symn can be written as a product of disjoint cycles Usually one omits cycles 04 of length 1 because they are the identity But let us include all these cycles Then the occurring cycles form precisely the ltagt orbits of 1 n under the natural action of Symn on 1 n i ai Therefore the decomposition of a into disjoint cycles is unique If we order the cycle lengths k1 2 kg 2 2 k2 1 then the sequence k1 k2 k7 is called the partition of n corresponding to a or the cycle type of a Note that k1 k7 n A partition of n is a sequence k1 k2 k of elements in N satisfying k1 2 kg 2 2 k and k1 k n For example 4 5 6 7 8 3 g 6 8 7 5gt124359678 has cycle type 4 2 2 1 f lfa E Symn has cycle type k1 k then 0a lcmk1 Wk verifyl 63 Proposition Let 07 6 Symn Then a and 739 are conjugate ltgt a and 739 have the same cycle type Proof It is easy to see that plta1akgtp1ltplta1gtpltakgtgt 63a 30 for every k cycle 117 7ak and every p E Symn Let a E Symn and write a 0391039T as a product of disjoint cycles of lengths 17 7 k7 Conjugation with any p E Symn yields again a product oap 1 palfl 70077071 of disjoint cycles of lengths 17 7 k by equation 63a Assume that the permutations a 011701277ak7b17b277bk2 and 739 oz17 7a 1b 17 7b 2 havethe same cycletype k17 27 7 k De nep E Symn by pa1 a 17pa2 1 27 Then7 by equation63a we obtain oap 1 739 Elk4 64 Remark For each n E N one obtains in the above way a bijection between the set of conjugacy classes of Symn and the set of partitions of 77 65 Example For n 5 we have the following partitions 5 4 1 32311221211111111 Thus7 Sym5 has seven conjugacy classes 66 Proposition a Symn 17 27 2737 7 n 717ngt b SymW lt1 2 123 ngt Proof a See Homework b Set a 17 27 771 and U 1727a Then also i17i 2 ai17 2a i E U for all i 17 7n 7 2 Thus7 U Symn by part a l 67 Theorem Symn is solvable if and only ifn g 4 Proof We already have seen that Symn is solvable for n g 4 Now let n 2 5 De ne U Symn as the subgroup generated by all 3 cylces Then U U ln fact7 let 2737 k E 17 771 be three distinct number Then M k WWWk7milji7m k 2391 i7k7ml for any two distinct numbers 7 m E 17 771 i7j7 this implies that l U is not solvable and also that Symn is not solvable 68 Remark It is not difficult to prove that Altn is a simple group for n 2 5 31 69 De nition Let n 2 3 The subgroup of Symn generated by a 1 23 n and 739 1n2n 71 is called the dihedral group D2 One usually extends this de nition to D4 V4 and D2 222 610 Proposition With the above notation one has for n 2 3 a 101 7117quot 2 and 7 07 0 b ngnl 2n and Dgn 1002 a 1739a73902739 a 1739 Proof a Easy veri cation b Since 0a n the elements 1 a a 1 are pairwise distinct and form the cyclic subgroup C a of D2 Since 0 1 is the only element in C that maps 1 to n we have 739 G Since 7390quot an lr and 7392 1 we can see that G C U 0739 and that G C 2 611 Remark Geometric interpretation of D2 For a natural number n 2 3 consider a regular n gon with center in the origin of the plane R2 Number the vertices of the n gon by the numbers 1 2 n counterclockwise From classical geometry we know that there are n rotations and n re ections about an axis that take the n gon to itself We also know that the composition of any two such geometric operations is again one of them So they form a group G of order 271 under composition Each element of the group is a linear automorphism of R2 which permutes the verticies Every element of G permutes the n vertices and this gives rise to a homomorphism f G a Symn Note that the rotation about the angle 27rn is mapped to the permutation a 1 2 3 n and that the re ection about the line that intersects the edge between vertex 1 and n perpendicularly at its center is mapped to 739 1n2n 7 1 This implies that fG contains D2 But since lGl 2n ngnl we obtain fG D2 Now the Fundamental Theorem of Homomorphisms shows that f is injective So 1 G a Dzn is an isomorphism For this reason we also say that D2 is the symmetry group of the regular n gon 612 Proposition Let G be a nite group H g G and N gea gHg l Then N 1 G and there exists a monomorphism GN a Symn with n G Proof G acts on GH by left translation 9aH gaH This gives rise to a group homomorphism p G a SymGH Symn lts kernel is the kernel of the action But this is equal to the intersection of all the stabilizers of elements of GH Obviously H has stabilizer H and therefore 32 gH has stabilizer gHg l It follows that the kernel of the action is N geG gHgil39 l 613 Theorem For every odd prime p there exist up to isomorphism exactly 2 groups of order 2p namely 22192 and Dgp Proof Since D217 is not abelian we have D2p 22192 lt suffices now to show that every non cyclic group G of order 2p is isomorphic to D217 By Let P a and Q b be Sylow p subgroups and Sylow 2 subgroups respec tively Since G P 2 P is normal in G If also Q is normal in G then otbofllf1 E P Q 1 and consequently ab ba This implies hat ab is an el ement of order 2p a contradiction Thus N0Q Q and the G set SylZG is given by QaQa 1a2Qa 2 ap 1Qa since Q aQa2Q a1 1Q are the p left cosets of Q in G We number the Sylow 2 subgroups in the above order Since the intersection of all the Sylow 2 subgroups is trivial the pre vious proposition yields an injective homomorphism f G 7 Symp and by the proof of the proposition we see that at 1 2 p 7 1 We still want to compute the image of I Since P is the only Sylow p subgroup every element of order p and order 1 is contained in P Since G is not cyclic there is no element of order 2p It follows that all the elements of Pb bP must have order 2 In particular baba 1 and consequently ba ap lb and bai a ib for all i 1 p 7 1 Now it is an easy veri cation that fba 1p2p 7 1 so that fG contains D217 Comparing orders we obtain fG D217 and f G 7 D21 is an isomorphism l 33 7 Direct and Semidirect Products 71 De nition Let H and N be groups We say that H acts on N by group automomhisms if H acts on N as a set and additionally hn1n2 hm 71a for all h E H and all 711712 6 N This condition is equivalent to saying that the image of the group homomorphism p H H SymN h H 717 resulting from the action of H on N is contained in Aut N in other words 717 E AutN for all h E Note that in this case h1 1 for all h E H We say that H acts trivially on N if hm n for all h E H and all n E N 72 Proposition Let H and N be groups and let H act on N by group automorphism The binary operation a hb k a hb h k 72a on the set N x H de nes a group structure with identity element 1 1 and H1 inverses ah 1 h fa1 1 for a h E N x H Proof Associativity the property of the identity element and inverse ele ments are easy veri cations and left as homework 73 De nition Assume the notation from the previous proposition The group constructed with the product 72a is called the semidirect product ofN and H and is denoted by N gt4 H If the action of H on N is not clear from the context one also writes N gt4p H where p H H AutN is the homomorphism corresponding to the action 74 Remark Assume the notation from the last proposition The function i N H N gt4 H a H a 1 is a monomorphism Moreover the function p N gt4 H ah H h is an epimorphism Note that imi kerp a 1 l a E N is a normal subgroup ofNgt4Hwhich is isomorphic to N such that N gt4 Hkerp E H by the fundamental theorem of homomorphisms In this sense the semidirect product is made up of two pieces namely the group N and the group H Note also that j H H N gt4 H h H 1 h is a group homomorhism and that the two subgroups imi and imj of N gt4 H satisfy imi imj 1 and imiimj N gt4 H If the action of H on N is trivial then the semidirect product is just the direct product of N and H 34 75 Example Let G be a group let N be a normal subgroup of G and let H be a subgroup of G with H O N 1 and HN G Then H acts by conjugation on N and the resulting semidirect product N x H is isomorphic to G Via the homomorphism N gt4 H a G 7171 gt gt G verify In this case we also say that G is the internal semidirect product of the normal subgroup N and the subgroup H or that N is a normal complement of H in G 76 Examples a For 3 g n E N Dzn is the semidirect product of the cyclic subgroup 1 2 ngt of order n and the cyclic subgroup 1 n2 n7 1 of order 2 b Sym4 is the semidirect product of the Klein 4 group V4 and the subgroup Sym3 c Alt4 is the semidirect product of the Klein 4 group V4 and the subgroup 123gt 77 De nition Let G be a group and let H and K be normal subgroups of G We say that G is the internal direct product of the normal subgroups H and K ifH K 1 and HK G Note that in this case hk kh for all h E H and all k E K since 71 k hkh lk l 6 HO K 1 Moreover the function H x K a G h k gt gt hk de ned on the external direct product of H and K is an isomorphism and for this reason