270 Chapter 5 Rings, Integral Domains, and Fields 270 Chapter 5 Rings, Integral Domains, and Fields

5.2 Integral5.2 DomainsIntegral Domains and Fields and Fields

In the precedingIn thesection preceding we definedsection we the defined terms the ring terms with ring unity, with unity, commutative commutative ring, ring,andandzerozero 270 Chapter 5 Rings, Integral Domains, and Fields 5.2divisors. IntegralAll Domains threedivisors. of and theseAll Fields three terms ! of these are usedterms inare defining used in defining an integral an integral domain. domain. Week 13

Definition 5.14DefinitionI Integral 5.14 IDomainIntegral Domain 5.2 LetIntegralD be a ring. Domains Then D is an andintegral Fields domain provided these conditions hold: Let D be a ring. Then D is an integral domain provided these conditions hold: In1. theD ispreceding a commutative section ring. we defined the terms ring with unity, commutative ring, and zero 1. D is a commutativedivisors.2. D hasAll a ring. unity three eof, and thesee 2terms0. are used in defining an integral domain. 2. D has a unity3. De,has and noe zero2 0 divisors.. Definition 5.14 I Integral Domain 3. D has no zero divisors. Note that the requirement e 2 0 means that an integral domain must have at least two Let D be a ring. Then D is an integral domain provided these conditions hold: Remark. elements. Note that the1. requirementD is a commutative e 2 0ring. means that an integral domain must have at least two elements. Example2. D has a1 unityThe e, ring andZe of2 all0. integers is an integral domain, but the ring E of all even in- tegers3. D ishas not no an zero integral divisors. domain, because it does not contain a unity. As familiar examples of integral domains, we can list the set of all rational numbers, the set of all real numbers, and Example 1 theThe set ofring allZ complexof all numbers—allintegers is an of integral these with domain, their usual but operations. the ring E of all even in-I tegers is not an integralNote domain,that the requirement because ite does2 0 means not contain that an integral a unity. domain As familiar must have examples at least two of elements. integral domains, we can list the set of all rational numbers, the set of all real numbers, and Example 2 The ring Z10 is a commutative ring with a unity, but the presence of zero the set of all complexdivisors numbers—allsuch as [2] and [5] of theseprevents withZ theirfrom usual being an operations. integral domain. Considered asI a Example 1 The ring Z of all integers10 is an integral domain, but the ring E of all even in- possible integral domain, the ring M of all 2 3 2 matrices with real numbers as elements fails tegers is not an integral domain, because it does not contain a unity. As familiar examples of on two counts: Multiplication is not commutative, and it has zero divisors. I Example 2 integralThe ring domains,Z10 is we a can commutative list the set of all ring rational with numbers, a unity, the but set the of all presence real numbers, of zero and the set of all complex numbers—all of these with their usual operations. I divisors such as [2]In and Example [5] prevents 4 of SectionZ10 5.1,from we being saw that an Z integraln is a ring domain. for every Considered value of n . as1 a. possible integralMoreover, domain,Zn theis a ring commutativeM of all 2ring3 since2 matrices with real numbers as elements fails Example 2 The ring Z10 is a commutative ring with a unity, but the presence of zero on two counts: Multiplication is not commutative,a ? b 5 ab and5 itba has5 zerob ? divisors.a I divisors such as [2] and [5] prevents Z10 from being an integral domain. Considered as a forpossible all a , integralb in Z domain,n. Since theZn3 ringhas4 M314ofas all the3 2 4unity,3 23 matricesZ4 n is3 an4 with integral3 4 real numbers domain asif elementsand only failsif it In Examplehason two4no ofzero counts: Section divisors. Multiplication 5.1,The following we is saw not theorem commutative, that Z ncharacterizesis a and ring it has these for zero everyZ divisors.n, and value it provides of n us. with1I. Moreover, Zn isa largea commutative 3class4 3 4 of finite ring integral since domains3 4 (that is, integral domains that have a finite number of elements).In Example 4 of Section 5.1, we saw that Z is a ring for every value of n . 1. a ? b 5 ab 5 ba 5 b ?n a Moreover, Zn is a commutative ring since Theorem 5.15 I The Integral Domain Z When n Is a Prime for all a , b in Zn. Since Zn3has4 314 as thea3 n ? 4unity,b 53 abZ4 n5is3 baan4 5integral3 b4 ? a domain if and only if it has no zero divisors.for all Thea , b followingin Z . Since theoremZ has 1 characterizesas the unity, Z theseis an Zintegraln, and domain it provides if and usonly with if it 3 4 3 4 For n . 1, Zn is ann integral3 4 domainn3 4 3 4if and3 only4 3if n4 nis a3 prime.4 3 4 a large class ofhas finite no zero integral divisors. domains The following(that is,theorem integral characterizes domains these that Z haven, and ait finiteprovides number us with of elements). Proofa large 3classFrom4 3 4 of the finite previous integral discussion, domains3 4 (that it is clearis, integral that we domains need only that provehave athat finite Zn numberhas no zeroof elements). divisors if and only if n is a prime. p ⇐ q Suppose first that n is a prime. Let a 2 0 in Zn,and suppose a b 5 0 for some b I Theorem 5.15 The Integralin Zn. DomainNow a b 5Zn0Whenimplies thatn Is ab a 5Prime0 ,and therefore,n ab. However, a 2 0 means Theorem 5.15 I The Integral Domain Zn When n Is a Prime that n a. Thus n ab and n a. Since n is3 4a prime,3 4 this implies that 3n43b,4 by Theorem3 4 2.16;3 4 3 43 4 3 4 3 4 3 4 0 3 4 3 4 For n . 1, Zn isFor ann integral. 1, Zn is domain an integral if anddomain only if andif n onlyis a if prime. n is a prime. >0 0 >0 0

