13 More on free abelian groups

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1 13 More on free abelian groups Recall. G is a free abelian group if G = i I Z for some set I Definition. Let G be an abelian group. A set B G is a basis of G if B generates G if for some x 1,...x k B and n 1,...,n k Z we have n 1 x n k x k =0 then n 1 = = n k = Theorem. An abelian group G has a basis iff G is a free abelian group. Proof. If B is a basis of G then the map f : x B Z G, f((n x ) x B )= x B n x x is an isomorphism (check!). Conversely, if we have an isomorphism f : i I Z G then for j I take δ j i I Z where δ j =(n i ) i I such that 1 if i = j n i = 0 otherwise Then B = {f(δ j )} i I is a basis of G (check!). 42

2 13.3 Proposition. If G is a free abelian group then any two basis of G have the same cardinality. Notation. If X is a set then X = cardinality of X Lemma. If {G i } i I is a family of abelian groups and H i is a subgroup of G i for i I then G 1 / H i = (G i /H i ) i I i I i I Proof. Exercise. Proof of Proposition Let B, B be two bases of G. Wewanttoshow: B = B. We have isomorphisms Z = G = Z y B x B Case 1. B,B -finitesets, B = m, B = n. Take 2G := {2a a G}. This is a subgroup G. Since G = x B Z,using Lemma 13.4 we obtain G/2G = x B Z/2Z Similarly, since G = x B Z we have G/2G = y B Z/2Z This gives: 2 m = x B Z/2Z = G/2G = y B Z/2Z =2 n so m = n. 43

3 Case 2. B -finiteset,b -infiniteset. As before this would give: Z/2Z = G/2G = Z/2Z x B x B This is however impossible since x B Z/2Z is a finite group and y B Z/2Z is an infinite group. Case 3. B,B -infinitesets. Check: if B is an infinite set then B = x B Z It follows that we have B = x B Z = G = y B Z = B 13.5 Definition. If G is a free abelian group then the rank of G is the cardinality of a basis of G. Note. By (13.3) rankofg does not depend on the choice of basis of G Theorem. Let G be a free abelian group of a finite rank n and let H be asubgroupofg. ThenH is a free abelian group and rank H rank G Note. This theorem is true also if G is a free abelian group of an infinite rank Lemma. If f : G H is an epimorphism of abelian groups and H is free abelian group then G = H Ker(f) 44

4 Proof. Recall (from homework): if we have homomorphisms f : G H, g: H G such that fg =id H then G = H Ker(f). If follows that we only need to construct the homomorphism g. Take a basis B of H. Sincef is onto, for every x B there is a x G such that f(a x )=x. Letg : H G be the unique homomorphism satisfying g(x) =a x for all x B. Thenfg(x) =x for all x B and so fg =id H. Proof of Theorem We can assume that G = Z n. We want to show: if H Z n then H is a free abelian group and rank H n. Induction with respect to n: If n =1then H = kz for some k 0 so H = {0} or H = Z. Next, assume that for some n every subgroup of Z n is a free abelian group of rank n, andleth Z n+1.takethehomomorphism We have: We have an epimorphism: f : Z n+1 Z, f(m 1,...,m n+1 )=m n+1 Ker(f) ={(m 1,...,m n, 0) m i Z} = Z n f H : H Im(f) Since Im(f H ) Z, thusim(f H ) is a free abelian group and so by Lemma 13.7 we have G = Im(f H ) Ker(f H ) We also have: Ker(f H )=Ker(f) H 45

5 It follows that that Ker(f H ) is a subgroup of Ker(f), andsinceker(f) is a free abelian group of rank n by the inductive assumption we get that Ker(f H ) is a free abelian group of rank n. Therefore H = Im(f H ) free abelian rank 1 Ker(f H ) free abelian rank n and so H is a free abelian group of rank n

6 14 Finitely generated abelian groups Goal. Describe all isomorphism types of finitely generated abelian groups Proposition. If G is an abelian group generated by n elements then G = F/H where F is a free abelian group of rank n and H is some subgroup of F. Proof. We have G = a 1,...a n for some a 1,...,a n G. Let{x 1,...,x n } be abasisoff.wehaveahomomorphism f : F G, f(x i )=a i Take H =Ker(f). Sincef is an epimorphism by the First Isomorphism Theorem (6.1) wehave G = F/H Recall. If rank F = n and H F then H is a free abelian group of rank n Theorem. Let F be a free abelian group of rank n and let H be a subgroup of F There exists a basis {x 1,...,x n } of F and integers d 1,...,d r > 0 such that d i d i+1 for i =1,...,r {d 1 x 1,...,d r x r } is a basis of H Theorem. If G is a finitely generated abelian group then G = (Z/d 1 Z)... (Z/d r Z) Z k for some k 0 and d 1,...,d r > 0 such that d i d i+1 for i =1,...,r. 47

7 Proof. By (14.1) wehave G = F/H for some free abelian group F of finite rank and H F. Let rank F = n. By Theorem 14.2 there is a basis {x 1,...,x n } of F such that {d 1 x 1,...,d r x r } is abasisofh for some d 1,...,d r > 0, d i d i+1. We have an isomorphism f : F Z Z n times where f(x 1 )=(1, 0,...,0), f(x 2 )=(0, 1,...,0),...,f(x n )=(0,...,0, 1) Notice that f(h) =d 1 Z... d r Z {0}... {0}.Thisgives G = (Z Z)/(d 1 Z... d r Z {0}... {0}) Using (13.4) weobtain where k = n r. G = (Z/d 1 Z)... (Z/d r Z) Z k Proof of Theorem Let {y 1,...,y n } be any basis of F and let {h 1,...,h m } H be any set generating H. Wehave h i = a i1 y 1 + a i2 y 2 + a in y n for some a ij Z. Considerthematrix a a 1n A =.. a m1... a mn Note: columns of A correspond to basis elements of F and rows of A correspond to generators of H. Consider the following operations on matrices: 48

