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13 اﻟﻤﺤﺎﺿﺮة ﺳﻨﺒﺤﺚ ﻧﻈﺮﯾﺔ ﻗﺎﺑﻠﯿﺔ اﻟﻘﺴﻤﺔ ﻓﻲ اﻟﺴﺎﺣﺔ اﻟﺘﻜﺎﻣﻠﯿﺔ وھﻲ ﺗﻌﻤﯿﻢ ﻟﺼﻔﺎت ﻣﺸﺎﺑﮭﺔ ﻓﻲ ﺣﻠﻘﺔ اﻹﻋﺪاد وﻧﺒﺪأ ﺑﺘﻌﺮﯾﻒ ﻣﻌﻨﻰ أن ﯾﻘﺴﻢ ﻋﻨﺼﺮ ﻣﺎ ﻋﻨﺼﺮ آﺧﺮ ﻓﻲ اﻟﺤﻠﻘﺔ أو أن ﯾﻜﻮن ﻋﻨﺼﺮ.اﻟﺼﺤﯿﺤﺔ .ﻣﻀﺎﻋﻒ ﻟﻌﻨﺼﺮ آﺧﺮ Definition 1.27 If a 0 and b are elements of a ring R, then a is said to divide b, in symbols a|b, provided that some c R exists such that b= ac. In case a does not divide b, we shall write a b. Other language for the divisibility property a|b is that a is a factor of b, that b is divisible by a, and that b is a multiple of a. .اﻟﻤﻼﺣﻈﺎت اﻟﺘﺎﻟﯿﺔ ﺗﺴﺘﻨﺘﺞ ﻣﻦ اﻟﺘﻌﺮﯾﻒ ﻣﺒﺎﺷﺮة Theorem 1.39 Let the elements a, b, c R. Then a) a|0, 1|a, a|a; b) a|1 if and only if a is invertible; c) if a|b, then ac| bc; d) if a|b and b|c, then a|c; e) if c|a and c|b, then c|(ax + by) for every x, yR. : ﻗﺴﻤﺔ اﻟﻌﻨﺎﺻﺮ ﻓﻲ اﻟﺤﻠﻘﺔ ﺗﺮﺗﺒﻂ ﺑﻌﻼﻗﺔ اﻻﺣﺘﻮاء ﺑﯿﻦ اﻟﻤﺜﺎﻟﯿﺎت ﺑﻤﻌﻨﻰa|b if and only if (b) (a). ذو ﻣﻌﻜﻮس )ﺑﺎﻟﻨﺴﺒﺔ إﻟﻰ ﻋﻤﻠﯿﺔ اﻟﻀﺮب( ﻓﺎن أي ﻋﻨﺼﺮ ﻓﻲ اﻟﺤﻠﻘﺔ ﯾﻘﺒﻞu ﻋﻨﺪﻣﺎ ﯾﻜﻮن اﻟﻌﻨﺼﺮ وﺑﺎﻟﺨﺼﻮص ﻓﻲ اﻟﺤﻘﻞ ﻓﺎن أي ﻋﻨﺼﺮ ﯾﻘﺒﻞ اﻟﻘﺴﻤﺔ ﻋﻠﻰ أي ﻋﻨﺼﺮ آﺧﺮ ﻏﯿﺮ. u اﻟﻘﺴﻤﺔ ﻋﻠﻰ .اﻟﺼﻔﺮ ﻟﺘﺠﺎوز ﻣﺸﺎﻛﻞ.( ﻟﯿﺲ ﻟﺪﯾﮫ أي ﻗﻮاﺳﻢ )ﻋﻮاﻣﻞ2 اﻟﻌﻨﺼﺮ، ﺑﺎﻟﻤﻘﺎﺑﻞ ﻓﻲ ﺣﻠﻘﺔ اﻷﻋﺪاد اﻟﺰوﺟﯿﺔ.اﻟﻤﻼﺣﻈﺔ أﻋﻼه ﻧﺤﺘﺎج إﻟﻰ اﻟﺘﻌﺮﯾﻒ اﻟﺘﺎﻟﻲ Definition 1.28 Two elements a, b R are said to be associated elements or simply associated if a = bu, where u is an invertible element. Example 1.33 In the case of the ring Z the only associates of an integer n Z are ± n, since ± 1 are the only invertible elements of R. Example 1.34 Consider the domain Z(i) of Gausian integers, a subdomain of the complex number field, whose elements form the set Z(i) = { a + bi | a, b Z; i 2 = -1}. Here the only invertible elements are ± 1 and ± i. So the class of associates determined by any Gausian integer a + bi consists of exactly four members : a + bi, -a- bi, -b +ai, b- ai. We have the following theorem: Theorem 1.40 Let a, b be nonzero elements of an integral domain R. Then the following statements are equivalent: 1. a and b are associates, 2. both a| b and b| a, 3. (a) = (b). Proof: In the class. اﻷﻋﺪاد ﺻﻮل إﻟﻰ ﻣﻦ اﺟﻞ ا ﻟ ﻮ . ﻧﺤﺘﺎج أوﻻ إﻟﻰ ﻣﻔﺎھﯿﻢ ﻣﺸﺎﺑﮭﮫ ﻟﻤﺎ ﻟﺪﯾﻨﺎ ﻓﻲ اﻷﻋﺪاد اﻟﺼﺤﯿﺤﺔ،اﻟﺼﺤﯿﺤﺔ إﻟﻰ ﻋﻮاﻣﻞ أوﻟﯿﺔ Definition 1.29 1. A nonzero element p R is called a prime if and only if p is not invertible and p| ab implies either p| a or else p| b. 2. A nonzero element q R is said to