Proof From theProof previousFrom thediscussion, previous discussion, it is clear it thatis clear we that need we needonly onlyprove prove that that Z nZnhashas no no Proof.zero divisors if zeroand divisors only if if n andis aonly prime. if n is a prime. p ⇐ q Suppose first that n is a prime. Let a 2 0 in Zn,and suppose a b 5 0 for some b p ⇐ q Suppose first that n is a prime. Let a 2 0 in Zn,and suppose a b 5 0 for some b in Zn. Now a b 5 0 implies that ab 5 0 ,and therefore,n ab. However, a 2 0 means in Zn. Now a thatb 5 n 0a. Thusimplies n ab thatand abn a.5 Since0 ,and n is3 4atherefore, prime,3 4 thisn impliesab. However, that 3n43b,4 bya Theorem23 4 0 means 2.16;3 4 that n a. Thus n ab and 3n43a.4 Since3 4 n is3 4a prime,33 44 this3 4 implies that0 3n43b,4 by Theorem3 43 4 3 4 2.16;3 4 3 43 4 >03 4 0 3 >0 4 3 4 0 0 3 4 3 4 >0 0 >0 0

1 5.2 Integral Domains and Fields 271

that is, b 5 0 . We have shown that if a 2 0 ,the only way that a b can be 0 is for b to be 0 . Therefore, Zn has no zero divisors and is an integral domain. ,p ⇐ ,q Suppose3 4 3 now4 that n is not a prime.3 Then4 3 n4 has divisors other 3than43 461 and3 64 n, so there3 4 are3 integers4 a and b such that 5.2 Integral Domains and Fields 271 n 5 ab where 15.2, a , Integraln and 1 Domains, b , n. and Fields 271 5.2 Integral Domains and Fields !Week 13 Thisthat is,meansb 5 that0 . Wea 2 have0 , shownb 2 0that, but if a 2 0 ,the only way that a b can be 0 is for b to that is, b 5 0 be. We0 . haveTherefore, shownZn thathas no if zeroa 2 divisors0 ,the and only is an way integral that domain. a b can be 0 is for b to ,p ⇐ ,q Suppose3 4 3 now4 3 4that3 n4 is3 not4 a3 aprime.4 b 53 Then4 ab3 n54 hasn divisors5 0 . other 3than43 461 and3 64 n, so there3 4 be 0 . Therefore, Zn has no zero divisors and is an integral domain. are3 integers4 a and b such that ,p ⇐ ,q Suppose3 4 3 now4Therefore, that n isa isnot a zero a prime. divisor3 Then 4in3 Z43n3 ,n44 andhasZ3 divisorsn 4is not3 an4 other integral3 4 3than 43domain. 461 and3 64 n, so there3 4 Combining the two cases, we see that n is a prime if and only if Z is an integral are3 integers4 a and b such that n 5 ab where 1 , a , n and 1 , b , n. n domain. 3 4 This meansn 5thatab a 2where0 , b 21 ,0 ,a but, n and 1 , b , n. One direct consequence of the absence of zero divisors in an integral domain is that the 3 4 3 4 3 4 3 a4 b 5 ab 5 n 5 0 . This means thatcancellation a 2 0 , lawb 2for multiplication0 , but must hold. Therefore, a is a zero divisor in3 Z43n,4 and Z3 n 4is not3 an4 integral3 4 domain. Combining the twoa b cases,5 ab we see5 thatn 5n is0 a. prime if and only if Z is an integral I3 4 3 4 3 4 3 4 n Theorem 5.16 domain.Cancellation3 4 Law for Multiplication Therefore, a isIf aa ,zerob, and divisorc are elementsin3 Z43n,4 and of anZ3 n integral4is not3 an4domain integral3 4 D such domain. that a 2 0 and ab 5 ac, then Combiningb 5 theOnec. two direct cases, consequence we see of that the absencen is a primeof zero divisors if and in only an integral if Zn domainis an integral is that the domain. 3 4 cancellation law for multiplication must hold. (p q) ⇒ r Proof Suppose a, b, and c are elements of an integral domain D such that a 2 0 and ab 5 ac. Now TheoremOne¿ direct5.16 IconsequenceCancellation of the Law absence for Multiplication of zero divisors in an integral domain is that the cancellation law for multiplication must hold.ab 5 ac ⇒ ab 2 ac 5 0 If a, b, and c are elements of an integral domain D such that a 2 0 and ab 5 ac, then ⇒ a(b 2 c) 5 0. b 5 c. Since a 2 0 and D has no zero divisors, it must be true that b 2 c 5 0, and hence b 5 c. I Theorem 5.16 Proof.