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Student Notes 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 andExample [5] prevents4 of SectionZ10 5.1,from we beingsaw 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 of zero counts: Section divisors. Multiplication 5.1, The following we issaw not theorem commutative,that Z ncharacterizesis a and ring it has thesefor zero every Z 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.5Since0 ,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 means that a 0 , b 0 , but n 5 ab 2where 21 , a , n and 1 , b , n.
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