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# Modern Algebra MAT 150C

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Lecture Notes For Mathematics 1500 Dr Tyler J Evans Spring 2001 Contents H O CO Rings and Fields lil De nition of a Ring i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 12 Integral Domains i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 13 Field of Quotients i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 14 Polynomial Rings i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 15 Factorization of Polynomials over a Field i i i i i i i i i i i i i i i i i i i i i i i i 16 Noncommutative Examples i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i Ideals and Quotient Rings 21 Ring Homomorphisms i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 22 ldeals i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 23 Quotient Rings i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 24 Prime and Maximal ldeals i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 25 Manufacturing Roots of Polynomials i i i i i i i i i i i i i i i i i i i i i i i i i i i Modules 31 Introduction to modules i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 32 Module homomorphisms i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 33 Free modules and bases i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 34 Some odds and ends about Ple i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 35 Torsion modules and order ideals i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 36 Structure theorem for nitely generated modules over Ple i i i i i i i i i i i i i i 40 40 41 4 Two Applications of the Structure Theorem 65 41 Structure theorem for nitely generated abelian groups i i i i i i i i i i i i i i i i 65 42 The Jordan canonical form i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i 68 Index 7 5 CHAPTER 1 MATHEMATICS 150C Spring 2001 1 Chapter 1 Rings and Fields 11 De nition of a Ring Up until now we have focused our attention primarily on sets with a single binary operation That is we have been studying groups However we have also considered sets with two binary operations namely elds The next topic of study rings deals with sets with two binary operations but we will weaken the eld axioms slightly with respect to the multiplication Here is the main de nition De nition 111 Ring A set R together with two binary operations and is a ring if R1 R is an abelian group We write 0 for the identity R2 The operation is associative R31 We have ab c ab ac for all ab c E R R31 We have a bc ac be for all a 125 6 R Before we give examples we invite the reader to think about the properties of a eld that have been omitted from the previous de nition For example if R is a ring the nonzero elements RX do not necessarily form a group under Example 112 Every eld is a ring As we have noted the converse is false In particular the familiar elds Q R and C are all rings with the usual addition and multiplication CHAPTER 1 MATHEMATICS 150C Spring 2001 2 Example 113 The integers Z form a ring under usual addition and multiplication This ring is not a eld however7 since n E Z has no multiplicative inverse if n i1 Example 114 If R is a ring7 the set MAR A aij Z aij E R71 S ivj S n of n X n matrices with entries in R is a ring with the usual addition and multiplication of n X n matrices over R This ring is also not a eld Example 115 If X is a set7 then the set of all functions f z X A R or C or any ring R is a ring under pointwise addition and multiplication of functions That is7 for g X A R we de ne f 9 and f9 by f 9W f1 995 MW fIyIA We will denote this ring by FX7 R We leave it as an exercise for the reader to verify that each of the above examples is indeed a ring Example 116 We de ne a multiplication in the group Zn by E E E We will see that this multiplication is well de ned and hence Zn is a ring7 called the ring of integers modulo n We will also see that Zn is a eld if and only if n is a prime We now want to see what elementary properties of rings that we can deduce only from the de nition For notation7 recall that 0 will denote the additive identity of R and if a E R 7a denotes the additive inverse for a Proposition 117 IfR is a ring then for all ab E R we have 1 0a a0 0 2 a7b 7a 7ab 5 7a7b ab CHAPTER 1 MATHEMATICS 150C Spring 2001 3 Proof Let ab E R be arbitrary 1 Using the left distributive law we have a0 a0 0 a0 a0 and therefore 0 a0 by the cancellation law in the group R Similarly we have 0 0a 2 Recalling that the additive inverse for an element in R is unique we note that ab a7b ab 712 a0 0 so that a7b 7ab Similarly we have 7a 7ab 3 Now using 2 we have 7a7b 7a7b 77ab Now 77ab is the element that when added to 7ab gives 0 Clearly ab satis es this property so that 7a 7b ab as desired l After our extensive work with groups and vector spaces the following de nition should feel very natural to the reader does it De nition 118 Ring homomorphism If R and S are ngs a map g0 R A S is called a ring homomorphism if for all ab E R we have both 1 a b 9001 b 2 900117 90009002 This de nition says that a ring homomorphism is a homomorphism of abelian groups R and S that also preserves the multiplication ln particular we can speak of the kernel and image of go as a group homomorphism Recall that the homomorphism g0 R A S as groups gives rise to a quotient group We will see soon that we also have a quotient ring77 in this situation as well Example 119 Let FR R be the ring of all functions f R A R as de ned above If a E R then a determines a ring homomorphism goa FR A R via the formula 90am faA We can verify that goa is a homomorphism immediately If fg E FR then by de nition we have soaf 9 f 5001 fa 901 soaf 90M 90am f9a fa9a soafsoa9A This homomorphism is called the evaluation at a homomorphism It will be very important when we study polynomial rings in detail lndeed solving a polynomial equation 101 0 amounts to nding a E R such that p E ker spa CHAPTER 1 MATHEMATICS 150C Spring 2001 4 Again to the experienced MAT 150AB student the following de nition comes as no surprise De nition 1110 Ring isomorphism A ring homomorphism go R A S is an isomorphism if go is hijective as a function If there exists an isomorphism go R A S we say that R and S are isomorphic Example 1 1 11 We note easily that the set 2Z of even integers form a ring under the usual addition and multiplication of integers Moreover the map go Z A 2Z de ned by gon 2n is easily seen to be a isomorphism of the abelian groups Z and 2Z However if nm E Z then seam 2mm y lt2ngtlt2mgt soltngtsoltmgt so that go is not an isomorphism of rings In fact the rings Z and 2Z are not isomorphic Did we prove this last statement here In many of the examples we have considered so far all but one of them in fact our rings have had a multiplicative identity as well as an additive identity That is we have seen an element 1 E R with la al a for all a E R We note that the singleton set 0 is a ring with 0 0 0 and 00 0 and in this case 0 is both and additive and multiplicative identity This is the only case where this can happen though Indeed if 0 is a multiplicative identity then by our proposition above a 0a 0 for all a E R We call this singleton ring the trivial ring We always exclude this example when we speak of rings with a multiplicative identity That is if 1 E R is a multiplicative identity then 1 f 0 De nition 1112 Commutative ring unity IfR is a ring such that ab 12a for all a b E R then we say R is a commutative ring fR has a multiplicative identity 1 E R then we say R is a ring With unity The multiplicative identity is called unity A ring with unity is also called a unital ring Proposition 1113 fR is a ring with unity l E R then 1 is the only unity Proof This follows from the general result that an identity for an associative binary operation is always unique I If R1R2 Rn are all rings then we can multiply elements of the product R1 gtlt gtlt Rn compo nentwise making the set R R1 gtlt gtlt Rn into a ring called the product ring Clearly the product R is commutative if each factor is commutative and if each Ri has unity 11 then 11 1n 6 R CHAPTER 1 MATHEMATICS 150C Spring 2001 5 is unity for R We leave the precise proofs of these statements as exercises for the reader Here is one more de nition De nition 1114 Unit7 eld7skew eld Let R be a ring with unity An element u E R is a unit if uv vu l for some v E R If every nonzero element of a ring R is a unit then R is called a division ring A eld is a commutative division ring A noncommutative division ring is sometimes called a skew eld We end the lecture with the notion of a subring Again7 the reader should be able to supply her or his own de nition ln particular7 a subset S of a ring R is a subring if S is itself a ring under the same operations Similarly you can de ne sub eld and subanything for that matter 12 Integral Domains One of the many useful algebraic facts about the ring of complex numbers C and its subrings RQ and Z is the so called zero product rule It states that the only way a product of two elements in the ring is equal to zero is for one of the elements to be zero We formalize this in the following de nition De nition 121 Divisor of zero IfR is a ring and 0 y a E R then a is a left divisor of zero if there exists an element 12 f 0 such that ab 0 In this case the element 12 is a right divisor of zero If R is a commutative ring7 then every left divisor of zero is also a right divisor of zero and vice versa In this case we simply say divisor of zero Our remarks above state that C has no divisors of zero Proposition 122 In the ring Zn the divisors of zero are precisely those elements m 6 Zn such that mn gt 1 Proof Let d m7 n and note that mZEn30 modn d d If d gt 17 then 0 nd mod n so that m is a zero divisor Conversely7 if d l and ma E 0 mod p then nlma so that necessarily nla This implies a E 0 mod p so that m is not a zero divisor l CHAPTER 1 MATHEMATICS 150C Spring 2001 6 Corollary 123 lfp is a prime the ring Z1 has no divisors of zero I De nition 124 Cancellation law for rings A ring R has the left cancellation law if for all a y 0 and all 127 c E R ab ac implies b c Similarly R has the right cancellation law if for all a f 0 and all bc E R ha ca implies b c lfR has both the left and right cancellation laws we say the cancellation law holds for R Theorem 125 The cancellation law holds for R and only if R has no left or right divisors of 267 0 Proof Suppose R has the cancellation law If a f 0 and ab 07 then ab a0 so that b 0 It follows that R has no left divisors of zero Similarly one can show that R has no right divisors of zero Conversely7 if R has no left or right divisors of zero and ab ac for some 0 f a and bc E R7 then 0 ab 7 ac al 7 c It follows that b 7 c 0 so that b c7 and hence R has the left cancellation law Similarly R has the right cancellation law I This brings us to the main de nition of the lecture De nition 126 Integral domain An integral domain is a commutative unital ring with no divisors of zero Example 127 All of the familiar rings C R Q and Z are integral domains Theorem 128 Every eld is an integral domain Proof Suppose F is a eld and ab 6 F satisfy ab 0 If a f 07 then a 1 E F and 0 a lab b so that a is not a divisor of zero I We remark that the proof of this theorem actually shows that if u E R is a unit7 then u is not a divisor of zero Theorem 129 Every nite integral domain is a eld Proof Let 1 a07 a17 7 an be all the nonzero elements of a nite integral domain D If aj is any one of the elements then each element in the list ajao7 ajah ajan is distinct by the cancellation law that holds in an integral domain It follows that 1 ajak for some h so that aj is a unit Therefore D is a commutative unital ring in which every nonzero element is a unit and hence D is a eld I CHAPTER 1 MATHEMATICS 150C Spring 2001 7 Corollary 1210 pr is prime then Z1 is a eld I De nition 1211 Characteristic ofa ring If R is a ring the Characteristic of R is the smallest positive integer n such that n a 0 for all a E R Here n a a a n times If no such positive integer exists we say R has Characteristic zero Example 1212 The characteristic of Zn is n The rings C R Q and Z all have characteristic zero Theorem 1213 IfR is a unital ring the R has characteristic n and only is the smallest positive integer such that n l 0 Proof Of course if R has characteristic n7 then n l 0 Conversely7 suppose n is the smallest positive integer that satis es n l 0 and let a E R be arbitrary Then naaaallanla00 so that R has characteristic n l 13 Field of Quotients We saw in the last lecture that there is no difference between nite integral domains and nite elds The integers Z provide an example of a integral domain that is not a eld The purpose of this lecture is to show that every integral domain can be regarded as being contained in some eld called the eld of quotients of the integral domain We will see that the eld Q of rational numbers is the eld of quotients for the integers lndeed7 the construction given below is just an abstract version of the construction of the rational numbers from the integers Nothing we do here is too dif cult7 but to be completely careful7 the construction is quite long We therefore will give a brief outline of what we propose to do Our goal is to construct the smallest eld F that contains a given integral domain D We will proceed in four steps 1 De ne the elements of F 2 De ne two binary operations and on F 3 Show that F7 is a eld 4 Show that there is an injective ring homomorphism D A F CHAPTER 1 MATHEMATICS 150C Spring 2001 8 We will use the result of step 4 to identify D with its image in F so that we can think of D as a subdomain of F We will leave some of the details to the reader as we proceed Step 1 Let D be an integral domain and let S be the subset of D X D de ned by Sab DgtltDzb9 0 The set S is too big to be our eld F7 so we cut it down a bit with the following lemma Lemma 131 The relation N de ned on S by a7 b N e7 d i c ad be is an equivalence relation on 5 Proof First7 since D is commutative we have ab ba for all a7 b E D so that ab N db Also7 if ab N e7 d7 then ad be so that eb da and hence e7 d N a7 b Finally if a7 b N e7 d and e7 d N e7 f7 then ad be and ef de Now7 using this and the commutativity in D7 we compute afd adf bef bde bed But d f 0 so that the cancellation law in D implies af be and hence a7 b N e7 I We will denote the equivalence class containing a7 b by a7 We now complete step 1 by de ning F S N to be the set of all equivalence classes in 5 under the relation N Step 2 We will now give the de nitions of and in F We will de ne these operations in terms of representatives of the class7 so that we will need to show that they are well de ned We state the precise result in the form of a lemma Lemma 132 The operations and de ned on F by the formulas ab ed ad bebd L u bl 67 dl 6067 W are well de ned binary operations on F Proof We begin by noting that since ab7 ed 6 F7 the pairs ab7 e7 d E 5 so that b 0 and d f 0 Since D is an integral domain7 it follows that bd 0 and hence ad be7 bd7 ae7 bd E S CHAPTER 1 MATHEMATICS 150C Spring 2001 9 Therefore the righthand sides of the de ning equations above both lie in F It remains to show that the operations are well de ned Suppose that ab E a 12 and 5d E c We must show that ad bc Jd E ad bcbd and ac Jd E acbdi We will do the rst one and leave the second to the reader We have by hypothesis a b Va and c d dc so that multiplying the rst by dd and the second by 1212 and adding gives a bdd cdbb Hadd dcbbi Using various properties in the integral domain we have ad bcbd bdad be so that ad bc Jd E ad bcbd as desired Similarly one shows that ac Jd E acbd and hence the operations