be irreducible (or nonfactorizable) if and only if q is not invertible and in every factorization q= bc with b, c R, either b or c is invertible. Lemma In an integral domain R, any prime element p is irreducible. Theorem 1.41 Let R be a principal ideal domain. A nonzero element p R is irreducible if and only if it is prime. Proof: In the class. Theorem 1.42 Let R be a principal ideal domain. If { I n }, n Z , is any infinite sequence of ideals of R satisfying I1 I 2 ... I n I n 1 ... , Then an integer m exists such that I n = I m for all n>m. Proof: In the class. A principal ideal of the ring R is said to be a maximal principal ideal if it is maximal ( with respect to inclusion) in the set of proper principal ideals of R. Lemma Let R be an integral domain. For a nonzero element p R the following hold: a) p is an irreducible element of R if and only if (p) is a maximal principal ideal; b) p is a prime element of R if and only if the principal ideal (p) R is prime. Proof: In the class. . ﺗﺘﻠﺨﺺ اﻟﺤﻘﺎﺋﻖ أﻋﻼه ﺑﻤﺎ ﯾﻠﻲ،ﻓﻲ اﻟﺴﺎﺣﺔ اﻟﺘﻜﺎﻣﻠﯿﺔ ذات اﻟﻤﺜﺎﻟﯿﺎت اﻟﺮﺋﯿﺴﯿﺔ Theorem 1.43 Let R be a principal ideal domain. The nontrivial ideal (p) is a maximal(prime) ideal of R if and only if p is an irreducible(prime) element of R. أن .ﻣﻌﻜﻮس ﯾﺘﺤﻠﻞ إﻟﻰ ﻋﻮاﻣﻞ أوﻟﯿﺔ Corollary Let a 0 be a noninvertible element of the principal ideal domain R. Then there exists a prime p R such that p| a. Proof . Since a is not invertible, the principal ideal (a) R. Thus by Theorem 1.28 a maximal ideal M of R exists such that (a) M. But the preceding result tells us that every maximal ideal is of the form M =(p), where p is a prime element of R ( in this setting, there is no distinction between prime and irreducible elements). Thus (a) (p), which is to say that p| a. 1.28 . 1.42ﺑﺎﻻﻋﺘﻤﺎد ﻋﻠﻰ ﻣﺒﺮھﻨﺔ .اﻟﺴﺎﺣﺔ اﻟﺘﻜﺎﻣﻠﯿﺔ إﻟﻰ ﻋﻮاﻣﻞ أوﻟﯿﺔ Definition 1.30 An integral domain R is a unique factorization domain (U.F.D.) in case the following two conditions hold: 1. every a R that is neither zero nor invertible can be factored into a finite product of irreducible elements; 2. if a= p 1 p 2 … p n = q 1 q 2 …q m are two factorizations of a into irreducible elements, then n=m, and there is a permutation of the indices such that p i and q (i ) are associates (i= 1, 2, …, n). .