Grossment College Solve the Equations Algebra Questions

0/4 Questions AnsweredHomework 5
Q1 Subring?
1 Point
Suppose S
= {(a, b) ∣ a − 2b = 0} ⊆ R × R. Is S a subring of
R × R? Prove your claim.
Save Answer
Q2 Important Subring
1 Point
> 1 be an integer. Suppose a ∈ Z.
Define Qa = {[ka]n ∣ k ∈ Z} ⊆ Zn .
(a) Prove that Qa is a subring of Zn .
(b) Suppose n = 12 and a = 8. List the elements of Q8 .
(c) Suppose n = 6 and a = 5. List the elements of Q5 .
Let n
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Save Answer
Q3 Matrix Example
1 Point
0 1
0 0
),O = (
) ∈ M2 (R).
0 0
0 0
Define L = {B ∣ BA = O} ⊆ M2 (R) and
R = {B ∣ AB = O} ⊆ M2 (R).
(a) Prove L is a subring of M2 (R).
(b) Is L = R? Prove your claim.
Let A
=(
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​
​
​
​
​
​
Save Answer
Q4 Unique Inverse
1 Point
Suppose R is a ring with identity and a is a unit.
(a) Prove that the inverse of a is unique. I.e. Suppose there are two
inverses b and c and show b = c.
(b) Prove that ∀a, b
solution.
∈ R, if a is a unit, then ax = b has a unique
Save Answer
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C H A P T E r
3
Rings
ALTERNATE ROUTE: If you want to cover groups before studying rings,
you should read Chapters 7 and 8 now.
We have seen that many rules of ordinary arithmetic hold not only in  but also in
the miniature arithmetics n. You know other mathematical systems, such as the
real numbers, in which many of these same rules hold. Your high-school algebra
courses dealt with the arithmetic of polynomials.
The fact that similar rules of arithmetic hold in different systems suggests
that it might be worthwhile to consider the common features of such systems.
In the long run, this might save a lot of work: If we can prove a theorem about one
system using only the properties that it has in common with a second system,
then the theorem is also valid in the second system. By “abstracting” the common core of essential features, we can develop a general theory that includes
as special cases , n, and the other familiar systems. Results proved for this
general theory will apply simultaneously to all the systems covered by the theory.
This process of abstraction will allow us to discover the real reasons a particular
statement is true (or false, for that matter) without getting bogged down in nonessential details. In this way a deeper understanding of all the systems involved
should result.
So we now begin the development of abstract algebra. This chapter is just
the first step and consists primarily of definitions, examples, and terminology.
Systems that share a minimal number of fundamental properties with  and n
are called rings. Other names are applied to rings that may have additional properties, as you will see in Section 3.1. The elementary facts about arithmetic and
algebra in arbitrary rings are developed in Section 3.2. In Section 3.3 we consider
rings that appear to be different from one another but actually are “essentially the
same” except for the labels on their elements.
43
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44
Chapter 3
3.1
Rings
Definition and Examples of Rings
We begin the process of abstracting the common features of familiar systems with this
definition:
Definition
A ring is a nonempty set R equipped with two operations* (usually written
as addition and multiplication) that satisfy the following axioms. For all a,
b, cPR:
1. If aPR and bPR, then a 1 bPR.
[Closure for addition]
2. a 1 (b 1 c) 5 (a 1 b) 1 c.
[Associative addition]
3. a 1 b 5 b 1 a.
[Commutative addition]
4. There is an element 0R in R such
that a 1 0R 5 a 5 0R 1 a for every
aPR.
[Additive identity
or zero element]
5. For each aPR, the equation
a 1 x 5 0R has a solution in R.†
6. If aPR and bPR, then abPR.
[Closure for multiplication]
7. a(bc) 5 (ab)c.
[Associative multiplication]
8. a(b 1 c) 5 ab 1 ac and
(a 1 b)c 5 ac 1 bc.
[Distributive laws]
These axioms are the bare minimum needed for a system to resemble  and n. But
 and n have several additional properties that are worth special mention:
Definition
A commutative ring is a ring R that satisfies this axiom:
9. ab 5 ba for all a, bPR.
Definition
[Commutative multiplication]
A ring with identity is a ring R that contains an element 1R satisfying this
axiom:
10. a1R 5 a 5 1R a for all aPR.
[Multiplicative identity]
*“Operation” and “closure” are defined in Appendix B.
†
Those who have already read Chapter 7 should note that Axioms 1–5 simply say that a ring is an
abelian group under addition.
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3.1
Definition and Examples of Rings
45
In the following examples, the verification of most of the axioms is left to the
reader.
EXAMPLE 1
With the usual addition and multiplication,
Z (the integers)   and   R (the real numbers)
are commutative rings with identity.
EXAMPLE 2
The set Zn, with the usual addition and multiplication of classes, is a commutative ring with identity by Theorem 2.7.
EXAMPLE 3
Let E be the set of even integers with the usual addition and multiplication.
Since the sum or product of two even integers is also even, the closure
axioms (1 and 6) hold. Since 0 is an even integer, E has an additive identity
element (Axiom 4). If a is even, then the solution of a 1 x 5 0 (namely – a) is
also even, and so Axiom 5 holds. The remaining axioms (2, 3, 7, 8, and 9)
hold for all ­integers and, therefore, are true whenever a, b, c are even.
Consequently, E is a commutative ring. E does not have an identity, however,
because no even integer e has the property that ae 5 a 5 ea for every even
­integer a.
EXAMPLE 4
The set of odd integers with the usual addition and multiplication is not a
ring. Among other things, Axiom 1 fails: The sum of two odd integers is
not odd.
Although the definition of ring was constructed with Z and Zn as models, there
are many rings that aren’t at all like these models. In these rings, the elements may not
be numbers or classes of numbers, and their operations may have nothing to do with
“ordinary” addition and multiplication.
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46
Chapter 3
Rings
EXAMPLE 5
The set T 5 {r, s, t, z} equipped with the addition and multiplication defined
by the following tables is a ring:
1
z
r
s
t
z
z
r
s
t
r
r
z
t
s
s
s
t
z
r
t
t
s
r
z
z
z
z
z
z
?
z
r
s
t
r
z
z
z
z
s
z
r
s
t
t
z
r
s
t
You may take our word for it that associativity and distributivity hold
(Axioms 2, 7, and 8). The remaining axioms can be easily verified from the
operation tables above. In particular, they show that T is closed under both
addition and multiplication (Axioms 1 and 6) and that addition is commutative (Axiom 3).
The element z is the additive identity—the element denoted 0R in Axiom 4. It behaves in the same way the number 0 does in Z (that’s why the notation 0R is used in the
axiom), but z is not the integer 0—in fact, it’s not any kind of number. Nevertheless,
we shall call z the “zero element” of the ring T.
In order to verify Axiom 5, you must show that each of the equations
r 1 x 5 z   s 1 x 5 z   t 1 x 5 z   z 1 x 5 z
has a solution in T. This is easily seen to be the case from the addition table; for
example, x 5 r is the solution of r 1 x 5 z because r 1 r 5 z.
Finally, note that T is not a commutative ring; for instance, rs 5 r and
sr 5 z, so that rs Þ sr.
EXAMPLE 6
Let M(R) be the set of all 2 3 2 matrices over the real numbers, that is, M(R)
consists of all arrays
a
a
c
b
b,   where a, b, c, d are real numbers.
d
Two matrices are equal provided that the entries in corresponding positions are equal;
that is,
a
a
c
For example,
a
1140058_CH0003.indd 46
b
r
b5a
d
t
4
23
s
b   if and only if   a 5 r, b 5 s, c 5 t, d 5 u.
u
0
2 12
b5a
1
1 24
0
b
1
but
a
1
5
3
3
b2 a
2
1
5
b.
2
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3.1
47
Definition and Examples of Rings
Addition of matrices is defined by
a
For example,
a
3
5
a
c
b
ar
b1a
d
cr
4
22
b1a
1
6
br
a 1 ar
b5a
dr
c 1 cr
7
314
b5a
0
516
7
22 1 7
b5a
110
11
Multiplication of matrices is defined by
For example,
a
2
0
a
b w
ba
d y
a
c
3 1
ba
24 6
x
aw 1 by
b5a
cw 1 dy
z
20
224
11
b.
228
5
b.
1
ax 1 bz
b.
cx 1 dz
25
21136
b5a
7
0  1 1 1 24 2 6
5a
b 1 br
b.
d 1 dr
2 1 25 2 1 3  7
b
0 1 25 2 1 1 24 2 7
Reversing the order of the factors in matrix multiplication may produce a different
answer, as is the case here:
a
1
6
25 2
ba
7 0
3
1  2 1 1 25 2 0
b5a
24
62170
5a
2
12
23
b.
210
1  3 1 1 25 2 1 24 2
b
6  3 1 7 1 24 2
So this multiplication is not commutative. With a bit of work, you can verify that
M(R) is a ring with identity. The zero element is the zero matrix
which is denoted 0 and X 5 a
2a
2c
a
a
c
a
0
0
0
b,
0
2b
b is a solution of
2d
b
0
b1X5a
d
0
0
b.
0
We claim that the multiplicative identity element (Axiom 10) is the matrix I 5 a
1
0
To prove this claim, we first multiply a typical matrix in M(R) on the right by I:
a
1140058_CH0003.indd 47
a
c
b 1
ba
d 0
0
a11b0
b5a
1
c11d0
a01b1
a
b5a
c01d1
c
0
b.
1
b
b.
d
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48
Chapter 3
Rings
Since multiplication is not commutative here, we also need to check left multiplication
by I as well:
a
1
0
0 a
ba
1 c
b
1  a 1 0c
b5a
d
0  a 1 1c
1b 1 0d
a
b5a
0b 1 1d
c
b
b.
d
This proves that I satisfies Axiom 10.* Consequently, I is called the identity matrix.
Note that the product of nonzero elements of M(R) may be the zero element; for
example,
a
4
2
6 23
ba
3
2
EXAMPLE 7
4 1 23 2 1 6  2
29
b5a
6
2 1 23 2 1 3  2
0
4 1 29 2 1 6  6
b5a
2 1 29 2 1 3  6
0
0
b.
0
If R is a commutative ring with identity, then M(R) denotes the set of all
2 3 2 matrices with entries in R. With addition and multiplication defined as
in Example 6, M(R) is a noncommutative ring with identity, as you can readily verify. For instance, M(Z) is the ring of 2 3 2 matrices with integer entries,
M(Q) the ring of 2 3 2 matrices with rational number entries, and M(Zn) the
ring of 2 3 2 matrices with entries from Zn.
EXAMPLE 8
Let T be the set of all functions from R to R, where R is the set of real
­numbers. As in calculus, f 1 g and fg are the functions defined by
( f 1 g)(x) 5 f  (x) 1 g(x)   and   ( fg)(x) 5 f  (x)g(x).