are well de ned I Step 3 In this step we must simply verify that the operations de ned in step 2 make F into a eld The details are largely boringi Except for multiplicative inverses each eld axiom follows directly from the corresponding property that holds in the integral domain Di We leave the details to the reader Step 4 It remains to show that D is isomorphic to a subdomain of F We accomplish this with the following lemma Lemma 133 The map 239 z D A F de ned by ia a1 is an injective ring homomorphism Proof For any two elements a b E D we have ia b a b1 and iaiba71b711a11a711 lta w so that ia b ia Similarly iab ab1 and 21002117 a71ll571l 04771 390 04771 and hence i is a ring homomorphismi Finally if ia ib then a1 12 1 so that a 1 1 b and hence a bi I This completes the construction of the eld of quotients for the integral domain D We summarize what we have proved as well as clarify in which sense F is the smallest eld containing D in the following theoremi CHAPTER 1 MATHEMATICS 150C Spring 2001 10 Theorem 134 If D is an integral domain then there exists a eld F and an injective ring ho momorphism i z D A F with the property that if E is any eld containing D then there exists an injective ring homomorphism 1 F A E with a for all a E D Proof We let F be the eld of quotients constructed above and let i z D A F be the map i z a gt gt a7 Then we have seen that i is an injective ring homomorphismi For notation7 if 0 f b 6 D7 then since D Q E and E is a eld7 b has a multiplicative inverse in E which we denote by b li Note that the element 12 1 may not be in Di Now we de ne 1 F A E by wab ab li To show that 1 is well de ned7 we note that if ab 6 a7 12 then ab ba so that ab 1 a b 1 in E Therefore 1 is well de ned We leave the proof that 1 is an injective ring homomorphism to the reader I Corollary 135 Every eld E containing an integral domain D contains the eld of quotients F of D Proof Referring to the proof of theorem 1347 if E contains D7 then E contains wF7 and is isomorphic to Fl l Corollary 136 Any two elds of quotients for an integral domain D are isomorphic Proof If E is another eld of quotients of D7 then the proof of theorem 134 can be suitably modi ed to show that there is an injective ring homomorphism go E A F with goa a for all a E Di You can check that go o w 1F and w a go 1 so that E is isomorphic to Fl l 14 Polynomial Rings Based on your previous algebraic experience7 you are probably completely willing to accept the idea that we can form polynomials with coef cients from an arbitrary ring Ri Moreover7 you are most likely willing to believe that we can add and multiply such polynomials using the usual rules so that in fact7 the set of all polynomials with coef cients in R form a ring RXi With that said7 we want to emphasize that we will be working with such polynomials from a slightly different point of view7 and there are many details about the constructions involved that we wish to discuss carefully To begin7 we will call X an indeterminate rather than a variable If our ring is the ring Z of integers7 one polynomial in ZX is 1X which we write simply as Xi ln solving polynomial equations7 the CHAPTER 1 MATHEMATICS 150C Spring 2001 11 reader is probably used to writing expressions like X l or X 2 However we will never write such things because 12 6 Z and X g Z Similarly we will never write X 4 0 because the polynomial X 4 is not the additive identity in the ring ZXi At this point the reader may feel we are being too formal in our discussion What we are actually trying to do is to develop the theory of solving polynomial equations77 purely algebraically and we want to avoid saying two things are equal in one context and not equal in another The rst step we need to take is to give a formal de nition of what a polynomial is This may seem easy a polynomial with coef cients in a ring R should be a formal sum aoa1XanX i Without saying something else this de nition is not good enough however After all surely we want the two distinct formal sums l 2X and l 2X 0X2 to denote the same polynomiali Surprising as it may seem we work around this difficulty by taking in nite sumsl Here is the main de nitioni De nition 141 Polynomial Let R be a ring A polynomial f with coe icients in R is an in nite formal sum 00 fZaiXia0a1Xquot39aanquot39 i0 where ai E R for all i E N and ai 0 for all but nitely many values ofi The elements ai E R are called the coef cients of the polynomial Iffor some i gt 0 ai 0 but aj 0 for all j gt i then i is the degree of f If no such i gt 0 exists then we say f has degree zero To simplify our notations for polynomials we agree that if ai 0 for i gt n then we will not write the terms aiXi so that a polynomial is written aoa1Xaani Moreover if R is unital we write Xi in place of lXii Finally we omit any term with ai 0 so that the polynomial l 0X X2 becomes 1 X An element of the ring R is called a constant polynomial De nition 142 Let R be a ring and let f 20 aiXi and g 20 biXi be two polynomials with coe cients in R We de ne the sum f 9 and the product fg by the formulas f 9 2ai biXi7 i0 CHAPTER 1 MATHEMATICS 150C Spring 2001 12 and 1 f9 Z awhile X11 i0 k0 We remark that if ai and bi are zero for all but nitely many values of i then same is true for Ci ai bi and di 220 akin7k so that f g and fg are polynomials with coefficients in Ri We remark that if R is not commutative we should not expect 220 akbik to equal 220 bkai ki The following theorem is routine although the notations involved make it seen difficult Theorem 143 fR is a 7ing the set RX of all polynomials with coe cients in R is a ring under polynomial addition and multiplication If R is commutative then RX is commutative If R is unital then RX is unital Proof We leave the veri cation that RX is an abelian group to the reader The veri cation of associativity of and the distributive laws is straight forward if not a little cumbersome We will write out associativityi Applying the ring axioms to ai bj ck E R we compute WdQ 50 n0 i0 i lt albjck X5 50 ijks i 5 a577 ibjcmgt X5 50 m0 j0 gm i Xm i0 m0 j0 7 lt n wgt and therefore multiplication is associativei Similarly you can show that the distributive laws hold and hence RX is a ring It is clear that if R is commutative then 220 akin7k 220 bkai zc so that fg gf and hence RX is commutative Finally ifl E R is unity the constant polynomial 1 is obviously unity for RXi l CHAPTER 1 MATHEMATICS 150C Spring 2001 13 Example 144 The ring ZX is the ring of polynomials with integer coef cients The ring is the ring of polynomials with rational coef cients Both of these rings are familiar from high school algebra Example 145 ln the ring Z2X we have X 12 X2 1 Moreover7 X 1 X 1 0 If R is a ring7 and X and Y are two indeterminates7 then we can form the ring RXY of polynomials in Y with coef cients in RX lt is fairly obvious7 but tedious to prove carefully7 that the ring RXY is canonically isomorphic to the ring That is7 every polynomial in Y whose coef cients are polynomials in X can be rewritten as a polynomial in X with coef cients in RY We use this isomorphism to identify RXY and RYX7 and we denote this ring by RXY Similarly we can de ne the ring of polynomials in n indeterminates X17 i Xn RX17 X2 Xn We will not work with polynomials with more than one indeterminate lt is a nice exercise to show that if D is an integral domain7 then so is DX ln particular7 if F is a field7 then the polynomial ring FX is an integral domain The eld of fractions for FX is called the eld of rational functions in X and is denoted by FX Similarly7 FX1 7Xn is the eld of fractions for FX17 Xn The eld FX17 Xn plays an important role in modern algebraic geometry The next goal of the lecture is to show how the problem of solving polynomial equations77 can be cast in the language of homomorphisms We begin with the a fundamental theorem about evaluation homomorphisms Theorem 146 Let F be a sub eld of a eld E and let a E E be arbitrary Then there is a unique Ting homomorphism gout FX A E such that goaX a and goaa a for all a E F Proof Given a E E7 we de ne goo FX A E by 00 00 goat EaXi Zaial i0 i0 The right hand side of this de ning equation is well de ned since ai 0 for all but nitely many values ofi E N and F C E The ring homomorphism property of good is an immediate consequence of our de nitions of polynomial addition and multiplication We leave the details of the computation to the reader Now7 if a 6 F7 then a is a constant polynomial and by the de nition of sou we have goaa a Finally7 again by definition7 we have goaX a CHAPTER 1 MATHEMATICS 150C Spring 2001 14 To show uniqueness we note that if 1 FX A E satis es a a for all a E F and a then for all polynomials f SalXi E FX we have M w galXi Zwltaigtwltxgti Zaiai sew so that 1 goal I We remark here that the previous theorem is valid with the same proof if F and E are commutative unital ringsi We will primarily be interested in the case where F and E are eldsi Although you would never guess from the simplicity of the proof of this theorem it is difficult to over estimate the importance of this theorem in eld theory It forms the basis for nearly every result in Galois theory the study of solving polynomial equationsi We complete the connection with solving polynomial equations with the following de nition De nition 147 Zero ofa polynomial Let F be a sub eld of a eld E and let a E E If f e F X we say that a is a zero of f if 040 0 We conclude this lecture with a remark It may seem to the reader that all we have done in this lecture is to take a simple idea solving polynomial equations and make it unnecessarily complicatedi In fact what we have done is to cast a familiar problem in the language of mappings homomorphismsi We can now use all the machinery we have developed and will continue to develop about mappings to solve these problems We will see that this point of view is very useful indeedi 15 Factorization of Polynomials over a Field One of the main problems in algebra is nding zeros of a given polynomial f E FX where F is a eld Suppose that F is a sub eld of a eld E and that f E FX factors in FX so that f gh with g h E FXi If a E E then using the evaluation homomorphism 00 we have 90040 soulyh soa9soah Since FX is an integral domain we see that a is a zero for f if a is a zero for g or a is a zero for h Therefore the problem of nding a zero for the polynomial f can be reduced to the problem of nding a zero for a factor of This is one reason why it is useful to understand factorization in Fm CHAPTER 1 MATHEMATICS 150C Spring 2001 15 The following theorem is the backbone for all of the work we will do in this lecture The reader is advised to compare this result with the division algorithm for Z whose importance ws established back in MAT 150A Before we state the theorem we remark that if f E FlX is a polynomial then degf denotes the degree of Theorem 151 Division algorithm for polynomials Let F be a eld and let f aanan71X 1ao y mem l bmilXMTl 120 be two elements of FX with an 12m 0 and Tn gt 0 Then theTe aTe unique elements 4 T E FlX such that f 94 T and degT lt Tn degg Proof First we will show existence Consider the set SfigssEFXi Using the Well Ordering Principal we can nd T E S with minimal degree so that f 94 T for some 4 E FXi We claim that degT lt mi lf degT 0 then we are done since Tn gt 0 Otherwise we have T ctX ct1Xquot1 co with 016 F and CE 7amp0 andtZ ll lfdegT th then f i 49 CthmXquotm9 T CthmXquotm9 1 1 and the second expression in lil is of the form T 7 01X terms of lower degree which is a polynomial of degree lower than t the degree of Ti However the polynomial in equation lil can be written in the form f 7 94 Ct17mXquotm7 so it is an element of Si But this contradicts the minimality of the degree of Ti Therefore we must have t degT lt mi This shows existence CHAPTER 1 MATHEMATICS 150C Spring 2001 16 For uniqueness suppose that f 94 T and f 94 T so that subtracting we have 94 7 4 T T A Since degT 7 T lt degg this can happen if q 7 q 0 and hence T 7 T 0 I We remark here that if F is any eld you can compute the polynomials q and T using polynomial long division just as you did for the ring RX in high school The following corollaries are also familiar from high school algebra courses We state them as corollaries to emphasize that they follow immediately from the division algorithm but they are important results on their own Corollary 152 fF is a eld and element a E F is a Toot of a polynomial f E FX and only ifX7 a is afactoT off Proof If f gX 7 a then 0040 goaga 7 a 0 so that a is a root of Conversely suppose a E F is a root and use the division algorithm 151 to write f X a T with degT lt 1 Again applying the evaluation homomorphism goa to f X 14 7 gives south 0 But degT 0 implies T E F is a constant polynomial so that T 0 and hence Xa is a factor of I We leave the proof of the next corollary to the reader Corollary 153 fF is a eld and f E FX has degTee n then f has at most n Toots in F l Example 154 Let fg E Z5X be de ned by f X473X32X24X71andg X272X3 Find qT E Z5X such that f gq T We will encounter polynomials that cannot be factored in a nontrivial manner at all De nition 155 Irreducible polynomial IfF is a eld a nonconstant polynomial f E FX is irreducible over F if f gh and g h E FX implies eitheT 9 0T h is a constant You will show in the homework that an element f E FX is a unit if and only if it is a constant Therefore a polynomial f E FX is irreducible if f is not a unit and f gh implies that either 9 or h is a unit This is taken as a de nition of irreducible in any ring for an abstract theory of factorization We will content ourselves to polynomial rings over elds We leave it as an exercise to show that degree 1 polynomials are irreducible Note that we have de ned the notion of a polynomial being iTTeducible oveT a eld F not just irreducible Indeed a polynomial f E FX may be irreducible over F but reducible not irreducible over E if F S E CHAPTER 1 MATHEMATICS 150C Spring 2001 17 Example 156 The polynomial X2 7 2 is irreducible over Q but over R we have X2 7 2 X 7 so that it is reducible over R The problem of determining if a polynomial f E FX is irreducible can be quite difficult The following theorem states that for low degree polynomials7 the problem is equivalent to nding zeros Theorem 157 If f E FX and degf S 3 then f is reducible over F and only f has a root in F Proof If f is reducible over F7 then f gh with gh E FX and deggdegh lt deg lt follows7 without loss of generality7 that degg 1 so that7 up to a factor in F7 9 X 7 a for some a E F It follows that a is a root of Conversely7 if a E F is a root of f7 then X 7 a is a factor so that f is reducible over F l The theory of factorization of polynomials over a eld is very similar to the theory of factorization of ordinary integers in the sense that the analog of the fundamental theorem of arithmetic holds in FX There is a wider class of rings7 unique factorization domains7 in which all elements can be factored uniquely into a product of irreducible elements Historically7 such rings were rst investigated in attempts to prove the famous Fermat s Last Theorem We will not undertake the study of factorization in rings in this course Rather7 we will content ourselves with the results of that theory as they apply in polynomial rings over elds As we proceed7 the reader is urged to keep in mind that irreducible polynomials are the analogs of prime integers This should make the following theorem come as no surprise Although we could give a proof of the theorem now7 it will be awkward with out some results about homomorphism so that we delay the proof until chapter 2 of these notes Before we state the theorem7 let us say that for g 6 FX7 that f divides y if g fh