Algebra and Trigonometry 26
Integral domains and
characteristic of a ring
Objectives
From this unit a learner is expected to achieve the following
- Familiarize with the concept of divisors of 0 in a ring.
- Study that the cancellation laws hold in a ring R if and only if R has no left or right divisors of 0.
- Learn the definition of integral domains and familiarize with some examples.
- 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 Using this idea, we can say that every integer divides 0. Also we note that in the ring of integers if and only if either or One uses this fact constantly, perhaps without realizing it. Suppose, for example, in order to solve the equation
we first factor the left side as follows:
.
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 observedthat 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 situationswhere the product bq is 0 even if neither b nor q is 0.
Example 1 Solve the equation in , the ring with the operations +12 ,the addition modulo 12,and 12, multiplication modulo 12.
Solution
Since and , we can factorize as follows:
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 .
But in , not only 0a = a0 = 0 for all a, 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 = 0 imply x 2 = 2 and x 3 = 6; so 4 and 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 = 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 0in 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.
Example2Consider the ring. 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 are exactly the numbers in that are not relatively prime to 12.
Theorem 1In the ring , the divisors of 0 are precisely those elements that are not relatively prime to n.
Proof
Recall that {0, 1, . . . , n 1} is ring under +n and n . Let m, where m 0, and m is not relatively prime to n, so that g.c.d. of m and n be d 1. Then
…(1)
Now is a divisor of implies is an integer, and hence is a multiple of and hence in , . Thus from Eq. (1), in Zn , while neither m nor is 0, so m is a divisor of 0.
We prove the converse using contra positive method of proving theorems. Suppose mis a non-zero element in Znthat is relatively prime to n. Then 0 < mn. If for sZn 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.
CorollaryIf 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, cR. 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 if and only if R has no left or right divisors of 0.
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, bR.
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 a1, then the solution x of ax = b is a1 b.
Notation In the case that R is commutative, in particular if R is a field, it is customary to denote a1 b and ba1 (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 a1 b or ba1.
- In a field F it is usual to define a quotient b/a, where 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.
Example3 Z is an integral domain.
Example4 The set of all even integers doesn’t form an integral domain as it doesn’t contain the multiplicative identity.
Theorem 3 is an integral domain.
Proof Every nonzero element in are relatively prime to p. Hence, by the corollary to Theorem 1, we know that has no divisors of 0. Also is a commutative ring with unity 1. Hence 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 implies either or ; implies or
- Theorem 2 and the definition of integral domain shows that the cancellation laws for multiplication hold in an integral domain.
Example5 is not an integral domain if n is not prime, because in that case contains divisors of 0.
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 . Then if ab = 0, we have
But then
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 aD, where a 0, there exists bD 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 is a field.
Proof
By Theorem 3, is an integral domain. Also, since contains finite number of elements, is a finite integral domain. Hence, by Theorem 5 , is a field.
Example 6Give an example of a field with three elements.
Solution under addition modulo 3 and multiplication modulo 3.
Example 7 Give an example of a ring which is not a field.
Solution
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 6The multiplicative inverse of a non-zero element in a field is unique.
Proof.
Let be the multiplicative identity of a field .
Let and be two inverses of a non-zero element
Then
and
Now
, by the associativity of multiplication
Thus inverse of an element in a field is unique
Theorem 7 If a and b are elements in a field F,
(i) .
In particular,
(ii)
(iii) for with
(iv) for with
Proof.
(i)We have
,
where 0 is the additive inverse.
Post multiplying both sides by b, we get
and this implies by the right distributive law that
Hence is the additive inverse of . That is,
(ii) is obtained from (i) by the commutativity of multiplication.
(iii)Let be the multiplicative identity of . We note that if the inverse of exists in F and is denoted by
Then
Also by the reversal law of inverses,
Since the above gives
.
This implies by right cancellation law that
.
(iv) Let d be the multiplicative inverse of . Then
…(2)
Using Property (i),
…(3)
From Equations (2) and (3), we obtain
Since is the additive inverse of and since -1 is the additive inverse of 1, the above implies
Pre-multiplying both sides by , we obtain
Hence
Since the above implies That is,
Example8 In a field , prove that either or .
Solution
Given
We note that is the additive inverse of Adding on both sides of the above equation, we obtain
, since
(by the distributive law)
(by commutativity of multiplication)
(by the distributive law)
Since a field is without divisors of 0, the last equation implies, either or i.e., either or .
Example9 In a field with multiplicative identity 1, if and then there exists a unique element such that
Solution
Since , exists and .
Left multiplying by we get
implies
(by the associativity of multiplication)
implies
since
implies
To prove uniqueness, suppose there are and such that and . This implies This implies (by the cancellation law in a field) that Hence the uniqueness is proved.
Example 10 A Gaussian integer is a complex number , with and being integers. It can be seen that the set of Gaussian integers is a commutative ring with unity. Since the product of two non-zero complex numbers cannot be zero, has no zero-divisors. It is therefore an integral domain. To check whether is a field, let be any non-zero element of . Then at least one of and is not zero. Since is a nonzero complex number and since the set of complex numbers form a field, we have
which is, in general, not a Gaussian integer as and are not necessarily integers. Therefore is not a field.
Definition A non commutative division ring is called a strictly skew field.
