product of cyclic groups

Zpe1 — · · · — Zper ,

r

1

where the pi are primes (not necessarily distinct) and the ei are positive

integers. This direct product of cyclic groups is unique up to the order of

the factors.

An alternative statement of this theorem is the following:

Theorem 8.44. A ¬nite abelian group (with more than one element) is

isomorphic to a direct product of cyclic groups

Zm1 — · · · — Zmt ,

where each mi > 1, and where for i = 1, . . . , t ’ 1, we have mi | mi+1 .

Moreover, the integers m1 , . . . , mt are uniquely determined, and mt is the

exponent of the group.

Exercise 8.29. Show that Theorems 8.43 and 8.44 are equivalent; that is,

show that each one implies the other. To do this, give a natural one-to-one

correspondence between sequences of prime powers (as in Theorem 8.43)

and sequences of integers m1 , . . . , mt (as in Theorem 8.44), and also make

use of Example 8.47.

Exercise 8.30. Using the fundamental theorem of ¬nite abelian groups

(either form), give short and simple proofs of Theorems 8.40 and 8.41.

We now prove Theorem 8.44, which we break into two lemmas, the ¬rst

of which proves the existence part of the theorem, and the second of which

proves the uniqueness part.

Lemma 8.45. A ¬nite abelian group (with more than one element) is iso-

morphic to a direct product of cyclic groups

Zm1 — · · · — Zmt ,

where each mi > 1, and where for i = 1, . . . , t ’ 1, we have mi | mi+1 ;

moreover, mt is the exponent of the group.

Proof. Let G be a ¬nite abelian group with more than one element, and let

m be the exponent of G. By Theorem 8.40, there exists an element a ∈ G of

order m. Let A = a . Then A ∼ Zm . Now, if A = G, the lemma is proved.

=

So assume that A G.

We will show that there exists a subgroup B of G such that G = A + B

and A © B = {0}. From this, Theorem 8.28 gives us an isomorphism of G

210 Abelian groups

with A — B. Moreover, the exponent of B is clearly a divisor of m, and so

the lemma will follow by induction (on the order of the group).

So it su¬ces to show the existence of a subgroup B as above. We prove

this by contradiction. Suppose that there is no such subgroup, and among

all subgroups B such that A © B = {0}, assume that B is maximal, meaning

that there is no subgroup B of G such that B B and A © B = {0}. By

assumption C := A + B G.

Let d be any element of G that lies outside of C. Consider the quotient

group G/C, and let r be the order of d + C in G/C. Note that r > 1 and

r | m. We shall de¬ne a group element d with slightly nicer properties

than d, as follows. Since rd ∈ C, we have rd = sa + b for some s ∈ Z and

b ∈ B. We claim that r | s. To see this, note that 0 = md = (m/r)rd =

(m/r)sa + (m/r)b, and since A © B = {0}, we have (m/r)sa = 0, which

can only happen if r | s. That proves the claim. This allows us to de¬ne

d := d ’ (s/r)a. Since d ≡ d (mod C), we see that d + C also has order r

in G/C, but also that rd ∈ B.

We next show that A©(B + d ) = {0}, which will yield the contradiction

we seek, and thus prove the lemma. Because A © B = {0}, it will su¬ce

to show that A © (B + d ) ⊆ B. Now, suppose we have a group element

b + xd ∈ A, with b ∈ B and x ∈ Z. Then in particular, xd ∈ C, and so

r | x, since d + C has order r in G/C. Further, since rd ∈ B, we have

xd ∈ B, whence b + xd ∈ B. 2

Lemma 8.46. Suppose that G := Zm1 — · · · — Zmt and H := Zn1 — · · · — Znt

are isomorphic, where the mi and ni are positive integers (possibly 1) such

that mi | mi+1 for i = 1, . . . , t ’ 1. Then mi = ni for i = 1, . . . , t.

Proof. Clearly, i mi = |G| = |H| = i ni . We prove the lemma by

induction on the order of the group. If the group order is 1, then clearly

all mi and ni must be 1, and we are done. Otherwise, let p be a prime

dividing the group order. Now, suppose that p divides mr , . . . , mt but not

m1 , . . . , mr’1 , and that p divides ns , . . . , nt but not n1 , . . . , ns’1 , where r ¤ t

and s ¤ t. Evidently, the groups pG and pH are isomorphic. Moreover,

pG ∼ Zm1 — · · · — Zmr’1 — Zmr /p — · · · — Zmt /p ,

=

and

pH ∼ Zn1 — · · · — Zns’1 — Zns /p — · · · — Znt /p .

=

Thus, we see that |pG| = |G|/pt’r+1 and |pH| = |H|/pt’s+1 , from which it

follows that r = s, and the lemma then follows by induction. 2

9

Rings

This chapter introduces the notion of a ring, more speci¬cally, a commu-

tative ring with unity. The theory of rings provides a useful conceptual

framework for reasoning about a wide class of interesting algebraic struc-

tures. Intuitively speaking, a ring is an algebraic structure with addition

and multiplication operations that behave like we expect addition and mul-

tiplication should. While there is a lot of terminology associated with rings,

the basic ideas are fairly simple.

9.1 De¬nitions, basic properties, and examples

De¬nition 9.1. A commutative ring with unity is a set R together with

addition and multiplication operations on R, such that:

(i) the set R under addition forms an abelian group, and we denote the

additive identity by 0R ;

(ii) multiplication is associative; that is, for all a, b, c ∈ R, we have

a(bc) = (ab)c;

(iii) multiplication distributes over addition; that is, for all a, b, c ∈ R, we

have a(b + c) = ab + ac and (b + c)a = ba + ca;

(iv) there exists a multiplicative identity; that is, there exists an element

1R ∈ R, such that 1R · a = a = a · 1R for all a ∈ R;

(v) multiplication is commutative; that is, for all a, b ∈ R, we have ab =

ba.

There are other, more general (and less convenient) types of rings ” one

can drop properties (iv) and (v), and still have what is called a ring. We

shall not, however, be working with such general rings in this text. There-

fore, to simplify terminology, from now on, by a “ring,” we shall always

mean a commutative ring with unity.

211

212 Rings

Let R be a ring. Notice that because of the distributive law, for any

¬xed a ∈ R, the map from R to R that sends b ∈ R to ab ∈ R is a group

homomorphism with respect to the underlying additive group of R. We call

this the a-multiplication map.

We ¬rst state some simple facts:

Theorem 9.2. Let R be a ring. Then:

(i) the multiplicative identity 1R is unique;

(ii) 0R · a = 0R for all a ∈ R;

(iii) (’a)b = a(’b) = ’(ab) for all a, b ∈ R;

(iv) (’a)(’b) = ab for all a, b ∈ R;

(v) (na)b = a(nb) = n(ab) for all n ∈ Z and a, b ∈ R.

Proof. Part (i) may be proved using the same argument as was used to prove

part (i) of Theorem 8.2. Parts (ii), (iii), and (v) follow directly from parts

(i), (ii), and (iii) of Theorem 8.20, using appropriate multiplication maps,

discussed above. Part (iv) follows from parts (iii) and (iv) of Theorem 8.3.

2

Example 9.1. The set Z under the usual rules of multiplication and addi-

tion forms a ring. 2

Example 9.2. For n ≥ 1, the set Zn under the rules of multiplication and

addition de¬ned in §2.3 forms a ring. 2