(pCancellationq) ⇒ r Proof LawSuppose for aMultiplication, b, and c are elements of an integral domain D such that a 2 0 and ab 5 ac. Now If a, ¿b, and c are elementsIt can be shown of an that integral if the cancellation domain Dlawsuch holds that in a commutativea 2 0 and ring,ab 5 thenac the, then ring cannot have zero divisors. The proofab 5 ofac this⇒ isab left2 asac an5 exercise.0 b 5 c. To require that a ring has no zero divisors is equivalent to requiring that a product of ⇒ a(b 2 c) 5 0. nonzero elements must always be different from 0. Or, stated another way, a product that is (p q) ⇒ r Proof Suppose0Since musta, ba have,2 and0 at and cleast areD hasone elements nofactor zero equal divisors, of toan 0. integral it must be domain true that bD2suchc 5 0,that and a hence2 0 band5 c. ab 5 ac. Now A field is another special type of ring, and we shall examine the relationship between a field and an integral domain. We begin with a definition. ¿ It can be shown that if the cancellation law holds in a commutative ring, then the ring cannot have zero divisors.ab 5 Theac ⇒ proofab of2 thisac is5 left0 as an exercise. 272 Chapter 5 Rings, Integral Domains, and Fields Definition 5.17 I FieldTo require that a ring has ⇒no zeroa(b divisors2 c) 5 is0. equivalent to requiring that a product of nonzero elements must always be different from 0. Or, stated another way, a product that is Since a 2 0 andLet0 must DFThehasbe have arational noring. at zero least Then numbers, divisors,one F is factor a thefield itrealequal providedmust numbers, to be0. these true and conditionsthat the complexb 2 hold:c 5numbers0, and are hence familiar b 5exam-c. A field is another special type of ring, and we shall examine the relationship between ples1. Fofis fields. a commutative We shall ring.see in Corollary 5.20 that if p is a prime, then Zp is a field. Other aand field less and familiar an integral examples domain. of fields We begin are found with ina definition. the exercises for this section. 2. F has a unity e, and e 2 0. It can be shownPart that of if the the relation cancellation between law fields holds and integralin a commutative domains is stated ring, in then the the following ring 3. Every nonzero element of F has a multiplicative inverse. cannot have zerotheorem. divisors. The proof of this is left as an exercise. Definition 5.17 I Field To require that a ring has no zero divisors is equivalent to requiring that a product of nonzeroTheorem elements 5.18 LetI mustFFieldsbe a always ring. and Then Integral be Fdifferentis a Domains field fromprovided 0. Or,these stated conditions another hold: way, a product that is 0 must have at least1. F oneis a commutativefactor equal ring. to 0. Every field is an integral domain. A field is another2. F has special a unity typee, and ofe 2ring,0. and we shall examine the relationship between a field and an integral domain. We begin with a definition. Proof. p ⇒ q Proof3. EveryLet nonzeroF be a field.element To ofprove F has that a multiplicativeF is an integral inverse. domain, we need only show that F has no zero divisors. Suppose a and b are elements of F such that ab 5 0. If a 2 0, then a21 [ F and Definition 5.17 I Field ab 5 0 ⇒ a21(ab) 5 a21 ? 0 Let F be a ring. Then F is a field provided these⇒ conditions (a21a)b 5 hold:0 eb 5 0 1. F is a commutative ring. ⇒ ⇒ b 5 0. 2. F has a unity e, and e 2 0. Similarly, if b 2 0, then a 5 0. Therefore, F has no zero divisors and is an integral 3. Every nonzerodomain. element of F has a multiplicative inverse.