we also write G H x K if G is the internal direct product of H and K 78 Examples a The Klein 4 group is the direct product of the subgroups ltlt1 2gtlt34gtgt and ltlt13gtlt24gtgt b If mn E N are relatively prime ie gcdm n 1 then Zng is the direct product of the subgroups 7712ng and nZng c The dihedral group D12 is the direct product of the subgroups 03gt of order 2 and 727 of order 6 The latter group is isomorphic to Sym3 and D6 35 8 Nilpotent Groups 81 De nition Let G be a group A normal series GG02G12G222Gl1 81a is called a central series if GiGi ZGGi1 for all i 0 l 7 1 If G has a central series we call G nilpotent 82 Remark Let G be a group For any two subsets X Y Q G we de ne a subgroup lX7Yl ltlmyyl l96 6 X4 6 Ygt a A normal series 81a of G is a central series if and only if lazy Gl Gm for all i 0l 7 1 In fact GiGi1 ltgt giGi1Gi1 Gi1 for all 6 G and 9139 E ltgt giGi1 Gi1 for all 6 G and 9139 E ltgt G G g 111 b Every abelian group A is nilpotent since A 2 1 is a central series c Every nilpotent group is solvable since in 81a GiGi ZGGi1 implies that GiGi is abelian 83 De nition For a group G on de nes the upper or ascending central series of G 1ZOltGgtlt21ltGgtlt recursively by Z0G 1 and Zi1GZiG ZGZiG fori 2 0 Moreover one de nes the lower or descending central series of G GZ0GgtZ1Ggt 7 recursively by Z0G G and Zi1G ZiGG for i 2 O 84 Remark Let G be a group a It is easy to show by induction that ZiG 1 G and ZiG 1 G for h C ar ar every i 6 No In particular these subgroups are normal in G a fact used implicitly in the de nition of the upper central series b One has Z1G G but in general 12 lt Z2G 36 c If ZnG 1 resp Z G G for some n E N then the lower resp upper central series is a central series of C using Remark 82a for the lower central series Thus G is nilpotent d If G is a p group then G is nilpotent7 since Z G G for some n E N by Theorem 510 85 Proposition Let G G0 2 G1 2 2 G1 1 be a central series ofa group G Then ZG g G g ZHG for alli O7 Z In particular for any 0 6 NO one has ZCG G ltgt ZCG 1 The minimal number 0 6 N0 with this property is called the nilpotency class ofG Proof Both inclusions are proved by induction7 one on i the other one on 1 7i Homework 86 Lemma Let G be a nite group P a Sylow p subgroup of G and U g G satisfying NaP U Then NaU U Proof Let g E N0U Then ng 1 U and hence7 P and ng 1 are Sylow p subgroups of U By Sylow s theorem there exists it E U such that uPu 1 ng l This implies that u lg E NaP U and thereforeg E U l 87 Theorem For every nite group G the following are equivalent i G is nilpotent i ZnG 1 for some n 6 N0 iii For all U lt C one has U lt N0U v Every maximal subgroup of G is normal in G v For every prime p and every P E SylpG one has P 1 G vi G is isomorphic to the direct product of its Sylow subgroups vii ZGN gt 1 for everyNlt1G viii Z G G for some n 6 N0 ix If m y E C have coprime orders then my ym AAAAAAAAA Proof All assertions are true for the trivial group We assume from now on that 1G gt 1 iii This is immediate from Proposition 85 37 iiiii Let U lt G and let i E N such that ZiG U but Zi1G i U Then for all z E Zi1G and all u E U we have 2712711171 211 6 Zi1G G ZiG U This implies that zuz l E U and that N0U 2 Zi1GU gt U iiiiv This is obvious ivv Let p be a prime and let P E SylpG Assume that P is not normal in G Then NaP lt G and there exists a maximal subgroup U of G with N0P U By Lemma 86 we have N0U U contradiction iv vvi Let p1 pT be the distinct prime divisors of G and let Pi be the unique Sylow p subgroup of G for 1 7 Consider the function fP1XXPgtG 1xgt gtm1mr We will show that f is an isomorphism First note that this is clear when r 1 and that both groups have the same order Next note that Pb Pj 1 fori 31 j from 1 7 since P and Pj are normal in G This implies that f is a group homomorphism Finally for every i 1 r we have P g imf This implies that divides limfl But then lGl divides limfl and f is surjective Comparing orders we see that f is an isomorphism vivii We may assume that G P1 x x P with pi groups P for pairwise distinct primes p1 p lf Q denotes a Sylow pi subgroup of N fori 1r then Q N Pi Pi and Q1 xx QT ltN Comparing orders we obtain Q1x XQT N Since N lt G there exists i0 6 1 r with Qio lt1 Pic By Theorem 510 we have ZPOQO RioQ for some intermediate subgroup Qio lt Rio Pic Therefore ZGNgtQ1xxRiOxxQ7Q1xxQ7gt1 viiviii This is clear from the de nition of the upper central series viiii This follows from Remark 84c viix Assume that G P1 x x P as before in vivii If x 1 Wm and y y1y7 have coprime orders then for every i 1 7 one of the elements mpg E P must be trivial since the order of z is the product of the orders of m i 1 r But then obviously 1 1 y ixv Let p be a prime divisor of lGl and let P E SylpG Then for any other prime divisor q of lGl and any Q E SylpG we have Q g 00P N0P This implies that lQl divides lNGPl But also lPl divides lNGP Altogether lGl divides lNGPl and we can conclude that N0P G 38 88 Proposition Each subgroup and factor group of a nilpotent group is again nilpotent Proof We can use statements ix and iv in Theorem 87 for nite groups Assume that G G0 2 G1 2 2 G1 1 is a central series of an arbitrary group G If H g G then it is easy to see with Remark 82a that HH G02H G122H Gl1 is a central series of H If N is a normal subgroup of G then it is easy to see with Remark 82a that GN GoNN 2 GlNN 2 2 GlNN NN is a central series of GN l 89 Proposition Let G be a group and let N g ZG be such that GN is nilpotent Then G is nilpotent Proof Let GN GON 2 GlN 2 2 GlN NN be a central series of GN Then G G0 2 G1 2 2 Gl N 2 1 is a central series of G l 39 9 Free Groups and Presentations 91 De nition Let X be a set not containing an element denoted by 1 and assume that for every element x E X the set X does not contain an element denoted by z l We de ne X 1 as the set of symbols m 1 where z runs through m Thus X O X 1 Q and the function X a X l z gt gt z l is a bijection A word with letters from X is a sequence 11012 with ai E XUX 1U1 such that there exists n E N with 1 an an an2 The word 1 1 1 is called the empty word A word 11012 13 is called reduced if for all i E N and z E X one has i ai m a 1 7 71 ii a z 1 Ll1 7 m and a 1 gt n1 92 Remark Every reduced word has the form lmy zfquot 1 1 with n 6 N0 m1 mn E X 61 en 6 i1 where ml m We abbre viate this word by mil mfg Two reduced words mil miquot and yfl pg are equal if and only if m 71 xi pi and q 6139 for all i 1 n For the empty word we write just 1 93 Proposition Let X be as in the above de nition The set of reduced words with letters in X is a group FX the free group on X under the following multiplication Let w mil miquot and v yfl be elements ofFX with n g m and let k E 0 n be determined by 34713 for allj O k 7 1 but mfg 7 ygfkl Then the product we is de ned by 7 5 xi1x6quotk k1y nm 1fkltn kykl 7 61 5n 51 5m i 5n1 6771 fk m1y1ymi yn1ym7 1 7nltmi 1 ifknm In the case n gt m the product is de ned in an obvious analogous way Proof The proof of associativity is a straightforward but lengthy case distinction and is left out The identity element is the empty word 1 and the inverse of m xiquot is zquot zf51 94 Example Let X 1 b c Then aab lcbcb lxbc labb aab lcbabb 40 95 Remark Let X be as in De nition 91 a For x E X the inverse of z 1 is equal to m 11 justifying the notation b We denote byi X a FX the function de ned by z 1 1 Note that the reduced word a mgquot is equal to the product im161 izn5quot This shows that FX is generated by the set l x E X 96 Theorem Universal property of the free group Let X be as in Def inition 91 For every group Gand every function f XH G there exists a unique group homomorphism f FX a G such that f oi 1 Proof Existence We set 1 and zflquot fm151fmn5quot Then it is easy to check that f is a homomorphism Moreover 351 fz1 x for all m e X Uniqueness Assume that also f FX a G is a homomorphism with foi 1 Then f f since f and f coincide on the generating set of l 97 De nition Let G be a group and let Y Q G be a subset The normal subgroup of G generated by Y is de ned as ltltYgtgt lt9y9 1 ly 6 K9 6 Ggt It is easy to see that is the intersection of all normal subgroups of G that contain Y and that it is also the smallest with respect to inclusion normal subgroup of G that contains Y 98 De nition a Let X be as in De nition 91 and let R be a subset of Then the group with generators X and relations R is de ned as the group and it is denoted by X l R b Let G be a group let X Q G and let FX a G be the unique homomorphism de ned by the inclusion function j X a G x gt gt m via the universal property of Let R Q FX be a subset We say that the relations R hold in G if R g kerG We write a X i R 98a if X