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Ring TheoryMassoud Malek

♣ Rings . A ring is a non-empty set R with two binary operations ( + , · ) , called addition and

multiplication, respectively satisfying :

Axiom 1. Closure ( + ) : ∀x, y ∈ R , x + y ∈ R .

Axiom 2. Commutative ( + ) : For every x, y ∈ R , x + y = y + x .

Axiom 3. Associative ( + ) : ∀x, y, z ∈ R , x + (y + z) = (x + y) + z .

Axiom 4. Neutral ( + ) : ∃ θ ∈ R , such that ∀x ∈ R, x + θ = θ + x = x . We denote θ by 0 .

Axiom 5. Inverse ( + ) : ∀ a ∈ R , the equation a+ x = θ has a solution.

Note that (R , + ) is an Abelian group.

Axiom 6. Closure ( ·) : ∀x, y ∈ R , x · y ∈ R

Axiom 7. Associative ( ·) : ∀x, y, z ∈ R, x · (y · z) = (x · y) · z .

Axiom 8. Ditributive ( + , ·) :

∀x, y, z ∈ R, x · (y + z) = (x · y) + (x · z) and (x + y) · z = (x · z) + (y · z) .

Definitions 1. A Ring is called a commutative ring, if it satisfies :

Axiom 9. Commutative ( · ) : ∀ x , y ∈ R , x · y = y · x

A ring may have an identity for the multiplication; satisfying :

Axiom 10. Identity ( · ) : ∃ ! e ∈ R , such that ∀x ∈ R, x · e = e · x = x . We denote e by 1 .

A ring with an identity, is called a ring with unity or a unital ring.

If a and b are nonzero but a b = 0 , we say that a is left zero divisor and b a right zero divisor.

For example,

if A =

[1 0

0 0

]and B =

[0 0

1 0

]. Then AB =

[0 0

0 0

], but neither A nor B is a zero matrix.

An Integral Domain is a commutative ring with identity with no zero divisors, i.e., if it satisfies the

following axiom:

Axiom 11. No zero divisors ( ·) : If for some x, y ∈ R , x · y = θ, then x = θ or y = θ .

The characteristic of a ring R , often denoted char (R ) , is defined to be the smallest positive number

n > 1 such that for every a ∈ R ,

a + a + a + · · · + a︸︷︷︸ = na = θ .

n times

If there is no such an n , then we say that R is of characteristic zero, and we write char(R) = 0 .

A ring, where R∗ = R−{θ} is group, i.e., R∗ satisfying:

An element u in a ring R with identity is called a unit, if there exists an element v ∈ R such that

u v = v u = 1 . In this case the element v is called the (multiplicative) inverse of u and is denoted

http://www.mcs.csueastbay.edu/~malek/Class/Ring.pdf

Rings Ring Theory 2

by u−1.

Axiom 12. Inverse ( ·) : ∀x ∈ R∗, ∃ ! x−1 ∈ R∗, such that xx−1 = x−1 x = e is called a Division

Ring or a Skew Field.

Finally, a commutative division ring is called a Field.

Rings

↙ ↘Commutative Rings Rings with Identity

↓ ↙ ↓Integral Domains Division Rings

↘ ↙Fields

Lemma 1. ∀ a ∈ R , the equation a+ x = θ has a unique solution.

Proof. We know that a + x = θ has at least one solution u by Axiom 5. If v is also a solution,

then a+ u = θ and a+ v = θ , so that

v = θ + v = ( a+ u ) + v = (u+ a ) + v = u+ ( a+ v ) = u+ θ = u

Therefore, u is the only solution.

In the ring Z6 , the solution of the equation 2 + x = 0 is 4, and so in this ring −2 = 4 .

Similarly, − 9 = 3 in Z12 because 3 is the solution of 9 + x = 0 .

Subtraction in a ring is now defined by the rule :

b− a means b+ (− a ) .

Note that by using the binary operation addition ( + ) , we just defined another operation called

subtraction (− ) .

Intuitively, in a ring we can do addition, subtraction, and multiplication without leaving the set,

while in a field (or skew field) we can do division (except by zero) as well. Hence Z is not a field.

The relationship among rings, integral domains, division rings, and Fields is shown in the next table :

Lemma 2. If a+ b = a+ c in a ring R , then b = c .