ﻓﯿﻤﺎ ﯾﻠﻲ ﺳﻨﺮى أن ﻛﻞ ﺳﺎﺣﺔ ﺗﻜﺎﻣﻠﯿﺔ ذات ﻣﺜﺎﻟﯿﺎت رﺋﯿﺴﯿﺔ ﺗﻤﺘﻠﻚ اﻟﺼﻔﺔ أﻋﻼه Theorem 1.44 If R is a principal ideal domain, then every element of R that is neither zero nor invertible has a factorization into a finite product of primes. Proof. Let a R be a nonzero noninvertible element. Then p 1 R, with p 1 prime and p 1 |a ( by the last corollary). So, a= p 1 a 1 for some 0 a 1 R, hence (a) (a 1 ). If (a) = (a 1 ), then a 1 = ra, (r R), hence a= p 1 a 1 = p 1 ra, or 1=p 1 r, a contradiction. This means that (a) (a 1 ). Repeat the procedure, now starting with a 1 , to obtain an increasing chain of principal ideals (a) (a 1 ) (a 2 ) … (a n ) …, with a n 1 =p n a n for some prime p n R. This process goes on as a n is not invertible. But, by Theorem 1.42, the above chain terminates, that is, a n must possess an inverse for some n, and (a) (a 1 ) (a 2 ) … (a n )=R . Therefore a= p 1 p 2 … p n 1 p n , where p n = p n a n , being an associate of prime, is itself prime. Corollary In a P.I.D. R, every nontrivial ideal is the product of a finite number of prime(maximal) ideals. وﻓﻲ اﻟﻤﺒﺮھﻨﺔ اﻟﻘﺎدﻣﺔ ﺳﻨﺜﺒﺖ وﺣﺪاﻧﯿﺔ،(ﻓﻲ اﻟﻤﺒﺮھﻨﺔ اﻟﺴﺎﺑﻘﺔ أﺛﺒﺘﻨﺎ وﺟﻮد اﻟﺘﺠﺰﺋﺔ )اﻟﺘﺤﻠﯿﻞ .اﻟﺘﺠﺰﺋﺔ Theorem 1.45 Every P.I.D. R is a U.F.D. Proof: In the class. ﻟ ﻜﻦ ﻣﻦ. P.I.D ﻟﯿﺲU.F.D. ﻏﯿﺮ ﺻﺤﯿﺢ وﺳﻨﻌﻄﻲ )ﻻﺣﻘﺎ( ﻣﺜ ﺎل ﻋﻠﻰ1.45 ﻋﻜﺲ اﻟﻤﺒﺮھﻦ (اﻟﻤﻔﯿﺪ أن ﻧﻨﺘﺒﮫ إﻟﻰ اﻟﺤﻘﯿﻘﺔ اﻟﺘﺎﻟﯿﺔ )واﻟﺘﻲ ﺳﻨﻮﺿﺤﮭﺎ ﻓﻲ اﻟﺼﻒ Remark. In a U.F.D. R, any irreducible element p R is necessarily prime. 14 اﻟﻤﺤﺎﺿﺮة . وﻧﺒﺪأ ﺑﺎﻟﺘﻌﺮﯾﻒ اﻟﺘﺎﻟﻲ،اﻟﺤﻠﻘﺎت ﯾﺴﻤﻰ اﻟﺴﺎﺣﺔ اﻻﻗﻠﯿﺪﯾﺔ Definition 1.31 An integral domain R is said to be Euclidean if a function : R→Z {0} (the Euclidean valuation) exists such that the following conditions are satisfied: 1. (a)= 0 if and only if a= 0; 2. for all a, b R, (ab) = (a) (b); 3. given a, b R, with b 0, there exist elements q, r R ( the quotient and the remainder) such that a= qb + r, where (r)< (b). As simple examples of Euclidean domains, we may take 1. any field F, with valuation defined by (a)= 1 for all nonzero a F and (0)= 0; 2. the ring Z of integers. With valuation defined by (a)= |a| for all a Z. 3. the Gaussian integers Z(i). with valuation defined by (a + bi) = a 2 + b 2 for all a + bi Z(i). . ﺗﻈﮭﺮ ﻓﻲ اﻟﻤﺄﺧﻮذة اﻟﺘﺎﻟﯿﺔ،ﻋﺪة ﺧﻮاص أوﻟﯿﺔ ﻟﻠﺴﺎﺣﺎت اﻻﻗﻠﯿﺪﯾﺔ Lemma. Let R be a Euclidean domain with valuation . Then 1. (1)= 1; 2. an element 0 a R is invertible if and only if (a)= 1; 3. if two elements a, b R are associates, then (a)= (b). Proof: In the class. ، r وq ، ﻓﻲ ﺧﻮارزﻣﯿﺔ اﻟﻘﺴﻤﺔ،ﺑﺸﻜﻞ ﻋﺎم .