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Algebra and Trigonometry
INTEGRAL DOMAINS AND
CHARACTERISTIC OF A RING
Objectives
From this unit a learner is expected to achieve the following
1. Familiarize with the concept of divisors of 0 in a ring.
2. Study that the cancellation laws hold in a ring R
or right divisors of 0.
3.
if and only if R has no left
Learn the definition of integral domains and familiarize with some examples.
4. Study the definitions of boolean ring and characteristic of a ring and study some
results.
Sections
1. Introduction
2. Divisors of 0
3. Cancellation Laws in a Ring R
4. Integral Domains
5. Some Results
6. Boolean Ring
7. Characteristic of a Ring
1. Introduction
In this session we discuss integral domains, a particular type of rings. We begin with the
definition and examples of divisors of 0. We illustrate that although rings are a direct
generalization of the integers, certain arithmetic facts in the ring of integers need not hold
in general rings.
We will see an application of divisors of 0 while discussing
multiplicative cancellation in a ring. We give the definitions of boolean ring and
characteristic of a ring.
2. Divisors of 0
We recall that a non-zero integer b divides an integer a if a = bq for some integer q. In particular, a non-zero integer b
divides 0 if there is another integer q such that bq = 0. Notice that if b is any integer at all, we can choose q = 0 and get
bq  b  0  0. Using this idea, we can say that every integer divides 0. Also we note that in the ring of integers bq  0 if and
only if either
b  0 or q  0.
One uses this fact constantly, perhaps without realizing it. Suppose, for example, in order to solve
the equation
x2  4 x  3  0
we first factor the left side as follows:
x2  4x  3  ( x  1)( x  3) .
Then, using the fact that product is 0 if and only if one of the factor is 0, one conclude
that either x  1 = 0 or x  3 = 0. Hence the possible values for x are 1 and 3. These
are the only solutions.
We have just observed that in our usual number system a product of two numbers
can only be 0 if at least one of them is 0. The next example illustrates that this most
important algebraic property of the ring of integers need not hold in general rings. In
other words, there are situations where the product bq is 0 even if neither b nor q is 0.
Example 1 Solve the equation x2  5x  6  0 in 12  {0, 1, , 11} , the ring with the
operations +12 , the addition modulo 12, and 12, multiplication modulo 12.
Solution
Since 2 12 3  5 and 2 12 3  6 , we can factorize x2  5x  6 as follows:
x2  5x  6  ( x  2)( x  3).
Clearly, the values of x for which either x  2 = 0 or x  3 = 0 are solutions of the
given equation, and hence obviously 2 and 3 are solutions of the equation in 12 .
, not only 0a = a0 = 0 for all a  12 , but also
(2)(6) = 6 +12 6 = 0; (3)(4) = 4 +12 4 +12 4 = 0 ;
similarly,
(6)(2) = (3)(8) = (8)(3)
= (4)(6) = (6)(4) = (4)(9) = (9)(4) = (6)(6) = (6)(8)
= (8)(6) = (6)(10) = (10)(6) = (8)(9) = (9)(8) = 0.
Using (2)(6) = 0, we can say that x  2x  3 = 0 imply x  2 = 2 and x  3 = 6; so 4 and
But in
12
9 are also solutions of the equation. Using (6)(2)=0, we have x  2 = 6 and x  3 = 2; so x = 8
and x = 5.
Using (4)(3) = 0, we can say that x  2x  3 = 0 imply x  2 = 4 and x  3 = 3;
so 6 is also a solution.
11 is another solution, since (11  2) (11  3) = (9)(8) = 0 in Z12.
The ideas seen in the above example are of such importance that we formalize them
in the following definition.
Definition If a and b are two nonzero elements of a ring R such that ab = 0, then
a and b are divisors of 0 (or 0 divisors ). In particular, a is left divisor of 0 and b is
a right divisor of 0.
Remarks

By the new definition, there is no divisor of 0 in the ring of integers although every non-zero integer divides 0 in the ring
of integers.