for some h E FX Theorem 158 Let p E FX be an irreducible polynomial fp divides the product rs with rs E FX then p divides r or p divides s l Corollary 159 fp E FX is irreducible and p divides r1 rn ri E FX then p divides ri for at least one i Proof We induct on n7 the case n 1 being trivial Suppose the theorem holds for some n 2 1 and suppose that p divides r1 rnrn1 Let s r1 rn and r rn1 so that p divides sr Therefore p CHAPTER 1 MATHEMATICS 150C Spring 2001 18 divides s or p divides r by theorem 1158 If p divides r rn1 we are done Otherwise p divides s r1 rn and hence p divides ri for some i S n by induction I Here is the main theorem we are after the analog of the fundamental theorem of arithmetic for polynomials over a eld We remind the reader that the units in the ring FX are precisely the nonzero constantsi Theorem 1510 fF is a eld then every nonconstant polynomial f E FX can be written as a nite product of irreducible polynomials and this product is unique up to the order of the irreducible factors and multiplication by a unit in F Proof We prove the existence of such a factorization by induction on deg lf degf 1 then f is irreducible Suppose that the theorem is true for all polynomials of degree less than or equal to n for some n 2 1 and suppose that deg f n 1 If f is irreducible we are done Otherwise we have f gh with deggdegh S n By induction both 9 and h factor into a product of irreducibles and hence f factors into a product of irreduciblesi This show existence It remains to show the uniqueness statementi Suppose then that pip2prf41q2qs are two factorizations of f into irreducible polynomialsi Now p1 obviously divides f and hence by the corollary to theorem 1518 p1 divides qi for some i Without loss of generality we say p1 divides ql so that 11 u1101 for some ul 6 FXi Since ql is irreducible we must have ul 6 F a unit Substituting ulpl for ql and canceling gives P2quot39Pr u142quot3945 Continuing inducting we have 1 u1u239 quotur4r1quot3945 Looking at the degree of each side we see that we must have r s so that the irreducible factors pi and qi were the same up to a unit I Example 1511 You can show that the polynomial f X4 3X3 2X 4 E Z5X factors into irreducibles as f X 7 13X 1 This is the same factorization up to units as f X7122X 723X3i CHAPTER 1 MATHEMATICS 150C Spring 2001 19 16 Noncommutative Examples In this lecture we discuss some noncommutative rings that will be very important in the rest of our course In MAT 150B we saw that groups arise naturally77 as groups of permutations In the same sense our rst example can be thought of as a natural example of a ring Let A be any abelian group and recall that a group homomorphism go A A A is also called an endomorphism of A We let EndA go A A A go is an endomorphism of A denote the set of all endomorphisms of A Theorem 161 If A is an abelian group the set EndA of all endomorphisms of A is a unital Ting under the operations and de ned for all 0111 6 EndA and all a E A by so WW 9001 01 WWW 9001101 Proof We will leave the veri cation that EndA is an abelian group under addition to the reader Moreover it is well know that function composition is an associative operation and obviously the identity map 1A A A A is an identity for this operation It remains to show that the distributive laws hold If go il 9 E EndA and a E A then using the homomorphism property and the de nitions of and we have 990 WW 98001 01 8000 00 984001 WWI and therefore 9go Ogo 91 Similarly you can show the right distributive law I Example 162 If V is a nite dimensional vector space over a eld F then the set EndFV goz V A V go is F linear is a subring of EndV We leave it as an exercise for the reader to show that choosing a basis 3 for V determines a ring isomorphism from EndF V to MatnF where n dimFV CHAPTER 1 MATHEMATICS 150C Spring 2001 20 Example 163 Let F be a eld and consider the abelian group FX which we denote with some possible confusion by FXi We consider three special elements of EndFXi First let A FX A FX be de ned by 00 00 A aZXZ gt gt ZaiXHli i0 i0 We will call A the shift operator77i We leave it to the reader to show that A is a group endomorphism A FX A FXi Note that A is not a ring homomorphismi Second let B FX A FX be de ned by B iaiXi gt gt iiaiXiili i0 i1 Again the reader can easily verify that B is a group homomorphismi Note that B is just formal differentiation of polynomials In your homework you will show that AB 7 BA 1 where 1 is the identity endomorphism 1 FX A FXi Finally if a E F multiplication by 1 de nes a group homomorphism FX A FXl gt gt If The subring W of EndFX generated by AB and the multiplications by all a in F is called the Weyl algebra We will study endomorphism rings EndA extensively Another important example of noncommutative rings are group rings Here is the de nition De nition 164 Group ring Let G 1 9192 i gn be a nite group and R a commuta tive unital ring Let RG denote the set of all formal sums n Zaigi i1 RG is called the group ring of C over R IfF is a eld FG is called the group algebra of C over F As usual we need to justify the terminology ring in group ring We will give the operations in the statement of the following theoremi Theorem 165 If G is a nite group and R is a commutative unital ring then the operations and de ned on RG de ned for all ab E RG by mm bigi ZWt 176913 CHAPTER 1 MATHEMATICS 150C Spring 2001 21 and Em j1 k1 make RG a unital ring 2 ajbk 9239 n 21 97 9kgz Proof By now the participating reader can see immediately that RG is an abelian group with identity EOgii The distributive laws follows at once since the multiplication is de ned by formally distributing and collecting like terms The associativity law is not difficult to prove but it is tedious to type We leave it as an exercise 1f 91 e is the identity element then the element of RG de ned by a1 1 and ai 0 for i gt 1 is unity l The unity element of RG is a member of the family of elements of RG de ned by ai 1 and aj 0 for j y ii This implies that RG contains an isomorphic copy of Cr In particular we see that RG will not be commutative if G is nonabeliani It can be shown that every unitary representation of a nite group G corresponds to a unitary representation of the group algebra CG This is the last bit of theory that we were missing to prove the orthogonality relations last quarter Our last example will be our rst example of a skew eld Let H R4 as an additive abelian group and let 1lt1000gt ilt0100gt jlt0010gt and klt0001gt so that we can write every element of H as a sum aa1a2iasja4k with a1a2a3a4 E R We de ne a multiplication on H by formally distributing and using the relations 1aa1aforallaEH ijk jkz39 kij ji7k kj7i and ik7ji These relations are easy to remember if you think of the so called cross product77 of the usual unit vectors ij k E Rgi Or if you prefer in the sequence 23127672313767 CHAPTER 1 MATHEMATICS 1500 Spring 2001 the product from left to right of two adjacent terms in the next term to the right and the product from right to left of two adjacent terms is the negative of the next term to the left Once again the distributive laws will hold in H by de nition and 1 is clearly unity The veri cation of the associativity of is a tedious chore that we leave to the reader If we put all of this together then we see tat H is a noncommutative unital ringi To show H is a skew eld we rst de ne for all a E H the length of a by W aia a ai and the conjugate of a by E a1 7 agi 7043 ia4ki We leave it as an exercise to show that lal 0 if a 0 and a5 M It follows that if a f 0 then a 1 Elalz Therefore every nonzero element of H is a unit so that H is a skew eld We call H the quaternionsi The symbol H is used in reference to Sir William Rowan Hamilton who discovered this division ring in 1843 We will look at some interesting properties of H in the exercises In particular you will show that the set UaEH lall is isomorphic to the special unitary group SUQ and the conjugacy class de ned by trA 0 the equator is the unit 2sphere in the space Spanij k of purely imaginary77 quaternionsi CHAPTER 2 MATHEMATICS 150C Spring 2001 23 Chapter 2 Ideals and Quotient Rings 21 Ring Homomorphisms Some of our discussion on ring theory up until now has been awkward due to the absence of a careful investigation of ring homomorphismsi As we have said any investigation of an algebraic object must include an investigation of the functions that preserve the algebraic structures whatever they may be Also we must look at the space of bers quotients of these maps because it also has the same structure Examples of this scheme of ideas include groups with group homomorphisms and quotient groups and vector spaces with linear maps and quotient spaces We will now investigate these notions for ringsi As we will see half the work is already done from what we know about the abelian groups R i We had to de ne the notion of ring homomorphism and isomorphism to discuss polynomial evaluation homomorphismsi We repeat the de nition here for easy reference De nition 211 Ring homomorphism If R and S are ngs a map g0 R A S is called a ring homomorphism if for all ab E R we have both 1 a b 9001 b 2 900117 9000900 We have seen several example of ring homomorphisms so lets get right to some theory In the proofs note how we use the fact that g0 R A S is a group homomorphismi CHAPTER 2 MATHEMATICS 150C Spring 2001 24 Theorem 212 Let go R A S be a ring homomorphism 1 If R S R is a subring then goR S S is a subring 2 If Squot S S is a suhiing then go 1S S R is a subring 5 fR is unital and golR 05 then goS is unital and golR 15 is unity Proof 1 We know that goR is a subgroup of 5 since go is a group homomorphism Moreover the equation goagob goab holds for all ab E R and ab 6 R so that goR is closed under multiplication and hence is a subring of S 2 Exercise 3 Note that for all a E R then writing 1 13 we have a 901a 94409001 and a 90011 9000900 Therefore if gol 0 it is a multiplicative identity and hence gol 15 by uniqueness of such an identity I Roughly speaking the theorem states that ring homomorphisms send subrings to subrings and unity to unity Here is a long over due de nition De nition 213 Kernel of a ring homomorphism If go R A S is a ring homomorphism then the kernel of go is the subset kergo a E R goa 0 Note that the kernel of a ring homomorphism go is just the kernel of go considered as a homomorphism of abelian groups in particular we have the following useful fact Proposition 214 A 7ing homomorphism go R A S is injective and only kergo l Recall that a ring homomorphism go is an isomorphism if go is bijective it should come as no surprise that the class of all rings is partitioned into isomorphism classes by the usual equivalence relation The same techniques that we used to show two groups are not isomorphic can be applied to rings in addition we can look for structural properties in rings such as being unital commutative etc Example 215 The ring Matg R of 2 X 2 real matrices is not isomorphic to C Note that Mat2R has zero divisors and the eld C does not CHAPTER 2 MATHEMATICS 150C Spring 2001 25 Example 216 The eld R is not isomorphic to the eld C Any isomorphism go C 7gt R maps 1 to 1 and hence 71 to 71 But then E R satis es W 902392 9071 71 which is absurd Here is the main theorem of the lecture Theorem 217 Let go R 7gt S be a ring homomorphism and let I kergo The operation de ned on the quotient group RI de ned by aIbI abI is a well de ned binary operation making RI into a ring If R is commutative then RI is commutative IfR is unital then RI is unital provided I f R Proof The whole issue here is to show that the operation is well de ned That is7 the associative and distributive laws will follow immediately from those laws in R Suppose then that a 7 a E I and b 7 b E I so that a ai1 and 12 bi2 for some ibig E I Then we have ab a i1b i2 ab aig i1bi1i2 Now7 aig agoi2 a0 0 and similarly ilb ilig 0 so that ai2i1bi1i2 E I It follows that a b 7 ab 6 I and hence abI a b JrI The reader can now easily verify the remaining ring axioms and that RI is commutative if R is commutative Finally7 1I is clearly unity ifl E R is unity l The ring RI is called the quotient ring 22 Ideals The purpose of this lecture is to abstract the properties of the kernel of a ring homomorphism and investigate quotient rings further Recall from the proof of the last theorem in the previous lecture that if go R 7gt S is a ring homomorphism7 a E kergo and b E R7 then ab 6 kergo and ba 6 kergo Moreover7 our work in group theory shows that kergo is a subgroup of the additive group R We abstract these ideas in the following de nition CHAPTER 2 MATHEMATICS 150C Spring 2001 26 De nition 221 Ideal IfR is a ring a subset I ofR is an ideal in R if 1 I7 S R is an additive subgroup of R7 2 For allaEI andalleR abEI andbaEI If we let bI ba a E I and similarly de ne Ib7 then an ideal in R is a subgroup I that satis es bICIandIbCIforalleR Example 222 If go R A S is a ring homomorphism7 then I kergo is an ideal in R Example 223 In the ring of integers7 every subgroup is an ideal This follows since if a E ltngt then a nh for some h E Z and hence for all b E Z we have ab nbh ba so that abba E Conversely7 if I is an ideal in Z then since I is a subgroup of the cyclic group Z we must have I for some n E Z To emphasize we are dealing with rings7 we will denote the ideal by We have shown in this example that every ideal in the ring Z has the form for some n Such rings are called principal ideal rings7 and we will study them in some detail In ring theory7 ideals play the same role as normal subgroups in the theory of groups as the following theorem shows Theorem 224 Let R be a ring and let I be an ideal in R then the operation de ned on the quotient group RI de ned by aIbI abI is a well de ned binary operation making RI into a ring If R is commutative then RI is commutative IfR is unital then RI is unital provided I f R Proof We can copy the proof for the case I ker go a ring homomorphism without any changes at all That is7 once again the whole issue here is to show that the operation is well de ned Suppose then that aia E I and bib E Iso that a ai1 and b bi2 for some i1i2 E I Thenwe have a b a 2 1b 2 2 ab dig i112 z 12 2i Now7 ai2i1bi1i2 E I since I is an ideal It follows that a b 7 ab 6 I and hence ab I a b I The reader can now easily verify the remaining ring axioms and that RI is commutative if R is commutative Finally7 l I is clearly unity if 1 E R is unity l CHAPTER 2 MATHEMATICS 150C Spring 2001 27 Continuing to parallel the theory of quotient groups we remark that at this point we know that kernels of ring homomorphisms are ideals and we can form quotient rings by any ideal The next step is to show that every ideal is the kernel of a ring homomorphism by showing that the quotient map R A RI is in fact a ring homomorphismi We state the precise result as a theoremi Theorem 225 IfI is an ideal in R then the quotient map 77 R A RI de ned by 77 z a gt gt a I is a 7ing homomorphism with ker 77 I Proof Given what we know from group theory it suf ces to show that 77 preserves the multiplication operationi So let ab E R and compute 77015 ab I 01 1b 1 77a77bA I We will look at the isomorphism theorems for quotient rings in the next lecturer Here are some useful facts about idealsi Proposition 226 IfI and J are ideals in a ring R then 1 I N J is an ideal 2 The setIJ ab a6 Ib6 J is an ideal 5 IfR is unital I R and only if I contains a unit Proof 1 We know I N J is a subgroup from MAT 150Ai If a E I N J and b E R then since both I and J are ideals we have ab 6 I and ab 6 Jr Similarly ba 6 I and ba 6 J so that abba E I N J and hence I N J is an ideal 2 Since R is abelian as a group under we know I J JI and hence I J is a subgroup