You can readily verify that T is a commutative ring with identity. The zero element is the function h given by h(x) 5 0 for all xPR. The identity element is the
function e given by e(x) 5 1 for all xPR. Once again the product of nonzero
elements of T may turn out to be the zero element; see Exercise 36.
We have seen that some rings do not have the property that the product of two
nonzero elements is always nonzero. But some of the rings that do have this property,
such as Z, occur frequently enough to merit a title.
Definition
An integral domain is a commutative ring R with identity 1R Þ 0R that
s­ atisfies this axiom:
11. Whenever a, bPR and ab 5 0R, then a 5 0R or b 5 0R.
*Checking a possible identity element under both right and left multiplication is essential. There
are rings in which an element acts like an identity when you multiply on the right, but not when you
multiply on the left. See Exercise 11.
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3.1
Definition and Examples of Rings
49
The condition 1R Þ 0R is needed to exclude the zero ring (that is, the single-element
ring {0R}) from the class of integral domains. Note that Axiom 11 is logically equivalent to its contrapositive.*
Whenever a Þ 0R and b Þ 0R, then ab Þ 0R.
EXAMPLE 9
The ring Z of integers is an integral domain. If p is prime, then Zp is an ­integral
domain by Theorem 2.8. On the other hand, Z6 is not an integral domain ­because
4  3 5 0, even though 4 Þ 0 and 3 Þ 0.
You should be familiar with the set Q of rational numbers, which consists of all
fractions ayb with a, bPZ and b Þ 0. Equality of fractions, addition, and multiplication are given by the usual rules:
a
r
5    if and only if   as 5 br
s
b
    
a
c
ad 1 bc
1 5
     
b
d
bd
a c
ac
 5
b d
bd
It is easy to verify that Q is an integral domain. But Q has an additional property that
does not hold in Z: Every equation of the form ax 5 1 (with a Þ 0) has a solution in
Q. Therefore, Q is an example of the next definition.
Definition
A field is a commutative ring R with identity 1R Þ 0R that satisfies this
axiom:
12. For each a Þ 0R in R, the equation ax 5 1R has a solution in R.
Once again the condition 1R Þ 0R is needed to exclude the zero ring. Note that
Axiom 11 is not mentioned explicitly in the definition of a field. However, Axiom 11
does hold in fields, as we shall see in Theorem 3.8 below.
EXAMPLE 10
The set R of real numbers, with the usual addition and multiplication, is a field.
If p is a prime, then Zp is a field by Theorem 2.8.
EXAMPLE 11
The set C of complex numbers consists of all numbers of the form a 1 bi,
where a, bPR and i2 5 21. Equality in C is defined by
a 1 bi 5 r 1 si   if and only if   a 5 r and b 5 s.
*See Appendix A for a discussion of contrapositives.
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50
Chapter 3
Rings
The set C is a field with addition and multiplication given by
(a 1 bi) 1 (c 1 di) 5 (a 1 c) 1 (b 1 d )i
(a 1 bi)(c 1 di) 5 (ac 2 bd) 1 (ad 1 bc)i.
The field R of real numbers is contained in C because R consists of all complex
­ umbers of the form a 1 0i. If a 1 bi Þ 0 in C, then the solution of the equation
n
(a 1 bi)x 5 1 is x 5 c 1 di, where
c 5 ay(a2 1 b2)PR   and   d 5 2by(a2 1 b2)PR (verify!).
EXAMPLE 12
Let K be the set of all 2 3 2 matrices of the form
a
a
2b
b
b,
a
where a and b are real numbers. We claim that K is a field. For any two matrices in K,
a
a
2b
a
a
2b
b
c
b1a
a
2d
b
c
b  a
a
2d
d
a1c
b5a
c
2b 2 d
d
ac 2 bd
b5a
c
2ad 2 bc
b 1d
b
a1c
ad 1 bc
b.
ac 2 bd
In each case the matrix on the right is in K because the entries along the main
diagonal (upper left to lower right) are the same and the entries on the opposite
diagonal (upper right to lower left) are negatives of each other. Therefore, K is
closed under addition and multiplication. K is commutative because
a
c
2d
d
a
ba
c 2b
b
ac 2 bd
b5a
a
2ad 2 bc
ad 1 bc
a
b5a
ac 2 bd
2b
Clearly, the zero matrix and the identity matrix I are in K. If
A5a
a
2b
b
c
ba
a 2d
d
b.
c
b
b
a
is not the zero matrix, then verify that the solution of AX 5 I is
X5 a
a/ d
b/ d
2b / d
b [ K,
a/ d
whe re d 5 a 2 1 b 2 .
Whenever the rings in the preceding examples are mentioned, you may assume
that addition and multiplication are the operations defined above, unless there is some
­specific statement to the contrary. You should be aware, however, that a given set (such
as Z) may be made into a ring in many different ways by defining different addition
and multiplication operations on it. See Exercises 17 and 22–26 for examples.
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3.1
Definition and Examples of Rings
51
Now that we know a variety of different kinds of rings, we can use them to produce
new rings in the following way.
EXAMPLE 13
Let T be the Cartesian product Z6 3 Z, as defined in Appendix B. Define
a­ ddition in T by the rule
(a, z) 1 (a9, z9) 5 (a 1 a9, z 1 z9).
The plus sign is being used in three ways here: In the first coordinate on the right-hand
side of the equal sign, 1 denotes addition in Z6; in the second coordinate, 1 denotes
­addition in Z; the 1 on the left of the equal sign is the addition in T that is being ­defined.
Since Z6 is a ring and a, a9PZ6, the first coordinate on the right, a 1 a9, is in Z6. Similarly
z 1 z9PZ. Therefore, addition in T is closed. Multiplication is defined similarly:
(a, z)(a9, z9) 5 (aa9, zz9).
For example, (3, 5) 1 (4, 9) 5 (3 1 4, 5 1 9) 5 (1, 14) and (3, 5)(4, 9) 5
(3 ? 4, 5 ? 9) 5 (0, 45). You can readily verify that T is a commutative ring with
identity. The zero element is (0, 0), and the multiplicative identity is (1, 1). What
was done here can be done for any two rings.
Theorem 3.1
Let R and S be rings. Define addition and multiplication on the Cartesian
product R 3 S by
(r, s) 1 (r9, s9) 5 (r 1 r9, s 1 s9) and
(r, s)(r9, s9) 5 (rr9, ss9).
Then R 3 S is a ring. If R and S are both commutative, then so is R 3 S. If both
R and S have an identity, then so does R 3 S.
Proof
•
Exercise 33.
Subrings
If R is a ring and S is a subset of R, then S may or may not itself be a ring under the
operations in R. In the ring Z of integers, for example, the subset E of even integers is
a ring, but the subset O of odd integers is not, as we saw in Examples 3 and 4. When
a subset S of a ring R is itself a ring under the addition and multiplication in R, then
we say that S is a subring of R.
EXAMPLE 14
Z is a subring of the ring Q of rational numbers and Q is a subring of the field
R of all real numbers. Since Q is itself a field, we say that Q is a subfield of R.
Similarly, R is a subfield of the field C of complex numbers.
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52
Chapter 3
Rings
EXAMPLE 15
The matrix rings M(Z) and M(Q) in Example 7 are subrings of M(R).
EXAMPLE 16
The ring K in Example 12 is a subring of M(R).
EXAMPLE 17
Let T be the ring of all functions from R to R in Example 8. Then the subset S
consisting of all continuous functions from R to R is a subring of T. To prove
this, you need one fact proved in calculus: The sum and product of continuous
functions are also continuous. So S is closed under addition and multiplication
(Axioms 1 and 6). You can readily verify the other axioms.
Proving that a subset S of a ring R is actually a subring is easier than proving ­directly
that S is a ring. For instance, since a 1 b 5 b 1 a for all elements of R, this fact is also true
when a, b happen to be in the subset S. Thus Axiom 3 (commutative addition) automatically holds in any subset S of a ring. In fact, to prove that a subset of a ring is actually a
subring, you need only verify a few of the axioms for a ring, as the next theorem shows.
Theorem 3.2
Suppose that R is a ring and that S is a subset of R such that
(i) S is closed under addition (if a, bPS, then a 1 bPS);
(ii) S is closed under multiplication (if a, bPS, then abPS);
(iii) 0RPS;
(iv) If aPS, then the solution of the equation a 1 x 5 0R is in S.
Then S is a subring of R.
Note condition (iv) carefully. To verify it, you need not show that the equation
a 1 x 5 0R has a solution—we already know that it does because R is a ring. You need
only show that this solution is an element of S (which implies that Axiom 5 holds for S).
Proof of Theorem 3.2
As noted before the theorem, Axioms 2, 3, 7, and 8 hold
for all elements of R, and so they necessarily hold for the elements of the
subset S. Axioms 1, 6, 4, and 5 hold by (i)–(iv).
•
EXAMPLE 18
The subset S 5 {0, 3} of Z6 is closed under addition and multiplication
(0 1 0 5 0; 0 1 3 5 3; 3 1 3 5 0; similarly, 0 ? 0 5 0 5 0 ? 3; 3 ? 3 5 3). By the
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3.1
Definition and Examples of Rings
53
definition of S we have 0PS. Finally, the equation 0 1 x 5 0 has solution
x 5 0PS, and the equation 3 1 x 5 0 has solution x 5 3PS. Therefore, S is a
subring of Z6 by Theorem 3.2.
EXAMPLE 19
a
Let S be the subset of M(R) consisting of all matrices of the form a
b
Then S is closed under addition and multiplication because
a
a
b
0
r
b1a
c
s
0
a1r
b5a
t
b1s
a
a
b
0 r
ba
c s
010
a1r
b5a
c1t
b1s
0
ar
b5a
t
br 1 cs
0
b
ct
0
b
c1t
0
b.
c
H S and
H S.
The identity matrix is in S (let a 5 1, b 5 0, c 5 1) and the solution of
a
a
b
0
0
b1x5a
c
0
0
b
0
Hence S is a subring by Theorem 3.2.
is x 5 a
2a
2b
0
b
2c
H S.
EXAMPLE 20
The set Z C”2D 5 5 a 1 b”2 0 a, b H Z 6 is a subring of R. You can easily verify
that
Aa 1 b”2B Ac 1 d “2B 5 ac 1 ad “2 1 bc”2 1 bd “2  “2
5 1 ac 1 2bd 2 1 1 ad 1 bc 2 “2 2
H Z 3 “2 4 .
So Z 3 “2 4 is closed under multiplication. See Exercise 13 for the rest of the proof.
■ Exercises
A. 1. The following subsets of Z (with ordinary addition and multiplication) satisfy
all but one of the axioms for a ring. In each case, which axiom fails?
(a) The set S of all odd integers and 0.
(b) The set of nonnegative integers.
2. Let R 5 {0, e, b, c} with addition and multiplication defined by the tables on
page 54. Assume associativity and distributivity and show that R is a ring with
identity. Is R commutative? Is R a field?
