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# REED MATHEMATICS 361 - Lecture Notes

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MATH 361 NUMBER THEORY SEVENTH LECTURE 1 The Unit Group of Z nZ Consider a nonunit positive integer Y n pep 1 The Sun Ze Theorem gives a ring isomorphism Y Z pep Z Z nZ The right side is the cartesian product of the rings Z pep Z meaning that addition and multiplication are carried out componentwise It follows that the corresponding unit group is Y Z pep Z Z nZ Thus to study the unit group Z nZ it suffices to consider Z pe Z where p is prime and e 0 Recall that in general Z nZ n so that for prime powers Z pe Z pe pe 1 p 1 and especially for primes Z pZ p 1 Here are some examples of unit groups modulo prime powers Z 2Z 1 20 0 Z Z Z 3Z 1 2 20 21 0 1 Z 2Z Z 4Z 1 3 30 31 0 1 Z 2Z Z 5Z 1 2 3 4 20 21 22 23 0 1 2 3 Z 4Z Z 7Z 1 2 3 4 5 6 30 31 32 33 34 35 0 1 2 3 4 5 Z 6Z Z 8Z 1 3 5 7 30 50 31 50 30 51 31 51 0 1 0 1 Z 2Z Z 2Z Z 9Z 1 2 4 5 7 8 20 21 22 23 24 25 0 1 2 3 4 5 Z 6Z 1 2 MATH 361 NUMBER THEORY SEVENTH LECTURE 2 Prime Unit Group Structure Abelian Group Theory Argument Proposition 2 1 Let G be any finite subgroup of the unit group of any field Then G is cyclic In particular the multiplicative group modulo any prime p is cyclic Z pZ Z p 1 Z That is there is a generator g mod p such that Z pZ 1 g g 2 g p 2 Proof We may assume that G is not trivial By the structure theorem for finitely generated abelian groups G Z d1 Z Z d2 Z Z dk Z k 1 1 d1 d2 dk Thus the polynomial equation X dk 1 whose additive counterpart is dk X 0 has d1 d2 dk roots in G forcing k 1 and consequently making G cyclic The proof tacitly relies on a fact from basic algebra Lemma 2 2 Let k be a field Let f k X be a nonzero polynomial and let d denote its degree thus d 0 Then f has at most d roots in k Proof If f has no roots then we are done Otherwise let a k be a root Write f X q X X a r X deg r 1 or r 0 Thus r X is a constant Substitute a for X to see that in fact r 0 and so f X q X X a By induction q has at most d 1 roots in k and we are done The lemma does require that k be a field not merely a ring For example the polynomial X 2 1 over the ring Z 24Z has for its roots 1 5 7 11 13 17 19 23 Z 24Z To count the generators of Z pZ consider any finite cyclic abelian group G Z nZ For any integer k the subgroup of Z nZ generated by k nZ is hk nZi kZ nZ gcd k n Z nZ hgcd k n nZi which clearly has order n gcd k n Especially each k nZ where k is coprime to n generates the full group and there are n such values of k In particular Z pZ has p 1 generators 3 Prime Unit Group Structure Elementary Argument From above a nonzero polynomial over Z pZ can not have more roots than its degree On the other hand Fermat s Little Theorem says that the polynomial f X X p 1 1 Z pZ X has a full contingent of p 1 roots in Z pZ For any divisor d of p 1 consider the factorization in consequence of the finite geometric sum formula p 1 d 1 f X X p 1 1 X d 1 X i 0 call X id g X h X MATH 361 NUMBER THEORY SEVENTH LECTURE 3 We know that f has p 1 roots in Z pZ g has at most d roots in Z pZ h has at most p 1 d roots in Z pZ It follows that g X X d 1 where d p 1 has d roots in Z pZ Now factor p 1 Y p 1 q eq For each factor q e of p 1 e Xq 1 Xq e 1 has q e roots in Z pZ 1 has q e 1 roots in Z pZ and so Z pZ contains q e q e 1 q e elements xq of order q e The order of an element is the smallest positive number of times that the element is multiplied by itself to give 1 Plausibly Y Y any product xq has order q eq p 1 q q and certainly there are p 1 such products In sum we have done most of the work of showing Proposition 3 1 Let p be prime Then Z pZ is cyclic with p 1 generators The loose end is as follows Lemma 3 2 In a commutative group consider two elements whose orders are coprime Then the order of their product is the product of their orders Proof We have ae bf 1 and so ab ef ae f bf e 1f 1e 1 Also we have e f 1 So for any positive integer d e ab d 1 1 ab d ae be d bed f ed f d and symmetrically e d Thus ef d 4 Odd Prime Power Unit Group Structure p Adic Argument Proposition 4 1 Let p be an odd prime and let e be any positive integer The multiplicative group modulo pe is cyclic Proof The structure theorem for finitely generated abelian groups and then the Sun Ze theorem combine to show that Z pe Z takes the form Z pe Z Ape 1 Ap 1 where An denotes an abelian group of order n By the Sun Ze Theorem it suffices to show that each of Ape 1 and Ap 1 is cyclic The natural epimorphism Z pe Z Z pZ maps Ape 1 to 1 in Z pZ since the orders of the two groups are coprime but the image is a quotient of the first and a subgroup of the second Consequently the restriction of the natural epimorphism to Ap 1 must be an isomorphism and thus Ap 1 is cyclic Furthermore this discussion has shown that Ape 1 is the kernel of the natural epimorphism Ape 1 a pe Z a 1 mod p 4 MATH 361 NUMBER THEORY SEVENTH LECTURE Working p adically consider two group isomorphisms p exp Zp pZp 1 pZp The exponential series converges on pZp because 1 1 p 1 0 making the summands decay X p ap n n n n pi n 1 …

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