In a commutative ring, every left divisor of 0 is also a right divisor of 0 and conversely. Thus there is no distinction
between left and right divisors of 0 in a commutative ring.
Example 2 Consider the ring 12 . We have seen in Example 1, that (2)(6) = 0. So both 2 and 6
are divisors of 0. Similarly, 3, 4, 8, 9 and 10 are other divisors 0. i.e., the elements 2, 3, 4, 6, 8, 9,
10 are all divisors of 0. Note that these divisors of 0 in 12 are exactly the numbers in 12 that
are not relatively prime to 12.
Theorem 1 In the ring
relatively prime to n.
Proof
n
, the divisors of 0 are precisely those elements that are not
Recall that n  {0, 1, . . . , n  1} is ring under +n and n . Let m  n , where m 
0, and m is not relatively prime to n, so that g.c.d. of m and n be d  1. Then
n
m
m      n
d   d 
…(1)
Now d is a divisor of m implies m is an integer, and hence  m  n is a multiple of
d
n and hence in
n
,  m  n  0 .
d
d
n
Thus from Eq. (1), m   0 in Zn , while neither
d 
n
is 0, so m is a divisor of 0.
d
We prove the converse using contra positive method of proving theorems. Suppose m
is a non-zero element in Zn that is relatively prime to n. Then 0 < m < n. If for s  Zn
we have ms = 0, then n divides the product ms of m and s as elements in Z, the
ring of integers. Since n has no factors > 1 in common with m, it must be that n
divides s, so s = 0 in Zn. Hence m is not a divisor of 0. This completes the proof.
m nor
Corollary If p is a prime, then Zp has no divisors of 0.
Proof By Theorem 1, a number m in Zp is a divisor of 0 if and only if m is not relatively
prime to p. Since p is a prime number, every non-zero element in Zp is relatively prime
to p. Thus, Zp has no divisors of 0.
3. Cancellation Laws in the Ring R
Let R be a ring, and let a, b, c  R. The cancellation laws hold in R if ab = ac, with a 
0, implies b = c, and ba = ca with a  0, implies b =c. These are multiplicative
cancellation laws.
Remark By saying that cancellation laws hold in a ring <R, +, .> we mean that
multiplicative cancellation laws hold in R. Always, being a group under the operation +,
the additive cancellation laws hold in any ring.
The importance of the concept of divisors of zero is given in the following theorem.
Theorem 2 The cancellation laws hold in a ring R
right divisors of 0.
if and only if R has no left or
Proof
Let R be a ring in which the cancellation laws hold. We have to show R has no left or
right divisors of 0. For this suppose
ab = 0 for some a, b  R.
We must show that either a or b is 0.
If a  0, then ab = 0 = a0 which implies by the multiplicative cancellation law that b
= 0;
similarly, if b  0, then ab = 0 = 0b which implies by the multiplicative cancellation
law that a = 0.
So there can be no left or right divisors of 0 if the cancellation laws hold.
Conversely, suppose that R has no left or right divisors of 0, and suppose that ab = ac
with a  0. Then
ab  ac = a (b  c) = 0.
Since a  0, and since R has no left divisors of 0, we must have b  c = 0, so b = c. A
similar argument shows that ba = ca, with a  0, implies that b = c.
This completes the proof of the theorem.
Corollary 1 Suppose that R is a ring with no divisors of 0. Then an equation ax = b,
with a  0, in R can have at most one solution x in R.
Proof If ax1 = b and ax2 = b, then ax1 = ax2 , and hence by the cancellation law
(which is possible by Theorem 2) it follows that x1 = x2 .
Corollary 2 If R has unity 1 and a is a unit in R with multiplicative inverse a1, then
the solution x of ax = b is a1 b.
Notation In the case that R is commutative, in particular if R is a field, it is customary
to denote a1 b and ba1 (they are equal by commutativity) by the formal quotient b/a.
Remarks
 The quotient notation given above must not be used in the event that R is not
commutative, for then one does not know whether b/a denotes a1 b or ba1.