is a generating set of G and if kerG In this case induces an isomorphism i FXltltRgtgt H G and we say that 98a is a presentation of G with generators X and relations R Altogether we obtain the commutative diagram 41 99 Remark a A group G can have many different presentations As sume that G X l R is a presentation of G and that a certain product of elements of X and their inverses is equal to 1 This product can be in terpreted as an element in FX and there it can be written as a product of conjugates of elements of R In other words7 every relation among the generators X of G can be derived in this way from the relations R b If G X l R is a presentation of G and if X 17mn and R 7617 7Tm are nite7 we also write G 1 Wm l n Mm instead of G ltm17mn l r1rmgt c As the following proposition shows7 knowing a presentation ofa group G helps to de ne homomorphisms from G to other groups 910 Proposition Let G X l R be a presentation of the group G Let H be also a group and let 1 X a H be a function such that for every element mil sz E B one has u zn 1 in Then there exists a unique group homomorphism f G a H such that x x for all x E X Proof The commutative diagram from De nition 98 can be extended to a commutative diagram 42 where f is de ned by the universal property of FX and f is de ned by the fundamental theorem of homomorphisms noting that R Q kerf results from the hypothesis of the proposition Since G X l R the homomor phism is an isomorphism Now we de ne f G a H by f f0 5 1 and obtain foj oi 10jfoj loioiOlOif0if as required Since G is generated by X the homomorphism f is uniquely determined by f oj f 911 Example Let n 2 3 be a natural number and let a 1 2 n E Sym and 739 1 n2 n713 712 E Symn Recall that the dihedral group Dzn was de ned as a 739 g Symn We will show that Dzn 07 l a 7392 707710 Here we suppress the function i X a FX by writing a 72707 10 E FX instead of ia i7 2iTiai7 1ia E Since a and 739 generate D2 and since the relations R a 7392 7 070 hold in D2 the universal property ofFX and the fundamental theorem of homomorphisms yield a group epimorphism a D2 mapping the coset of a resp 739 to a resp 739 Using the equation TUltltRgtgt 0 17Rgtgt we can rewrite every element wRgtgt E as aleRgtgt with kl E Z Using the equations a Rgtgt and 72Rgtgt we can rewrite UleRgtgt o39iTjRgtgt with 0 g 2 g n 71 and 0 g j g 1 This implies that has at most 271 elements Since the map a Dzn is surjective it must be an isomorphism This means by de nition that G X l R Now that we know that G X l R we can de ne a homomorphism from D2 to any group H by simply nding elements st E H such that s 1 t2 1 and tst ls 1 Then by the previous proposition there exists a unique group homomorphism D2 a H with maps a to s and 739 to t The proposition shows that HomD2nH a st e H x H l s t2 25525713 1 gt H gta7 gtr is a bijection 43 Chapter II Rings 10 Rings Basic De nitions and Properties 101 De nition A Ting is a set R together with two binary operations l R x R a R ab gt gt ab and R x R a R ab gt gt ab called addition and multiplication respectively such that i R is an abelian group ii R is a monoid and iii abc abac and abc acbc for all a b c C R distributivity lf additionally a I I a holds for all a b C R then the ring R is called a commutative ring 102 Remark Let R be a ring The identity element with respect to ad dition is denoted by 0 and called the zero element the identity element with respect to multiplication is denoted by 1 and called the one element or identity element Usually we write ab instead of a I and a 7 1 instead of a 71 for ab C B To save parentheses one has the rule that multipli cation has priority over addition For instance a be always means a be for a b c C R The following facts for a b C R are immediate consequences of the axioms i 0a 0 a0 ii 7ab 7ab a7b iii 7a7b ab 103 Examples a 0 is a ring with 0 1 the trivial ring If one has 0 1 in some ring R then R is the trivial ring b Z C Q C R C C are rings N is not a ring c If R i C I are rings then the cartesian product Hid R is again a ring with componentwise addition and multiplication The O element is the tuple with entry OR in the i th component and the 1 element is the tuple with 1R in the i th component d If R is a ring one de nes its opposite ring Rquot as the same set R with the same addition as R but with multiplication gien by a b ba Note that R R e Let R be a ring and n C N Then MatnR the set of n x n matrices with entries in R is a ring with the usual matrix addition and multiplication 44 f For every ring R the polynomial ring RX in the indeterminate X with coef cients in R given by Ftlea0a1Xa2X2aanln N0a0an R is a ring with the usual addition and multiplication of polynomials Note We do not View polynomials as functions from R to R but as abstract sums of the above form Two polynomials pX a0 a1X 01an and qX b0b1X mem are equal ifan only ifbl a for all i 6 No where we set ai 0 ifigt nand bi 0ifigtm lfpX a0a1XanX with an 7 0 we call an the leading coe cient of pX and degpX n the degree of pX The O polynomial has no leading coef cient lts degree is foo by de nition Note that degpXqX degpX degqX provided that ab 0 implies a 0 or b 0 for all a b E R If pX E RX has leading coef cient 1 we call pX a manic polynomial One could introduce RX also as the set of all sequences 010011012 in R with only nitely many non zero entries together with componentwise addition and multiplication given by a0 a1 b0 b1 0001 02 where oi aobi 111371 aibo for i 6 No The element X corresponds to the sequence 0 10 One could also introduce RX as the set of functions 1 No a R which are non zero on only nitely many elements together with the usual addition of functions and the multiplication given by f 220 fkgi 7 k for i 6 N0 also called convolution g More generally one introduces the polynomial ring RX1 Xn in n indeterminates X1 Xn with coef cients in R in a similar way Ab stractly one can introduce this ring again as the set of functions 1 N3 a R with 7 0 for only nitely many 1 i1in 6 N3 More intu itively one sets X3 X11 Xf for g i1in 6 N3 and de nes RX1 Xn as the set of formal nite sums of elements of the form rXi with r E R and 1 6 N3 and the usual addition and multiplication One can also de ne polynomial rings RX l i E I for any set of indeter minates Xi i E I h If A is an abelian group then its endomorphism ring EndA HomA A is a ring with pointwise addition f ga fa 9a for fg E EndA and a E A and composition as multiplication Verify i If R is a ring and G is a monoid then the monoid m9 RG is the set of all formal sums 2960 rgg with g E G and r9 6 R such that all but nitely many coef cients r9 are 0 Addition is de ned by Z T99 l 599 09 l 599 960 960 96 45 and multiplication is de ned by Z 799 Shh Z rgsh gh Z 7132939 960 heG gheGgtltG 960 Ragga If G is a group we call RG the group ring of C over R Note that the polynomial ring RX can also be interpreted as the monoid ring over the monoid N0 under addition over the ring R 104 De nition Let R be a ring a An element u E R is called a unit of R or invertible in R if there exists v E R with uv 1 U Ll In this case7 the element v is uniquely determined It is called the inverse of u and is denoted by u l The set of invertible elements of R forms a group under multiplication It is called the unit group of R and is denoted by RX or UR b Assume that R is commutative An element a E R is called a zero divisor if there exists I E R 0 such that ab 0 105 Examples a If R is a commutative non trivial ring then 0 is a zero divisor b In the ring Z x Z the element 27 0 is a zero divisor c Let R be a commutative ring The unit group of the ring MatnR is denoted by GLMR and called the general linear group Note A is invertible in MatnR if and only if detA is a unit in R In fact7 one direction follows from the product formula for determinants and the other direction follows from the explicit formula for the inverse7 involving the inverse of the determinant d For every group G and every ring R7 the elements of G form a subgroup of RGX 106 De nition Let R be a ring a R is called an integral domain if it is commutative7 if 1 7 0 and if 0 is the only zero divisor of R b R is called a division ring if 1 7 0 and if RX R c R is called a eld if it is a commutative division ring 107 Examples a Every eld is an integral domain b Z is an integral domain c If R is an integral domain then RX is an integral domain and X RX RX 46 108 De nition A subset S of a ring R is called a subring if S together with the restricted addition and multiplication is again a ring lf additionally 13 C S then we call S a unitary subring of R 109 Remark Let R be a ring a A subset S of R is a unitary subring of R if