from our work in MAT 150Ai See the section on product of groups in the 150A notesi Now since both I and J are ideals ifab E IJ and c E R then ca 6 I and cl 6 J so that cab cacb E IJi Similarly a bc E I J so that I J is an ideal 3 Suppose that R is unitali If I R the l E I and l is a unit so I contains a unit Conversely suppose that u E I and u is a unit Then there exists an element 1 E R with on l and hence l on E I since I is an ideal But then for all a E 7 a a1 6 I and hence I Ri l CHAPTER 2 MATHEMATICS 150C Spring 2001 28 We note that for any ring R the subgroups R and 0 are always ideals Part 3 of this proposition gives us the following corollary Corollary 227 IfF is a eld then F and 0 are the only ideals in F l De nition 228 Simple ring A nontrivial ring is called simple ifR and 0 are the only ideals in R The corollary states that all elds are simple There are simple rings that are not elds as the following example shows Example 229 Let F be a eld and let R MatnF be the ring of n X n matrices over F for some n gt 1 We claim that R is simple To show this suppose that 0 a I is an ideal in R Since I f 0 there exists a nonzero element A E I Since A f 0 we must have at least one entry say aiojo 0 Using the matrix units eij we note that for any h 6 12 n we have ekio Aejok aiojo 61616 and hence aiojoekk E I since A E I and I is an ideal It follows that n aiojOIn Zaiojoekk E I 161 But aiojo 0 so that aiojOIn is invertible a unit It follows that I R and hence R is simple De nition 2210 Principal ideal Let R be a commutative unital ring and let a E R The reader can show that the set a raerR is an ideal of R We call a the principal ideal generated by a IfR has the property that every ideal in R is principal then R is called a principal ideal ring PIR Example 2211 Since every ideal in Z is of the form n Z is a PlR Example 2212 If R is a simple commutative unital ring then R is a PlR This follows since R only has two ideal 0 and R and 0 0 and R 1 so that both are principal We end this lecture with an important result about polynomial rings over a eld Theorem 2213 IfF is a eld the polynomial ring FX is a PIR CHAPTER 2 MATHEMATICS 150C Spring 2001 29 Proof We know FX is a commutative unital ring so that it suf ces to show that every ideal I in F X is principali The zero ideal is always principal so let 0 f I be an ideal in FX and choose a nonzero polynomial f E I with minimum degreei Without loss of generality we may assume the leading coef cient of f is 1 otherwise we multiply f by the inverse of its leading coef cient getting another polynomial in I with the same degree and leading coef cient 1 We claim Ii Clearly C I since f E I and Iis an ideal lfdegf 0 then f E FlX is aunit and henceI FX l is principal Otherwise degf gt 1 and given any 9 E I we can use the division algorithm to write 9 fq T where qT E FX and either T 0 or degT lt deg Now f E I so that fq E I since I is an ideal Therefore T g 7 94 E I and hence T 0 by the minimality of the degree of It follows that g fq E and hence I C We have already shown that C I and hence I and I is principal I 23 Quotient Rings In this lecture we will state and prove the ring theoretic analog of the rst isomorphism theoremi We will also give some examples to show how this theorem is used analyze the structure of quotient ringsi We begin with a general lemma about lifting homomorphismsi Lemma 231 Lifting lemma Suppose that R S and R we Tings go R A S and 7r R A R aTe Ting homomoTphisms and Tr is suTjective Then theTe exists a unique Ting homomoTphism 1 R A S such that 1 c 7r go and only kerTr Q kergo MoTeoveT 1 is suTjective if and only if go is suTjective and 1 is injective and only if kerTr kergo Proof By hypothesis we are given the diagram R903 R with 7r R A R surjective and kerTr Q ker so We de ne 1 R A S as follows Given a E R choose a E R with 7ra a 7r is onto and de ne iMa a We must show that 1 is well CHAPTER 2 MATHEMATICS 150C Spring 2001 30 de ned7 independent of the choice of the preimage of a Suppose then that b E R satis es 7rlz a Then a 7 b E ker7r and hence a 7 b E kergo so that a h and hence 1 is well de ned Clearly 1 is a ring homomorphism since it is de ned by go and go and Tr and both of these maps are ring homomorphisms Specifically7 if a7 b E R and a7 12 E R satisfy 7ra a and 7rlz 12 then 7ra b a b and 7rab a b and hence we have7 by de nition7 Wa b 9001 5 a 9007 Wa WV WW 80W 8000800 MELWUZl Now7 if 1W R 7gt 5 also satis es 1W 0 7r go then for all a E R7 we have WW WWW 8001 WWW WM so that 1 1W and hence 1 is unique Now7 if go is surjective and c E S then choose a E R with a c so that ib7ra a c and hence 1 is surjective lf ker7r kergo and iMa ile then if 7ra a and 7rlz 12 then a 7 b a 7 h iMa 7 ile 0 so that a 7 b E ker go Therefore a 7 b E ker7r so that a b and hence 1 is injective l Corollary 232 First isomorphism theorem If go R 7gt S is a ring homomorphism and I ker go then RI is isomoijohic to imgo Proof We apply the lifting lemma to the diagram R imgogt5 RI where 7r R 7gt RI is the quotient map That is7 go is onto its image and kergo I ker7r so that the lifting lemma 231 gives an isomorphism 1 RI 7 im go I Corollary 233 Second isomorphism theorem Let R be a ring and let I and J be two ideals in R Then I J is an ideal in J I is an ideal in I J and IJI g JI J CHAPTER 2 MATHEMATICS 150C Spring 2001 31 Proof Clearly 1 N J is an ideal in J and 1 is an ideal in 1 J Let go be the composition J L 1 quot 1 JIi Clearly ker g0 1 J so that we may use the rst isomorphism theorem 232 to conclude J 1 J E imgoi Now if a 12 1 6 1 J1 then since a E 1 we have a b I b I and hence b b 1 a b 1 so that go is surjectivei This completes the proof I Corollary 234 Third isomorphism theorem Let R be a ring and let 1 and J be two ideals in R such that 1 S J S R Then J1 is an ideal in RI and RIJI EFLJr Proof First consider the diagram RJ Since 1 ker 77 and J ker 77 the lifting lemma implies that we have a surjective map g0 RI gt RJ commuting with the quotient maps 77 and 77 Moreover if a1 E RI then a1 a J by construction Therefore we know that kergo a1 RIzae J 771J J1i It follows that J1 is an ideal in RI it is a kernel and we have the diagram R 71 R1 7RIJI 90 RJ where 77 RI gt R1J1 is the quotient mapi Since go is surjective and kergo ker 77 the rst isomorphism theorem 23 implies that R1J1 E RJi I We conclude the lecture with some examples of identifying quotient ringsi CHAPTER 2 MATHEMATICS 150C Spring 2001 32 Example 235 If F is a eld then FXX is isomorphic to F To see this we note that the evaluation homomorphism goo FX A F is trivially surjective since goo a a for all a E F Moreover we see that f E ker goo iff 0 iff f E X by Corollary 152 so that ker goo The result now follows directly from the rst isomorphism theorem Example 236 The ring RXX2 1 is isomorphic to C To see this we consider the evaluation homomorphism goi RX A C Again W is clearly surjective since for all a ii 6 C a b E R we have g0ia 12X a bi Moreover X2 1 6 ker goi since i2 1 0 and hence X2 1 Q ker 90139 Now since RX is a PlR we have ker goi for some f E RX Using the division algorithm in RX we can write f X2 1M T where q T E RX and either T 0 or degT lt 2 Applying W to this equality shows that galT 0 1f degT 1 this implies that i E R a contradiction Therefore T E R must be a constant and hence galT 0 implies T 0 This shows that f E X2 1 and hence kergoi Q X2 1 We have shown that ker goi X2 1 and hence the rst isomorphism theorem implies that RXX2 1 2 C We will see that both of these examples are quite general 24 Prime and Maximal Ideals The purpose of this lecture is to look more closely at the lattice of ideals in a ring R Historically many of these notions were rst investigated abstractly in the study of factorization in rings The terminology involved here re ects these origins We recall here that every ring R has at least two ideals the improper ideal R and the trivial ideal A ring is simple if these are the only two ideals The quotients by these ideals are not interesting at all RR has only one element and RO is obviously isomorphic to R We will therefore turn our attention to ideals I in R such that 0 f I f R Such an ideal is called a proper nontrivial ideal Let us investigate the conditions under which the quotient ring RI is a integral domain or a eld The answer is not as straight forward as you might expect De nition 241 Maximal ideal If R is a Tiny and ideal M is maximal if M y R and if N is any ideal afR satisfying M Q N then either N M 07 N R CHAPTER 2 MATHEMATICS 150C Spring 2001 33 Another way to state this very important de nition is as follows An ideal M is maximal if M is a proper ideal in R and R is the only ideal of R that properly contains M One useful fact about maximal ideals in commutative unital rings is given in the following theorem Theorem 242 IfR is a commutative unital ring and M is an ideal in R then M is maximal RM is a eld Proof Suppose that M is maximal We know RM is a commutative unital ring by theorem 224 It therefore suffices to show that every nonzero element of RM is a unit To this end let a M E RM with a M f M ie a M is nonzero in Now let N a M so that N is an ideal in R by proposition 2262 Moreover we clearly have M Q N and a E N But aM M so that a g M It follows that N properly contains M and hence N R by maximality Since 1 E R we can write 1 ramfor some r E R ande M and hence we have 1MramMraMrMaM This shows that a M is a unit in RM and hence RM is a eld Conversely suppose that RM is a eld Then RM is unital and hence M f R If N is an ideal of R such that M C N C R proper inclusions then the third isomorphism theorem implies that NM is an ideal of RM But RM is a eld so that the only two ideals of RM are RM improper and MM trivial In the rst case we have N R and in the second we have N M Therefore M is maximal I This theorem is false if you omit commutativity ie you do not even get a division ring For example we have shown that 0 is a maximal ideal in the matrix ring R MatnF but R 0 E R has zero divisors We now turn to the question when is RI an integral domain The zero divisor condition for cosets reads a I b I I implies that a I I or b I I This motivates the following de nition De nition 243 Prime ideal An ideal N y R in a commutative ring R is prime if for all ab E R ab 6 N implies that either a E N or b E N Theorem 244 If R is a commutative unital ring and N f R is a proper ideal in R then N is prime i c RN is an integral domain Proof Again since N y R we know RN is a commutative unital ring by theorem 224 It therefore remains to show that RN has no divisors of zero Suppose then that aNb N N CHAPTER 2 MATHEMATICS 150C Spring 2001 34 in RN Then ab N N so that ab 6 N But N is a prime ideal so that either a E N or b E N It follows that a N N or b N N so that RN has no divisors of zero Conversely if RN is an integral domain then it is a unital ring so that N f R Moreover if ab E Rand ab 6 N then aNbN abN N so that aN Nor bN N It follows that a E N or b E N and hence N is a prime ideal I Corollary 245 fR is a commutative unital Ting then eveTy maximal ideal is pTime Proof If M is maximal then RM is a eld hence RM is an integral domain and hence M is prime l De nition 246 Divides prime element7 irreducible element LetR be a commutative uni tal Ting If ab E R we say a divides b if b aT foT some T E R We wTite all a divides b An element p E R is called prime ifp is not a unit and plab implies pla 0T plb An element T is irreducible if T is not a unit and T ab implies that eitheT a 0T 12 is a unit We now specialize to the class of Ple that have no divisors of zero Such rings are called principle ideal domains PID We remark that if F is a eld then FX is a PlD Theorem 247 IfR is a PID and 0 f p E R then p is a maximal ideal and only ifp E R is iTTeducible Proof Suppose that p is maximal Then p f R and hence p is not a unit by proposition 2263 If p ab then ab 6 Now p is a maximal ideal and hence p is a prime ideal by corollary 245 so that we have a E p or b E If a E p then a pT for some T E R and hence p ab pr This implies that l Tb since R is a integral domain and hence I is a unit Similarly if b E p then a is a unit and hence p is irreducible Conversely if p is irreducible then p is not a unit and hence p f R To show that p is maximal suppose that p C a Q l where the rst inclusion is proper Then p E a so that p ab for some I E R But p is irreducible so that either a or b is a unit lf 1 is a unit then bu l for some u E R and hence pu abu a This implies that a E p and hence a E p contrary to assumption Therefore we must have a a unit and hence a R by proposition 2263 This shows that p is a maximal ideal I CHAPTER 2 MATHEMATICS 150C Spring 2001 35 Theorem 248 IfR is a PID and 0 f p E R then p is a prime ideal if and only ifp E R is a prime element Proof If 0 f p is a prime ideal7 then p f R so that p is not a unit If ab 6 R and plab then ab 6 p so that either a E p or b E p and hence pla or plb This shows that p is a prime elementi Conversely7 if p is a prime element7 then p is not a unit and hence p Ri If a b E R and ab 6 1o7 then plab and hence pla or plb It follows that a E p or b E p and hence p is a prime ideal I Corollary 249 fR is a PID then every irreducible element is prime Proof If R is a PD and p is irreducible7 then theorem 247 implies that p is a maximal ideal7 and hence p is a prime ideal by corollary 245 It then follows from theorem 248 that p is a prime element I This corollary patches a hole in our study of polynomial rings Namely7 recall that we gave no proof of the following theorem theorem 158 Theorem 2410 Let p E FX be an irreducible polynomial If p divides the product rs with rs E FX then p divides r or p divides s l Proof Note that if F is a eld7 then FX is a PlR by theorem 22137 and FX is an integral domain by problem 4 of homework 2 Therefore FX is a PD and we can apply the previous corollary to the irreducible element f E FX to conclude that f is a prime elementi I We conclude this lecture with an interpretation of theorem 247 for polynomial rings Theorem 2411 fF is a eld and f E FX then f is irreducible over F if and only is a eld Proof If f is irreducible7 then is maximal by theorem 2 47 and hence is a eld by theorem 242 Conversely7 if is a eld7 then is a maximal ideal by 242 and hence f is irreducible7 again by theorem 247 l 25 Manufacturing Roots of Polynomials We have now developed enough machinery tackle the problem of nding roots for polynomialsi As we will see7 if we are willing to enlarge the eld of coefficients7 we can always nd a root for any CHAPTER 2 MATHEMATICS 150C Spring 2001 36 nonconstant polynomial As you read this section notice how much use we make of the evaluation homomorphism We begin by establishing the terminology commonly used in this business De nition 251 Field extension If E and F are elds with F Q E we say that E is an extension of F A sequence of extensions is called a tower We leave it as an exercise for the reader to show that if E is an extension of F then E is a vector space over F The dimension dimF is called the degree of the extension and is written E F If E F lt 00 then we say that E is a nite extension of F We are ready for our rst main theorem Before we give it we caution the reader that we are going to do some things that for the purest will be entirely beyond the pale Speci cally if go F A E is an injective homomorphism from one eld into another then we will often choose to identify F with its image goF Q E and hence we will consider E to be an extension of F There are some set theoretic dangers with such shenanigans but the economics gained are worth the risks Of course if the danger is acute we will avoid such identi cations Theorem 252 If F is a eld and f E FX is