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54
Chapter 3
Rings
1
0
e
b
c
?
0
e
b
c
0
0
e
b
c
0
0
0
0
0
e
e
0
c
b
e
0
e
b
c
b
b
c
0
e
b
0
b
b
0
c
c
b
e
0
c
0
c
0
c
3. Let F 5 {0, e, a, b} with operations given by the following tables. Assume
associativity and distributivity and show that F is a field.
1
0
e
a
b
?
0
e
a
b
0
0
e
a
b
0
0
0
0
0
e
e
0
b
a
e
0
e
a
b
a
a
b
0
e
a
0
a
b
e
b
b
a
e
0
b
0
b
e
a
4. Find matrices A and C in M(R) such that AC 5 0, but CA 2 0, where 0 is the
zero matrix. [Hint: Example 6.]
5. Which of the following six sets are subrings of M(R)? Which ones have an identity?
(a) All matrices of the form a
(b) All matrices of the form a
(c) All matrices of the form a
(f) All matrices of the form a
a
0
a
c
(d) All matrices of the form a
(e) All matrices of the form a
0
0
a
a
a
0
a
0
r
b with rPQ.
0
b
b with a, b, cPZ.
c
b
b with a, b, cPR.
0
0
b with aPR.
0
0
b with aPR.
a
0
b with aPR.
0
6. (a) Show that the set R of all multiples of 3 is a subring of Z.
(b) Let k be a fixed integer. Show that the set of all multiples of k is a subring of Z.
7. Let K be the set of all integer multiples of “2, that is, all real numbers of the
form n”2 with nPZ. Show that K satisfies Axioms 1–5, but is not a ring.
8. Is the subset {1, –1, i, –i} a subring of C?
9. Let R be a ring and consider the subset R* of R 3 R defined by R* 5 {(r, r) | rPR}.
(a) If R 5 Z6, list the elements of R*.
(b) For any ring R, show that R* is a subring of R × R.
1140058_CH0003.indd 54
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3.1
55
Definition and Examples of Rings
10. Is S 5 {(a, b) | a 1 b 5 0} a subring of Z 3 Z? Justify your answer.
11. Let S be the subset of M(R) consisting of all matrices of the form a
a
b
(a) Prove that S is a ring.
1
(b) Show that J 5 a
every A in S). 0
a
b.
b
1
b is a right identity in S (meaning that AJ 5 A for
0
(c) Show that J is not a left identity in S by finding a matrix B in S such that
JB 2 B.
For more information about S, see Exercise 41.
12. Let Z[i] denote the set {a 1 bi | a, bPZ}. Show that Z[i] is a subring of C.
13. Let Z C”2D denote the set {a 1 b”2 | a, bPZ}. Show that Z C”2D is a subring
of R. [See Example 20.]
14. Let T be the ring in Example 8. Let S 5 { fPT | f (2) 5 0}. Prove that S is a
subring of T.
15. Write out the addition and multiplication tables for
(a) Z2 3 Z3
(b) Z2 3 Z2
1 1
0
b and 0 5 a
1 1
0
such that AB 5 0.
16. Let A 5 a
(c) Z3 3 Z3
0
b in M(R). Let S be the set of all matrices B
0
(a) List three matrices in S. [Many correct answers are possible.]
(b) Prove that S is a subring of M(R). [Hint: If B and C are in S, show that
B 1 C and BC are in S by computing A(B 1 C) and A(BC).]
17. Define a new multiplication in Z by the rule: ab 5 0 for all a, b,PZ. Show that
with ordinary addition and this new multiplication, Z is a commutative ring.
18. Define a new multiplication in Z by the rule: ab 5 1 for all a, b,PZ. With
ordinary addition and this new multiplication, is Z is a ring?
19. Let S 5 {a, b, c} and let P(S) be the set of all subsets of S; denote the
elements of P(S) as follows:
S 5 {a, b, c};
A 5 {a};
D 5 {a, b};
B 5 {b};
E 5 {a, c};
C 5 {c};
F 5 {b, c};
0 5 [.
Define addition and multiplication in P(S) by these rules:
M 1 N 5 (M 2 N) ∙ (N 2 M)   and   MN 5 M ∙ N.
Write out the addition and multiplication tables for P(S). Also, see Exercise 44.
B. 20. Show that the subset R 5 {0, 3, 6, 9, 12, 15} of Z18 is a subring. Does R have
an identity?
21. Show that the subset S 5 {0, 2, 4, 6, 8} of Z10 is a subring. Does S have an
identity?
1140058_CH0003.indd 55
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56
Chapter 3
Rings
22. Define a new addition ! and multiplication ( on Z by
a ! b 5 a 1 b 2 1   and   a ( b 5 a 1 b 2 ab,
where the operations on the right-hand side of the equal signs are ordinary
addition, subtraction, and multiplication. Prove that, with the new operations
! and (, Z is an integral domain.
23. Let E be the set of even integers with ordinary addition. Define a new
multiplication * on E by the rule “a * b 5 abY2” (where the product on the
right is ordinary multiplication). Prove that with these operations E is a
commutative ring with identity.
24. Define a new addition and multiplication on Z by
a ! b 5 a 1 b 2 1   and   a ( b 5 ab 2 (a 1 b) 1 2.
Prove that with these new operations Z is an integral domain.
25. Define a new addition and multiplication on Q by
r ! s 5 r 1 s 1 l   and   r ( s 5 rs 1 r 1 s.
Prove that with these new operations Q is a commutative ring with identity. Is
it an integral domain?
26. Let L be the set of positive real numbers. Define a new addition and
multiplication on L by
a ! b 5 ab   and   a # b 5 alogb.
(a) Is L a ring under these operations?
(b) Is L a commutative ring?
(c) Is L a field?
27. Let S be the set of rational numbers that can be written with an odd
denominator. Prove that S is a subring of Q but is not a field.
28. Let p be a positive prime and let R be the set of all rational numbers that can
be written in the form rYpi with r, iPZ, and i $ 0. Note that Z # R because
each nPZ can be written as nYp0. Show that R is a subring of Q.
29. The addition table and part of the multiplication table for a three-element ring
are given below. Use the distributive laws to complete the multiplication table.
r
r
s
t
1
r
s
t
s
s
t
r
t
t
r
s
?
r
s
t
r
r
r
r
?
w
x
y
z
w
w
w
w
w
s
r
t
t
r
30. Do Exercise 29 for this four-element ring:
1
w
x
y
z
1140058_CH0003.indd 56
w
w
x
y
z
x
x
y
z
w
y
y
z
w
x
z
z
w
x
y
x
w
y
y
w
z
w
w
w
y
6/16/12 9:43 AM
3.1
Definition and Examples of Rings
k
31. A scalar matrix in M(R) is a matrix of the form a
0
number k.
57
0
b for some real
k
(a) Prove that the set of scalar matrices is a subring of M(R).
(b) If K is a scalar matrix, show that KA 5 AK for every A in M(R).
(c) If K is a matrix in M(R) such that KA 5 AK for every A in M(R), show
a b
1 0
that K is a scalar matrix. [Hint: If K 5 a
b, let A 5 a
b. Use the
c d
0 0
fact that KA 5 AK to show that b 5 0 and c 5 0. Then make a similar
0 1
argument with A 5 a
b to show that a 5 d.]
0 0
32. Let R be a ring and let Z(R) 5 {aPR | ar 5 ra for every rPR}. In other
words, Z(R) consists of all elements of R that commute with every other
element of R. Prove that Z(R) is a subring of R. Z(R) is called the center of
the ring R. [Exercise 31 shows that the center of M(R) is the subring of scalar
matrices.]
33. Prove Theorem 3.1.
34. Show that M(Z2) (all 2 3 2 matrices with entries in Z2) is a 16-element
noncommutative ring with identity.
35. Prove or disprove:
(a) If R and S are integral domains, then R 3 S is an integral domain.
(b) If R and S are fields, then R 3 S is a field.
36. Let T be the ring in Example 8 and let f, g be given by
f 1×2 5 e
0
x22
if x # 2
22x
     g ( x ) 5 e
if x . 2
0
if x # 2
if x . 2.
Show that f, gPT and that fg 5 0T. Therefore T is not an integral domain.
37. (a) If R is a ring, show that the ring M(R) of all 2 3 2 matrices with entries in
R is a ring.
(b) If R has an identity, show that M(R) also has an identity.
38. If R is a ring and aPR, let AR 5 {rPR | ar 5 0R}. Prove that AR is a subring
of R. AR is called the right annihilator of a. [For an example, see Exercise 16 in
which the ring S is the right annihilator of the matrix A.]
39. Let Q(“2) 5 (r 1 s”2 | r, sPQ}. Show that Q(“2) is a subfield of R.
[Hint: To show that the solution of (r 1 s”2)x 5 1 is actually in Q(“2),
multiply 1y(r 1 s”2) by (r 2 s”2)y(r 2 s”2).]
40. Let d be an integer that is not a perfect square. Show that Q(“d) 5
{a 1 b”d | a, bPQ} is a subfield of C. [Hint: See Exercise 39.]
1140058_CH0003.indd 57
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58
Chapter 3
Rings
41. Let S be the ring in Exercise 11.
(a) Verify that each of these matrices is a right identity in S:
1
2
±
1
2
1
2
.7
≤, a
1
.3
2
x
(b) Prove that the matrix a
y
x 1 y 5 1.
.7
2
b, and a
.3
21
2
b.
21
x
b is a right identity in S if and only if
y
x x
b is not a left identity in S.
y y
42. A division ring is a (not necessarily commutative) ring R with identity
1R ≠ 0R that satisfies Axioms 11 and 12 (pages 48 and 49). Thus a field is a
commutative division ring. See Exercise 43 for a noncommutative example.
Suppose R is a division ring and a, b are nonzero elements of R.
(c) If x 1 y 5 1, show that a
(a) If bb 5 b, prove that b 5 1R. [Hint: Let y be the solution of bx 5 1R and
note that by 5 b2 y.]
(b) If u is the solution of the equation ax 5 1R, prove that u is also a solution
of the equation xa 5 1R. (Remember that R may not be commutative.)
[Hint: Use part (a) with b 5 ua.]
43. In the ring M(C), let
15a
1
0
0
i
b   i 5 a
1
0
0
0
b     j 5 a
2i
21
1
0
b     k 5 a
0
i
i
b
0
The product of a real number and a matrix is the matrix given by this rule:
ra
t
y
u
rt
b5a
w
ry
ru
b
rw
The set H of real quaternions consists of all matrices of the form
al 1 bj 1 cj 1 dk 5 aa
5a
5a
1
0
a
0
0
i
b 1ba
1
0
0
bi
b 1 a
a
0
a 1 bi
2c 1 di
0
0
b 1ca
2i
21
0
0
b 1 a
2bi
2c
c 1 di
b,
a 2 bi
1
0
b1da
0
i
c
0
b1a
0
di
i
b
0
di
b
0
where a, b, c, and d are real numbers.