and only if 13 C S and aibCSandabCSforalla7bCS b S 730 l r C R is a subring of the ring R x R lt s identity element is 10 S is not a unitary subring of R x R c Every non trival subring of a eld is a unitary subring and an integral domain d The intersections of unitary subrings of R is again a unitary subring of R 1010 De nition Let R be a commutative ring and let S be a unitary subring of R For any subset Y of R we denote by SY the intersection of all subrings of R that contain S and Y It is the smallest subring of R containing S and Y It is equal to the set of elements ofthe form 19yl7 7y where n EN pX C SX1Xn7 and y17yn C Y 1011 Examples a Z C Q C R C C are unitary subrings of each other Q R and C are elds b abi l a7 b E Z is a subring of C called the ring of Gaussian integers It is an integral domain c a bi l ab 6 Q is a subring of C It is a eld7 called the eld of Gaussian numbers In fact7 1 a b abi 77a2b2 77a2b22 for a bi 7 0 d a bxgi l ab 6 Z is a subring of C and an integral domain e For each ring R7 its center ZR r C R l rs sr for all s C R is a unitary commutative subring of R f The set of 2 x 2 matrices of the form 3 wee is a unitary subring of the ring Mat2C verifyl It is a non commutative division ring In fact7 for a non zero matrix as above7 we have a6 bb 7 0 47 and we can check easily that 1 a 7b aab5 b a is its inverse This division ring is denoted by H and called the quaternion division ring discovered by Hamilton in 1843 If one sets 1 710 72 0 7 01 kA7H702 quot0172quot07 7Jquot7107 quot23 107 then H is a 4 dimensional subspace with basis 1 i j k of the 8 dimensional real vector space Mat2C lts center is equal to R 1 the eld of real numbers Note that i1 ii ij ik is a subgroup of the unit group of H isomorphic to the quaternion group Q8 verifyl g Every ring R is a unitary subring of the polynomial ring RX 48 11 Homornorphisms Ideals and Factor Rings 111 De nition Let R and S be rings A function f R H S is called a ring homomorpism if fabfafb and fabfafb for all a7 1 E R The notions mono 7 epi 7 iso 7 endo 7 and automorphism are de ned as for groups R and S are called isomorphic notation R E S as for groups if there exists an isomorphism f R H S If 1 RH S is a ring homomorphism then its kernel is de ned as kerf a E R fa 0 112 Remark Let R and S be rings and let 1 R H S be a ring homomor phism a We do not require that f1 1 and this does not follow from the axioms lf f1 1 then we call 1 a unitary ring homomorphism If f is surjective then f is unitary If f is unitary and if z E RX then fz E Sgtlt and MA MW b The composition of two unitary ring homomorphisms is again a unitary ring homomorphism If f is bijective then also f 1 is a ring homomorphism The automorphisms of R form a group AutR7 the au tomorphism group of the ring R 113 Examples a Complex conjugation is a ring automorphism of C b Let R be a ring The map r R H R x R7 r H r707 is a ring homomorphism7 but it is not unitary unless R is trivial c For any ring R7 a E R and n E Z we de ne aa l la7 ifrigt07 na 0 ifri07 ifrilt07 7 7017a7a as we did for groups now applied to the additive group R7 The func tion Z H R7 71 H it 17 is a unitary ring homomorphism in fact the unique unitary ring homomorphism from Z to R lts image is contained in ZR lts kernel must be of the form kZ for a unique k 6 No We call k the characteristic of R It is denoted by charR d Let S be a ring7 R Q S a unitary subring with R Q ZS For every a E S one has the evaluation map ea RX H S7 pX H pa It is a unitary ring homomorphism lf pa 0 we call a a zero7 or a root7 ofpX is S 49 114 De nition Let R e a ring and let I C R be an additive subgroup of RieOEIandmiyEIforallzweI a I is called a left ideal if rm 6 I for all r E R and all z E I b I is called a right ideal if xi 6 I for all r E R and all z E I c I is called an ideal or sometimes for emphasis a two sided ideal if it is a left and right ideal 115 Proposition Let R be a ring and let I be an additive subgroup ofR The following are equivalent i I is an ideal ofR ii Thegroup RI7 is a ring under rIsI rsI for r s E R iii There exists a ring S and a ring homomorphism f R a S with I kerf Proof iii We know that RI7 is an abelian group The multipli cation from ii is well de ned Let 77773 s E R with r I r I and sIs I Thenthere exist 71 6 Iwith r rz and s sy It follows that TSI rmsyI rsrymszyI TSI since rymsmy 6 I7 as I is an ideal of R It is easy to check that RI becomes a monoid with identity element 1 I under this multiplication Distributivity for RI follows immediately from the distributivity of R iiiii Let V R a RI7 r gt gt r I7 be the natural epimorphism of the additive groups It is easy to check that it is a ring homomorphism Obviously7 kerf I iiii Let f R a S be a ring homomorphism with kerf I For every r E R and every x E I we have rm fzfr O Thus7 I is a left ideal of R Similarly7 I is a right ideal of R l 116 De nition Let R be a ring and let I be an ideal The ring RI with the multiplication de ned as in Proposition 115ii is called the factor ring of R with respect to the ideal I Note that the canonical epimorphism V R a RI is a unitary ring homomorphism 117 Examples Let R be a ring a 0 and R are always ideals of R If R 7 0 and if these are the only ideals then R is called simple Warning This de nition of a simple ring is not the usual convention For a ring to be simple one usually additionally requires that the ring is artinian7 a notion that we do not introduce in this class Our de nition of simple is often called quasi simple in the literature 50 b Let I be a left or right ideal of R and assume that I contains a unit a of R Then I R In fact in the case of I being a left ideal we have 1 u lu E I and then r r 1 E I for all r E R A similar argument is used if I is a right ideal c Let A be a set and let Ia Oz 6 A be a collection of left ideals of R Then aeA ID is again a left ideal of R Similar statements hold for right ideals and for ideals d Let X Q R be a subset The left ideal ofR generated by X is de ned as the intersection of all left ideals of R that contain X It is the smallest left ideal containing X and is denoted by RX Similarly one de nes the right ideal and the ideal of R generated by X They are denoted by X R and by X respectively One has RXnm1mmnln ll0 nrn R 1mn X Rx1nznmln ll0 nrn R 1mn X x X nz131rnxnsnlnENO nrn31sn R 1mn X le 0 then RX XR X If R is commutative then X RX XR lf additionally X consists of only one element then zR Rm is called a pn39ncz39pal ideal of R e Let Ia Oz 6 A be a collection of left ideals of R Then their sum is de ned by Z Iama1man ln 6 N004104n E Ama E Ia1man 6 Ian 04614 It is again a left ideal in fact the left ideal generated by UaeA Ia One de nes sums of right ideals and of ideals the same way They are again right ideals and ideals respectively f Let I and J be ideals Their product I J or just IJ is de ned by IJ1y1mnynln N0 17n617 y177yn Jl It is again an ideal For ideals I JK ofR one has IJK I IK IJ Q I J IJKIJK IJKIK JK IJK IJIK g The ideals of Z are precisely the subsets kZ with k 6 NO The factor ring of Z with respect to kZ is ZkZ with multiplication m kZn kZ mm kZ In the ring Z we have the following identities between ideals k l gcdk7l7 k m l 16mk7l7 kW kl 51 118 Theorem Fundamental theorem of homomorphisms for rings Let R and S be rings let 1 R a S be a ring homomorphism and letI be an ideal ofR with I Q kerf There exists a unique ring homomorphism RI a S such that 0 1 f where V R a RI denotes the natural epimorphism Moreover imq imf and kerq kerfI In particular Rkerf E imf as rings Proof By the fundamental theorem of homomorphisms for groups there exists a unique group homomorphism RI a S for the underlying addi tive groups such that 0 1 1 For T r E R we have Ir I rr I n frfr TIr I Thus is a ring homo morphism The statement about the image and the kernel of only depends on the additive group structure and therefore follows from the group version of the theorem 1 119 Theorem 1st isomorphism theorem for rings Let R be a ring let S Q R be a unitary subring and let I be an ideal of R Then S I 5m l s E Sz E I is a unitarysubring ofR S I is an ideal ofS and there exists a ring isomorphism gtSSOIHSII With gtsS IsIforalls S Proof For 3 s E S and zm E I one has szs z 33H sz zs zz E S I Moreover 13 E S Q S I and S I is an additive subgroup of R Thus S I is a unitary subring of R Since I is an ideal in B it is also an ideal in SI Form 6 S Iand s E Swe have sz 6 S I and also ms 6 S O I Since S O I is an additive subgroup of S it follows that S O I is an ideal of S Finally the rst isomorphism theorem for groups yields a group isomorphism j SS O I a S II with gts S O I s I For 33 6 S we have gts S Is S