a nonconstant polynomial then there exists an extension eld E ofF and an element a E E such that a is a root of That is fa 0 Proof Given a nonconstant f E FX theorem 1510 implies that f is a product of irreducible polynomials If p is one such irreducible factor and E 2 F is an extension of F with a root 1 for p then clearly a is also a root of Therefore we may assume that f E FX is irreducible over F Now theorem 242 implies that is a maximal ideal so that E is a eld by theorem 2411 Let go F A E be the restriction of the quotient map FX A to the constant polynomials Then go is a ring homomorphism since the restriction of a homomorphism to a subring is a homomorphism Now if a E F then goa a if a E if a fg for some 9 E FX But f is irreducible so that degf 2 l and hence degfg 2 1 unless g 0 But dega 0 unless a 0 so that we must have a 0 and hence go is injective If we identify F with its image goF S E then E is an extension of F CHAPTER 2 MATHEMATICS 150C Spring 2001 37 Now let a E E be de ned by a X If f a0 a1X anX then we compute using the evaluation homomorphism goo FlX A E WU 004010 a1X 39 39 39 aan a0 a190aX quot39 an80aXn aoaiaana a0 a1ltX f anX f a0a1Xfquot39aanf aoa1XanX f ff fl Therefore f E ker good and hence a E E is a root of f as desired l Example 253 Let f X2 1 6 RlX Then we have seen that E C and the iso morphism is given explicitly by g gt gt In particular we have X gt gt i so that the construction in the previous theorem when applied to RlX and X2 1 produces the extension C of R and the root i E C De nition 254 Algebraic and transcendental elements IfE is an extension eld of a eld F and element a E E is called algebraic over F if kergoa y 0 where gout FX A E is the evaluation homomorphism If ker 90a 0 then we say a is transcendental over F IfE C and F Q then an algebraic element a E C over Q is called an algebraic number Similarly we have transcendental numbers In other words a E E is algebraic over F if a is the root of some nonzero polynomial f E FlX Since FlX is a PD theorem 2213 we know that ker goo for some f E FlX Moreover the proof of theorem 2213 shows that we may assume that f is the smallest degree monic polynomial in FlX for which a is a root If a E E is transcendental over F then goo FlX A E is injective and hence an isomorphism onto its image in E We summarize all of these ideas in the following theorem Theorem 255 Let E be an extension eld of a eld F and let a E E Then a is transcendental over F and only if the evaluation homomorphism gout FX A E is an isomorphism onto its CHAPTER 2 MATHEMATICS 150C Spring 2001 38 image If a is algebraic over F then there exists a unique monic irreducible polynomial p E FX such that ker 90a Proof First as we have already remarked a E E is transcendental over F iff ker goo 0 if good is injective if good is an isomorphism onto its image in E This proves the rst statement of the theoremi Now ifa is algebraic then we have also remarked that since FX is a PD we have 0 ker goo p for some monic p E FX of minimal degree Moreover ker goo FX since F gZ ker good so that degp 2 1 It remains to show that p is irreduciblei If p rs for some rs E FX then pa 0 implies that rasa 0 so that either ra 0 or 8a 0 since E is an integral domain it is a eld By the minimality of the degree of p one of r or 8 must be a constant and hence p is irreducible and the proof is complete I De nition 256 Irreducible polynomial for a If E is an extension of F and a E E is al gebraic over F the irreducible polynomial for a over F is the unique monic polynomial p of minimal degree satisfying pa 0 We denote the irreducible polynomial for a over F by irraF The degree of irra F is the degree of a over F and is denoted by dega F Example 257 lfi xil E C then clearly irrilR X2 1 since i is not a root of an linear polynomial over R and X2 1 is monici Therefore degi R 2 Let us try to summarize what we have done so far If E is an extension eld of a eld F and a E E we consider the evaluation homomorphism goo FX A E We have two cases Algebraic case If a E E is algebraic over F then the kernel of good is the maximal 24 ideal irraF in FX and hence FX irraF is a eld 242 The rst isomorphism theorem implies that this eld is isomorphic to the image 90aFXl S E which we denote by Fai We leave it as an exercise to show that the image Fa goaFX is the smallest sub eld of E containing F and 1 We denote this eld by Fa and call it F adjoin 1 Therefore if a E E is algebraic over F Fa Fa Transcendental case This time sou FX A Fa is an isomorphism 255 and hence Fa is not a sub eld of E It is a subring and hence an integral domaini Now theorem 1 34 implies that E contains the eld of fractions of Fai We leave it as an exercise to show that this eld of fractions is the smallest sub eld of E containing F and 1 That is the eld of fractions of Fa is F a CHAPTER 2 MATHEMATICS 150C Spring 2001 39 De nition 258 Simple extension An extension E of F is simple if E Fa for some a E E We conclude this lecture with a theorem about the dimension of simple extensions Fa in the case that a is algebraic Theorem 259 Suppose E Fa is a simple extension of F with a E E algebraic over F If dego 7 F n then Fa F n so that in particular simple algebraic extensions are nite Proof Let p irraF Since a is algebraic over F7 we know Fa Fa imgoa We claim that 17 1 an l is a basis for Fa over F7 and the result follows immediately First7 if B E FOz then B goag for some 9 E FX Using the division algorithm7 we write 9 pq r where degr lt degp or r 0 This shows that B 30049 goo ng r south If r 07 then B 0 E SpanFlaan 1 Otherwise we have r a0 alX an71Xn 1 with all aj E F and hence 90040 a0 a1a aniian l so that 104 an l spans Fa Moreover7 if aoaia an71a 1 0 with all aj 6 F7 then f a0 a1X an71Xn 1 E p so that f 0 by the minimality of the degree of p This shows that 104 an l is linearly independent over F and hence a basis for F a over F l CHAPTER 3 MATHEMATICS 150C Spring 2001 40 Chapter 3 Modules 31 Introduction to modules This lecture introduces our next topic modules Loosely speaking7 a module is like a vector space7 except that the scalars will only be assumed to come from a commutative ring As we will see7 this will change some familiar results from linear algebra dramatically From now on7 R will always be a commutative unital ring unless stated otherwise Here is the main de nition De nition 311 Rmodule If R is a commutative unital ring on obelian group M written additively is an Rmodule we are given afunction R X M A M denoted T m gt gt Tm satisfying for allmn E M 7 78 6 R M1 1m m M2 Tm n Tm Tn M3 7 sm Tm sm M4 Tsm Tsm ln particular7 we note that the axioms M1M4 are precisely the axioms for M to be a vector space over R in the case that R is a eld In fact7 we see at once that if R is a eld7 then an R module M is just a vector space over R CHAPTER 3 MATHEMATICS 150C Spring 2001 41 Example 312 Every ring R is an R module under the map R X R A R given by multiplication in the ring The axioms M1M4 are readily veri ed they are the ring axioms This module is called the regular Rmodule Example 313 If R is a ring and n 2 1 is an integer7 the product group R is an Rmodule under the map r7 rhi zn gt gt rzhi rzn We leave the proof of this as an exercise One way to get started thinking about modules is to look for some over familiar rings We have remarked that modules over elds are vector spaces7 so we know all about those The next familiar ring is the ring of integers Z As we will see7 we know all about Z modules as well Theorem 314 Every abelian group M is a Zmodule in a unique way Proof De ne Z X M A M by a7 m gt gt a mi Then M1 is trivial7 and M3 and M4 are the laws of exponents in the abelian group M written additively of course Finally7 M2 is the identity mn man which is valid in an abelian group This shows that M is a Zmodulei Furthermore7 since the module axioms are all exponent laws that must be valid in any abelian group7 we see that this module structure is unique I We will make the last statement of the theorem more precise when we have de ned the notion of isomorphic R modules OK7 by now the notation r7 m gt gt rm must be driving your group action sensors wildl It is no accident As we have remarked before7 just as groups were meant to act on sets7 rings were meant to act as endomorphisms of abelian groups We have looked at ideas like this two times now7 so we will leave the proof of the following theorem to you Go do it Theorem 315 Let M be an abelian group and let R be a commutative unital ring Then M is an Rmodule and only if there exists a unital ring homomorphism p R A EndM I With this theorem at hand7 we see that what we are doing is studying the representation theory of commutative unital rings 32 Module homomorphisms Since a module is similar to a vector space in the sense that there is a ring of scalars that operates on an abelian group7 the reader can probably formulate his or her own de nition of a module homomorphismi CHAPTER 3 MATHEMATICS 150C Spring 2001 42 De nition 321 Module homomorphism If R is a commutative unital ring and M and N are Rmodules then a mapping go M A N is an Rmodule homomorphism if for all zy E M and all 7 E R we have 1 901 y I y 2 907 1 Tgoz That is an Rmodule homomorphism is a homomorphism of abelian groups 1 that is compatible With the action of R Note that is R is a eld so that M and N are just vector spaces over R then an Rmodule homomorphism is What we have called a linear transformation Similarly homomorphisms of Z modules are simply homomorphisms of abelian groups Example 322 Consider the abelian group Z as a Z module The map f Z A Z given by 2n is easily seen to be a Z module homomorphism Note that f is not a ring homomorphism Along With homomorphisms the subobjects in this business are called submodules You can guess that they are de ned to subsets of the module that are themselves modules under the same operations This amounts to requiring that we have a subgroup of M that is invariant under the action of R Here is the de nition De nition 323 Submodule Let be a R commutative unital ring and let M be an Rmodule A subset N Q M ofM is a submodule of M if 1 N is a subgroup of the group M 2 TnENforallTERzmdallnEN We often refer to condition 2 of the de nition by saying that N is stable under R or that N is Rinvariant Example 324 1 If M is an Rmodule then 0 and M are submodules called the trivial sub module and the improper submodule respectively 2 The submodules of the regular module R are just the ideals of R 5 The submodules of a Z module M are just the subgroups of the abelian group M CHAPTER 3 MATHEMATICS 150C Spring 2001 43 Now7 the rest of the lecture can be summed up in one sentence Namely7 all of the usual results about homomorphisms7 subobjects and their interaction hold for Rmodules We will list the results here7 but we omit the proofs You will be asked to supply the detailed proofs in your homework exercises Theorem 325 Let be a R commutative unital ring M and N be Rmodules and let go M A N be a Rmodule homomorphism 1 If M S M is a submodule then M S N is a submodule 2 If N S N is a submodule then go 1N S M is a submodule 5 The kernel kergo and image imgo are submodules of M and N respectively where ker go denotes the kernel ofgo as a group homomorphism 4 go is injective i c kergo Of course an isomorphism of Rmodules is a bijective module homomorphism We say that two R modules M and N are isomorphic if there exists an isomorphism go M A N Theorem 326 Let R be a commutative unital 7ing and let M be an Rmodule Then if N is a submodule of M the operation de ned on the quotient group MN de ned by rz N rz N is a well de ned operation R X MN A MN making MN into an Rmodule l The module MN is called the quotient module of M by N The canonical map M A MN is a surjective R module homomorphism with kernel equal to N Proposition 327 IfK and N are submodules of an Rmodule M then 1 K N is a submodule of M 2 The setKNknk Kn N isasubmodule ofM CHAPTER 3 MATHEMATICS 150C Spring 2001 44 We say that K and N are independent if K N 0 We say that K and M generate M if K N M We write M K 69 N if K and N are independent and generate M Note that in this case7 every element of M cam be written uniquely as a sum h n with h E K and m E M cf Proposition 296 of the 150A notes De nition 328 Simple module A nontrivial Rmodule M is called simple if M and 0 are the only submodules in M Lemma 329 Lifting lemma Suppose that MN and M are Rmodules go M A N and 7r M A M are module homomorphism and Tr is surjective Then there exists a unique module homomorphism 1 M A M such that 1 c 7r go if and only if ker7r Q kergo Moreover 1 is surjective and only ifgo is surjective and 1 is injective and only if ker7r kergo l Corollary 3210 First isomorphism theorem Ifgo M A N is an Rmodule homomorphism and K ker go then MK is isomorphic to imgo as an Rmodule l Corollary 3211 Second isomorphism theorem Let M be an Rmodule and let K and N be two submodules of M Then K N N is a submodule of N K is a submodule of K N and KNK g NK N Corollary 3212 Third isomorphism theorem Let M be an Rmodule and let K and N be two submodules ofM such that K S N S M Then NK is a submodule of MK and MKNK g MN We conclude this lecture with a remark on logical ef ciency Theoretically we could have developed the notion of module as soon as we knew that EndM is a ring for all abelian groups M We could have then stated and proved the lifting lemma and the isomorphism theorems for modules The same results for rings would then follow at once by applying the module results to the regular R module R In fact7 many of the notions we have looked at for rings can be de ned for R modules7 and the particular case of the regular module coincides with the ring theoretic notions For example7 a ring R is simple if it is simple as the regular R module and so on CHAPTER 3 MATHEMATICS 150C Spring 2001 45 33 Free modules and bases We continue to assume that R is a commutative unital ring throughout this section The reader will no doubt agree that matrices play an invaluable role in linear algebra The same is true for the theory of R modules for any commutative ring R not just elds In particular one can go back to the de nition of a matrix over QR and C given in section 112 of the MAT 150A notes still online and replace the entries with entries in any ring Ri Such a matrix is called an R matrix The de nitions of matrix addition and multiplication are unchanged Moreover the proof of theorem 121 in the same section goes through for R matrices unchangedi We can also take the same de nition of determinant for an n X n Rmatrixi The same proof used over elds shows that detAB detA detB for n X n R matrices A and Bi Therefore A is invertible if and only if detA 6 is a unit in the ring Ri Now the concepts of spanning sets and independent sets transfer immediately to R modulesi Namely we have the following de nition De nition 331 Let M be an Rmodule and let m1i i i mn be an ordered set of elements of M Then an Rlinear combination of the m is an element m E M of the form mT1m1T2m2quot39Tnmn where each ri E R The elements ri E R are called the coef cients of the linear combination If S m1m2i i i mn is an ordered set of vectors in M the set of all linear combinations of the vectors in S is called the span of S and is denoted RS Therefore RS mmr1m1r2m2rnmnri E R If S C M and RS M then we say 5 generates M If S is nite and generates M we say that M is nitely generated The reader should show that if S is any subset of an Rmodule M then R5 is always a submodule of Mr More generally if I is any ideal in R then S is a submodule of Mr Finally 5 is a submodule of M if RS Si All nice exercises All of the modules that we will encounter will be nitely generated If R is a eld