(a) Prove that
i2 5 j2 5 k2 5 –1
ij 5 –ji 5 k
jk 5 –kj 5 i
ki 5 –ik 5 j.
(b) Show that H is a noncommutative ring with identity.
1140058_CH0003.indd 58
6/16/12 9:43 AM
3.2
Basic Properties of Rings
59
(c) Show that H is a division ring (defined in Exercise 42). [Hint: If M 5 a1 1
bi 1 cj 1 dk, then verify that the solution of the equation Mx 5 1 is the
matrix ta1 2 tbi 2 tcj 2 tdk, where t 5 ly(a2 1 b2 1 c 2 1 d 2).]
(d) Show that the equation x2 5 21 has infinitely many solutions in H.
[Hint: Consider quaternions of the form 01 1 bi 1 cj 2 dk, where
b2 1 c 2 1 d 2 5 1.]
44. Let S be a set and let P(S) be the set of all subsets of S. Define addition and
multiplication in P(S) by the rules
M 1 N 5 1 M 2 N 2 < 1 N 2 M2 and MN 5 M d N. (a) Prove that P(S) is a commutative ring with identity. [The verification of additive associativity and distributivity is a bit messy, but an informal discussion using Venn diagrams is adequate for appreciating this example. See Exercise 19 for a special case.] (b) Show that every element of P(S) satisfies the equations x2 5 x and x 1 x 5 0P(S). C. 45. Let C be the set R 3 R with the usual coordinatewise addition (as in Theorem 3.1) and a new multiplication given by (a, b)(c, d) 5 (ac 2 bd, ad 1 bc) Show that with these operations C is a field. 46. Let r and s be positive integers such that r divides ks 1 1 for some k with 1 # k < r. Prove that the subset {0, r, 2r, 3r, . . . , (s 2 1)r} of Zrs is a ring with identity ks 1 1 under the usual addition and multiplication in Zrs. Exercise 21 is a special case of this result. Application: Applications of the Chinese Remainder Theorem (Section 14.2) may be covered at this point if desired. 3.2 Basic Properties of Rings When you do arithmetic in , you often use far more than the axioms for an integral domain. For instance, subtraction appears regularly, as do cancelation and the various rules for multiplying negative numbers. We begin by showing that many of these same properties hold in every ring. Arithmetic in Rings Subtraction is not mentioned in the axioms for a ring, and we cannot just assume that such an operation exists in an arbitrary ring. If we want to define a subtraction 1140058_CH0003.indd 59 6/16/12 9:43 AM 60 Chapter 3 Rings operation in a ring, we must do so in terms of addition, multiplication, and the ring axioms. The first step is Theorem 3.3 For any element a in a ring R, the equation a 1 x 5 0R has a unique solution. Proof •  e know that a 1 x 5 0R has at least one solution u by Axiom 5. If v is W also a solution, then a 1 u 5 0R and a 1 v 5 0R, so that v 5 0R 1 v 5 (a 1 u) 1 v 5 (u 1 a) 1 v 5 u 1 (a 1 v) 5 u 1 0R 5 u. Therefore, u is the only solution. We can now define negatives and subtraction in any ring by copying what happens in familiar rings such as . Let R be a ring and aPR. By Theorem 3.3 the equation a 1 x 5 0R has a unique solution. Using notation adapted from , we denote this unique solution by the symbol “2a.” Since addition is commutative, 2a is the unique element of R such that a 1 (2a) 5 0R 5 (2a) 1 a. In familiar rings, this definition coincides with the known concept of the negative of an element. More importantly, it provides a meaning for “negative” in any ring. E xample 1 In the ring 6, the solution of the equation 2 1 x 5 0 is 4, and so in this ring 22 5 4. Similarly, 29 5 5 in 14 because 5 is the solution of 9 1 x 5 0. Subtraction in a ring is now defined by the rule b 2 a means b 1 (2a). In  and other familiar rings, this is just ordinary subtraction. In other rings we have a new operation. EXAMPLE 2 In 6 we have 1 2 2 5 1 1 (22) 5 1 1 4 5 5. In junior high school you learned many computational and algebraic rules for dealing with negatives and subtraction. The next two theorems show that these rules are valid in any ring. Although these facts are not particularly interesting in themselves, it is essential to establish their validity so that we may do arithmetic in arbitrary rings. Theorem 3.4 If a 1 b 5 a 1 c in a ring R, then b 5 c. 1140058_CH0003.indd 60 6/16/12 9:43 AM 3.2 Proof • Basic Properties of Rings 61 Adding 2 a to both sides of a 1 b 5 a 1 c and then using associativity and negatives show that 2a 1 (a 1 b) 5 2a 1 (a 1 c) (2a 1 a) 1 b 5 (2a 1 a) 1 c 0R 1 b 5 0R 1 c b 5 c. Theorem 3.5 For any elements a and b of a ring R, (1) a ? 0R 5 0R 5 0R ? a. In particular, 0R ? 0R 5 0R. (2) a(2b) 5 2ab and (2a)b 5 2ab. (3) 2(2a) 5 a. (4) 2(a 1 b) 5 (2a) 1 (2b). (5) 2 (a 2 b) 5 2a 1 b. (6) (2a)(2b) 5 ab. If R has an identity, then (7) (21R)a 5 2a. Proof • (1) Since 0R 1 0R 5 0R, the distributive law shows that a ? 0R 1 a ? 0R 5 a(0R 1 0R) 5 a ? 0R 5 a ? 0R 1 0R. Applying Theorem 3.4 to the first and last parts of this equation shows that a ? 0R 5 0R. The proof that 0R ? a 5 0R is similar. (2) By definition, 2ab is the unique solution of the equation ab 1 x 5 0R, and so any other solution of this equation must be equal to 2ab. But x 5 a(2b) is a solution because, by the distribution law and (1), ab 1 a(2b) 5 a[b 1 (2b)] 5 a[0R] 5 0R. Therefore, a(2b) 5 2ab. The other part is proved similarly. (3) By definition, 2(2a) is the unique solution of (2a) 1 x 5 0R. But a is a solution of this equation since (2a) 1 a 5 0R. Hence, 2(2a) 5 a by uniqueness. (4) By definition, 2(a 1 b) is the unique solution of (a 1 b) 1 x 5 0R, but (2a) 1 (2b) is also a solution, because addition is commutative, so that (a 1 b) 1 [(2a) 1 (2b)] 5 a 1 (2a) 1 b 1 (2b) 5 0R 1 0R 5 0R. 1140058_CH0003.indd 61 6/16/12 9:43 AM 62 Chapter 3 Rings Therefore, 2(a 1 b) 5 (2a) 1 (2b) by uniqueness. (5) By the definition of subtraction and (4) and (3), 2(a 2 b) 5 2(a 1 (2b)) 5 (2a) 1(2(2b)) 5 2a 1 b. (6) (2a)(2b) 5 2(a (2b)) [By the second equation in (2), with 2b in place of b] 5 2(2ab) [By the first equation in (2)] 5 ab [By (3), with ab in place of a] (7) By (2), (21R)a 5 2(1Ra) 5 2(a) 5 2a. When doing ordinary arithmetic, exponent notation is a definite convenience, as is its additive analogue (for instance, a 1 a 1 a 5 3a). We now carry these concepts over to arbitrary rings. If R is a ring, aPR, and n is a positive integer, then we define a n 5 aaa ? ? ? a   (n factors). It is easy to verify that for any aPR and positive integers m and n, aman 5 am1n and   (am)n 5 amn. If R has an identity and a 2 0R, then we define a0 to be the element 1R. In this case, the exponent rules are valid for all m, n $ 0. If R is a ring, aPR, and n is a positive integer, then we define na 5 a 1 a 1 a 1 ? ? ? 1 a. (n summands) 2na 5 (2a) 1 (2a) 1 (2a) 1 ? ? ? 1 (2a). (n summands) Finally, we define 0a 5 0R. In familiar rings this is nothing new, but in other rings it gives a meaning to the “product” of an integer n and a ring element a. EXAMPLE 3 Let R be a ring and a, bPR. Then (a 1 b)2 5 (a 1 b)(a 1 b) 5 a(a 1 b) 1 b(a 1 b) 5 aa 1 ab 1 ba 1 bb 5 a2 1 ab 1 ba 1 b2. Be careful here. If ab 2 ba, then you can’t combine the middle terms. If R is a commutative ring, however, then ab 5 ba and we have the familiar pattern (a 1 b)2 5 a2 1 ab 1 ba 1 b2 5 a2 1 ab 1 ab 1 b2 5 a2 1 2ab 1 b2. For a calculation of (a 1 b)n in a commutative ring, with n . 2, see the Binomial Theorem in Appendix E. It’s worth noting that subtraction provides a faster method than Theorem 3.2 for showing that a subset of a ring is actually a subring. 1140058_CH0003.indd 62 6/16/12 9:43 AM 3.2 Basic Properties of Rings 63 Theorem 3.6 Let S be a nonempty subset of a ring R such that (1) S is closed under subtraction (if a, bPS, then a 2 bPS); (2) S is closed under multiplication (if a, bPS, then abPS). Then S is a subring of R. Proof • We show that S satisfies conditions (i)2(iv) of Theorem 3.2 and hence is a subring. The conditions will be proved in this order: (ii), (iii), (iv), and (i). (ii) Hypothesis (2) here is identical with condition (ii) of Theorem 3.2. Hence, S satisfies condition (ii). (iii) Since S is nonempty, there is some element c with cPS. Applying (1) (with a 5 c and b 5 c), we see that c 2 c 5 0R is in S. Therefore, S ­satisfies condition (iii) of Theorem 3.2. (iv) If a is any element of S, then by (1), 0R 2 a 5 2a is also in S. Since 2a is the solution of a 1 x 5 0R, condition (iv) of Theorem 3.2 is satisfied.   (i) If a, bPS, then 2b is in S by the proof of (iv). By (1), a 2 (2b) 5 a 1 b is in S. So S satisfies condition (i) of Theorem 3.2. Therefore, S is a subring of R by Theorem 3.2. Units and Zero Divisors Units and zero divisors in n were introduced in Section 2.3. We now carry these concepts over to arbitrary rings. Definition An element a in a ring R with identity is called a unit if there exists uPR such that au 5 1R 5 ua. In this case the element u is called the (multiplicative) inverse of a and is denoted a21. EXAMPLE 4 The only units in  are 1 and 21. EXAMPLE 5 By Theorem 2.10, the units in 15 are 1, 2, 4, 7, 8, 11, 13, and 14. For instance, 2 ? 8 5 1, so 221 5 8 and 821 5 2. 