اﻻﻗﻠﯿﺪﯾﺔ ﯾﻤﻜﻦ اﻟﺒﺮھﻨﺔ ﻋﻠﻰ وﺣﺪاﻧﯿﺘﮭﻤﺎ Theorem 1.46 The quotient and remainder in condition (3) of the definition of a Euclidean domain R are unique if and only if (a+ b) max{ (a), (b)} for all a, b R. Proof: In the class. :ﻣﻼﺣﻈﺎت ﻣﺜﺎل ﻋﻠﻰ ﺳﺎﺣﺔ اﻗﻠﯿﺪﯾﺔ ﻻ ﺗﺤﻘﻖ،( ﺳﺎﺣﺔ اﻷﻋﺪاد اﻟﺼﺤﯿﺤﺔ )ﻣﻊ اﻟﻘﯿﻤﺔ اﻟﻤﻄﻠﻘﺔ داﻟﺔ اﻟﻘﯿﻤﺔ.1 .1.46 اﻟﺸﺮط ﻓﻲ اﻟﻤﺒﺮھﻨﺔ ( ) .2 ،اﻟﻘﺴﻤﺔ وﻧﺴﺘﻨﺞ ذﻟﻚ ﻣﻦ ﺧﻼل إﺛﺒﺎت أن ﻛﻞ ﺳﺎﺣﺔ، ﻋﻠﻰ ﻏﺮار ﻣﺎ ﻣﻮﺟﻮد ﻓﻲ اﻷﻋﺪاد اﻟﺼﺤﯿﺤﺔ . .ﻣﺜﺎل Theorem 1.47 Every Euclidean domain is a P.I.D. Proof: In the class. Corollary. Every Euclidean domain is a U.F.D. Problems 5. 1. If u is a nonzero element of an integral domain R, prove that the following statements are equivalent: a) u and 1 are associates; b) u is an invertible element of R; c) the principal ideal (u)= R. 2. Assuming that R is a Euclidean domain with valuation , prove the statements below: a) For nonzero a, b R, if a| b and (a)= (b), then a and b are associates. [ Hint: Show that b|a.] b) For nonzero a, b R, (ab) > (a) if and only if b is not an invertible element.[ Hint: Use the " division algorithm" to write a= q(ab) + r.] c) If n is any integer such that (1) + n 0, then the function : R-{0}→ Z defined by (a) = (a) + n is also a Euclidean valuation on R. 3. In the domain Z(i), let a= -4 + I and b= 5 + 3i. Show that : a) the Euclidean valuation defined by (r + si) = r 2 + s 2 does not satisfy (a + b) max{ (a), (b)}; b) the element a has the two representations a= (-1 + i)b + (4- i) = (-1)b + (1 + 4i) and hence uniqueness fails in the "division algorithm" for Z(i). 4. Let n 1 be a square-free integer ( an integer that is not divisible by square of any positive integer >1), let Q be the set of rational numbers, and let Q(n) = { a+ b n | a, b Q}. a)Prove that for each square-free integer n, Q(n) is a field, it is a subfield of the field of complex numbers C. b)For each = a+ b n in Q(n), define = a- b n (called the conjugate of ), and define N( ) = (called the norm of ) . Prove that: 1. N( )=0 if and only if =0; 2. N( )=N( )N( ); 3. N(1)=1. c) Let Z( n ) = { a+ b n | a, b Z}. Show that Z( n ) is an integral domain and the following two statements are hold: 1- N( )= ±1 if and only if is invertible in Z( n ); 2- if N( )=±p, where p is a prime number, then is an irreducible element of Z( n ).