gtss SO I 33 I s Is I gts S I gts S I Thus 1 is a ring homomorphism and the theorem is proved l 1110 Theorem Correspondence theorem and 2nd isomorphism theorem for rings Let R be a ring let I be an ideal ofR and let V R a RI denote the natural epimorphism a The function S gt gt SI de nes a bijection between the unitary sub rings of R containing I and the unitary subrings of RI Its inverse is T gt gt V 1T 52 b The function J gt gt JI de nes a bijection between the left resp right resp two sided ideals of R containing I and the left resp right resp two sided ideals of RI c IfJ is an ideal ofR containingI then RIJI RJ as rings Proof This follows immediately from the correspondence theorem and the second isomorphism theorem for groups after verifying that images and preimages of left right ideals resp unitary subrings under a ring epi morphism are again left right ideals resp unitary subrings The latter is an easy homework problem 1111 Theorem Chinese Remainder Theorem Let R be a ring let n E N let I1 In be ideals off and denote by 39y the function 39yR7gtRI1XXRIn Tgt gtTIlTIn a 39y is a unitary ring homomorphism with ker39y 121 Ik In partic ular RI1 In is isomorphic to a unitarysubring ofRI1 x xRIn b IfIk I1 R for all kl E 1 n with k 1 then 39y is surjective In particular RI1 O In RI1 x x RIn c IfIkIl R for all kl E 1 n with k 7 l and R is commutative then I1 In I1 In In particular RI1 I E RI1 x x RIn Proof a This is an easy veri cation b For each pair kl with kl E 1n and k 7 l we can nd elements akl E Ik and bk 6 I such that akl bk 1 For every I E 1 n set 31 alla21quot39anl a product with n 7 1 factors there is no factor all Then 31 E Ik for every k 7 l and s E 1 I1 The latter follows from s 1 7 b111 7 bgl 1 7 bnl To show the surjectivity of 39y let r1 rn E R be arbitrary and set r r131r252 er By the above properties of the elements 81 we have for every k E 1 n the equation TIk T181Ik Tn8n1k Tk8k1k Tk1k8k1k Tk Ik1 Ik Tk Ik Thus yT T1 I1 Tn In c Obviously I1In Q I1 O In It suffices to show the other inclusion We do this by induction on n The case n 1 is trivial We also need to consider the case n 2 for later use Let akl and bid be as in the proof of part b for kl E 1 2 with k 7 I Then for every r 6 I1 Ig we have T Ta12 512 mm 7512 112 7512 E 112 53 Finally if n 2 3 and akl and bk are as before then we have 1 aln b1nquot an71n bn71n 1 1771 In which implies that 1 In71 In R Note that by induction we have 1 In1 1 In1 Using this and the case n 2 applied to the two ideals 1 In1 and In we obtain 1 In71 In Q 1In1 ln Q 1 In1In and the theorem is proved l 1112 Corollary Let m1 Wm E N be pairwise coprime and set m m1mr Then 27712 gt 2m12 X X 2m72 am2gt gtam12 am72 is an isomorphism In particular for any a1 aT E 2 there exists a E 2 such that a E ai mod mi for alli E 1 r Moreover the integer a is uniquely determined modulo m Proof Follows immediately from Theorem 1111 l 1113 Example Let n p 19 be the prime factorization of n E N Then 2712 2pf12 X X 2piT2 as rings and in particular as additive groups So for example 2122 E 242 x 232 1114 De nition The Euler Ap functz39on 4p N a N is de ned by Mn KZnZW 1115 Remark Let n E N a For a E 2 one has a 712 E 2712gtlt ltgt there exists I E 2 with a n2b 712 1 712 ltgt there exists I E 2 with ab 712 1 712 ltgt there exist I c E 2 with 1 ab no ltgt gcdan 1 54 Thus Lplt7lgt Ha E 1 n lgcdan b If n m1 m with pairwise coprime natural numbers m1 m then the isomorphism in the Chinese Remainder Theorem induces an iso morphism of the unit groups ZnZX Zm1Z x x Zm2x 2le x x ZmTZX Thus Lpn ltpm1 4pm In particular if n p 19 is the prime factorization of n then Lpn 4ppi14pp c For a prime p e E N andi E 1 p5 we have gcdip5 gt 1 if and only ifp l i This implies that Lpp5 p5 7p5 1 p5 1p 7 1 Altogether if n p 19 is the prime factorization of n then Mn 291171091 71p5 1pr 71 e Obviously we have ZnZ is eld ltgt 4pn n 7 1 ltgt n is a prime 1116 Theorem Fermat For every n E N and a E Z one has gcdan 1 ltgt aw E 1 mod n In particular for a prime p and a E Z one has pla ltgt 0177121 modp Proof If gcda n 1 then a nZ e 2712X and a 72 1 nZ since the group lZnZgtlt has order Conversely if aw E 1 mod n then there exists q E Z with WW 7 gm 1 This implies gcdan 1 l 1117 De nition Let R be a ring An ideal M of R is called maximal if M 7 R and if there exists no ideal I ofR with M C I C R Note that by the correspondence theorem we have for every ideal M of R M is a maximal ideal of R if and only if RM is a simple ring Similarly one de nes maximal left and right ideals 1118 Proposition Let R be a commutative and simple ring Then R is a eld Proof Let a E R As R is simple we have aR B This implies that there exists I E R with ab 1 Thus a is a unit of B By de nition R is a eld I 55 1119 De nition Let R be a commutative ring An ideal P of R is called aprime idealifP 7 Randifonehas for all 011 6 R ab 6 P a a E P or b E P 1120 Proposition Let R be a commutative ring and let I be an ideal of a I is a maximal ideal ofR ltgt RI is a eld b I is a prime ideal ofR ltgt RI is an integral domain Proof a I is maximal if and only if RI is simple By Proposition 1118 if RI is simple then RI is a eld The converse follows immediately from the de nitions 10 RI 7 0 since I 7 B Let ab E R with aIbI 0 Then ab 6 I Since I is a prime ideal we havea E I or b E I Thus aI 0 or b I 0 s Since RI 7 0 we have I 7 B Let 011 6 R with ab 6 I Then 0 ab I a Ib I Since RI is an integral domain we obtain aI00rbI0Thusa Iorb I l 1121 Examples a In a commutative ring R every maximal ideal is a prime ideal but in general not every prime ideal is maximal For instance 0 is a prime ideal of Z but not a maximal ideal 10 Let n 6 NO The ideal n2 is a maximal ideal of Z if and only if n is a prime The ideal n2 is a prime ideal if and only if n is a prime or n O 1122 De nition a A partially ordered set X g is a set X together with a relation satisfying i m g m iimltyandyltzmyand iiimltyandyltzxltz for alll myz E X A partially ordered set X g is called totally orderedif for every x y E X onehasm yory z 10 Let Xlt be a partially ordered set and let Y Q X An element x E X is called an upper bound of Y if y g z for all y E Y c Let X g be a partially ordered set An element x E X is called maximal in X if for every element y E X one has x g y a z y 1123 Lemma Zorn s Lemma Let X be a non empty partially ordered set and assume that every totally ordered subset ofX has an upper bound in X Then X has a maximal element 56 1124 Remark We do not prove Zorn s Lemma It is equivalent to the axiom of choice which is used as one of the axioms in set theory Zorn s Lemma is for example used in the proof that every vector space has a basis We illustrate its use in the following theorem 1125 Theorem Let R be a ring and let I be a left resp right resp two sided ideal ofR with I 7 R Then there exists a maximal left resp right resp two sided ideal M of R with I g M In particular every non zero ring has maximal left right and two sided ideals Proof We only prove the statement about left ideals The others are proved in the same way Consider the set X of all left ideals J ofR with I Q J C B This set is non empty as I E X and partially ordered by inclusion Let Y be a totally ordered subset of X By Zorn s Lemma it suffices to show that there exists an upper bound for Y in X Let J1 denote the union of all elements J of Y Then it is easy to see that J1 is again an ideal Obviously7 it contains all elements of Y7 so it is an upper bound in X if we can show that J1 E X Since I Q J17 it suffices to show that J1 7 R Assume that J1 R Then 1 6 J1 Since J1 is the union of the left ideals J 6 Y7 there exists J E Y such that 1 E J But this is a contradiction since every element J E X satis es J 7 R 57 12 Divisibility in Integral Domains Throughout this section B denotes an integral domain Thus7 R is a com mutative ring with 1 7 0 and ab 0 implies a 0 or b 0 for all 071 6 R 121 De nition Let a and b be elements of R a One says that a divides b notation a l b if there exists 0 E R with ac b 10 One says that a and b are associated notation a N b if a l b and b l a 122 Remark Let a719 07b17bmm177mn E R and let u E RX Then the following rules are easily veri ed from the de nitions aalaandal0 b Ola ifand only ifa0 d a l u ifand only ifa E RX e lfalbandblcthena c lfa b17a l bn thena l x1b1mnbn g a l b if and only if bR Q aR h N is an equivalence relation on R 123 Proposition For 071 6 R the following are equivalent i a N 1 ii aR 1913 iii There exists u E RX with au 1 Proof ltgt ii This is immediate from Remark 122g iiiii First note that if a 0 then also I 07 since I 6 LR and then a 1 b We assume from now on that a 7 0 Since aR bR