Proof. By adding − a to both sides of the equation a+ b = a+ c and then using associativity and

negatives show that

− a+ ( a+ b ) = − a+ ( a+ c )

(− a+ a ) + b = (− a+ a ) + c

θ + b = θ + c

b = c .

Theorem 1. The characteristic of a finite ring is not zero.

Proof. Let a ∈ R , then since the ring is closed under addition,

a , (a + a) = 2 a , (a + a + a) = 3 a, · · · , (a + a + a · · · + a) = na , . . .

Rings Ring Theory 3

are also in R . For a finite ring, the sequence of sums can not be infinite, thus it must have repetitions;

so for some integer k , k a = θ.

Theorem 2. In an integral domain R , if char (R ) 6= 0 , then it must be prime.

Proof. Suppose char (R ) = n = p q , where p and q are positive integers greater than one, then

( p 1 ) ( q 1 ) = n× 1 = 0 ; so either p 1 = 0 or q 1 = 0 , a contradiction with the minimality of n .

Lemma 3. (Cancelation) In an integral domain, if a 6= 0 and a b = a c , then b = c .

Proof. From a b = a c , we have a ( b− c ) = 0 . Since a 6= 0 , we must have b = c .

Remark 1. The equation x 2 − 5x+ 6 = 0 has more solutions in Z 12 than in Z . Here is how :

The fact that Z 12 is not an integral domain, implies that

0 = 2× 6 = 6× 2 = 3× 4 = 4× 3 = 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 .

By factoring out the equation x 2 − 5x+ 6 = 0 into (x− 2 ) (x− 3 ) = 0 , we conclude that x1 = 2

and x2 = 3 are solutions to the equations in both Z and Z 12 . Notice that (x−3 ) and (x−2 ) are

consecutive numbers. Also note that 3 × 4 = 0 and 8 × 9 = 0 in Z 12 . We conclude that x 3 = 6

and x 4 = 11 are also the solutions to the equation x 2 − 5x+ 6 = 0 .

Theorem 3. A finite ring R of order n ≥ 2 , with no zero divisors has unity.

Proof. Let S = { a1 , a2 , a3, . . . , an−1 } be the set of all non-zero elements of the ring and let

T = { a1 a1 , a2 a1 , a3 a1 , . . . , an−1 a1 } . T must be equal to S , if not, then for some r and s ,

ar a1 = as a1; hence ( ar − as ) a1 = θ . Since the ring has no zero divisors, ar − as = θ or ar = as.

Thus T = S and a1 = ak a1, for some integer k . From aka1 = (ak)2a1, we conclude that a 2

k = ak ;

this element is the unity of R .

Theorem 4. Let R be a ring with more than one element. Suppose for all x ∈ R , there exists a

unique y ∈ R such that x = x y x . Then

(i) R has no zero divisors,

(ii) if x 6= 0 and x = x y x , then y = y x y for all x, y ∈ R ,

(iii) R has an identity,

(iv) R is a division ring.

Proof. (i) Let x be a nonzero element of R and x z = 0 for some z ∈ R . Now by the hypothesis,

there exists a unique y ∈ R such that x = x y x . Thus,

x ( y − z )x = x y x− x z x = x y x .

Hence, by the uniqueness of y , y − z = y , so z = 0 . This proves that R has no zero divisors.

(ii) Let x 6= 0 and x y x = x . Then

x ( y − y x y ) = x y − x y x y = x y − x y = 0.

Because R has no zero divisors and x 6= 0 , so y − y x y = 0 , thus y x y = y .

Rings Ring Theory 4

(iii) Let 0 6= x ∈ R . Then there exists a unique y ∈ R such that x y x = x . Let e = y x . If

e = 0 , then x = x y x = 0 , which is a contradiction. Therefore, e 6= 0 . Also,

e2 = y x y x = y (x y x ) = y x = e .

Let z ∈ R . Then

( z e− z ) e = z e2 − z e = z e− z e = 0 .

Thus, by (i), either z e− z = 0 or z e = z . Similarly, e ( e z − z ) = 0 implies that e z = z . Hence,

e is the identity of R .

(iv) By (iii), R contains an identity element e . To show R is a division ring, it remains to be shown

that every nonzero element of R has an inverse in R . Let x be a nonzero element in R . Then

there exists a unique y ∈ R such that x y x = x . Thus, x y x = x e , i.e., x ( y x− e ) = 0 . Because

R has no zero divisors and x 6= 0 , y x− e = 0 , so y x = e . Similarly, x y x = e x implies x y = e .