Example 11Show that the set of all matrices of the form
where and are real numbers, is a strictly skew-field under matrix addition and multiplication.
Solution
It can be easily verified that is a ring. The matrix
is the zero element of
The matrix is the unit of
Now let
be any non-zero element in Then not all of are 0. We have det . Since and are real and not all of them are 0, det Therefore the matrix is invertible . In fact,
which is easily seen to be an element of Thus the non-zero elements of form a group under multiplication. We can show by examples that the commutative law of multiplication does not hold in . Therefore is a strictly skew field.
6. Boolean Ring
Definition A ring is called Boolean if for all .
Example 12 If is a Boolean ring, prove that
for all ;
(b) implies and
(c) is commutative.
Solution
(a) Since for all , and since , we have
, by the right-distributive law.
by the cancellation law of addition,
(b) If then, since (by Part (a)), we have , and hence by the cancellation law of addition.
(c) For any we have . That is,
, since and
using the commutativity of addition.
, by the cancellation law of addition.
, by part (b).
Thus is commutative.
7. Characteristic of a Ring
Let R be any ring. We now examine whether there is a positive integer n such that for all , where means a + a+ + a for n summands. i.e.,
,
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
for all
- Consider the ring < , +n,n). In this case
.
as 0 is the remainder obtained when the usual sum is divided by n. This shows that
for
It can be seen that for
for
. . .
for
. . .
Definition If for a ring R a positive integer n exists such that
for
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
for
Consider the ring of positive integers Z. Since there is no positive
integer n such that aZ
Z has characteristic 0. Similarly, Q, R, Call have characteristic 0.
Theorem 8 If R is a ring with unity 1, then R has characteristic if and only if n is the smallest positive integer such that
Proof Let R be a ring with unity 1. By the definition, if R has characteristic n > 0, then
aR,
so in particular
Conversely, suppose that n is the smallest positive integer such that. Then for any aR, we have
.
Hence the theorem is proved.
Summary
In this session we have seen the definitions of divisors of 0 and integral domain. We have seen 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 have seen the definitions of boolean ring and characteristic of a ring.
Assignments
- Find all solutions of the equation in Z6.
- Let X be a non-empty set and R = the set of all subset of X. For A, B R, define A + B = (A B ) (B A) and A . B = A B. Show that <R, +, .> is a commutative ring with unit element, but not an integral domain in general.
- Show that 1 and p 1 are the only elements in the field Zp that are their own multiplicative inverse. (Hint: Consider the equation x2 1 = 0)
- If are the members of a field, then show that if
- Show that the characteristic of an integral domain is either 0 or a prime p.
Quiz
1. All the solutions of the equation in are ______
(a). 1 and 2
(b). 0 and 1
(c). 2 and 3
(d). none of these
Ans. (d)
2. A division ring contains exactly ______number of idempotent elements.
(a). 1
(b). 2
(c). 3
(d). 4
Ans. (b)
3. Pick the true statement.
(a).A divisor of zero in a commutative ring with unity can have no multiplicative inverse.
(b). is a subfield of .
(c). has zero divisors if n is not prime.
(d).All the above are false.
Ans. (a)
FAQ
1. Is the following statement true? If there are no left 0 divisors in a ring R then there cannot be right 0 divisors in R.
Ans. Yes. This follows immediately from the definition of divisors of 0.
2. Is the following statement true? Whenever there is a left 0 divisor, then there is a right 0 divisor.
Ans.Yes . 0 divisors will always come in pairs. So whenever you have a left 0 divisor, you automatically have a right 0 divisor. Sometimes left and right 0 divisors turns to be the same, for example, in , corresponding to the left 0 divisor 2, the right 0 divisor is 2 itself.
3. Is the following statement true? The number of distinct 0 divisors in a ring is an even number.
Ans. No. For example, in there is only one 0 divisor and is 2.
4. Can 0 be a 0 divisor?
Ans. The definition clearly states that 0 divisors must not be 0.
5. Is the following statement true? An integral domain always contain 0 and 1.
Ans. Yes. This follows from the definition.
Glossary
Divisors of 0: 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.
Cancellation Law in a Ring: Let R be a ring, and let a, b, cR. The cancellation laws hold in R if ab = ac, with a 0, implies b = c, and ba = ca with a 0, implies b =c.
Integral Domain: An integral domain D is a commutative ring with unity containing no divisors of 0.
Gaussian integer: A Gaussian integer is a complex number , with and being integers.
Boolean Ring: A ring is called Boolean if for all .
Characteristic of a ring:If for a ring R a positive integer n exists such that
for
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.
REFERENCES
Books
- I.N. Herstein, Topics in Algebra, Wiley Easten Ltd., New Delhi, 1975.
- K. B. Datta, Matrix and Linear Algebra, Prentice Hall of India Pvt. Ltd., New Delhi, 2000.
- P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, First Course in Linear Algebra, Wiley Eastern, New Delhi, 1983.
- P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, Basic Abstract Algebra (2nd Edition), CambridgeUniversity Press, Indian Edition, 1997., New Delhi, 1983.
- S.K. Jain, A. Gunawardena and P.B. Bhattacharya, Basic Linear Algebra with MATLAB,KeyCollege Publishing (Springer-Verlag), 2001.
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