so that an Rmodule M is just a vector space then M is nitely generated if and only if it is nite dimensionali Now we turn to independence Again the de nition transfers to R modules without change CHAPTER 3 MATHEMATICS 150C Spring 2001 46 De nition 332 A subset S m1m2mn of an Rmodule M is called linearly inde pendent if the equation rimi r2m2 rnmn 0 with ri E R implies that ri 0 for all i 127 7n If S is not linearly independent we say it is dependent De nition 333 Basis IfM is an Rmodule a subset S Q M is called a basis if S is a linearly independent spanning set Example 334 The standard basis77 e17 7e116 is easily seen to be a basis for the module Rni Just as with vector spaces7 if ml7 7mn is a basis for an R module M7 then every element m E M can be written as an R linear combination of the mi in a unique way One place we we see that the theory of arbitrary R modules differs dramatically from vector spaces7 is that not every R module will have a basis In fact7 most R modules will not have a basis if R is not a eld De nition 335 Free module A nitely generated Rmodule M is free if there exists an R module isomorphism go R A M for some n E N The integer n is called the rank of the free module M To avoid some pathology in the future we say that the trivial module 0 is free of rank 0 and the empty set 0 is a basis Proposition 336 A nitely generated Rmodule M has a basis and only it is free Proof Let B ml7 7mn denote an ordered set of elements of M and de ne a map 1 R A M by ibXBXm1iiimn zlm1znmn In This map is easily seen to be an R module homomorphismi Moreover7 1 is surjective if 13 generates M and 1 is injective if B is independent Therefore 3 is a basis for M if 1 R A M is an isomorphism if M is free I Example 337 The proposition shows that a free Z module is isomorphic to the Z module Z for some n Therefore every free Z module is in nite It follows that no nite Zmodule nite abelian group is free So we see at last an example of a nitely generated module that is not free7 any nite abelian group will do CHAPTER 3 MATHEMATICS 150C Spring 2001 47 De nition 338 An Rmodule M is cyclic it is geneTated by a single element That is M RI TI T E R foT some I E M IfN is any Rmodule and z E N then RI S N is the cyclic submodule generated by 1 Proposition 339 IfF is a fTee Rmodule with basis 11 In then F R11 69 EB Rzn Proof We know that Rzi R11 Rzi1 Rzi1 Rzn 0 since 11 zn is independent Moreover7 R11 Rzn F since 11 zn generates F Therefore F Rhea69sz I it is a surprising fact that for some rings7 the cyclic submodule R1 for z E M is not free even if M is free of rank ll This is not the case for Ple7 as the following proposition shows We will use this lemma later on as a base step for an induction Proposition 3310 If R is a PID and M is a fTee Rmodule 0f Tank 1 then eveTy nonzeTo submodule ofM is fTee 0f Tank 1 Proof By hypothesis7 there exists I E M such that the map 1 R A M given by TI is an isomorphism Let 0 f N S M be a non zero submodule of M so that 1N is a non zero submodule of R That is7 1N is a non zero ideal of R7 and hence 1N a for some 0 f a E R since R is a PD Therefore N Tax T E R Therefore the map g0 R A N given by 907 Tax is an onto homomorphism of R modules Moreover7 907 0 iff Tax 0 iff iMTa 0 iff Ta 0 since 1 is injective But R is an integral domain and a f 0 so that we must have T 0 and hence go is injective as well This shows that g0 R A N is an isomorphism of R modules and hence N is free of rank 1 by de nition I We remark here that7 just as for vector spaces7 Rhomomorphisms M A N of free R modules M and N can be given by multiplication by R matrices once basis are chosen Moreover7 change of basis matrices work just as they do for vector spaces7 and a change of a basis matrix is necessarily invertible These ideas are discussed in detail on page 455 in Artin7s text We will make little use of these facts7 so we do not expound on them here We can now state the main goal of the current chapter We want to classify all nitely generated modules M over a PD R To put the problem in perspective7 consider the case that R is a eld so that an R module is a vector space over R Since every nitely generated vector space has a CHAPTER 3 MATHEMATICS 150C Spring 2001 48 basis see Lemma 3311 of the 150A notes every nitely generated vector space is free Moreover Theorem 3314 of the 150A notes implies that any two vector spaces over R of the same dimension are isomorphic Therefore ifR is a eld we see that every nitely generated R module is isomorphic to R and thus we have classi ed nitely generated modules over elds We want to do the same thing for Ple We already know that not every R module is free so the answer will be different However since a change of basis matrix is invertible any two free R modules of the same rank n are isomorphic We end this lecture with an important general fact about free modules as well as the rst step toward the structure theorem we desire Theorem 3311 IfM is an Rmodule and 11 In E M then there is a unique homomorphism g0 R A M with ei z for all i In particular if M is generated by the Ii go is surjective Proof Let 11 In E M and de ne g0 R A M by 801 leuan l Tlxl l quot397 n1n Then the reader can check that go is a Rmodule homomorphism and that ei z for all i If 1 R A M also satis es iMei Ii then for all r Eriel E R we have WT ZWMED ZTiIi ZNM T i1 i1 i1 and hence g0 1 is unique Finally if the I generate M then every I E M has the form I E rizi so that if r Erie we see that r z and hence go is surjective l The theorem implies that every nitely generated R module M is the image of a free module of nite rank Theorem 3312 Let R be a PID and let F be a free Rmodule of rank n Then if G S F is a submodule of F then G is free of rank m with m S n Proof Let 11zn be a basis for F and for each 1 S k S n let Fk SpanRzlzk and Gk G N Fk By induction we will prove that each Gk is free of rank less than or equal to k If G1 0 then G1 is free of rank 0 If G1 0 then G1 is a submodule of R11 and hence G1 is free of rank 1 by proposition 3310 Let k gt 1 and suppose that Gk1 is free of rank less than or equal to k 7 1 If Gk C1671 then we are done Otherwise let I b E R there exists 16 F1671 and z bzk E Gk CHAPTER 3 MATHEMATICS 150C Spring 2001 49 If 7rc F 7gt Rzk is the usual coordinate projection restricted to Gk then I b E R bzk 6 im Therefore I is an ideal in R say I ai Since Gk C1611 we must have a f 0 Therefore there is some y E Gk and y E F1611 with y y azkl We claim that Gk Gk1 EB Ryl Let 2 E Gki Then there exist 2 E F1611 and c E R with 2 2 czki Therefore c E a and hence c ad for some a E Ri This means that 2 7 dy 2 7 dy E Gk1 and hence Gk Gk1 Ryi Now ry E Gk1 Q F1611 iffl razk 0 iff ra 0 if r 0 Therefore Gk1 Ry 0 and hence Gk Gkil EB Ryl By induction Gk1 is free of rank less than or equal to k 7 1 so that Gk is free of rank less than or equal to hi I Corollary 3313 If R is a PID and M is an Rmodule generated by m elements then ifN is a submodule of M N is generated by less than or equal to m elements Proof Suppose that M is generated by m elements and take a surjective homomorphism go Rm 7 M theorem 3311 Now go 1N is a submodule of the free module Rm by theorem 325 so that go 1N is free of rank n S m by theorem 33112 If y1luyn is a basis for go 1N then easily yl i i i goyn is a generating set for N and hence N is generated by less than or equal to m elements I 34 Some odds and ends about PIDS In this lecture we list and prove some useful facts about Ple that we will need in our classi cation theoreml We could have covered this material before we began studying modules Until further notice let R be a PD De nition 341 If R is a commutative unital ring and a1iuan E R a greatest common divisor GCD of the ai is an element a E R such that dlai for all i and e E R is another element satisfying elai for all i then eld A least common multiple LCM of the ai is an element m E R such that aim for all i and if n E R is another element in R with ailn for all i then We say that the elements a1 1 i i an are relatively prime if the GOD of the ai is 1 Before we state the following proposition we recall that for a commutative ring R and ab E R a E b if and only if blai Proposition 342 Let R be a PID and let a1i i i an E R CHAPTER 3 MATHEMATICS 150C Spring 2001 50 1 The generator of the ideal a1i i i an is a GOD of the ai 1 The generator of the ideal al N an is a LCM of the ai 93 fd and d are GCDs of the ai then d du for some unit u E R Ifm and m are LCMs of the ai then m mu for some unit u E R Fquot a1i i i an d if and only ifd is a GOD of the ai 91 a1i i i an R if and only if the ai are relatively prime 7 If ab E R are relatively prime and 12 Q a then a R Proof 1 Since R is a PD we know a1 i i i an d for some d E Ri Now each ai E a1i i i an and hence each ai E d so that dlai for all ii If e E R and elai for all i then ai eri for some ri E R for all i Now d 6 a1 i i i an so that d slal snan for some 316 R and hence d Siai Snan 81r16 SnTne 81r1 SnTn57 so that eldi This shows that d is a GCD of the ail 2 Again R is a PD so that the ideal al N an is principle generated by m for some m E Ri Moreover m E ai for all i so that aim for all i Now if ailn for all i then for all i we have n airi for some ri E R and hence n 6 al N an It follows that n E and hence This shows that m is a LCM of the ail 3 1f d and d are GCDs of the ai then dld and d ldi Therefore d E d and d E d so that d d i It follows from you midterm exam problem that d d u for some unit u E Ri g Proceed as in 01 1f d a1 i i i an then d is a GCD of the ai by Now let d be any GCD of the ail Then by 3 d do for some unit u E R and hence d du Therefore d a1 i i i an 6 This is immediate the ai are relatively prime iff the GCD is 1 iff a1 i i i an 1 Ri 7 If a and b are relatively prime then by 6 we have 1 az by for some zy E Ri Since 12 Q a azby E a and hence 1 E a so that a Ri CHAPTER 3 MATHEMATICS 150C Spring 2001 51 I Now since R is commutative it is easy to see that al an a1 an so that we have the following corollary Corollary 343 IfR is a PID and a1an E R then a1gtnnltangt a1quot39an Proof We know that ifm and d are an LCM and GCD for the ai respectively then a1 0 an m and a1 an Now dlai for all i and ailm for all i so that dlm and hence m E d so that Q I Now if we let a a1 an be the product of all the ai then aila for all i and hence mla This implies that a E and hence a Q This proves the following corollary Corollary 344 IfR is a PID and a1an E R then aagtglta1gtnnltangt We do not want to go into the details here but for completeness we remark that one can show that every PlD has the unique factorization into products of irreducible elements such as the integers Z and polynomials F X over a eld Such rings are called Unique factorization domains UFD Every PD is a UFD Using this it is not hard to show that the inclusion in the previous corollary is an equality if and only if the ai are relatively prime Let use brie y sketch the details for the case n 2 and leave the general induction proof to the reader Lemma 345 fR is a UFD and ab E R are relatively prime then ab a N Proof Since R is a UFD there are irreducible elements p1 pn E R and natural numbers e1enf1fn 2 0 such that a 10 105quot and b Pf1quot39107fzquot If we let ai minei and maxei for all i then the reader can easily show that Din dPi 1pn CHAPTER 3 MATHEMATICS 150C Spring 2001 52 is a GCD for a and b and m pr an is an LCM for the air Moreover since trivially ai i ei for all i we see that ab dmi If a and b are relatively prime then d is a unit and hence ab The result now follows I Since every PD is a UFD lemma 345 is valid for any PlD Bi 35 Torsion modules and order ideals Recall that our present goal is to classify all nitely generated modules over a PD We have remarked that not all such modules are free so the situation is more complicated than that of modules over a eld vector spaces As a rst step in our classi cation we will see that every nitely generated module M over a PD R does have a possibly 0 direct summand which is free The complementary summand consists of elements of nite orderlli More precisely we make the following de nition De nition 351 An Rmodule M is torsion free iffor all z E M and all a E R ax 0 implies either a 0 or z 0 Note that any nitely generated free module over an integral domain is torsion free This follows since if M is nitely generated and free then M E R so that z E M has the form r1 i i i rn E Rni Now if R is an integral domain then ari 0 implies that a 0 or ri 0 But az 0 iff ari 0 for all i so that we see that either a 0 or ri 0 for all i and hence I 0 The converse of this fact is only partially truer Proposition 352 fR is a PID then a nitely generated Rmodule M is free and only ifM is torsion free Proof We have just remarked that a nitely generated free module over an integral domain is torsion free For the converse assume that 0 y M is torsion free If M 0 then it is trivially freer Let 11iuzn be a generating set for M so that M R11 R1 and each I 7 0 Since M is torsion free if 0 f z E M then is independent so that R1 is free of rank 1 If we choose a maximal independent subset from 11 i i i zn say 11 1m reenumerating if necessary then the submodule FRzlRzm CHAPTER 3 MATHEMATICS 150C Spring 2001 53 is free of rank m since each Rzi is free of rank 1 and the z are independent Now for each i gt m since 11Hzm is maximal independent there exists a nonzero element ai E R with aizi 6 Fl Let a am1am2 on and note that am 6 F for all z E M homework exercisei Now R is an integral domain so that a f 0 Therefore since M is torsion free the map Au M A F given by Aaz am is injective and hence an isomorphism onto its image imAa aM ax z E But Au is a R module homomorphism homework exercise so that aM is a submodule of the nitely generated free module F Therefore theorem 3312 implies that aM is nitely generated and free But aM M so that M is nitely generated and free I Now let M be any R module and let I 6 Mi The map R A M de ned by a gt gt am is an R module homomorphismi lts image is the cyclic module generated by 1 De ne the order ideal 075 of z to be the set O aERzaz0i Note the order ideal of z is just the kernel of the map a gt gt am and hence it is an an ideal If R is a PD and z E M has a non zero order ideal then 0 f 01 a for some non zero a E R which is unique up to multiplication by a unit We call a the order of z 6 Mi De nition 353 IfM is on Rmodule an element 1 E M is torsion if 0 y 0 We let M z E M I is torsion A module M is a torsion module ifM Mt We note that a module is torsion free if M 0 This is because M is torsion free if 0 f a E R and 0 f z E M implies ax 0 if 0 f 1 implies 075 0 if 0 E M is the only torsion elementi Lemma 354 For every R module M M is a submodule ofM that is a torsion module Moreover the quotient MMt is torsion free Proof Suppose that Ly 6 ME and 0 ab E R satisfy az by 0 Since R is a PD we have ab 0 Moreover abz y abz aby 0 0 0 so that z y E M Also for all r E R arz arz raz raz r0 0 so that rm 6 M as well This shows that M is a submodule of M By de nition each element of M is torsion so that Mt ME and hence M is a torsion module Finally suppose that z M E MMt and 0 f a E R satis es ax M M Then am 6 Mt so that Mar 0 for some 0 f b E Ri Now R is a PD so that ab 0 and abz 0 so CHAPTER 3 MATHEMATICS 150C Spring 2001 54 that z E M This shows that M is the only torsion element of MM and hence MM is torsion free I Not surprisingly we call M the torsion submodule of Mr We can now take the next step in out classi cationi Theorem 355 IfM is a nitely generated module over a PID and M is the torsion submodule of M then there is a nitely generated free submodule Mf S M such that MMMfi That is every nitely generated module over a PID