Therefore, x y = e = y x . Hence, R is a division ring.

Exercises.

1. Let R be a ring and a, , b ∈ R . Then find ( a+ b ) (a− b) and ( a+ b ) 3.

2. An element e of a ring R is said to be idempotent if e 2 = e . A nilpotent element z satisfies

zn , for some N∗ .

(a) Find all idempotent and nilpotent elements in the ring of 3× 3 real matrices.

(b) Find all idempotents and nilpotent elements in Z 24 .

3. Find all 3× 3 real matrices with the property A3 = A .

4. Show that a ring has only one zero element.

5. If a b is a zero divisor in a ring R , prove that a or b is a zero divisor.

6. Let a , b ∈ R and let n be a positive integer. Under what conditions ( a b)n = an bn ?

7. Show that M2 (Z2 ) , the set of all 2× 2 matrices with entries in Z2 is a non-commutative ring

with identity. What is the order of this ring?

8. Prove that in a division ring, if a2 = a , then a = 1 .

9. Show that in an integral domain the only idempotent elements are 0 and 1.

10. Let R be the set of all 3× 3 matrices of the form

a 0 0

b a 0

c b a

, where a , b , c ∈ Z

(a) Show that R is commutative. (b) Find all the unit and nilpotent elements of R .

The Pigeon-Hole Principle . If n pigeons are placed in m holes, and if n > m

then some holes receives at least two pigeons.

Did you know that if you pick five numbers from the integers 1 to 8, then two of them must add up

to nine.

We immediately make use of this idea in proving our next result.

Theorem 5. Every finite integral domain is a field.

Proof. We need to show that every nonzero element of D has an inverse. For each a ∈ D ∗ , define

Rings Ring Theory 5

the map

λ a : D ∗ 7→ D ∗ with λ a ( d ) = a d ∀ d ∈ D ∗.

The map λ a is injective because D is an integral domain. If D is finite, then by the Pigeon-Hole

Principle, the map is surjective as well, so for some d ∈ D ∗, λ a ( d ) = 1 . Therefore, d a has a

left inverse. Since D is commutative, d must also be a right inverse for a . Consequently, D is a

field.

Corollary 1. If p is a prime number, then Jp the set of integers modulo p is a field.

Proof. Since Jp has only a finite number of elements, we only need to show that it is an integral

domain. Suppose for some a , b ∈ Jp , a b = 0 . Then p must divide the ordinary integer, a b . Since

p is prime, it must divide either a or b . But either a ≡ 0 (mod p) or b ≡ 0 (mod p) , hence in

Jp , one of these is zero.

Definitions 2. Given a ring R , let R o (read as R opposite), be the same set R . Define addition

( + ) and multiplication ( * ) on R o as

a + b = a + b and a ∗ b = b · a, ∀a , b ∈ R .

Under these operations , R o is a ring called the ring opposite to R or the opposite ring of R . It

is clear that (R o ) o = R .

Remark 2.

(1) R = R o if and only if R is commutative.

(2) R has unity if and only if R o has unity. In fact the unity of R is the same as that of R o .

♥ Examples.

I. The most familiar examples of rings are the integers Z and the sets Q , R , and C of rational,

real, and complex numbers, respectively, all with respect to the usual addition and multiplication.

All of these are commutative rings, since multiplication is commutative. The set of integers is an

integral domain and the other three are actually fields.

II. We can build new rings from old

1. For n > 1 , Zn is a commutative ring with unity and characteristic n . If n = p q, then Zncontains zero divisors; for example, p · q = 0 . But if p is a prime number, then Zp is a field.

2. The set

Zn1 × Zn2 × Zn3 × · · · × Znr

is a ring with the identity e = (1 , 1 , 1 , . . . , 1 ) .

3. The complex plane C may be defined as R [ i ] = {a + b i : a , b ∈ R}. By analogy with the

definition of the complex plane, we define the set

Z [ i ] = { a + b i : a , b ∈ Z }, Z [√

2 ] ={a + b

√2 : a , b ∈ Z

}, Q [

√2 ] =

{a + b

√2 : a , b ∈ Q

}.