there exist 0 d E R such that b ac and a bd Substituting the rst expression for b in the last equation we obtain a acd This implies a 7 Led 0 and a1 7 Cd 0 Since a 7 0 and since R is an integral domain7 we obtain 1 Cd In particular 0 E RX and iii holds iiii Obviously7 au 1 implies a l I Since u is a unit in R we also have a bu l and therefore I l a 124 Remark a In Z an integer a is always associated to a and 7a and no other element b ieRlleRgtlt and aERlaNO0 58 c The group RX acts on R by left multiplications u7a gt gt ua The orbits of this group action are the equivalence classes with respect to N d For a177an E B one has a177anm1a1mnanl177mn Ra1RanR e FordERanda177an Ronehas dla177dlan ltgt a1 EdR77an dR ltgt a177an dR In this case the element at is called a common divisor of a17 7 an Every unit is a common divisor of a17 7a If these are the only common divisors of a17 7 an then we call a17 7a coprime The elements a17 7a are called pairwise coprime if a7 and aj are coprime whenever i 7 j For instance7 767 15 and 10 are coprime in Z7 but they are not pairwise coprime f FormERand a177an Ronehas a1 lm77anlm ltgt mealR77m anR ltgt mREalR w anR In this case we call in a common multiple of a17 7 an 125 De nition Let a17 7a 6 R a A common divisor d of a17 7an is called a greatest common divisor of a17 7 an if every common divisor of a17 7a divides d The set of all greatest common divisors of a17 7a is denoted by gcda17 7an b A common multiple in of a17 7a is called a least common multiple of a17 7an if m divides every common multiple of a17 7an The set of all least common multiples of a17 7a is denoted by lcma17 7an 126 Remark A greatest common divisor resp least common multiple of a17 7an E B does not exist in general But if there exists one then gcda17 7a resp lcma17 7an is a single full associated class For instance7 in Z we have gcd6715 737 3 and lcm6715 7307 30 127 Example We want to get acquainted with the ring R a bx5i l a7b E Z One tool to study it is the norm map N R a N0 de ned by Na b i a bx5ia 7 b i la bx5il2 a2 512 127a One has Nrs NrNs for all r7 3 E R Thus7 for r7t E R7 we have i l t in R s No Nt in Z 12710 59 First we show that RX 711 Clearly 1 and 71 are units in R Con versely assume that u E RX Then by 127b NuNu 1 N1 1 and Nu 1 or Nu 71 But a2 5b2 1 has precisely two solutions namely ab i10 corresponding to u i1 and a2 5b2 71 has no solution We want to determine all common divisors of r 6 and s 21 Assume that t a bxSi is a common divisor of r and s Then since the norm map is multiplicative Nt a2 5b2 is a divisor of Nr 36 and also a divisor of Ns 24 Thus Nt is a divisor of 12 This implies that a2 5b2 6 1 2346 12 Obviously 23 and 12 are not of the form a2 5132 Moreover Nt1 ltgt te i1 Nt4 ltgt te i2 Nt 6 ltgt t6 i15ii175i Clearly i1i2i1 are common divisors of 6 and 21 On the other hand 1 7 f 21 since 22522541578454 12 f2f2f2 f7775i R39 17 6 6 3 3 Thus i1 i2 and i1 are the common divisors of r and 3 But none of them is a greatest common divisor since Ni1 1 Ni2 4 and Ni1 xSi 6 128 De nition Let p E R 0 and assume that p is not a unit in R a p is called irreducible if for all ab E B one has p ab a E RX or b E RX b p is called a prime element if for all 011 6 B one has p l ab p l a or p l b 129 Remark a lfp E R is irreducible resp a prime element in R then every element of R that is associated to p is as well irreducible resp a prime element b If p is a prime element then p is irreducible In fact assume that p ab with a b E R Since p is a prime element it follows that p l a or p l 1 Without loss of generality assume that p l a Then a pc for some 0 E R It follows that p ab pcb This implies that be 1 and that b E RX as desired 60 1210 Example a In Z we have for all integers p p is a prime element ltgt p is irreducible ltgt lpl is a prime number b In R the element p 1 is irreducible In fact7 assume that 1 rs Then 6 N1 NrNs Since 2 and 3 are not in the image of N7 we obtain Nr 1 or Ns 1 But this means r i1 6 RX or s i1 6 RX as desired However7 p is not a prime element7 since p l 1917 6 2 3 but p l 2 and l 3 Np 67 N2 4 and N3 9 1211 Proposition Let p E R Then p is a prime element in R if and only ipr is a prime ideal ofR Proof Since p is not a unit7 we have pR 7 R Assume that 071 6 R are such that ab 6 pR Then p l ab Since p is a prime element7 we obtain p l a or p l b This implies a E pR or b E pR as desired a Since pR 7 R7 the element p is not a unit in R Assume that at7 b E R are such that p l ab Then ab 6 pR7 and since pR is a prime ideal7 we obtain a E pR or b E pR But this means p l a or p l 67 as desired 1212 Theorem Assume that as 6 N0 and 111315191 7197 and q1 7q5 are prime elements in R such that p1 pT q1q9 Then T s and after a suitable renumbering of the elements ql 7q5 one has pi N qi for alli1r Proof We prove the result by induction on r Let r 0 Then ql qS 1 If s gt 0 then ql is a unit in contradiction to being a prime element Therefore7 s 0 Now let T 2 1 and assume that the statement is true for all smaller values of r Since p1pT q1qS7 we have p l q1q9 Since p is a prime element7 we have p l qi for some i 17 73 After renumbering the elements ql 7q5 we may assume that p l qs Then there exists it E R with qs 14197 As qs is also irreducible by Remark 1297 we have u E RX and p is associated to qs Now we have p1 pT ql q911upr and after canceling p on both sides we have 191 4971 qi q972q971u Thus7 by induction7 r71 3717 and after reordering the elements ql q5127q511u W8 haVe P1 N my 7 19772 N q39riZ and 19771 N qrilu N qril D 61 13 Unique Factorization Domains UFD Princi pal Ideal Domains PID and Euclidean Do mains Throughout this section B denotes an integral domain 131 De nition R is called a unique factorization domain or UFD if ev ery non zero element a E R with a RX can be written as a product of prime elements Note that in this case it can be written uniquely in the sense of Theorem 1212 132 Theorem The following are equivalent i R is a unique factorization domain ii Every non zero element a E R with a RX can be written as a product of irreducible elements and if p1 pT q1qS with irreducible elements 191prq1qS E R then r s and p N q after a suitable renumbering of the elements ql qs iii Every non zero element a E R with a RX can be written as a product of irreducible elements and every irreducible element of R is a prime element Proof iiii This is obvious iiiii This is immediate from Theorem 1212 iiii Since every prime element is irreducible it suf ces to show that every irreducible element a of R is a prime element By i there exist prime elements p1 p of R with a p1 pT Since a is irreducible we have T 1 and a p1 is a prime element iiiii We need to show that every irreducible element p of R is a prime element Assume that p ab for ab E R Then pc ab for some 0 E R We may assume that a 7 0 since otherwise p l a and we are done and that b 7 0 Also we may assume that a RX since otherwise pca 1 b and p l b so that we are done and 1 RX Then also 0 7 0 RX since if c is a unit then p c lab can be written as a product of two non units contradicting the fact that p is a prime element Thus we can write a p1 pk b q1ql c r1 rm as product of irreducible elements and obtain pr1rm p1 qu1ql The uniqueness part of ii implies that p N p or p N qj for somei E 1l orj E 1 l Thusplaorplb l 62 133 Remark Assume that R is a UFD and let 73 be a set of representatives for the prime element of R with respect to the equivalence relation N Then every element a E R 0 has a unique prime factorization aqueF 12673 with u E RX 517 6 No such that 517 0 for all but nitely many 19 E 73 Let also I E R 0 and let I 121117673 pf be its prime factorization Then the following statements follow easily from the unique prime factorization aalb ltgt ep fpforallpep b Hpep pmh ewfp is a greatest common divisor of a and b c Hpep pma epfpl is a least common multiple of a and b d a and b are coprime if and only if7 for every p E 73 one has 517 0 f 7 e lf 1 E gcdab and m E lcmab then ab N dm f If a and b are coprime and c E R satis es a l c and b l c then ab l c 134 Example a Z is a UFD and we may choose for 73 the set of positive prime numbers b is not a UFD7 since 1 is irreducible but not a prime element 135 De nition R is called a pn39ncz39pal ideal domain or PlD if every ideal I of R is principal7 ie7 I a aR for some element a E R 136 Example Z is a PlD7 since every ideal of Z is of the form nZ for some 77 E No 137 Proposition Assume that R is a PID and let p E R Then the following are equivalent i p is a maximal ideal ii p is a prime ideal iii p is a prime element iv p is irreducible Proof i ii This was shown in Example 1121 ii ltgt iii This is the statement