1140058_CH0003.indd 63 6/16/12 9:43 AM 64 Chapter 3 Rings EXAMPLE 6 1 Every nonzero element of the field R is a unit: If a Þ 0, then a  a 5 1. The same thing is true for every field F. By definition, F satisfies Axiom 12: If a Þ 0F, then the equation ax 5 1F has a solution in F. Hence, Every nonzero element of a field is a unit. EXAMPLE 7 a c easily verify, A matrix a a a c d b ad 2 bc b± d 2c ad 2 bc b b in M(R) such that ad 2 bc Þ 0 is a unit because, as you can d 2b ad 2 bc 1 ≤ 5a a 0 ad 2 bc d 0 ad 2 bc b and ± 1 2c ad 2 bc 2b ad 2 bc a ≤a a c ad 2 bc b 1 b5a d 0 0 b. 1 In particular, each of these matrices is a unit: A5a 3 7 2 b, 5 B5a 4 22 3 b, 5 Units in a matrix ring are called invertible matrices. C5a 13 5 0 b. 6 EXAMPLE 8 Definition Let F be a field and M(F) the ring of 2 3 2 matrices with entries in F. If a b A5a b PM(F) and ad 2 bc ≠ 0F, then ad 2 bc is a unit in F by Example 6. c d 1 The computations in Example 7, with replaced by (ad 2 bc)21, show that A is ad 2 bc d 1 ad 2 bc 2 21 2b 1 ad 2 bc 2 21 an invertible matrix [unit in M(F)] with inverse a b. 2c 1 ad 2 bc 2 21 a 1 ad 2 bc 2 21 An element a in a ring R is a zero divisor provided that (1) a ≠ 0R. (2) There exists a nonzero element c in R such that ac 5 0R or ca 5 0R. Note that in requirement (2), the element c is not unique: many elements in the ring may satisfy the equation ax 5 0R or the equation xa 5 0R (Exercise 6). Furthermore, 1140058_CH0003.indd 64 6/16/12 9:43 AM 3.2 Basic Properties of Rings 65 in a noncommutative ring, it is possible to have ac 5 0R and ca ≠ 0R (Exercise 4 in Section 3.1). EXAMPLE 9 Both 2 and 3 are zero divisors in 6 because 2 ? 3 5 0. Similarly, 4 and 9 are zero divisors in 12 because 4 ? 9 5 0. For a zero divisor A in a matrix ring, it is possible to find a matrix C such that AC 5 0 and CA 5 0. E xample 1 0 a b b in M(F) such that ad 2 bc 5 0F is a c d zero divisor because, as you can easily verify, Let F be a field. A nonzero matrix a a a c b d ba d 2c 2b 0F b5a a 0F 0F d b and a 0F 2c 2b a ba a c b 0F b5a d 0F In particular, each of these matrices is a zero divisor in the given ring: A5a 3 9 2 b in M 1  2 , 6 B5a 43 22 28 4 b in M 1  2 , and C 5 a 12 2 0F b. 0F 1 b in M 1 6 2 . 5 EXAMPLE 11 Every integral R domain satisfies Axiom 11: If ab 5 0R, then a 5 0R or b 5 0R. In other words, the product of two nonzero elements cannot be 0. Therefore, An integral domain contains no zero divisors. Finally, we present some useful facts about integral domains and fields. Theorem 3.7 Cancelation is valid in any integral domain R: If a ≠ 0R and ab 5 ac in R, then b 5 c. Cancelation may fail in rings that are not integral domains. In 12, for instance, 2 ? 4 5 2 ? 10, but 4 ≠ 10. Proof of Theorem 3.7 If ab 5 bc, then ab 2 bc 5 0R, so that a(b 2 c) 5 0R. Since a ≠ 0R, we must have b 2 c 5 0R (if not, then a is a zero divisor, contradicting Axiom 11). Therefore, b 5 c. 1140058_CH0003.indd 65 • 6/16/12 9:43 AM 66 Chapter 3 Rings Theorem 3.8 Every field F is an integral domain. Proof •  ince a field is a commutative ring with identity by definition, we need S only show that F satisfies Axiom 11: If ab 5 0F, then a 5 0F or b 5 0F. So suppose that ab 5 0F. If b 5 0F, there is nothing to prove. If b ≠ 0F, then b is a unit (Example 6). Consequently, by the definition of unit and part (1) of Theorem 3.5, a 5 a1F 5 abb2l 5 0F b21 5 0F. So in every case, a 5 0F or b 5 0F. Hence, Axiom 11 holds and F is an integral domain. The converse of Theorem 3.8 is false in general ( is an integral domain that is not a field), but true in the finite case. Theorem 3.9 Every finite integral domain R is a field. Proof • Since R is a commutative ring with identity, we need only show that for each a ≠ 0R, the equation ax 5 1R has a solution. Let a1, a2, . . . , an be the distinct elements of R and suppose at ≠ 0R. To show that at x 5 1R has a solution, consider the products at a1, at a2, at a3, . . . , at an. If ai ≠ aj, then we must have at ai ≠ at aj (because at ai 5 at aj would imply that ai 5 aj by cancelation). Therefore, at a1, at a2, . . . , at an are n distinct elements of R. However, R has exactly n elements all together, and so these must be all the elements of R in some order. In particular, for some j, at aj 5 1R. Therefore, the equation atx 5 1R has a solution and R is a field. ■ Exercises A. 1. Let R be a ring and a, bPR. (a) (a 1 b)(a – b) 5 ? (b) (a 1 b)3 5 ? (c) What are the answers in parts (a) and (b) if R is commutative? 2. Find the inverse of matrices A, B, and C in Example 7. 3. An element e of a ring R is said to be idempotent if e 2 5 e. (a) Find four idempotent elements in the ring M(R). (b) Find all idempotents in 12. 1140058_CH0003.indd 66 6/16/12 9:43 AM 3.2 Basic Properties of Rings 67 4. For each matrix A find a matrix C such that AC 5 0 or CA 5 0: A5 a 6 2 9 b; 3 A5 a 5 22 21 0 b; 4 A5 a 1/ 2 3 1/ 4 b. 3/ 2 5. (a) Show that a ring has only one zero element. [Hint: If there were more than one, how many solutions would the equation 0R 1 x 5 0R have?] (b) Show that a ring R with identity has only one identity element. (c) Can a unit in a ring R with identity have more than one inverse? Why? 6. (a) Suppose A and C are nonzero matrices in M(R) such that AC 5 0. If k is any real number, show that A(kC) 5 0, where kC is the matrix C with every entry multiplied by k. Hence the equation AX 5 0 has infinitely many solutions. (b) If A 5 a 1 3 2 b, find four solutions of the equation AX 5 0. 6 7. Let R be a ring with identity and let S 5 {n1R u nP}. Prove that S is a subring of R. [The definition of na with nP, aPR is on page 62. Also see Exercise 27.] 8. Let R be a ring and b a fixed element of R. Let T 5 {rb u rPR}. Prove that T is a subring of R. a 4b 9. Show that the set S of matrices of the form a b, with a and b real b a numbers is a subring of M(R). 10. Let R and S be rings and consider these subsets of R 3 S: R 5 {(r, 0S) u rPR} and S 5 {(0R, s) u sPS}. (a) If R 5 3 and S 5 5. What are the sets R and S? (b) For any rings R and S, show that R is a subring of R 3 S. (c) For any rings R and S, show that S is a subring of R 3 S. 11. Let R be a ring and m a fixed integer. Let S 5 {rPR u mr 5 0R}. Prove that S is a subring of R. 12. Let a and b be elements of a ring R. (a) Prove that the equation a 1 x 5 b has a unique solution in R. (you must prove that there is a solution and that this solution is the only one.) (b) If R is a ring with identity and a is a unit, prove that the equation ax 5 b has a unique solution in R. 13. Let S and T be subrings of a ring R. In (a) and (b), if the answer is “yes,” prove it. If the answer is “no,” give a counterexample. (a) Is S ∙ T a subring of R? (b) Is S ∙ T a subring of R? 1140058_CH0003.indd 67 6/16/12 9:43 AM 68 Chapter 3 Rings 14. Prove that the only idempotents in an integral domain R are 0R and 1R. (See Exercise 3.) 15. (a)­ If a and b are units in a ring R with identity, prove that ab is a unit whose inverse is (ab)21 5 b21a21. (b)­ Give an example to show that if a and b are units, then a21b21 need not be the multiplicative inverse of ab. 16. Prove or disprove: The set of units in a ring R with identity is a subring of R. 17. If u is a unit in a ring R with identity, prove that u is not a zero divisor. 18. Let a be a nonzero element of a ring R with identity. If the equation ax 5 1R has a solution u and the equation ya 5 1R has a solution y, prove that u 5 y. 19. Let R and S be rings with identity. What are the units in the ring R 3 S? 20. Let R and S be nonzero rings (meaning that each of them contains at least one nonzero element). Show that R 3 S contains zero divisors. 21. Let R be a ring and let a be a nonzero element of R that is not a zero divisor. Prove that cancelation holds for a; that is, prove that (a) If ab 5 ac in R, then b 5 c. (b) If ba 5 ca in R, then b 5 c. 22. (a) If ab is a zero divisor in a ring R, prove that a or b is a zero divisor. (b) If a or b is a zero divisor in a commutative ring R and ab ≠ 0R, prove that ab is a zero divisor. 23. (a) Let R be a ring and a, bPR. Let m and n be nonnegative integers and prove that (i) (m 1 n)a 5 ma 1 na. (ii) m(a 1 b) 5 ma 1 mb. (iii) m(ab) 5 (ma)b 5 a(mb). (iv) (ma)(nb) 5 mn(ab). (b) Do part (a) when m and n are any integers. 24. Let R be a ring and a, bPR. Let m and n be positive integers. (a) Show that a ma n 5 a m1n and (a m)n 5 a mn. (b) Under what conditions is it true that (ab)n 5 a nb n? 25. Let S be a subring of a ring R with identity. (a) If S has an identity, show by example that 1S may not be the same as 1R. (b) If both R and S are integral domains, prove that 1S 5 1R. B. 26. Let S be a subring of a ring R. Prove that 0S 5 0R. [Hint: For aPS, consider the equation a 1 x 5 a.] 27. Let R be a ring with identity and b a fixed element of R and let S 5 {nb u nP}. Is S necessarily a subring of R? [Exercise 7 is the case when b 5 1R.] 1140058_CH0003.indd 68 6/16/12 9:44 AM 3.2 Basic Properties of Rings 69 28. Assume that R 5 {0R, 1R, a, b} is a ring and that a and b are units. Write out the multiplication table of R. 29. Let R be a commutative ring with identity. Prove that R is an integral domain if and only if cancelation holds in R (that is, a ≠ 0R and ab 5 ac in R imply b 5 c). 30. Let R be a commutative ring with identity and bPR. Let T be the subring of all multiples of b (as in Exercise 8). If u is a unit in R and uPT, prove that T 5 R. 31. A Boolean ring is a ring R with identity in which x2 5 x for every xPR. For examples, see Exercises 19 and 44 in Section 3.1. If R is a Boolean ring, prove that (a) a 1 a 5 0R for every aPR, which means that a 5 2a. [Hint: Expand (a 1 a)2.] (b) R is commutative. [Hint: Expand (a 1 b)2.] 