It is certainly not true that every integral domain is a field. For example, the set Z of all integers forms an integral domain, and the integers 1 and 21 are the only elements of Z that have multiplicative inverses. It is perhaps surprising, but an integral domain with a finite 2 number of elements is always a field. This is the other part of the relationship between a field and an integral domain.

Theorem 5.19 I Finite Integral Domains and Fields

Every finite integral domain is a field.

p ⇒ q Proof Assume that D is a finite integral domain. Let n be the number of distinct ele- ments in D; say,

D 5 d1 , d2 , c, dn ,

where the di are the distinct elements of5 D. Now let a6be any nonzero element of D, and consider the set of products

ad1 , ad2 , c, adn .

These products are all distinct,5 for a 2 0 and ad6r 5 ads would imply dr 5 ds,by Theorem 5.16, and the di are all distinct. These n products are all contained in D, and no two of them are equal. Hence they are the same as the elements of D, except possibly for order. This means that every element of D appears somewhere in the list

ad1 , ad2 , c, adn . 272 Chapter 5 Rings, Integral Domains, and Fields

The rational numbers, the real numbers, and the complex numbers are familiar exam- ples of fields. We shall see in Corollary 5.20 that if p is a prime, then Zp is a field. Other and less familiar examples of fields are found in the exercises for this section. Part of the relation between fields and integral domains is stated in the following theorem. 5.2 Integral Domains and Fields 273 Theorem 5.18 I Fields and Integral Domains