of Proposition 1211 iii iv This was shown in Remark 129b iv i We have p 7 R7 since p RX Now assume that I is an ideal of R with p Q I 7 R Since R is a PlD7 there exists a E R with 63 I a It follows that p ab for some I C R Since I 7 R a is not a unit Since p is irreducible the element 1 must be a unit in R Since p ab we obtain p a I This shows 138 Theorem HR is a PID then R is a UFD Proof Assume that R is not a UFD Then by Proposition 137 and The orem 132 there exists a non zero element a1 6 R RX such that an is not a product of irreducible elements In particular a1 is not irreducible itself Thus we can write a1 agbg with non zero elements a2 2 C RRX Since an is not a product of irreducible elements the same must be true for Lg or 2 After interchanging a2 with 2 we may assume that a2 is not a product of irreducible elements Note that since I is not a unit in R we have 04 C a2 We can repeat the above with a2 in place of an to obtain an element 13 which is not a product of irreducible elements such that a2 C a3 We can iterate this process to obtain an in nite strictly ascending chain 04 C a2 C 13 C Note that I UnEan is an ideal of R Since R is a PlD there exists a C I such that I a But by the construction of I there exists n C N such that a C an This implies a Q an and we obtain the contradiction a Q an C an Q I a Thus R is a UFD l 139 Theorem Bezout Let R be a PID and let a1 an d C RThen d6gcda1an ltgt da1an In particular ifd C gcda1 an then there exist 71 7 C R such that dna1 rnan Proof There exists a C R with an an a From Remark 124e it follows that a C gcda1 n Since also 1 C gcda1 an we have a N 1 Thus 1 a a1 an s This follows immediately from Remark 124e l 1310 Corollary Let R be a PID and let an an E R Then 11 an are coprime ifand only ifthere exist 71 rn C R with 1 r1a1 rnan Proof This is immediate from Theorem 139 l 64 1311 De nition R is called a Euclidean domain if there exists a function N R a N0 with N0 0 such that for any two elements ab E R with b 7 0 there exist qr E R with a bq r and with r 0 or Nr lt Nb In this case N is called a Euclidean norm for R 1312 Example a The ring Z is a Euclidean domain with N Z a No a gt gt lal b For every eld F the polynomial ring FX is a Euclidean domain with N FX a N0 de ned by N0 0 and NpX degpX for pX E FX See Homework problem c The ring of Gaussian integers R a bi l 411 6 Z is a Euclidean domain with N gt gt N0 04 a bi gt gt a2 b2 046 l04l2 In fact let 043 6 R with B 7 0 We can write 043 z yi with zy E Q Choose mn E Z such that lz 7 ml g 12 and ly 7 nl g 12 and set 39ymni Randp047339y R Obviously 0439y pand NP la 7 vlz l lzlg Vlz NWWE m y 7 71W N 7702 y i 702 lt N 2 52 lt NW lt NW since NW 7 O 1313 Theorem HR is a Euclidean domain then R is a PID More pre cisely ifN is a Euclidean norm for R and ifI 7 0 is an ideal ofR then I b for every I E I 0 with Nb minNa l a E I Proof lt suffices to show the second statement Since I E I one has 1 Q I Conversely let a E I Since I 7 0 there exist qr E R with a qb r and with r 0 or Nr lt Nb Note that r aiqb E I lfr 7 0 then Nr lt Nb contradicting our choice of 1 Thus r 0 and a qb E I l 1314 Remark Euclidean Algorithm Assume that R is a Euclidean do main and that N is a Euclidean norm for R For ab E R 0 one can compute a greatest common divisor d of a and b and also elements r s E R with d NH 51 using the following so called Euclidean Algorithm by nding 65 elements 71 6 N0 q17qn1 E R and r1 776 E R with aqibT17 71 7307 N01 ltNb7 b W1 T27 T2 7 07 N02 lt NT17 7 1 quz T37 r3 7 0 NT3 lt N02 Tn72 annil TV Tn 7g 07 lt NTn717 Tnil qn1rn Then Tn is a greatest common divisor of a and b and can be expressd as a linear combination of a and b using the above equations If already 71 0 then b is a greatest common divisor of a and b More explicitly Compute triples uhvhwi E RE7 i 017 recur sively as follows Step 1 Set 140120100 1070 and 141121101 0171 Step 2 If ui vi wi with w 7 0 is computed then nd elements qi and M with veal1 qiwi n such that n 0 or Nn lt Nwi and set Ui17 Ui17 wi1gt 3 Ui717vi717wi717 qiltui7 Uiy UM lf le 0 then W E gcda7 b and W uia vib lf le 7 0 then repeat Step 2 66 14 Loca za on Throughout this section let R denote a commutative ring 141 Theorem Let S Q R be a multiplicatively closed subset ofR contain ing 1 a The relation on R x S de ned by T17 81 N T2 82 C ltgt 3t 6 S 2 718275 738175 is an equivalence relation The equivalence class of T7 5 is denoted by g and the set of equivalence classes is denoted by S lR One has 2 g for all T E R and all 572 E S b S lR is a commutative ring under T1 T2 7 T182 T281 81 82 39 8182 and T1 T2 7 T1T2 81 82 39 8182 with O element and 1 element c The function Li R a S lR T gt gt is a unitary ring homomorphism with the property LS Q S lRX Proof a Re exivity and symmetry clearly hold To show transitivity7 let T1781 N T2782 and T2782 N T3783 Then there exist thtg E S with Tlsgtl Tgsltl and Tgsgtg T332252 Multiplying the rst equation by sgtg and the second equation by sltl yields Tlsgtlsgtg Tgsltlsgtg Tgsgtgsltl This implies T1781 N 763337 since Sgtltg E S Moreover7 g since Ttsl T3751 with 1 E S b This is shown in the following steps7 each ofwhich is a straightforward veri cation i Addition is well de ned ii Multiplication is well de ned iii Addition is associative v is an additive identity element 77 is an additive inverse of i Addition is commutative vii Multiplication is associative iii is a multiplicative identity element ix Multiplication is commutative ltH lt AAA2AAAA 67 X Distributivity holds c First note that i1 1 which is the 1 element of S lR Moreover we have for 737 E R and s E S mmg mwww 7 1 f 1 LW i 1 1 1 WMTL 1 s 1 1 LS Ei gi39 This completes the proof l 142 Remark a S lPt is called the localization ofR with respect to S b It is not dif cult to show that i is injective if and only if S does not contain any zero divisors of R check This is for example true when R is an integral domain and 0 S In this case we can View R as a unitary subring in S lPt by viewing the element r as Also the notation g is justi ed by noting that g 1 1 f r 3 1 under this identi cation c lfO E S then S lPt In fact g g for all 733 E R x S since r10 s 0 0 Conversely if 0 e S then S lR 7g 0 since 1 7g g In fact0137 11sforalls S 143 Examples a If R is an integral domain and S R 0 then S lPt 1rs E Rs 7 0 is a eld In fact ifg 7 then r 7 0 and therefore r E S It follows that g 3 1 so that g is invertible with inverse In this case S lPt is called the eld of fractions or quotient eld of R It is denoted by QuotR Note that QuotR contains R as a unitary subring if one identi es the elements of R with elements in QuotR via i R a QuotR b For any given element 1 E R the set S 1 1 f2 f3 is multi plicatively closed In this case the ring S lPt is also denoted by Rf c For any prime ideal P of B one can choose S to be R P the complement of P in R In this case the ring S lPt is also denoted by Rp The set Pp 1 r E P 3 P is an ideal of Rp such that every element in Rp Pp is a unit In fact let g E Rp such that g Pp Then T P and f is an inverse of Moreover 1 Pp In fact assume that 1 g with r 6P ands P Thenthere existst RPsuchthat1st1rt This implies st rt 6 P but 3 P and t P a contradiction to P being a prime ideal Therefore Pp does not contain a unit and Pp Rp RpX 68 A ring T such that M T TX is an ideal is called a local ring In this case M is a maximal ideal and the unique maximal ideal of T So RP is a local ring explaining the terminology of localization 144 Theorem Universal property of Li R a S lR Let 1 E S Q R be multiplicatively closed let T be a ring and let 1 R a T be a unitary ring homomorphism with fS Q TX Then there exists a unique unitary ring homomorphism f S lR a T such that f f0 L R 4 T S lR Moreover frfs 1 for all r E R and s E S Proof Uniqueness If f S lR a T is a unitary ring homomorphism with Mr M then f M for an 7 e R argd re ew f 1 fs 1 for all s e 5 Hence g a f f frfs 1 for all r E R and all s E S This shows the uniqueness of f and the last statement of the theoremL Existence We de ne frfs 1 for r E R and s E S First we show that f is well de ned Assume that g g with r E R and s E S Then there exists it E S with rs t r st and we obtain frfs ft fr fsft Since ft is invertible in T this implies frf s fr fs Multiplying both sides by fs 1fs 1 and noting that imf is commuta tive we obtain frfs 1 fr fs 1 Next we show that f is additive and multiplicative Let r r E R and let 3 s E S Then noting again that imf is commutative we have 7 78 TS 9 T ms T sfss 1 mm fr f8f8 1fs 1 frf8 1 m w rl Ag 1H and frrfss 1 f7 f7 fs 1fs 1 ilt gtilt gt Finally f is unitary since f1f1 1 1T This completes the proof l 145 Remark Assume that R is