is a direct sum of a torsion module and a free module of nite rank Proof Let M be the torsion submodule and let 7r M A MM be the quotient mapi By lemma 354 MM is torsion freer Since M is nitely generated MM is nitely generated and hence MMt is free by proposition 352 This means that we can choose a basis 11ML7I2MLHizmME for MM We can then de ne a R module homomorphism 1 MM A M by de ning iMthMt z for l S i S m and extending linearly theorem 3311 Then clearly 7r 0 1 lMMt so that 1 is injectivei If we let Mf im LJ then Mf is a submodule of M and 1 MM A Mf is an isomorphismi it follows that Mf is nitely generated and free We claim that M M ED Mfr If I 6 M N Mf then 7rz ME and I y ME for some y aizi with ai 6 Bi But Mt 1 Wlt lt9 Mt yML so that y E M But 11 M 12 MHizm M is a basis in MM so that we must have ai 0 for all i and hence y 0 so that z 0 This shows that M N Mf 0 Finally if z E M then 7rz Eaizi M Let f Eaizi so that f E im LJ Mfr Moreover if t z 7 f then m we 7W m 7 7r w ltme 7M 7 m 7 Eatmm 7 so 7 so o and hence t E kerrr M Clearly I t f and hence M M Mf and the proof is complete I CHAPTER 3 MATHEMATICS 150C Spring 2001 55 Why is this the rst step in our classi cation We know that a nitely generated free R module is isomorphic to R where n is the rank of the module Just as for vector spaces any two free modules of the same rank are isomorphic We alluded earlier to change of basis matrices which are still valid over commutative rings These are the desired isomorphisms between two Rmodules of a given rank Therefore nite rank free modules are classi ed by their rank Having this in our hands the previous theorem implies that we can complete out classi cation of nitely generated modules over Ple if we can classify all nitely generated torsion modules over a PD This is what we now set out to do 36 Structure theorem for nitely generated modules over PIDs Let R be a PD and let 73 be a complete set of representatives of all prime elements of R That is we consider two primes pp in R to be equivalent if p 10 for some unit This is an equivalence relation on the set of all prime elements nice exercise and we choose one prime from each class For the ring Z this amounts to choosing say the positive primes ln FX we choose irreducible monic polynomials For each prime p E 73 and each R module M we let Mp z E M 075 10 for some n 2 0 That is Mp is the set of all elements of M that have order a power of p Lemma 361 Mp is a submodule of M for all primes p E 73 Proof Exercise l The following decomposition theorem holds for arbitrary torsion modules not just nitely generated ones Lemma 362 fR is a PID and M is a torsion module over R then M 63AM 1767 Proof Let 0 f z E M Since 11 z 075 f R But R is a PD so that 075 a for some 0 f a E R We know a f 0 since M is a torsion module Now a f R so that a is not a unit and hence we can write 2 e ap11pnn CHAPTER 3 MATHEMATICS 150C Spring 2001 56 where pi is irreducible for each i This is because R is a PID and hence a UFD Since R is a PD corollary 2 49 implies that each pi is a primer For each i l i i i n let 8 e n 4139 P rift1101 win so that a qipf i Now since ax 0 we have qiz E for all i Moreover we leave it as an exercise to show that 11 i i i qn are relatively primer Using this we can write 1Tiqiann for some Ti 6 R and hence I 11 qulzannz EMp1Mpni This shows that the Mp p E 73 generate Mi Now suppose that m i i i pn q E 73 are distinct and that IE Mp1quot39MPn M4 Since I E Mq then 075 gm for some m 2 0 Also since I E Mp1 Mpn z T111 TnIn where Ti 6 R and there exist ei 2 0 such that in 0 for all ii If we let a 10 3910quot then clearly ax 0 so that a 6 gm and hence a Q Since the p1 i i pn q E 73 are distinct primes they are relatively prime and hence a and qm are relatively primer By proposition 3427 this means that gm R and hence 075 R so that z 0 Therefore the Mp are independent and the proof is complete I If we combine theorem 3 55 and lemma 362 then we see that if M is a nitely generated module over a PID R then M 69 M p 69 M f 17673 where 73 is a complete set of all primes in R and Mf is a free module over R of nite rank Therefore to complete our classi cation we need only classify nitely generated torsion modules of the form That is nitely generated modules M with the property that pnz 0 for all z E M for some prime p E R and n 6 N For such a module M given I E M we have 075 a for some a E Ri But pnz 0 so that p E a and hence alpni But since p E R is prime this implies that a pT for some 0 S 7 S n The element CHAPTER 3 MATHEMATICS 150C Spring 2001 57 1oT is called the order of 1 Note that if 1oT is the order of I then pTz 0 but pr lz 0 Also 1oT is the order of I if and only R pr 2 RI De nition 363 Let M be a nitely generated module with pnM 0 for some p E 73 and some n E N The socle of M is SocM z E M pr 0 We leave it as an exercise for the reader to show that SocM is a submodule of M Lemma 364 IfM is a nitely generated module with pnM 0 for some p E 73 and some n E N and M M1 69 M2 then SocM SocM1 EB SocM2 Proof Let I E SocM so that pr 0 Since M M169M2 there exist unique y 6 M1 and 2 6 M2 with z y 2 Now 0 pr pypz and py 6 M1 and pz 6 M2 It follows that py p2 0 by uniqueness so that y E SocM1 and 2 E SocM2 so that SocM SocM1 SocM2 Now for any module M SocM C M so that if z E SocM1 SocM2 then I 6 M1 MQ so that z 0 Therefore SocM1 and SocM2 are independent and the proof is complete I Lemma 365 IfM is a nitely generated module with pnM 0 for some p E 73 and some n E N then SocM is a Rpmodule via the operation a 10 I H M Proof First if z E SocM then pr 0 so that for all a E R paz apz a0 0 and hence am 6 SocM The map de ned in the statement of the lemma clearly satis es the module axioms provided it is well de ned lf a p 12 p then a7 b E p so that am 7 121 a7 bz phz 0 Therefore the map is well de ned and SocM is a module over R I Now R is a PD so that the prime element p is irreducible and hence p is a maximal ideal It follows that Rp is a eld and hence SocM is a vector space over Rp Since SocM is a submodule of M and M is nitely generated corollary 3313 implies that SocM is nitely generated as an R module and hence as an R pmodule We have shown the following important fact Soc is a nite dimensional vector space over the eld R De nition 366 IfM is an Rmodule the elements y1ym are weakly independent if Tiyi39Tmym 0 implies that riyi 0 for all i CHAPTER 3 MATHEMATICS 150C Spring 2001 58 Lemma 367 IfM is an Rmodule and y1 ym are weakly independent then Ry1 ym Ry1Rym Proof We have Ry1 i ym Ryi 39 Rym by de nition If I 6 RM Ryi R9171 Ri1yi1 RM then there exist elements aj E R l S j S m with aiyi a191 042219271 ai1yi1 amym and hence alyl l quotquotaiyi l quot39amym0 Therefore ajyj 0 for all j since the yj are weakly independent and hence I aiyi 0 l OK7 we need one more technical lemma before we can give our next big step in the decomposition theorem I know it is easy to feel fatigued by now7 but stick with me it will all pay off Lemma 368 Suppose that M is a torsion module with pnM 0 and pn lM a 0 for some n 2 1 Let 0 f z E M with pn lz 0 and let M MRz Let 313m E M be weakly independent elements of M Then for each i there exists a representative yi of E such that yi has the same order as E Moreover the elements I yl ym are weakly independent Proof If M 07 there is nothing to prove7 so let E E M have order p8 for some e 2 1 If y is a representative of E in M7 then pay 6 RI and hence pay am for some a E R Since p is a prirne7 we can write a 17 where p X c If s 2 n7 then pay 0 so that the order of y is also p8 if pry 0 s for some r lt e7 then pr 07 a contradiction If s lt n7 then p501 has order p and hence y has order 1084quot We must have e n 7 s S n why and hence e S s Moreover7 the element has order p8 and y 7 y 7 psiecz psiecz 6 RI so that y 7 ps ecz also represents Q We have shown that if y 6 M7 we can nd a representative y E M ofgwith the same order as 3 Therefore for each i 1 7m7 we can choose a representative yi of E with the same order as E CHAPTER 3 MATHEMATICS 150C Spring 2001 59 We will now show that 17 M7 i i i 7 gm are weakly independenti Suppose that a7 a17 i i i 7 am 6 R and aza1y1amym 0 Therefore in the quotient M7 we have alylamvm 0 But the 7 are weakly independent so that aim U for all ii If p81 is the order of 377 then aim 5 implies ai 6 p81 and hence p81 diVides air That is7 ai pelhi for all i with hi 6 Bi Now my pe hiyi 0 since pe is also the order of yin This in turn implies that am 0 and hence I7 y17i i i 7 gm are weakly independent as desired I We now have the machinery in place to prove the next step in out decompositioni Recall that we are working on modules M over Ple with the property that pnM 0 and pn lM 0 for some prime p 6 Bi Lemma 369 IfM is a nitely generated module with pnM 0 for some p E 73 and some n 6 N then there exist unique natural numbers M g RPm 39 691 3Pnkl Proof We will induct on d dimRpSocM7 the case d 0 being triViali Given 0 f z 6 M7 the remarks before de nition 3 63 imply that the order of I has the form pr with 1 S r S n Pick 0 f 11 E M so that the order of 11 is 10 7 with n1 maximal Note that n1 is uniquely determined by Mr Let M MRzl denote the quotient modulei Since M is nitely generated7 M is nitely generated Moreover7 pnM 0 implies pnM 0 We claim that dimRpSocM lt dimRpSocMi To show this suppose that 317 i i i 7m 6 Socm is a basis for SocM over Rp We claim that the 7 are weakly independent over Ri For suppose that 131 39 amm U REL Then a1 pm am mm U CHAPTER 3 MATHEMATICS 150C Spring 2001 60 which implies that ai E p for all 239 because the E are linearly independent over Rp Therefore for all i ai pki for some 16 E R and hence ME kiPE39 6 for all 239 since 1 E SocMi This shows that the 1 are weakly independent over R establishing our claimi Now pnl lzl 0 and clearly pnl lzl E SocM so that we have an element 1 pnl lzl 6 R11 SocMi Now lemma 31618 implies that we can nd representatives yi E M for each i with the same order respectively That is yi E SocM for each 1 S i S mi Moreover the same lemma 31618 implies that the set z y1iii ym is weakly independent over R1 We leave it as an exercise for the reader to show that this implies that z y1i i i ym are linearly independent over Rp and hence dimRpSocM 2 721 1 so that dimRpSocM lt dimRpSocM as claimed Now by induction there exist unique natural numbers n2 2 2 nk 2 1 with M E Rpm EB EB RPnkl For each i 21 1116 we have Rpnl RE for some El 6 Mi Therefore the order of T is p and the E are weakly independent by lemma 3617 Therefore by lemma 3618 there are representatives 121 1 i 1k 6 M with the same orders 10 respectively and 11 121 1 i 1k are weakly independenti Therefore lemma 31617 implies that Rzluizk Rzl B GBRzzci But given I E M we know 5 E M is a R linear combination of 52111519 Say 5 ijji Then if y 3991 it is easy to see that z 7 y E kerM A M R11 so that z 7 y T111 for some 7 1 6 R and hence 11 i i i 1k generate Mi Therefore MR11 B BRzki Also we know n1 is unique and n1 2 n by the maximality of n1 so that Finally since the order of z is 10 we have R1 2 Rpnl so that nally M g Rpm EB GER10 CHAPTER 3 MATHEMATICS 150C Spring 2001 61 This brings us to our rst major characterization of nitely generated modules over a PD R1 The statement of those theorem may seem a bit overwhelming but we have already done all of the work to prove it so the proof will be a nice recap of the material above Theorem 3610 IfM is a nitely generated module over a PID R then there exist unique primes pl 1 i 1 pm 6 73 and for each such prime pi there exist unique natural numbers nilZni22quot392nikgt1 and there is a unique integer r 2 0 such that m k MngEB GB RPi U i1 j1 where Mf S M is free of rank r Proof Given M we know by theorem 31515 that M M f 69 M where Mf is a free module of rank r for some unique r 2 0 and M is the torsion submodule of Mr Moreover lemma 31612 implies that we have M 69 Md 1767 Since M is nitely generated M is nitely generated by corollary 3313 Therefore Mtp 0 for all but a nite number of distinct primes pl 1 i 1 pm 6 73 so that M 69 Mt i1 This shows that M Mf 69 A i1 Now for all i 1 i i i m we have pl1Mp 0 for some i E N and hence for each 1 S i S m we apply lemma 31619 to conclude that there exist unique natural numbers nil 2 rim 2 2 nikl 2 1 with MAM g RP 1EBEB RPihkll Putting all of this together we then have that m k MMf GB RPi U i1 j1 CHAPTER 3 MATHEMATICS 150C Spring 2001 62 as desired l The uniqueness statement of the theorem implies that the prime powers p together with the rank 7 completely characterize Mi We call the prime powers p the elementary divisors of Mr There fore theorem 3610 states that a nitely generated module over a PD is completely determined by the rank of the free part Mf S M and the elementary divisors of the torsion submodule M Next7 we want to reassemble the p power modules into another valuable decomposition for Mr The key fact is contained in the following lemma Lemma 3611 Let R be a PID and let plpiqpn E 73 be distinct pn39mes If e17i i r en E N and z pil 10quot then RI g RPil EB 39 39 39 GER102quot Proof Let n R A Rp denote the quotient map and de ne g0 7r1 gtlt gtlt 7Tn R A Rpil gtlt by a 7rlai i i Jr a Then go is a surjective ring homomorphism with kergo p N Therefore the rst isomorphism theorem 3210 implies that RPi1 quot39 10quot g RPi1 quot 3969 RPffl Since the pi are distinct primes7 they are relatively prime in pairs so that p N pi1 by lemma 345 Therefore E Rpi1 EB l OK7 if M is nitely generated and p1 i i 7pm 6 73 are the primes present in the elementary divisors of M7 we organize the exponents nij in the following not necessarily square array P11n112quot392n1k1 P21n212quot392n2k2 Pminm12quot392nmkm Then for eachhliiikmaxkliiikm7 let MA MA nmh 4hp1 172 quot39pm where we put a l for any p with nih not in the array Easily we see that for 2 S h S k we have thqh1 and hence qh71 S qh This proves all but the uniqueness statement in the follow theorem our nal goal CHAPTER 3 MATHEMATICS 150C Spring 2001 63 Theorem 3612 Fundamental Theorem of Finitely Generated Modules over a PID Let M be a nitely generated module over a PID R Then there is a unique integer r 2 0 and a unique chain of non trivial ideals 41 S 42 S E 4k in R with MngRqiRq2R4k where Mf is a free R module of rank r Proof Given M we use theorem 3610 to write in k M g Mf EB 63 RAP i1 j1 where r 2 0 is unique m i i i pm 6 73 are distinct and for each such prime pi there exist unique natural numbers nilZni22quot392nik 21 Using the notation given before the statement of the theorem lemma 3611 implies that RQh E RPilh EB RPSM EB 39 39 39 EB RP TW A Therefore we have that m k maxk1 m k 69 gatepr ea QBRltp gtgteRltqn We have already remarked that qh71 S qh for 2 S h S h maxhii This shows that MngR41R42quot3969R4k with 41 S 42 S S qk and each qh non triviali It remains to establish the uniqueness For this we may assume that M is a torsion modulei We induct on the number of primes that appear as factors in the decomposition 362 If there is only one prime p then M Mp and hence the uniqueness is settled by lemma 369 If there is more than one prime with say p one of them then by lemma 362 we have M Mp 69 N where N is the sum of the non zero Mq appearing in decomposition of Mr Now N is a nitely generated torsion module with one less prime in its factorization so that by induction there are unique ideals 41 S 4 S S 412 CHAPTER 3 MATHEMATICS 150C Spring 2001 in N With N E 30106 13012 69 EBB 112 Moreover7 by lemma 3 697 there are unique natural numbers n1 2 2 nh 2 1 such that Mun E RPm EB Rpm EB GER10 Then for M7 it is Clear that we have L11 971an 42 45pm and so on The elements qi E R are called the invariant factors for M i CHAPTER 4 MATHEMATICS 150C Spring 2001 65 Chapter 4 