These sets are integral domains under the usual addition and multiplication. The set Z [ i ] is called

the ring of Gaussian integers.

The ring Z [√

2 ,√

3 ] is equal to{a + b

√2 + c

√3 + d

√6 : a , b , c , d ∈ Z

}. To see why, observe

that the fact that rings are closed under multiplication and addition and√

2√

3 =√

6 , we must

have every element of the form a + b√

2 + c√

3 + d√

6 in the ring.

Rings Ring Theory 6

The ring of Gaussian integers modulo 3

Z 3 [ i ] = { a + b i : a , b ∈ Z 3 } = { 0 , 1 , 2 , i , 1 + i , 2 + i , 2 i , 1 + 2 i , 2 + 2 i }

is a field. Here is its multiplicative table :

Z3 [ i ] 1 2 i 1 + i 2 + i 2 i 1 + 2 i 2 + 2 i

1 1 2 i 1 + i 2 + i 2 i 1 + 2 i 2 + 2 i

2 2 1 2 i 2 + 2 i 1 + 2 i i 2 + i 1 + i

i i 2 i 2 2 + i 2 + 2 i 1 1 + i 1 + 2 i

1 + i 1 + i 2 + 2 i 2 + i 2 i 1 1 + 2 i 2 i

2 + i 2 + i 1 + 2 i 2 + 2 i 1 i 1 + i 2 i 2

2 i 2 i i 1 1 + 2 i 1 + i 2 2 + 2 i 2 + i

1 + 2 i 1 + 2 i 2 + i 1 + i 2 2 i 2 + 2 i i 1 + 2 i

2 + 2 i 2 + 2 i 1 + i 1 + 2 i i 2 2 + i 1 2 i

What are the additive and multiplicative tables of Z2 [ i ] ?

4. If R is a ring, then the set R [X ] of all polynomials with coefficients in R , under the usual

addition and multiplication of polynomials is a ring.

5. Mn (R ) , the set of n× n (n > 1 ) real matrices is a non-commutative ring with identity.

6. Consider the commutative rings

S1 =

{[a 0

0 0

]: a ∈ R

}and S2 =

{[0 0

b 0

]: b ∈ R

}.

The ring S1 is an integral domain but every element of S2 is a zero divisor.

7. We can also build new rings from old by taking external direct sums and defining both addition

and multiplication component wise. Note that the unity element of the direct sum

R = R l ⊕R 2 ⊕ · · · ⊕ Rn

is the n-tuple ( 1 , . . . , 1 , 1 ) , where the k-th entry is the unity of R .

III. The set F (R ) of all functions f from R to R . is a commutative ring.

IV. A Boolean ring [ in honor of the English mathematician George Boole ( 1815− 1864 ) ] is a ring

in which every element is idempotent; that is, a 2 = a for all a . The set

Z 2 × Z 2 × · · · × Z 2

is a boolean ring.

Lemma 4. Let B be a Boolean ring. Then char (B ) = 2 .

Proof. Let x ∈ B . Then x 2 = x and (x + x ) 2 = (x + x ) which implies that x 2 + 2x + x 2 =

x + x ; hence 2x = 0 . Thus, char (B ) = 2 . It follows then that x = −x for all x ∈ B .

Lemma 5. A Boolean ring is commutative.

Proof. Let x , y ∈ B . Since B is Boolean, then x 2 = x and y 2 = y . Now,

x + y = (x + y )2 = x 2 + x y + y x + y 2 = x + x y + y x + y.

Rings Ring Theory 7

Therefore, x y = −y x . Using the previous lemma, we have x = −x for every x in B . Hence,

x y = y x .

A boolean field is isomorphic to Z 2 .

Example. Let X be any non-empty set and P (X ) be the power set of X , i.e., set of all subsets

of X . Define addition and multiplication on P (X ) as follows :

Addition is the symmetric difference of sets, namely,

A B = A∆B = (A ∪ B ) − (A ∩ B ) = (A− B ) ∪ (B − A ) .

Multiplication is the intersection of sets, i.e., A · B = A ∩ B .

Under these operations, P (X ) is a commutative ring with unity and it is a Boolean ring..

V. A quaternion is a number in the form:

a1 + b i + c j + d k,

where a , b , c , d ∈ R and 1, i, j, k are entities related to each other in the following way:

i 2 = j 2 = k 2 = i j k = −1 , i j = − j i = k, j k = −k j = i, k i = − i k = i .