an integral domain and that f R a F is an injective unitary ring homomorphism from R to a eld F By the universal property of L R a QuotR there exists a unique unitary ring homomorphism f QuotR a F such that f0 L f R4F QuotR Since f is unitary and QuotR is a eld f is injective Note that the image of f is the set of elements of the form frfs 1 with r E R and s E R These elements form a sub eld E of F which is isomorphic to QuotR under If F E we also call 1 R a F a quotient eld of R In this case f QuotR a F is an isomorphism such that the above diagram commutes In particular if R is already contained in a eld F we can apply the above reasoning to the inclusion map f R a F and setting E rs l l r s E R s 7 0 we obtain a eld E and an isomorphism f l E a QuotR whose restriction to R is equal to the embedding L R a QuotR The eld E is called the quotient eld of R in F 146 Example Let R be the ring of Gaussian integers viewed as a sub eld of the complex numbers Then the eld of Gaussian numbers a bi l ab E Q is the quotient eld of R in C In fact every element of can be written as a quotient of an element of R and a non zero integer Conversely every quotient rs l with r s E R and s 7 0 lies in 1 i 7 7 7 a 4 QL since for s 7 a 2 we have s 7 m 7 mi 70 15 Polynomial Rings Throughout this section B denotes UFD and K denotes its eld of fractions We View R as a subset of K via the canonical embedding i R a K r gt gt 151 Proposition The polynomial ring AX over a commutative ring A has the following universal property Given a ring B a unitary ring homo morphism p A a B with 4pA Q ZB and an element 1 E B there exist a unique unitary ring homomorphism 4Z7 AX a B such that X b and a 4pa for all a E A Moreover one has Laoan a1X a0 Lpanbn Lpa1b Lpa0 151a Proof The uniqueness statement is clear since every element of AX is a sum of products of elements from A and powers of X We de ne the homomorphism 4 as in 151a It is a straightforward computation that 4 is a unitary ring homomorphism Clearly 4 satis es X b and a 4pa for all a E A 152 De nition A polynomial fX E RX is called primitive if the co efficients of fX are coprime For example 6X2 15X 7 10 is primitive in 153 Proposition Assume that R is a UFD For every 1 E KX 0 there exists an element a E K 0 and a primitive polynomial g E RX such that f ag If also 1 bh with b E K and a primitive polynomial h E RX then there exists it E UR with h ug and b au l Proof Write f 272 X X with r E R 0 7 316 R and set 3 30813n Then 3 7 0 and sf 6 RX Thus we can write sf t0t1XtnX with t E B Let 1 E gcdt0tn Thend 0 and tie tquot are coprime and g if is primitive in RX Thus 1 ag with a g Now for b and h as in the statement of the proposition we can write I f with et E R t 7 0 Then tdg sch Since 9 and h are primitive this implies td 35 Thus td use for some it E UR and we obtain a iiiubandhb 1fua 1fug l s 154 Lemma Gauss7 Lemma 177771855 Let fg E RX Then 1 andg are primitive if and only if fg is primitive 71 Proof 7 Assume that fg is not primitive and let p E R be a prime element dividing all the coefficients of fg Let 1 R a Rp denote the canonical epimorphism Then by Proposition 151 the map 17 RX a RpX ZiaiXi gt gt ZizaiXi is a unitary ring homomorphism with 17fg O This implies 17fz7g 0 Since Rp is an integral domain also RpX is Therefore 17f 0 or 179 0 But if 17f 0 then every coefficient of f is divisible by p and f is not primitive and similarly for g This is a contradiction lt773 Assume that f is not primitive and let p E R be a prime element dividing all the coefficients of 1 Then p divides all the coefficients of fg This is a contradiction Symmetrically one handles the case that g is not primitive 155 Theorem For f E RX the following are equivalent i f is irreducible in RX ii Either f is constant and irreducible in R or f is primitive in R and irreducible in Proof iii We have 1 7 0 and 1 URX UR lf degf 0 then clearly f is irreducible in R since URX UR So assume that degf gt 0 Then 1 UKX K 0 and clearly f is primitive in RX We show that f is irreducible in So let 971 6 KX with f gh We need to show that g or b is a constant polynomial By Proposi tion 153 there exist a b E K and primitive polynomials 91 In E RX such that g agl and h bhl We have 1 1 gh abglh1 with glhl prim itive by Gauss7 Lemma 154 Now the uniqueness part of Proposition 153 implies that ab u for some unit it of B This implies f ug hl Since 1 is irreducible in RX we can conclude that ugl or bl is a unit in RX and therefore constant This implies that g or b is a constant polynomial iii Assume that f is constant and irreducible in R Then clearly f is irreducible in RX Next assume that f is irreducible in KX and primitive in RX and assume that f gh for some 971 6 RX Note that degfX gt 0 since 1 is irreducible in For the same reason 9 or b is a unit in KX and therefore constant Thus 9 E R or b E R say 9 E R then 9 divides all the coefficients of 1 Since 1 is primitive g E UR URX l Xl 156 Corollary Let f E KX be a not constant and write 1 a g with 0 7 a E K and primitive g E RX according to Proposition 153 Then 1 is irreducible in if and only ifg is irreducible in RX 72 Proof By Theorem 155 since 9 is primitive and deggX gt O we have 9 is irreducible in RX ltgt g is irreducible in KX ltgt f is irreducible in KX 157 Theorem HR is a UFD then RX is a UFD Proof By Proposition 132 and Theorem 155 it suffices to show that i every 1 E RX which is non zero and not a unit is a product of irreducible elements and ii whenever r1rmp1pn 51 kq1ql with irreducible ele ments r1 rm 31 3k of R and primitive elements p1 pnq1 ql of RX which are irreducible in KX then m k n l r is associate to 3139 in R and pj is associate to qj in RX after reordering if necessary Proof of i If f E R then the assertion is clear Assume degf gt 0 Let 1 be a greatest common divisor of the coefficients of f and write 1 dg with g E RX primitive We may write 1 f1fT with f1fT irreducible in KX since KX is a Euclidean domain For i 1 r write fi aigi with 0 7 L E K and 9 E RX primitive according to Proposition 153 Then 1 a1 11791 97 with 91 97 primitive in RX by Gauss7 Lemma 154 By Proposition 153 there exists a unit u of R such that g uglgg 97 and ud a1a7 With f1 fT also 149192 gT are irreducible in KX and then by Theorem 155 also in RX Now write 1 as a product of irreducible elements of R if d is not a unit of R lf 1 is a unit in R then f dug gg 97 In both cases we are done Proof of ii By Proposition 153 and Gauss7 Lemma 154 there exists a unit u ofR with r1rm slsku 1 and p1pn q1qlu Since R is a UFD it follows that m k and n is associate to 3139 in R for all i 1 771 after reordering if necessary Since KX is a UFD n l and pj is associate to qj in KX for j 1 n after reordering if necessary Thus pj zjqj with some non zero z E K for all j 1 n Since pj and qj are primitive in RX Proposition 153 implies that zj is a unit in R Therefore pj is associate to qj in RX for each j 1 n 158 Remark The statements R is a PlD RX is a PlD77 and R is a Euclidean domain RX is a Euclidean domain77 are false in general A 73 counterexample to both is the ring R Z since Z is a Euclidean domain but ZX is not a PlD In fact the ideal 2X is not a principal ideal of ZX verifyl 159 Theorem Eisenstein s irreducibility criterion 182371852 Assume that fX aan a1X a0 6 RX is primitive and not constant and assume that p E R is a prime element such that pla07pla17 7Plan717plan andpzlao Then 1 is irreducible in RX Proof Obviously f 7 0 and f is not a unit in RX Assume that f gh withg kakb1Xb0 e RX and h chlchco RX and bk 7 0 7 cl Sincep l a0 b000 and p2 l am we have p l 0 or p l 00 but not both Without loss of generality assume that p l 0 and p l 00 Since p l an bkcl we have plbk and pie Therefore there existsj E 1 k such that p b0 p bjil and p bj Since aj bjCO bjilcl bOCj with c 0 ifi gt Z we obtain p l aj This impliesj 71 But j g k and k 1 n then implies that k n and l 0 Thus h E R Since 1 is primitive also h is by Gauss7 Lemma 154 This implies that h is a unit in R and also in RX Thus 1 is irreducible in RX 1510 Remark Let p E N be a prime number and let i E 1p71 Then the binomial coefficient is divisible by p because it is equal to 1 71154 1511 Theorem Let p E N be a prime number Then 1X1XXp 1 is irreducible in ZX and QX 11X is called the p th cyclotomic poly nomial Proof By Proposition 151 we obtain ring homomorphisms 6 ZX 7 ZX fX 7 fX 1 and 6 ZX 7 ZX fX 7 fX 71 It is easy to see that they are inverses of each other and therefore unitary ring isomorphisms In order to see that 1 is irreducible in ZX it suffices to show that 64 is irreducible in ZX Note that 117 X 7 1 X17 7 1 74

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