In a field F it is usual to define a quotient b/a, where a  0, as the solution x in
F of the equation ax = b. In particular, the multiplicative inverse of a nonzero
element a of a field is 1/a.
4. Integral Domains
Definition An integral domain D is a commutative ring with unity containing no
divisors of 0.
Example 3 Z is an integral domain.
Example 4 The set of all even integers doesn’t form an integral domain as it doesn’t
contain the multiplicative identity.
Theorem 3
p
is an integral domain.
Proof Every nonzero element in
Theorem 1, we know that
unity 1. Hence
p
p
p
are relatively prime to p. Hence, by the corollary to
has no divisors of 0. Also
p
is a commutative ring with
is an integral domain.
Remarks

Since an integral domain contains no 0 divisors, if the coefficients of a
polynomial are from an integral domain, one can solve a polynomial equation
in which the polynomial can be factored into linear factors in the usual fashion
by setting each factor equal to 0. For example, in 3 , ( x  1 )x( 2 ) 0
implies either x  1  0 or x  2  0 ; implies x  1 or x  2.

Theorem 2 and the definition of integral domain shows that the cancellation
laws for multiplication hold in an integral domain.
Example 5
n is not an integral domain if n is not prime, because in that case
contains divisors of 0.
n
Theorem 4 Every field is an integral domain.
Proof
Let F be a field and let a, b  F such that a  0 so that 1  F . Then if ab = 0, we
a
have
1
1
  ab     0  0.
a
a
But then
 1  
1
0    ab     a  b  1b  b.
a
 a  
We have shown that ab = 0 with a  0 implies that b = 0 in F, so there are no divisors
of 0 in F. A field is a commutative division ring, hence, F is a commutative ring with
unity. This completes the proof of the theorem.
The next theorem provides us some more fields of finite order.
Theorem 5 Every finite integral domain is a field.
Proof
Let D be a finite integral domain. We have to show that every non-zero element in
D has a multiplicative inverse. Being an integral domain D certainly contains 0 and 1.
Let
0, 1, a1, . . . , an
be all the elements of D. We need to show that for a  D, where a  0, there exists b
 D such that ab = 1. Now consider
a1, aa1, . . . , aan.
We claim that all these elements of D are distinct, for aai = aaj implies that for ai = aj,
by the cancellation laws that hold in an integral domain. Also, since D has no 0 divisors,
none of these elements is 0. Hence by counting, we find that
a1, aa1, . . . , aan
are the elements
1, a1, . . . , an
in some order, so that
either a1 = 1 , i.e. , a =1, or
aai = 1 for some i.
Thus a has a multiplicative inverse. Hence, when
a = 1 its inverse is 1 and when a 
1 and a  0, then the inverse is some ai.
Corollary If p is a prime, then
Proof
By Theorem 3,
elements,
p
p
p
is a field.
is an integral domain. Also, since
p
contains finite number of
is a finite integral domain. Hence, by Theorem 5 ,
p
is a field.
Example 6 Give an example of a field with three elements.
Solution 3  0, 1, 2 under addition modulo 3 and multiplication modulo 3.
Example 7 Give an example of a ring which is not a field.
Solution
4  0, 1, 2, 3 under addition modulo 4 and multiplication modulo 4 is not a field as the
nonzero element 2 doesn’t have a multiplicative inverse.
The Hierarchy of Algebraic Structures
The hierarchy of algebraic structures is displayed in the following figure. We note that