32. Let R be a ring without identity. Let T be the set R 3 . Define addition and multiplication in T by these rules: (r, m) 1 (s, n) 5 (r 1 s, m 1 n). (r, m)(s, n) 5 (rs 1 ms 1 nr, mn). (a) Prove that T is a ring with identity. (b) Let R consist of all elements of the form (r, 0) in T. Prove that R is a subring of T. 33. Let R be a ring with identity. If ab and a are units in R, prove that b is a unit. 34. Let F be a field and A 5 a a c b b a matrix in M(F). d (a) Prove that A is invertible if and only if ad 2 bc ≠ 0F. [Hint: Examples 7, 8, and 10 and Exercise 17.] (b) Prove that A is a zero divisor if and only if ad 2 bc 5 0F. 35. Let A 5 a a c b b be a matrix with integer entries. d (a) If ad 2 bc 5 ±1, show that A is invertible in M(). [Hint: Example 7.] (b) If ad 2 bc ≠ 0, 1, or 21, show that A is neither a unit nor a zero divisor in M(). [Hint: Show that A has an inverse in M(R) that is not in M(); see Exercise 5(c). For zero divisors, see Exercise 34(b) and Example 10.] 36. Let R be a commutative ring with identity. Then the set M(R) of 2 3 2 matrices with entries in R) is a ring with identity by Exercise 37 of Section 3.1. a b If A 5 a b PM(R) and ad 2 bc is a unit in R, show that A is invertible in c d 1 M(R). [Hint: Replace by (ad 2 bc)21 in Example 7.] ad 2 bc 37. Let R be a ring with identity and a, bPR. Assume that a is not a zero divisor. Prove that ab 5 1R, if and only if ba 5 1R. [Hint: Note that both ab 5 lR and ba 5 1R imply aba 5 a (why?); use Exercise 21.] 1140058_CH0003.indd 69 6/16/12 9:44 AM 70 Chapter 3 Rings 38. Let R be a ring with identity and a, bPR. Assume that neither a nor b is a zero divisor. If ab is a unit, prove that a and b are units. [Hint: Exercise 21.] 39. (a) If R is a finite commutative ring with identity and aPR, prove that a is either a zero divisor or a unit. [Hint: If a is not a zero divisor, adapt the proof of Theorem 3.8, using Exercise 21.] (b) Is part (a) true if R is infinite? Justify your answer. 40. An element a of a ring is nilpotent if a n 5 0R for some positive integer n. Prove that R has no nonzero nilpotent elements if and only if 0R is the unique solution of the equation x2 5 0R. The following definition is needed for Exercises 41243. Let R be a ring with identity. If there is a smallest positive integer n such that n1R 5 0R, then R is said to have characteristic n. If no such n exists, R is said to have characteristic zero. 41. (a) Show that  has characteristic zero and n has characteristic n. (b) What is the characteristic of 4 3 6? 42. Prove that a finite ring with identity has characteristic n for some n . 0. 43. Let R be a ring with identity of characteristic n . 0. (a) Prove that na 5 0R for every aPR. (b) If R is an integral domain, prove that n is prime. C. 44. (a) Let a and b be nilpotent elements in a commutative ring R (see Exercise 40). Prove that a 1 b and ab are also nilpotent. [You will need the Binomial Theorem from Appendix E.] (b) Let N be the set of all nilpotent elements of R. Show that N is a subring of R. 45. Let R be a ring such that x3 5 x for every xPR. Prove that R is commutative. 46. Let R be a nonzero finite commutative ring with no zero divisors. Prove that R is a field. 3.3 Isomorphisms and Homomorphisms If you were unfamiliar with roman numerals and came across a discussion of integer arithmetic written solely with roman numerals, it might take you some time to realize that this arithmetic was essentially the same as the familiar arithmetic in  except for the labels on the elements. Here is a less trivial example. E xample 1 Consider the subset S 5 {0, 2, 4, 6, 8} of 10. With the addition and multiplication of 10, S is actually a commutative ring, as can be seen from these tables:* *The reason the elements of S are listed in this order will become clear in a moment. 1140058_CH0003.indd 70 6/16/12 9:44 AM 3.3 1 0 6 2 8 4 0 0 6 2 8 4 6 6 2 8 4 0 2 2 8 4 0 6 8 8 4 0 6 2 4 4 0 6 2 8 71 Isomorphisms and Homomorphisms ? 0 6 2 8 4 0 0 0 0 0 0 6 0 6 2 8 4 2 0 2 4 6 8 8 0 8 6 4 2 4 0 4 8 2 6 A careful examination of the tables shows that S is a field with five elements and that the multiplicative identity of this field is the element 6. We claim that S is “essentially the same” as the field 5 except for the labels on the elements. You can see this as follows. Write out addition and multiplication tables for 5.* To avoid any possible confusion with elements of S, denote the elements of 5 by 0, 1, 2, 3, 4. Then relabel the entries in the 5 tables according to this scheme: Relabel 0 as 0,   relabel 1 as 6,   relabel 2 as 2, relabel 3 as 8,   relabel 4 as 4. Look what happens to the addition and multiplication tables for 5: By relabeling the elements of 5, you obtain the addition and multiplication tables for S. Thus the operations in 5 and S work in exactly the same way—the only difference is the way the elements are labeled. As far as ring structure goes, S is just the ring 5 with new labels on the elements. In more technical terms, 5 and S are said to be isomorphic. In general, isomorphic rings are rings that have the same structure, in the sense that the addition and multiplication tables of one are the tables of the other with the elements suitably relabeled, as in Example 1. Although this intuitive idea is adequate for small finite systems, we need a rigorous mathematical definition of isomorphism that agrees with this intuitive idea and is readily applicable to large rings as well. There are two aspects to the intuitive idea that rings R and S are isomorphic: ­relabeling the elements of R and comparing the resulting tables with those of S to verify that they are the same. Relabeling means that every element of R is paired with a unique element of S (its new label). In other words, there is a function f:R S S that *The 5 tables (in congruence class notation) are shown in Example 2 of Section 2.2. 1140058_CH0003.indd 71 6/16/12 9:44 AM 72 Chapter 3 Rings assigns to each rPR its new label f (r)PS. In the preceding example, we used the relabeling function f: 5 S S, given by f(0) 5 0   f(1) 5 6   f (2) 5 2   f(3) 5 8   f(4) 5 4. Such a function must have these additional properties: (i) Distinct elements of R must get distinct new labels: If r 2 r9 in R, then f(r) 2 f (r9) in S. (ii) Every element of S must be the label of some element in R:* For each sPS, there is an rPR such that f(r) 5 s. Statements (i) and (ii) simply say that the function f must be both injective and surjective, that is, f must be a bijection.† In order for a bijection (relabeling scheme) f to be an isomorphism, applying f to the addition and multiplication tables of R must produce the addition and multiplication tables of S. So if a 1 b 5 c in the R-table, we must have f (a) 1 f(b) 5 f (c) in the S-table, as indicated in the diagram: However, since a 1 b 5 c, we must also have f(a 1 b) 5 f (c). Combining this with the fact that f(a) 1 f(b) 5 f(c), we see that f (a 1 b) 5 f(a) 1 f(b). This is the condition that f must satisfy in order for f to change the addition tables of R into those of S. The analogous condition on f for the multiplication tables is f(ab) 5 f(a) f(b). We now can state a formal definition of isomorphism: Definition A ring R is isomorphic to a ring S (in symbols, R ù S) if there is a function f:R S S such that (i) f is injective; (ii) f is surjective; (iii) f(a 1 b) 5 f(a) 1 f(b) and f(ab) 5 f(a) f(b) for all a, bPR. In this case the function f is called an isomorphism. *Otherwise, we couldn’t possibly get the complete tables of S from those of R. † Injective, surjective, and bijective functions are discussed in Appendix B. 1140058_CH0003.indd 72 6/16/12 9:44 AM 3.3 Caution: Isomorphisms and Homomorphisms 73 In order to be an isomorphism, a function must satisfy all three of the conditions in the definition. It is quite possible for a function to satisfy any two of these conditions but not the third; see Exercises 4, 25, and 32. EXAMPLE 2 In Example 12 on page 50, we considered the field K of all 2 × 2 matrices of the form a a 2b b b, a where a and b are real numbers. We claim that K is isomorphic to the field C of complex numbers. To prove this, define a function f:K S C by the rule fa a 2b b b 5 a 1 bi. a To show that f is injective, suppose fa a 2b b r b5fa a 2s s b. r Then by the definition of f, a 1 bi 5 r 1 si in C. By the rules of equality in C, we must have a 5 r and b 5 s. Hence, in K a a 2b b r b5a a 2s s b, r so that f is injective. The function f is surjective because any complex number a 1 bi is the image under f of the matrix a a 2b b b a in K. Finally, for any matrices A and B in K, we must show that f(A 1 B) 5 f(A) 1 f(B) and f(AB) 5 f(A) f(B). We have f ca a 2b b c b1a a 2d d a1c bd 5 f a c 2b 2 d b1d b a1c 5 1a 1 c2 1 1b 1 d2i 5 1 a 1 bi 2 1 1 c 1 di 2 5fa 1140058_CH0003.indd 73 a 2b b c b1fa a 2d d b c 6/16/12 9:44 AM 74 Chapter 3 Rings and f ca a 2b b c ba a 2d Therefore, f is an isomorphism. d ac 2 bd bd 5 f a c 2ad 2 bc ad 1 bc b ac 2 bd 5 1 ac 2 bd 2 1 1 ad 1 bc 2 i 5 1 a 1 bi 2 1 c 1 di 2 5fa a 2b b c bfa a 2d d b. c It is quite possible to relabel the elements of a single ring in such a way that the ring is isomorphic to itself. EXAMPLE 3 Let f:C S C be the complex conjugation map given by f (a 1 bi) 5 a – bi.