Quaternions were first discovered by William Rowan Hamilton in 1843. The set of all quaternions

is usually denoted by H in honor of its discoverer. This set with the binary operations, addition

and multiplication forms a division ring, where the operation are defined as follows:

Let

x1 = a1 1 + b1 i + c1 j + d1 k and x2 = a2 2 + b2 i + c2 j + d2 k

be two quaternions, then

x1 + x2 = (a1 + a2)1 + (b1 + b2)i + (c1 + c2)j + (d1 + d2)k;

andx1 x2 = (a1a2 − b1b2 − c1c2 − d1d2)1 + (a1b2 + b1a2 + c1d2 − d1c2)i

+ (a1c2 − b1d2 + c1a2 + d1b2)j + (a1d2 + b1c2 − c1b2 + d1a2)k.

For example, if x = 21− i + 3j + k and y = 31 + 2i− j− k , then x + y = 51 + i + 2j and

x y=(6 + 2 + 3 + 1)1 + (4−3−3 + 1)i + (−2−1 + 9 + 2)j + (−2 + 1−6 + 3)k=121−i + 8j−4k .

In fact H is a division ring; the inverse of

a + b i + c j + dk is ( a 2 + b 2 + c 2 + d 2 )−1 ( a− b i− c j− dk ) .

Construction from Complex Pairs. A quaternion can be defined as an ordered pair (x , , y ) ,

where x , , y ∈ C , on which the operations of addition and multiplication are defined as follows:

Let x1 , x2, y1 , y2 be complex numbers. Then

(x1 , y1) + (x2, y2) = (x1 + x2, y1 + y2)

(x1 , y1 ) (x2, y2 ) := (x1 x2 − y2 y1, x1 y2 + y1 x2 ),

Rings Ring Theory 8

where x1 and y1 are the complex conjugates of x1 and y1 respectively.

Let x = ( 2− i , 3 + i ) and y = ( 2 + 2 i , −1− i ) , then

x y = ( 2− i, 3 + i ) ( 2 + 2 i,−1− i )

= ( ( 2− i ) ( 3 + 2 i )− (−1− i ) ( 3− i ) , ( 2 + i ) (−1− i ) + ( 3 + i ) ( 3 + 2 i ) )

= ( 12− i , 8− 4 i ) .

Construction from Complex Matrices. The quaternions can be represented as:

1→

[1 0

0 1

], i→

[i 0

0 −i

], j→

[0 1

−1 0

], k→

[0 i

i 0

].

It is shown that a general element x of H has the form:[a + bi c + di

−c + di a− bi

]

It is much easier to use the matrix form of quaternions to perform the multiplication. For example,

the product of the quaternions x = 21− i + 3j + k and y = 31 + 2i− j− k is obtain as follows:[2− i 3 + i

−3 + i 2 + i

] [3 + 2i −1− i1− i 3− 2i

]=

[12− i 8− 4i

−8− 4i 12− bi

]

The inverse of the matrix

[2− i 3 + i

−3 + i 2 + i

]gives us x−1 .

x−1=(21− i + 3j + k)−1→

[2− i 3 + i

−3 + i 2 + i

]−1

=1

15

[2 + i −3− i3− i 2− i

]→ x−1=

1

15(21 + i− 3j− k).

Consider the products

( a1 + b1 i+ c1 j + d1 k ) ( a2 + b2 i+ c2 j + d2 k ) and ( 7 + 2 i− 3 j + 4 k ) ( 4 + 5 i+ 2 j + 3 k ).

One may use the following method to obtain the result :

1 i j k

a2 b2 c2 d2a1 b1 c1 d1

step 1 − d1 d 12 d1 c

22 d1 b

32 d1 a

42

step 2 − c1 c 22 c1 d

12 c1 a

42 − c1 b 3

1

step 3 − b1 b 32 b1 a

42 − b1 d 1

2 b1 c22

step 4 a1 a42 a1 b

32 a1 c

22 a1 d

12

S1 S2 S3 S4

1 i j k

4 5 2 3

7 2 − 3 4

step 1 − 12 1 − 8 2 20 3 16 4

step 2 6 2 − 9 1 − 12 4 15 3

step 3 − 10 3 8 4 − 6 1 4 2

step 4 28 4 35 3 14 2 21 1

12 26 16 56

Thus ( a1 + b1 i+ c1 j + d1 k ) ( a2 + b2 i+ c2 j + d2 k ) = S1 + S2 i+ S3 j + S4 k

and ( 7 + 2 i− 3 j + 4 k ) ( 4 + 5 i+ 2 j + 3 k ) = 12 + 26 i+ 16 j + 56 k .