in the hierarchy of algebraic structures, an integral domain belongs between a
commutative ring with unity and a field.
5. Some Results
Theorem 6 The multiplicative inverse of a non-zero element in a field is unique.
Proof.
Let e be the multiplicative identity of a field F .
Let a and a be two inverses of a non-zero element a  F 
Then
a  a  a  a  e and a  a  a  a  e
a  e  a  (a  a )  a
Now
 a  (a  a) , by the associativity of multiplication
 a  e  a.
Thus inverse of an element in a field is unique
Theorem 7 If a and b are elements in a field F,
(i) (a)  b  (a  b), for a, b  F .
In particular,
(1)  b  b,
(ii) a  (b)  (a  b),
(iii) (a )  a
1 1
for
b  F.
for a, b  F.
for a  F with a  0
(iv) (a)1  a 1 for a  F with a  0
Proof.
(i)We have
a  a  0 ,
where 0 is the additive inverse.
Post multiplying both sides by b, we get
( a  a)  b  0  b
and this implies by the right distributive law that
( a)  b  a  b  0
Hence (a)  b is the additive inverse of a  b . That is,
(a)  b  (a  b).
(ii) is obtained from (i) by the commutativity of multiplication.
(iii) Let e be the multiplicative identity of F . We note that if a  0, the inverse of a
exists in F and is denoted by a1.
Then
a  a1  e
Also by the reversal law of inverses,
(a 1 )1  a 1  e
Since e  a  a 1 , the above gives
(a 1 ) 1  a 1  a  a 1 .
This implies by right cancellation law that
(a 1 ) 1  a .
(iv) Let d be the multiplicative inverse of a . Then
(a)  d  1.
…(2)
Using Property (i),
(a)  d  (a  d ).
…(3)
From Equations (2) and (3), we obtain
(a  d )  1.
Since ( a  d ) is the additive inverse of a  d and since -1 is the additive inverse of 1,
the above implies
a  d  1.
1
Pre-multiplying both sides by a , we obtain
a1  (ad )  a1  (1).
Hence
(a1  a)  d  a1.
Since a1  a  e, the above implies d  a1. That is,
(a)1  a 1.
Example 8 In a field F , prove that a2  b2  either a  b or a  b .
Solution
Given
a2  b2 .
We note that b2 is the additive inverse of b2 . Adding (b 2 ) on both sides of the above
equation, we obtain
a 2  (b 2 )  b 2  (b 2 )
 a 2  (b 2 )  0
 a  a  b  b  0 , since b2  b  b.
 a a  a b  a b  bb  0
 a  (a  b)  (a  b)  b  0 (by the distributive law)
 a  (a  b)  b  (a  b)  0 (by commutativity of multiplication)
 (a  b)  (a  b)  0 (by the distributive law)
Since a field is without divisors of 0, the last equation implies, either a  b  0 or
a  b  0. i.e., either a  b or a  b .
Example 9 In a field F with multiplicative identity 1, if a, b  F and a  0 then there
exists a unique element x such that a  x  b.
Solution
Since a  0 , a1 exists and a1  F .
Left multiplying ax  b by a1 we get
a 1  (ax)  a 1  b
implies
(a 1  a ) x  a 1  b (by the associativity of multiplication)
implies
1  x  a 1  b, since a1  a  1
implies
x  a1  b.
To prove uniqueness, suppose there are x1 and x2 such that ax1  b and ax2  b . This
implies ax1  ax2 . This implies (by the cancellation law in a field) that x1  x2 . Hence the
uniqueness is proved.
Example 10 A Gaussian integer is a complex number a  ib , with a and b being
integers. It can be seen that the set Z (i) of Gaussian integers is a commutative ring with
unity. Since the product of two non-zero complex numbers cannot be zero, Z (i) has no
zero-divisors. It is therefore an integral domain. To check whether Z (i) is a field, let
z  a  ib be any non-zero element of Z (i) . Then at least one of a and b is not zero.
Since z is a nonzero complex number and since the set of complex numbers form a field,
we have
z 1 