8 1) interchange of two rows 2) multiplication of a row by ( 1) 3) addition of a multiple of one row to another row. These operations are called elementary row operations for matrices of integers. Elementary column operations are defined analogously. Notice that: application of an elementary row operation to the matrix A corresponds to replacing of the set of generators of H by another set of generators of H; application of an elementary column operation to the matrix A corresponds to passing to a new basis of F and rewriting the generators of H in terms of this new basis. Key step. Starting with any matrix of integers A and applying a sequence of elementary row and column operations we can obtain a matrix of the form D 0 B = 0 0 where 0 s denote zero matrices (of appropriate dimensions) and D is a square diagonal matrix d d D = d r for some d 1,...,d r > 0 such that d i d i+1 for all i. Note. The matrix B is called the Smith normal form of the matrix A. Let {x 1,...,x n } be the basis of F corresponding to columns of the matrix B. In terms of this basis a set of generators of H is given by {d 1 x 1,...,d r x r, 0,...,0}. It follows that {d 1 x 1,...d r x r } is a basis of H. 49

9 Key step. How to compute the Smith normal form of a matrix A. Step 1. Produce a matrix of the form a A 1 = 0. A 1 0 This can be done as follows. (1a) By interchanging rows and columns if necessary we can make sure that a 11 =0. Also, by multiplying the first row by ( 1) we can get a 11 > 0. (1b) By adding multiples of the first row to the other rows (and multiples of the first column to the other columns) we can make all other entries in the first row and the first column positive and smaller than a 11. (1c) If all these other entries of the first row and column are 0 we are done. If some entry is non-zero then by replacing rows (or columns) we can move that entry to the (1, 1)-position. Then we go back to (1b). (1d) After a finite number of iterations we get a matrix of the form of A 1. Step 2. Given a matrix A 1 as above make sure that the entry a 11 divides all entries of A 1. This can be done as follows. (2a) If all entries of A 1 are already divisible by a 11 we are done. (2b) If some entry a ij is not divisible by a 11 add the i-th row to the first row. Then go back to (1b). (2c) After a finite number of iterations we get a matrix of the form of A 1 with a 11 dividing all entries of A. Step 3. We are done with the first row and and the first column. Next we apply the same steps recursively to reduce A 1. 50

10 14.4 Lemma. Let G be a group and let a 1,...,a k G be elements such that a i = n i. Assume that for i =1,...,k we have gcd(n i, j=i n i )=1 Then a 1... a k = n 1... n k. Proof. Exercise Corollary. If m>0 is an integer and m = p n p n k k are distinct primes then Z/mZ = (Z/p n 1 1 Z)... (Z/p n k k Z) where p 1,...,p k Proof. It is enough to show that the group (Z/p n 1 1 Z)... (Z/p n k k Z) contains an element of order n. ThisfollowshoweverfromLemma Theorem. If G is a finitely generated abelian group then G is isomorphic to a finite direct sum groups Z and Z/p n Z where p is a prime. Proof. This follows directly from (14.3) and(14.5) Theorem. For any finitely generated abelian group G there are unique (up to order) integers p n 1 1,...,p ns s > 1 where p 1,...,p s are primes, and a unique integer k 0 such that G = (Z/p n 1 1 Z)... (Z/p ns Z) Zk k 14.8 Note. The integers p n 1 1,...,p ns s group G. are called the elementary divisors of the 51

11 14.9 Lemma. If p is a prime number and 0 <n 1 n s 0 <m 1 m r be integers such that (Z/p n 1 Z) (Z/p ns Z) = (Z/p m 1 Z) (Z/p mr Z) Then s = r and n i = m i for all i. Proof. Exercise. Proof of Theorem Assume that and G = (Z/p n 1 1 Z)... (Z/p ns s Z) Z k (1) G = (Z/q m 1 1 Z)... (Z/q mr r Z) Z l (2) where k, l 0, p i, q j are primes and and n 1,m j > 0. We want to show that k = l, r = s and (after reordering) p n i i = q m i i for all i. Define G t := {a G a < } This is the torsion subgroup of G. Fromtheisomorphism(1) weobtain: while from the isomorphism (2) weget If follows that G t = (Z/p n 1 1 Z)... (Z/p ns s Z) G t = (Z/q m 1 1 Z)... (Z/q mr r Z) G/G t = Z k and G/G t = Z l Using Proposition 13.3 we obtain from here that k = l. Next, we want to show that the family of integers {p n i i same as the family {q m j j j =1,...,r}. i =1,...,s} is the 52

12 For a prime p define G p := {a G a = p l for some l>0} From the isomorphisms (1) and(2) weget (Z/p n i i Z) = G p = (Z/q m j j Z) p i =p Using Lemma 14.9 we obtain that the family {p n i i p i = p} is the same as {q n j j q j = p}. Sincethisholdsforallprimesp we are done. q j =p 53

Math 210B: Algebra, Homework 4

Math 210B: Algebra, Homework 4 Ian Coley February 5, 2014 Problem 1. Let S be a multiplicative subset in a commutative ring R. Show that the localisation functor R-Mod S 1 R-Mod, M S 1 M, is exact. First,