SELECTED CHAPTERS FROM ALGEBRA I. R. Shafarevich Preface

72 I. R. Shafarevichfollows: if M = M 1 + M 2 is a partition of M, thenthesetU(M) isinaone-to-onecorrespondence with the set U(M 1 ) U(M 2 ). Denote the number n(U(M)) byv(M)|this is the required number of all subsets. Applying Theorem 1 we deducethat(3) v(M 1 + M 2 )=v(M 1 )v(M 2 ):The equality (3) reduces the evaluation of v(M) to the evaluation of v(M 1 )and v(M 2 ) for the sets M 1 and M 2 with smaller number of elements. In orderto obtain the nal result, consider the partition of M not into two, but into anarbitrary number of subsets. We can dene this concept inductively, saying thatM = M 1 + + M r if M = (M 1 + + M r;1 )+M r , where the expressionM 1 + + M r;1 is taken to be already dened. In fact, when we say that M =M 1 ++M r , this means that M 1 ,...,M r are subsets of M and that every elementof M belongs to one and only one of the subsets M 1 , ... , M r . For example, if Mis the set of all positive integers, then M = M 1 + M 2 + M 3 , where M 1 is the subsetof all numbers divisible by 3,M 2 is the subset of all numbers of the form 3r +1and M 3 is the subset of all numbers of the form 3r +2.From (3), for nite sets M i we obtain, by induction(4) v(M 1 + + M r )=v(M 1 ) v(M r ):If n(M) =n, then there exists the \tiniest" partition of M into n subsets M i ,each having only one element, i.e. M = M 1 + + M n . If M = fa 1 ...a n g,thenM i = fa i g. The one element set M i has two subsets: the empty set ? and M iitself (see Table 1, rst row). Hence, v(M i )=2and applying formula (4) to thepartition M = M 1 + + M n weobtainthatv(M) =2 n , as stated in Theorem 2.The question of the number of all subsets of a given set appears in connectionwith certain problems regarding numbers. For example, consider the followingquestion: in how many ways can a positive integer n be written as a product oftwo relatively prime factors? Let n = ab, wherea and b are relatively prime and letn = p 11 pr rbe the canonical prime factorization. Then a and b are divisors of nand, as we saw in Section 3 of Chapter I, each one of them has the form p 11 pr rwhere 0 6 i 6 i . But since a and b are relatively prime, then if some p i divides a,then it cannot divide b and hence appears in a with degree i . Therefore, in orderto obtain the required factorization n = ab, it is necessary to choose an arbitrarysubset N of the set M = fp 1 ...p r g and to equate a to the product of p iiforp i 2 N. Then a divides n and n = ab is the required factorization. According toTheorem 2, the number of all factorizations of n into products of two relativelyprimer factors is 2 r ,wherer is the number of dierent prime factors of n.It should be noted that in the above evaluation we considered the factorizationn = ab and n = ba to be dierent. In fact, if a, and hence the factorizationn = ab, corresponds to the subset N fp 1 ...p r g, then b corresponds to thesubset consisting of those p i 2 M which do not belong to N, i.e. which belongto the complement N of N. Therefore, in our evaluation we corresponded thefactorizations n = ab and n = ba to two dierent subsets N and N. Hence, if we