* The function f satisfies and f 3 1 a 1 bi 2 1 1 c 1 di 2 4 5 f 3 1 a 1 c 2 1 1 b 1 d 2 i 4 5 1 a 1 c 2 2 1 b 1 d 2 i 5 1 a 2 bi 2 1 1 c 2 di 2 5 f 1 a 1 bi 2 1 f 1 c 1 di 2 f 3 1 a 1 bi 2 1 c 1 di 2 4 5 f 3 1 ac 2 bd 2 1 1 ad 1 bc 2 i 4 5 1 ac 2 bd 2 2 1 ad 1 bc 2 i 5 1 a 2 bi 2 1 c 2 di 2 5 f 1 a 1 bi 2 f 1 c 1 di 2 . You can readily verify that f is both injective and surjective (Exercise 17). Therefore f is an isomorphism. EXAMPLE 4 If R is any ring and iR:R S R is the identity map given by iR(r) 5 r, then for any a, bPR iR (a 1 b) 5 a 1 b 5 iR(a) 1 iR(b)   and   iR(ab) 5 ab 5 iR(a)iR(b). Since iR is obviously bijective, it is an isomorphism. Our intuitive notion of isomorphism is symmetric: “R is isomorphic to S” means the same thing as “S is isomorphic to R”. The formal definition of isomorphism is not *The function f has a geometric interpretation in the complex plane, where a 1 bi is identified with the point (a, b): It reflects the plane in the x-axis. 1140058_CH0003.indd 74 6/16/12 9:44 AM 3.3 Isomorphisms and Homomorphisms 75 symmetric, however, since it requires a function from R onto S but no function from S onto R. This apparent asymmetry is easily remedied. If f:R S S is an isomorphism, then f is a bijective function of sets. Therefore, f has an inverse function g:S S R such that g + f 5 iR (the identity function on R) and f + g 5 iS.* It is not hard to verify that the function g is actually an isomorphism (Exercise 29). Thus R ù S implies that S ù R, and symmetry is restored. Homomorphisms Many functions that are not injective or surjective satisfy condition (iii) of the definition of isomorphism. Such functions are given a special name. Definition Let R and S be rings. A function f:R S S is said to be a homomorphism if f(a 1 b) 5 f(a) 1 f(b) and f(ab) 5 f(a)f(b) for all a, bPR. Thus every isomorphism is a homomorphism, but as the following examples show, a homomorphism need not be an isomorphism because a homomorphism may fail to be injective or surjective. EXAMPLE 5 For any rings R and S the zero map z:R S S given by z(r) 5 0S for every rPR is a homomorphism because for any a, bPR z(a 1 b) 5 0S 5 0S 1 0S 5 z(a) 1 z(b) and z(ab) 5 0S 5 0S ? 0S 5 z(a)z(b). When both R and S contain nonzero elements, then the zero map is neither ­injective nor surjective. EXAMPLE 6 The function f:Z S Z6 given by f(a) 5 [a] is a homomorphism because of the way that addition and subtraction are defined in Z6: for any a, bPZ f(a 1 b) 5 [a 1 b] 5 [a] 1 [b] 5 f (a) 1 f (b) and f(ab) 5 [ab] 5 [a][b] 5 f (a)f (b). The homomorphism f is surjective, but not injective (why?). *See Appendix B for details. 1140058_CH0003.indd 75 6/16/12 9:44 AM 76 Chapter 3 Rings EXAMPLE 7 The map g:R S M(R) given by g1r2 5 a 0 2r is a homomorphism because for any r, sPR g1r2 1 g1s2 5 a 5a and g1r2g1s2 5 a 0 2r 0 0 b1a r 2s 0 21r 1 s2 0 2r 0 b r 0 0 b5a s 2r 2 s 0 b 5 g1r 1 s2 r1s 0 0 ba r 2s 0 0 b5a s 2rs 0 b r1s 0 b 5 g 1 rs 2 . rs The homomorphism g is injective but not surjective (Exercise 26). Caution: Not all functions are homomorphisms. The properties f(a 1 b) 5 f(a) 1 f(b)   and   f (ab) 5 f(a)f(b) fail for many functions. For example, if f:R S R given by f(x) 5 x 1 2, then f(3 1 4) 5 f(7) 5 9   but   f (3) 1 f(4) 5 5 1 6 5 11 so that f (3 1 4) 2 f (3) 1 f(4). Similarly, f(3 ? 4) 2 f(3) f(4) because f(3 ? 4) 5 f (12) 5 14,   but   f (3)f(4) 5 5 ? 6 5 30. Theorem 3.10 Let f:R S S be a homomorphism of rings. Then (1) f(0R) 5 0S. (2) f(2a) 5 2f(a) for every aPR. (3) f(a 2 b) 5 f(a) 2 f(b) for all a, bPR. If R is a ring with identity and f is surjective, then (4) S is a ring with identity f(1R). (5) Whenever u is a unit in R, then f(u) is a unit in S and f(u)21 5 f(u21). 1140058_CH0003.indd 76 6/16/12 9:44 AM 3.3 Proof Isomorphisms and Homomorphisms (1) f(0R) 1 f(0R) 5 f (0R 1 0R) f(0R) 1 f(0R) 5 f (0R) f(0R) 1 f(0R) 5 f (0R) 1 0S f(0R) 5 0S • 77 [ f is a homomorphism.] [0R 1 0R 5 0R in R] [ f(0R) 1 0S 5 f(0R) in S] [Subtract f(0R) from both sides.]. (2) First, note that f(a) 1 f(2a) 5 f (a 1 (2a)) [ f is a homomorphism.] 5 f (0R) [a 1 (2a) 5 0R] 5 0S [Part (1)]. Therefore, f(2a) is a solution of the equation f(a) 1 x 5 0S. But the unique solution of this equation is 2f(a) by Theorem 3.3. Hence f(2a) 5 2f(a) by uniqueness. (3) f(a 2 b) 5 f (a 1 (2b)) 5 f (a) 1 f(2b)) 5 f (a) 1 (2f (b)) 5 f (a) 2 f(b) [Definition of subtraction] [ f is a homomorphism.] [Part (2)] [Definition of subtraction]. (4) We shall show that f (1R)PS is the identity element of S. Let s be any element of S. Then since f is surjective, s 5 f (r) for some rPR. Hence, s ? f(lR) 5 f(r)f (lR) 5 f(r ? 1R) 5 f (r) 5 s and, similarly, f(1R) ? s 5 s. Therefore, S has f(1R) as its identity element. (5) Since u is a unit in R, there is an element v in R such that uv 5 1R 5 vu. Hence, by (4) f(u)f(v) 5 f (uv) 5 f(1R) 5 1S. Similarly, vu 5 1R implies that f(v)f(u) 5 1S. Therefore, f (u) is a unit in S, with inverse f(v). In other words, f (u)21 5 f (v). Since v 5 u21, we see that f(u)21 5 f(v) 5 f(u21). If f:R S S is a function, then the image of f is this subset of S: Im f 5 {sPS u s 5 f(r) for some rPR} 5 { f(r) u rPR}. If f is surjective, then Im f 5 S by the definition of surjective. In any case we have: Corollary 3.11 If f:R S S is a homomorphism of rings, then the image of f is a subring of S. Proof 1140058_CH0003.indd 77 • Denote Im f by I. I is nonempty because 0S 5 f(0R)PI by (1) of Theorem 3.10. The definition of homomorphism shows that I is closed under multiplication: If f(a), f(b)PI, then f(a) f(b) 5 f(ab)PI. Similarly, I is closed under subtraction because f (a) 2 f(b) 5 f(a 2 b)PI by Theorem 3.10. Therefore, I is a subring of S by Theorem 3.6. 6/16/12 9:44 AM 78 Chapter 3 Rings Existence of Isomorphisms If you suspect that two rings are isomorphic, there are no hard and fast rules for finding a function that is an isomorphism between them. However the properties of homomorphisms in Theorem 3.10 can sometimes be helpful. EXAMPLE 8 If there is an isomorphism f from Z12 to the ring Z3 3 Z4, then f(1) 5 (1, 1) by part (4) of Theorem 3.10. Since f is a homomorphism, it has to satisfy f(2) 5 f(1 1 1) 5 f(1) 1 f (1) 5 (1, 1) 1 (1, 1) 5 (2, 2) f(3) 5 f(2 1 1) 5 f(2) 1 f (1) 5 (2, 2) 1 (1, 1) 5 (0, 3) f(4) 5 f(3 1 1) 5 f(3) 1 f (1) 5 (0, 3) 1 (1, 1) 5 (1, 0). Continuing in this fashion shows that if f is an isomorphism, then it must be this bijective function: f(1) 5 (1, 1) f (4) 5 (1, 0) f (7) 5 (1, 3) f (10) 5 (1, 2) f(2) 5 (2, 2) f (5) 5 (2, 1) f (8) 5 (2, 0) f (11) 5 (2, 3) f(3) 5 (0, 3) f (6) 5 (0, 2) f (9) 5 (0, 1) f (0) 5 (0, 0). All we have shown up to here is that this bijective function f is the only possible isomorphism. To show that this f actually is an isomorphism, we must verify that it is a homomorphism. This can be done either by writing out the tables (tedious) or by observing that the rule of f can be described this way: f([a]12) 5 ([a]3, [a]4), where [a]12 denotes the congruence class of the integer a in Z12, [a]3 denotes the class of a in Z3, and [a]4 the class of a in Z4. (Verify that this last statement is correct.) Then f([a]12 1 [b]12) 5 f([a 1 b]12) [Definition of addition in Z12] 5 ([a 1 b]3, [a 1 b]4) [Definition of f ] 5 ([a]3 1 [b]3, [a]4 1 [b]4) [Definition of addition in Z3 and Z4] 5 ([a]3, [a]4) 1 ([b]3, [b]4) [Definition of addition in Z3 3 Z4] 5 f([a]12) 1 f ([b]12) [Definition of f ]. An identical argument using multiplication in place of addition shows that f([a]12[b]12) 5 f([a]12)f([b]12). Therefore, f is an isomorphism and Z12 ù Z3 3 Z4. Up to now we have concentrated on showing that various rings are isomorphic, but sometimes it is equally important to demonstrate that two rings are not isomorphic. To do this, you must show that there is no possible function from one to the other satisfying the three conditions of the definition. 1140058_CH0003.indd 78 6/16/12 9:44 AM 3.3 Isomorphisms and Homomorphisms 79 EXAMPLE 9 Z6 is not isomorphic to Z12 or to Z because it is not possible to have a surjective function from a six-element set to a larger set (or an injective one from a larger set to Z6). To show that two infinite rings or two finite rings with the same number of elements are not isomorphic, it is usually best to proceed indirectly. EXAMPLE 10 The rings Z4 and Z2 3 Z2 are not isomorphic. To show this, suppose on the ­contrary that f :Z4 S Z2 3 Z2 is an isomorphism. Then f (0) 5 (0, 0) and f(1) 5 (1, 1) by Theorem 3.10. Consequently, f (2) 5 f (1 1 1) 5 f (1) 1 f (1) 5 (1, 1) 1 (1, 1) 5 (0, 0). Since f is injective and f (0) 5 f (2), we have a contradiction. Therefore, no isomorphism is possible. Suppose that f:R S S is an isomorphism and the elements a, b, c, . . . of R have a particular property. If the elements f (a), f (b), f (c), . . . of S have the same property, then we say that the property is preserved by isomorphism. According to parts (1), (4), and (5) of Theorem 3.10, for example, the property of being the zero element or the identity element or a unit is preserved by isomorphism. A property that is preserved by isomorphism can sometimes be used to prove that two rings are not isomorphic, as in the following examples. EXAMPLE 11 In the ring Z8 the elements 1, 3, 5, and 7 are units by Theorem 2.10. Since being a unit is preserved by isomorphism, any isomorphism from Z8 to another ring with identity will map these four units to four units in the other ring. Consequently, Z8 is not isomorphic to any ring with less than four units. In ­particular, Z8 is not isomorphic to Z4 3 Z2 because there are only two units in this latter ring, namely (1, 1) and (3, 1) as you can readily verify. EXAMPLe 12 None of Q, R, or C is isomorphic to Z because every nonzero element in the fields Q, R, and C is a unit, whereas Z has only two units (1 and 21). EXAMPLE 13 Suppose R is a commutative ring and f:R S S is an isomorphism. Then for any a, bPR, we have ab 5 ba in R. Therefore, in S f(a)f (b) 5 f (ab) 5 f (ba) 5 f (b)f (a). 