Two Applications of the Structure Theorem 41 Structure theorem for nitely generated abelian groups Our rst application of the structure theorem 3612 Will be the case R Zr Here7 a nitely7 generated Z module is just a nitely generated abelian group Theorem 411 fA is a nitely generated abelian group then there is a unique integer r 2 0 and a unique possibly empty set of integers mlpu mc with mj gt 1 for all j and mklmk71l lml such that AEZTEBZM1 GBZm2 Elan69ka Proof Since Z is PH and a nitely generated abelian group A is a nitely generated Z module7 the structure theorem 3612 applies and hence there is a unique integer r 2 0 and a unique chain of non triVial ideals m1 S m2 S S mg in Z With AEZTZm1Zm2EBZmki An ideal in Z is nontrivial if and only if the generator mj gt 1 and S mj1 if and only if mj1lmji l CHAPTER 4 MATHEMATICS 150C Spring 2001 66 Note that if in addition the group A is nite then necessarily r 0 and hence A is a nite direct sum of cyclic groups and the cyclic factors are uniquely determined up to order by the invariant factors mji Moreover the product m m1 mk of the invariant factors is the order of Al We illustrate these ideas with an example Example 412 Find all abelian groups of order 24 up to isomorphismi There are exactly three possibilities for the invariant factors m1 24 m1 127212 2 and m1 6mg 27213 2 Therefore there are exactly three abelian groups of order 24 up to isomorphismi They are respectively Z24 Z12 EB Z2 and Z5 69 Z2 69 Zgi In practice if the order of A is a large number it can be difficult to determine the invariant factors mj for Al However we do have the alternate form of the structure theorem 3610 which decomposes A using the elementary divisorsi In practice it is easy to determine all of the elementary divisors of A in fact if you look back at the proof of theorem 3612 we used the elementary divisors to nd the invariant factors We will illustrate this by example as well but rst we restate theorem 3610 for nitely generated Z modulesi Theorem 413 fA is a nitely generated abelian group then there is a unique integer r 2 0 and n not necessarily distinct prime integers p1i i i pn such that AgZTGBpri quot39Zpni The prime powers p are uniquely determined by A and this direct sum decomposition is unique up to the order of the factors Proof By theorem 3610 there exist a unique r 2 0 and unique positive primes pl 1 i i pm 6 Z such that for each such prime pi there exist unique natural numbers 7111271122 quot27 gt1 such that If we let n 16 then CHAPTER 4 MATHEMATICS 150C Spring 2001 67 is a sum of n cyclic groups of prime power order The uniqueness follows from the uniqueness of the nij I Since the product of the elementary divisors of a nite group must be the order of the group we can quickly determine all possible elementary divisors if a nite abelian group A by looking at the various ways to form elementary divisors from the prime factorization of Here is an example Example 414 Determine all abelian groups of order 24 up to isomorphism First note that 24 23 3 Therefore the only possible collections of elementary divisors are 233 22 23 and 2 2 2 3 The corresponding abelian groups are Z3 69 Z3 Z4 69 Z2 69 Z3 and Z2 69 Z2 69 Z2 69 Z3 respectively Theorem 3610 implies that any abelian group of order 24 is isomorphic to one of these 3 groups and no two of these three are isomorphic The reader should compare this example with the invariant factor decomposition given earlier When reconciling the two decompositions it will be useful to recall that for two integers ab Z 69 Z1 Zab if a and b are relatively prime Example 415 Determine all abelian groups of order 1500 up to isomorphism First note that 1500 22 3 53 Therefore the only possible collections of elementary divisors are 22353 22353 223525 223525 223555 and 223555 Each of these families determines a group of order 1500 For example 2 2 353 determines Z2 69 Z2 69 Z3 69 Z125 Theorem 3610 implies that any abelian group of order 1500 is isomorphic to one of these 6 groups and no two of these 6 are isomorphic Example 416 From the proof of theorem 3612 we can determine the possible invariant factors for an abelian group of order 1500 from the elementary divisors For example for the elementary divisors 22 3 5 5 5 we form the array of exponents 22 31 5212121 so that the elementary divisors are m1 22 3 5m2 5 and 7213 5 Using invariant factors the corresponding abelian group of order 1500 is Z50Z5EBZ5 CHAPTER 4 MATHEMATICS 150C Spring 2001 68 and 5560i Using the elementary divisors7 the corresponding abelian group of order 1500 is Z4Z3Z5Z5Za But 4 35 are relatively prime in pairs so that Z4 69 Z3 69 Z5 2 Z50 and the decomposition is the same 42 The Jordan canonical form Our second application of the structure theorem 3612 concerns a linear operator T V A V on a nite dimensional vector space At rst7 it is not obvious how the structure theorem applies to this situation at all The next proposition will tip our hand h Proposition 421 Let V be a nite dimensional vector space over a eld lF and let T V A V be a linear operator on V Then I V is an Mtmodule with the operation Mt gtlt V A V given by aw H 1 fa 2 The restriction of the Mt action on V to lF is just the given vector space structure on V and hence V is nitely generated as a Mtmodule 5 tv Tv for all v E V 4 An additive subgroup W S V is a Mt submodule if and only if W is a Tinvariant lFsubspace of V Proof 1 Exercise 2 The restriction to lF is the scalar multiplication in V over lF by de nition Now7 dimpV is nite so that every element v E V is a lF linear combination of some nite set of vectors v17 i i i vn C Vi But every such combination is a Mtlinear combination as well so that V is nitely generated as a Mtmodulei 3 The de nition 4 If W S V is an Mtsubmodule7 then W is stable under lF constant polynomials as well as t By the de nition of the Mt action on V7 this implies that W is a lF subspace of V and TW Q Wi CHAPTER 4 MATHEMATICS 150C Spring 2001 69 Therefore W is a Tinvariant lF subspace of Vi Conversely if W is a Tinvariant lF subspace of V then W is stable under lF and t It follows that W is invariant under any polynomial 6 Mt and hence W is a Mtsubmodule of Vi I Now the polynomial ring Mt is a PD and the proposition states among other things that given a linear operator T V A V we have a nitely generated Mtmodule V and hence the structure theorem 3612 applies Namely we have the following interpretation of 3612 Theorem 422 If V is a nite dimensional vector space over a eld lF and T V A V is a linear operator on V then there exist n not necessarily distinct irreducible monic polynomials p1 i i i pn 6 Mt such that rmV g FltlPi1 39 39 39 EB ll Bl102quot ale tmodules The prime powers p are uniquely determined by T and V Proof We have remarked that the structure theorem 3612 applies to the Mtmodule V and so we have a decomposition of V into a direct sum of a free Mtmodule and cyclic submodules of prime power order with the appropriate uniqueness statementi However since dimpV lt 00 the free part in the decomposition is necessarily 0 Mt is in nite dimensional as a lF module I Let 1 V A Mtpi1 EB EB Mtpfquot be the Mtmodule isomorphism of theorem 4 22 and let W1 TllFltlPil for 1 S i S n Then proposition 4i2il4 implies that Wi is a lF subspace of V necessarily nite dimensional Moreover since the Wi are 39 A t as Mt d l they are 39 A t as lF subspaces as well It follows that if we choose a basis Bi for Wi as an lF subspace for all i then B 81PM En is a basis for Vi That is pVW1 B BWn as lF vector spacesi Moreover proposition 4i2i14 also implies that Wi is Tinvariant for all i so that the matrix of T with respect to the basis 3 has block formi Now x an i between 1 and n and for notational brevity let W Wi and 10 tk akiltkil a1t a0 Note that 1 generates E W as an Mtmodulei Let we il 1l E W and let wi Tiw0 for i 2 1 Then wi E W for all i and ti for all i the latter since 1 is an Mtmodule homomorphism and Tiw0 tiwo by de nition CHAPTER 41 MATHEMATICS 150C Spring 2001 70 Now7 given w 6 W7 we know gt for some 9 6 Mt and since degf k we may assume that degg S k 7 11 Say 90 bk71tk71 39 39 39 l1175 be with bj E F Therefore we have 1671 V 1671 w 1gt f 7 BMW 1 7 ijwj j0 j0 and hence the set 10011 1 71111671 spans W as an lF module Moreover7 if bowo bk71wk71 0 for some scalars bj E F then gt where 91 be bl tk l 6 Mt But deggt S k 7 1 so that gt E if g 0 and hence bj 0 for all 1 S j S k 711 This shows that 100111 101671 is a basis for W as an lF module We note that dimpW k degf degpf Note that by construction7 wi1 Twi forOSiSk7land 1671 1 wk tk f 7 ZELM f 1671 wk7 Zajwj 1 j0 It follows that the matrix of le with respect to the basis 100111 7111z 1 for W has the form and hence 0 0 0 fag 1 0 0 7a1 0 1 0 7112 1 411 0 0 1 7ak1 We have essentially proven the following theoremi CHAPTER 4i MATHEMATICS 150C Spring 2001 71 Theorem 423 Let V be a nite dimensional vector space over a eld lF and let T V 7gt V be a linear operator on V Then there is a basis 3 for V in which the matrix ofT is in block form where each of the blocks has the form 41 Moreover the scalars in the nontrivial column of each such block are uniquely determined by T Proof Given T V 7gt V we let Wi be as above so that Bi wiopiqwiklil is a basis for W where h degpg i On Wi the restriction of T has the form 41 and the nontrivial column consists of the nonleading coef cients of the monic polynomial 101 and are therefore uniquely determined by theorem 422 Finally we have seen that B 31PM Bn is a basis for V and the matrix for T with respect to B is in block form with the ith block being the matrix of leli This completes the proof I When we write T in the above basis the operator is said to be in rational canonical formi It is the best form available for an arbitrary operator over a lF vector space when lF is any eld If lF C is the eld of complex numbers we can do better than the rational canonical formi If lF C then every irreducible polynomial has the form t7 a for some a E C Therefore using the notation above we have p t 7 aiel so that W E CMO a e and dimpWZ eii Again we x an i between 1 and n and write W Wia ai f t 7 aiel and e h for brevity We again de ne we d 1l but this time we set wiT7aIiw0 for i 2 1 Note that t 7 ai because 1 is a CMmodule homomorphismi Moreover 0 iff wi 0 since 1 is an isomorphismi We have wl T7aIw0 Hi wk1T7aIwk2 wk T7aIwk1 T7aIkw0 0 where wk 0 since Solving each equation gives Two wi awe my ka72 wk71 awk727 ka71 awkir We leave it as an exercise to show that w0iiiwk1 is a C linearly independent set Since dimcW h w0iiiwk1 is a basis for W and the above computation shows that the ma CHAPTER 4 MATHEMATICS 150C Spring 2001 72 trix for the restriction of T to W in this basis is a0 00 1a 00 0100i 4 2 001a These matrices are called Jordan blocks The previous remarks constitute a proof of the following theoremi We will write out the proof to tidy up Theorem 424 Let V be a nite dimensional complex vector space and T V A V an operator on V then there is a basis 3 for V in which the matrix for T is block form with Jordan blocks Proof Since every irreducible polynomial over C is linear theorem 42 implies that V W1 69 69 Wn where each Wi is isomorphic to CMt 7 aieli For each i we construct the basis for Wi as above Then the matrix of T is block form and the blocks have the form 42 l The matrix of T in the basis of the theorem is referred to as the Jordan form of T It is uniquely determined by T up to a permutation of the basis vectors since the elementary divisors of the CMmodule de ned by T are uniquely determined by T Note that the Jordan form of T is lower triangular and hence the diagonal entries that is those complex numbers ai E C appearing in the decomposition of V are the eigenvalues of T The number of appearances of any given eigenvalue a for T is exactly the multiplicity of the root a for the characteristic polynomial p of Ti Let J be a Jordan block with diagonal entry a E C Clearly a is the only eigenvalue for J and the eigenspace for a is 1 dimensional aI 7 T has nullity 1 Conversely if an operator has all eigenvectors scalar multiples of some xed vector then it can only have one eigenvalue else independent eigenvectorsi Moreover its Jordan form cannot have more than one Jordan block else it has at least 2 eigenvectors in a basis It follows that for an operator T the number of Jordan blocks with diagonal entry a in the Jordan form of T is exactly the dimension of the eigenspace for the eigenvalue a E C This dimension is an integer between 1 and the multiplicity of a in the characteristic polynomial p of T If these multiplicities are not too big we can use this to nd the Jordan formi Here is an example CHAPTER 4 MATHEMATICS 150C Spring 2001 73 Example 425 Determine the Jordan form of the matrix 110 A010 011 The characteristic polynomial is t 7 137 and by inspection the matrix 13 7 A has rank 1 Therefore the eigenspace is 2 dimensional so that there are two Jordan blocks in the Jordan form Necessarily the Jordan form is 100 717110 001 If a is a 16fold root in the characteristic polynomial of T7 the number of Jordan blocks in the Jordan form with diagonal entry a is an integer between 1 and Is Here are the possibilities for small k we omit 0 entries k1a a a 162 7 1a a a a a 63 1a 7 1a 7 a 7 1a a a a a a a a 1a 1a 1a 1a a 7641 7 7 7 7 1a 1a a a a la a la a a Note that for k S 37 the number of Jordan blocks with diagonal entry a ie the dimension of the null space of T 7 a 1 is completely determines the portion of the Jordan form with diagonal entry a When Is 4 however7 there are two possibilities for two 2 Jordan blocks It can be shown that the operator T 7 a 12 distinguishes the cases We won t worry about the details Example 426 What is the Jordan form of a matrix whose characteristic polynomial is t 7 3W 7 7gt4 CHAPTER 4 MATHEMATICS 150C Spring 2001 74 and such that the space of eigenvectors for the eigenvalue 3 is 2 dimensional and the space of eigenvectors for the eigenvalue 7 is 3 dimensional The information given means that for a 37 we have 2 Jordan blocks in a 3 X 3 matrix and for a 7 we have 3 Jordan blocks in a 4 X 4 matrix The only possibility is Index algebraic element 37 basis for modules 46 characteristic 7 coefficients of a polynomial ll commutative ring 4 conjugate 22 cyclic module 47 degree of a polynomial ll divides 34 division ring 5 divisor of zero 5 elementary divisors 61 endomorphism 19 evaluation homomorphism 3 extension nite 36 simple 39 eld 5 eld extension 36 free module 46 greatest common divisor 49 group algebra 20 group ring 20 homomorphism evaluation 3 of modules 42 of rings 3 23 ideal improper 32 maximal 32 prime 33 principle 28 proper nontrivial 32 trivial 32 independent 46 submodules 44 indeterminate 10 invariant factors 63 irreducible l6 irreducible element 34 isomorphic rings 4 isomorphism INDEX MATHEMATICS 15OC Spring 2001 of modules 43 of rings 4 Jordan blocks 71 Jordan form 71 kernel of a ring homomorphism 24 least common multiple 49 length 22 linear combination for modules 45 maximal ideal 32 order 57 order ideal 53 polynomial ll constant 11 prime element 34 prime ideal 33 principle ideal 28 principle ideal domain 34 principle ideal ring 28 product ring 4 quaternions 22 quotient modules 43 quotient ring 25 rank 46 76 rational canonical form 70 regular module 41 relatively prime 49 ring 1 commutative 4 division 5 unital 4 With unity 4 ring homomorphism 3 23 simple extension 39 ring 28 44 skew eld 5 socle 57 span 45 submodule 42 improper 42 trivial 42 torsion 53 torsion free 52 torsion module 53 torsion submodule 54 transcendental element 37 trivial ring 4 unit 5 unital ring 4 unity 4 Weyl algebra 20 zero

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