In particular, theEvery unity field e is is onean integral of these domain. products. That is, adk 5 e for some dk. Since mul- tiplication is commutative in D, we have dka 5 adk 5 e, and dk is a multiplicative inverse of a. Thusp ⇒Dq is Proofa field. Let F be a field. To prove that F is an integral domain, we need only show that F has no zero divisors. Suppose a and b are elements of F such that ab 5 0. If a 2 0, then a21 [ F and ab 5 0 ⇒ a21(ab) 5 a21 ? 0 Corollary 5.20 I The Field Zn When n Is a Prime ⇒ (a21a)b 5 0 5 Zn is a field if and only if n is a prime. ⇒ eb 0 ⇒ b 5 0. This follows at once from Theorems 5.15, 5.18, and 5.19. Proof Similarly, if b 2 0, then a 5 0. Therefore, F has no zero divisors and is an integral domain. We have seen that the elements of a ring form an abelian group with respect to addition. A similar comparisonIt is can certainly be made not true for that the every nonzero integral elements domain isof a afield. field. For It example, is readily the seen set Z thatof all integers forms an integral domain, and the integers 1 and 21 are the only elements of Z that the nonzero elements form an abelian group with respect to5.2 multiplication. Integral Domains The and definition Fields of273 a field can thushave be reformulatedmultiplicative inverses.as follows: It is perhaps A field surprising,is a set of but elements an integral in domain which with equality, a finite number of elements is always a field. This is the other part of the relationship between a field addition, and multiplicationand an integral aredomain. defined such that the following conditions hold. 5.2 Integral DomainsIn particular, and Fields the unity! e is one of these products. That is, adk 5 e for some dk. SinceWeek mul- 13 1. F forms antiplication abelian groupis commutative with respect in D, weto addition.have dka 5 adk 5 e, and dk is a multiplicative inverse Theorem2. The nonzero 5.19 ofI a.elementsFiniteThus D Integralis of a field.F form Domains an abelian and Fields group with respect to multiplication. 3. The distributiveEvery finitelaw x integral(y 1 z domain) 5 xy is1 a field.xz holds for all x, y, z in F.

CorollaryThep last⇒ 5.20q example IProofThe inAssume Field this Zsection thatn WhenD ispointsn a finiteIs a out Prime integral that some domain. of Letourn mostbe the familiar number rings of distinct do not ele- form integral domains.ments in D; say, Zn is a field if and only if n is a prime.

D 5 d1 , d2 , c, dn , ProofFor n $This2, follows each of at theonce rings from Theorems 5.15, 5.18, and 5.19. Example 3 where the di are the distinct elements of5 D. Now let a6be any nonzero element of D, and consider the set of products We haveM seenn(Z that), theM elementsn(Q), ofM a ringn(R form), anM abeliann(C) group with respect to addition.

A similar comparison can be made forad 1the , ad nonzero2 , c, ad elementsn . of a field. It is readily seen that is not an integralthe domain, nonzero elements since multiplication form an abelian in group each with of them respect is to not multiplication. commutative. The definitionIt is also of These products are all distinct,5 for a 2 0 and ad6r 5 ads would imply dr 5 ds,by true that each ofa fieldthem can contains thus be zeroreformulated divisors as if follows: n $ 2 A. For fieldnis5 a set2, theof elements product in which equality, Theorem 5.16, and the di are all distinct. These n products are all contained in D, and addition, and multiplication are defined such that the following conditions hold. no two of them are equal.10 Hence00 they are the same00 as the elements of D, except possibly for order.1. F Thisforms means an abelian that every group element with respect of 5D appears to addition. somewhere in the list 10 11 00 2. The nonzero elements of F form an abelian group with respect to multiplication. ad1 , ad2 , c, adn . illustrates this statement.3. The distributive Similar lawexamples x(y 1 z )can5 xyeasily1 xz beholds constructed for all x, y, forz in nF. . 2. I The last example in this section points out that some of our most familiar rings do not form integral domains. B R B R B R Exercises 5.2 Example 3 For n $ 2, each of the rings True or False Mn(Z), Mn(Q), Mn(R), Mn(C) Label each of the following statements as either true or false. is not an integral domain, since multiplication in each of them is not commutative. It is also 1. An integraltrue domain that each contains of them at contains least 2 zero elements. divisors if n $ 2. For n 5 2, the product 2. Every field is an integral domain. 10 00 00 5 3. Every integral domain is a field. 10 11 00 4. If a set S isillustrates not an integral this statement. domain, Similar then examples S is not cana field. easily be constructed for n . 2. I

B R B R B R Exercises 5.2 True or False Label each of the following statements as either true or false. 1. An integral domain contains at least 2 elements. 2. Every field is an integral domain. 3. Every integral domain is a field. 4. If a set S is not an integral domain, then S is not a field.

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