1140058_CH0003.indd 79 6/16/12 9:44 AM 80 Chapter 3 Rings Hence, S is also commutative because any two elements of S are of the form f(a), f(b) (since f is surjective). In other words, the property of being a commutative ring is preserved by isomorphism. Therefore, no commutative ring can be isomorphic to a noncommutative ring. ■ Exercises A. 1. Let f:6 S 2 3 3 be the bijection given by 0 S (0, 0),   1 S (1, 1),   2 S (0, 2),   3 S (1, 0),   4 S (0, 1),   5 S (1, 2). Use the addition and multiplication tables of Z6 and 2 3 3 to show that f is an isomorphism. 2. Use tables to show that 2 3 2 is isomorphic to the ring R of Exercise 2 in Section 3.1. 3. Let R be a ring and let R* be the subring of R 3 R consisting of all elements of the form (a, a). Show that the function f:R S R* given by f(a) 5 (a, a) is an isomorphism. 4. Let S be the subring {0, 2, 4, 6, 8} of 10 and let 5 5 5 0, 1, 2, 3, 4, 6 (notation as in Example 1). Show that the following bijection from Z5 to S is not an isomorphism: 0 h 0 1 h 2 2 h 4 3 h 6 4 h 8. 5. Prove that the field R of real numbers is isomorphic to the ring of all 2 3 2 0 0 matrices of the form a b, with aPR. [Hint: Consider the function f given 0 a 0 0 by f 1 a 2 5 a b. 4 0 a 6. Let R and S be rings and let R be the subring of R 3 S consisting of all elements of the form (a, 0S). Show that the function f:R S R given by f(a) 5 (a, 0S) is an isomorphism. 7. Prove that R is isomorphic to the ring S of all 2 3 2 matrices of the form a 0 a b, where aPR. 0 a 8. Let  1 "2 2 be as in Exercise 39 of Section 3.1. Prove that the function f: 1 "2 2 S  1 "2 2 given by f 1 a 1 b"2 2 5 a 2 b"2 is an isomorphism. 9. If f : S  is an isomorphism, prove that f is the identity map. [Hint: What are f(1), f(1 1 1), . . . ?] 10. If R is a ring with identity and f:R S S is a homomorphism from R to a ring S, prove that f(1R) is an idempotent in S. [Idempotents were defined in Exercise 3 of Section 3.2.] 1140058_CH0003.indd 80 6/16/12 9:44 AM 3.3 Isomorphisms and Homomorphisms 81 11. State at least one reason why the given function is not a homomorphism. (a) f:R S R and f(x) 5 "x. (b) g:E S E, where E is the ring of even integers and f(x) 5 3x. (c) h:R S R and f(x) 5 2x. a b (d) k: S , where k(0) 5 0 and ka b 5 if a Þ 0. a b 12. Which of the following functions are homomorphisms? (a) f:Z S Z, defined by f (x) 5 2x. (b) f:Z2 S Z2, defined by f (x) 5 2x. 1 (c) g:Q S Q, defined by g 1 x 2 5 2 . x 11 2a 0 (d) h:R S M(R), defined by h 1 a 2 5 a b. a 0 (e) f:Z12 S Z4, defined by f([x]12) 5 [x]4, where [u]n denotes the class of the integer u in Zn. 13. Let R and S be rings. (a) Prove that f:R 3 S S R given by f((r, s)) 5 r is a surjective homomorphism. (b) Prove that g:R 3 S S S given by g((r, s)) 5 s is a surjective homomorphism. (c) If both R and S are nonzero rings, prove that the homomorphisms f and g are not injective. 14. Let f:Z S Z6 be the homomorphism in Example 6. Let K 5 {aPZ 0 f (a) 5 [0]}. Prove that K is a subring of Z. 15. Let f:R S S be a homomorphism of rings. If r is a zero divisor in R, is f (r) a zero divisor in S ? B. 16. Let T, R, and F be the four-element rings whose tables are given in Example 5 of Section 3.1 and in Exercises 2 and 3 of Section 3.1. Show that no two of these rings are isomorphic. 17. Show that the complex conjugation function f:C S C (whose rule is f(a 1 bi) 5 a – bi) is a bijection. 18. Show that the isomorphism of 5 and S in Example 1 is given by the function whose rule is f([x]5) 5 [6x]10 (notation as in Exercise 12(e)). Give a direct proof (without using tables) that this map is a homomorphism. 19. Show that S 5 {0, 4, 8, 12, 16, 20, 24} is a subring of 28. Then prove that the map f :7 S S given by f([x]7) 5 [8x]28 is an isomorphism. 20. Let E be the ring of even integers with the * multiplication defined in Exercise 23 of Section 3.1. Show that the map f :E S  given by f(x) 5 xy2 is an isomorphism. 21. Let * denote the ring of integers with the ! and ( operations defined in Exercise 22 of Section 3.1. Prove that  is isomorphic to *. 1140058_CH0003.indd 81 6/16/12 9:44 AM 82 Chapter 3 Rings 22. Let  denote the ring of integers with the ! and ( operations defined in Exercise 24 of Section 3.1. Prove that  is isomorphic to Z. 23. Let C be the field of Exercise 45 of Section 3.1. Show that C is isomorphic to the field C of complex numbers. 24. (a) Let R be the set R 3 R with the usual coordinatewise addition, as in Theorem 3.1. Define a new multiplication by the rule (a, b)(c, d) 5 (ac, bc). Show that R is a ring. (b) Show that the ring of part (a) is isomorphic to the ring of all matrices in a 0 M(R) of the form a b. b 0 a 0 25. Let L be the ring of all matrices in M(Z) of the form a b. Show that the b c a 0 function f :L S Z given by f a b 5 a is a surjective homomorphism but b c not an isomorphism. 26. Show that the homomorphism g in Example 7 is injective but not surjective. 27. (a) If g:R S S and f:S S T are homomorphisms, show that f + g:R S T is a homomorphism. (b) If f and g are isomorphisms, show that f + g is also an isomorphism. 28. (a) Give an example of a homomorphism f:R S S such that R has an identity but S does not. Does this contradict part (4) of Theorem 3.10? (b) Give an example of a homomorphism f :R S S such that S has an identity but R does not. 29. Let f :R S S be an isomorphism of rings and let g:S S R be the inverse function of f (as defined in Appendix B). Show that g is also an isomorphism. [Hint: To show g(a 1 b) 5 g(a) 1 g(b), consider the images of the left- and right-hand side under f and use the facts that f is a homomorphism and f + g is the identity map.] 30. Let f :R S S be a homomorphism of rings and let K 5 {rPR u f (r) 5 0S}. Prove that K is a subring of R. 31. Let f :R S S be a homomorphism of rings and T a subring of S. Let P 5 {rPR u f(r)PT}. Prove that P is a subring of R. 32. Assume n ; 1 (mod m). Show that the function f :m S mn given by f([x]m) 5 [nx]mn is an injective homomorphism but not an isomorphism when n  2 (notation as in Exercise 12(e)). 33. (a) Let T be the ring of functions from R to R, as in Example 8 of Section 3.1. Let u:T S R be the function defined by u( f ) 5 f (5). Prove that u is a surjective homomorphism. Is u an isomorphism? (b) Is part (a) true if 5 is replaced by any constant cPR? 34. If f :R S S is an isomomorphism of rings, which of the following properties are preserved by this isomorphism? Justify your answers. (a) aPR is a zero divisor. 1140058_CH0003.indd 82 6/16/12 9:44 AM 3.3 Isomorphisms and Homomorphisms 83 (b) aPR is idempotent.* (c) R is an integral domain. 35. Show that the first ring is not isomorphic to the second. (a) E and Z (b) R 3 R 3 R 3 R and M(R) (c) 4 3 14 and 16 (d) Q and R (e)  3 2 and  (f) 4 3 4 and 16 36. (a) If f :R S S is a homomorphism of rings, show that for any rPR and nPZ, f(nr) 5 nf (r). (b) Prove that isomorphic rings with identity have the same characteristic. [See Exercises 41–43 of Section 3.2.] (c) If f :R S S is a homomorphism of rings with identity, is it true that R and S have the same characteristic? 37. (a) Assume that e is a nonzero idempotent in a ring R and that e is not a zero divisor.* Prove that e is the identity element of R. [Hint: e2 5 e (Why?). If aPR, multiply both sides of e2 5 e by a.] (b) Let S be a ring with identity and T a ring with no zero divisors. Assume that f :S S T is a nonzero homomorphism of rings (meaning that at least one element of S is not mapped to 0T). Prove that f (1S) is the identity element of T. [Hint: Show that f (1S) satisfies the hypotheses of part (a).] 38. Let F be a field and f :F S R a homomorphism of rings. (a) If there is a nonzero element c of F such that f(c) 5 0R, prove that f is the zero homorphism (that is, f (x) 5 0R for every xPF ). [Hint: c21 exists (Why?). If xPF, consider f(xcc21).] (b) Prove that f is either injective or the zero homomorphism. [Hint: If f is not the zero homomorphism and f(a) 5 f(b), then f(a 2 b) 5 0R.] 39. Let R be a ring without identity. Let T be the ring with identity of Exercise 32 in Section 3.2. Show that R is isomorphic to the subring R of T. Thus, if R is identified with R, then R is a subring of a ring with identity. C. 40. For each positive integer k, let kZ denote the ring of all integer multiples of k (see Exercise 6 of Section 3.1). Prove that if m 2 n, then mZ is not isomorphic to nZ. 41. Let m, nPZ with (m, n) 5 1 and let f :mn S m 3 n be the function given by f([a]mn) 5 ([a]m, [a]n). (Notation as in Exercise 12(e). Example 8 is the case m 5 3, n 5 4.) (a) Show that the map f is well defined, that is, show that if [a]mn 5 [b]mn in mn, then [a]m 5 [b]m in m and [a]n 5 [b]n in n. (b) Prove that f is an isomorphism. [Hint: Adapt the proof in Example 8; the difference is that proving f is a bijection takes more work here.] 42. If (m, n) 2 1, prove that mn is not isomorphic to m 3 n. *Idempotents are defined in Exercise 3 of Section 3.2. 1140058_CH0003.indd 83 6/16/12 9:44 AM 1140058_CH0003.indd 84 6/16/12 9:44 AM