1
1
(a  ib)
a  ib


 2
z a  ib (a  ib) (a  ib) a  b2
a
 b 
 i 2
,
2
a b
 a  b2 
2
which is, in general, not a Gaussian integer as a /(a 2  b 2 ) and b /(a 2  b 2 ) are not
necessarily integers. Therefore Z (i) is not a field.
Definition A non commutative division ring is called a strictly skew field.
Example 11 Show that the set M of all 2  2 matrices of the form
 a  ib
 c  id

c  id 
a  ib 
where a, b, c and d are real numbers, is a strictly skew-field under matrix addition and
multiplication.
Solution
It can be easily verified that M is a ring. The matrix
0  i0 0  i0 
 0  i 0 0  i 0 


is the zero element of M .
0  i0
1  i 0
 0  i 0 1  i 0 


The matrix
Now let
 a  ib
A
 c  id
is the unit of M .
c  id 
a  ib 
be any non-zero element in M . Then not all of a, b, c, d are 0. We have det
A  a 2  b 2  c 2  d 2 . Since a, b, c and d are real and not all of them are 0, det A  0.
Therefore the matrix A is invertible . In fact,
A1 
1  a  ib c  id 
det A c  id a  ib 
which is easily seen to be an element of M . Thus the non-zero elements of M form a
group under multiplication. We can show by examples that the commutative law of
multiplication does not hold in M . Therefore M is a strictly skew field.
6. Boolean Ring
Definition A ring R is called Boolean if x2  x for all x  R .
Example 12 If R is a Boolean ring, prove that
(a) 2 x  0 for all x  R ;
(b) x  y  0 implies x  y; and
(c) R is commutative.
Solution
(a) Since x2  x for all x  R , and since x  x  R , we have
( x  x) 2  ( x  x)  ( x  x) ( x  x)  x  x
 ( x  x) x  ( x  x) x  x  x , by the right-distributive law.
 ( x2  x2 )  ( x2  x2 )  x  x
 ( x  x)  ( x  x )  ( x  x )  0
 x  x  0, by the cancellation law of addition,
 2x  0
(b) If x  y  0, then, since x  x  0 (by Part (a)), we have x  y  0  x  x , and hence
y  x, by the cancellation law of addition.
(c) For any x, y  R, we have ( x  y )2  x  y . That is,
( x  y) ( x  y)  x  y  ( x  y) x  ( x  y) y  x  y

( x 2  yx)  ( xy  y 2 )  x  y


x  yx  xy  y  x  y , since x2  x and y 2  y.
( x  y)  ( xy  yx)  ( x  y)  0, using the commutativity of addition.


xy  yx  0 , by the cancellation law of addition.
xy  yx ,
by part (b).
Thus R is commutative.
7. Characteristic of a Ring
Let R be any ring. We now examine whether there is a positive integer n such that
n  a  0 for all a  R , where n  a means a + a+ + a for n summands. i.e.,
na  a  a  ...  a ,
n summands
where + is the additive operation in the ring.
 If the ring under consideration is , the ring of integers, it can be seen that there
is no positive integer n such that
n  a  0 for all a 

Consider the ring <
n
, +n, n). In this case
n  a  a n a n . . . n a  0 .
n summands
as 0 is the remainder obtained when the usual sum a  a  . . .  a  n  a is divided by n.
n summands
This shows that
n  a  0 for a 
It can be seen that
2n  a  0 for a 
n
.
n
3n  a  0 for a 
n
...
kn  a  0 for a 
n
...
Definition If for a ring R a positive integer n exists such that
n  a  0 for a  R,
then the least such positive integer is the characteristic of the ring R. If no such positive
integer exists, then R is of characteristic 0.
Example 13 The ring Zn is of characteristic n, since n is the least positive integer such
that
n  a  0 for a  n .
Consider the ring of positive integers Z. Since there is no positive
na  0  a  Z
integer n such that
Z has characteristic 0. Similarly, Q, R, C all have characteristic 0.
Theorem 8 If R is a ring with unity 1, then R has characteristic n  0 if and only if n
is the smallest positive integer such that n 1  0.
Proof
Let R be a ring with unity 1. By the definition, if R has characteristic n > 0, then
n  a  0  a  R,
so in particular n 1  0.
Conversely, suppose that n is the smallest positive integer such that n  1  0 . Then
for any a  R, we have
n  a  a  a  . . .  a  a(1  1 
 1)  a(n 1)  a0  0 .
n summands
Hence the theorem is proved.