† The powers indicate the orders of the products.

Rings Ring Theory 9

Proposition 1. If a b = b a in a ring, then ( a + b )n =n∑k=1

(n

k

)an−k b k is valid.

Proof. In ( a+ b )n = ( a+ b ) ( a+ b ) . . . ( a+ b ) , we may use the fact that a b = b a , in order to

obtain the desired result.

Properties of Rings. Let a , b , and c belong to a ring R . Then

1. a 0 = 0 a = 0 .

2. a ( 2 b ) = ( 2 a ) b = 2 ( a b ) .

3. a ( b − c ) = a b − a c and ( b = c ) a = b a − c a .

Furthermore, if R has a unity element 1 , then

4. (−1 ) a = − a.

5. (−1 ) (−1 ) = 1 .

♣ Subrings. A subgroup S of (R, + ) which is closed under multiplication is called a subring of

(R, + , · ). Clearly, a ring is a subring of itself. The subset { 0 } of R is a subring and is called the

trivial subring. Any subring different from the trivial subring and the ring itself is called a proper

subring.

Here is a formal definition of a subring :

Let R be a ring and S a subset of R . Then S is a subring of R if and only if the following

conditions are satisfied.

1. S 6= ∅ ;

2. r s ∈ S for all r , s ∈ S ;

3. r − s ∈ S for all r , s ∈ S .

Although it is true for rings, as it is for groups, that the multiplicative identity is unique, it is not

generally true that a subring S of a ring R has the same unity as the whole ring. Of course, if the

element 1 of R happens to be in S , then it is necessarily the unity of S , and in this case we say

that S is a unitary subring of R .

Consider

S =

{[a 0

0 0

]: a ∈ R

}a subring of M =

{[a c

b d

]: a , b , c , d ∈ R

}.

The matrix I 2 =

[1 0

0 1

]is the unit of the ringM, but the matrix

[1 0

0 0

]is a unity for the subring

S . The only identity of the subring T =

{[a 0

b c

]: a , b , c ∈ R

}is I 2 .

Theorem 6. Let S ⊂ R such that ∀x , y ∈ S, x−y ∈ S , then S is a subring of R , if it is closed

under multiplication.

Proof. By choosing x for y, we show that x− x = θ ∈ S . Now if x = θ, then −y must also in S .

Finally, since −(−(y) ) = y , we conclude that S is closed under addition.

♥ Examples.

1. The set of even numbers 2Z is a subring of Z with no multiplicative identity.

2. The set of Gaussian integers Z [ i ] is a subring of the complex numbers C .

Rings Ring Theory 10

3. The set S = { 0̄ , 2̄ , 4̄} is a subring of the ring Z 6 . Note that although 1̄ is the unity in Z 6 ,

but 4̄ is the unity in S .

4. The set of n× n upper triangular (or lower triangular) matrices is a subring of Mn(R ) .

5. The set of n× n diagonal matrices is a commutative subring of Mn(R ) .

6. The sets

S1 =

{[a 0

b 0

]: a, b ∈ R

}, S2 =

{[0 c

0 d

]: c, d ∈ R

},

T1 =

{[a b

0 0

]: a, b ∈ R

}, T2 =

{[0 0

c d

]: c, d ∈ R

}.

are subrings of M2 (R ) .

7. The set C(R ) of all continuous functions f from R to R is a subring of F(R ) .

8. From the fact that the Quaternion ring H can be expressed in matrix form, where each quaternion

is represented as an element of M2 (C ) . It follows that H is a subring of the matrix space M2 (C ) .

Exercises.

1. The ring R = {0 , 2 , 4 , 6 , 8 } under addition and multiplication modulo 10 has a unity. Find

it.

2. Solve the equation x 2 − 7x+ 10 = 0 in Z 12

3. Let R be a ring. The center of R is the set

Z ( R ) = { z ∈ R | a z = z a , ∀ a ∈ R} .

Prove that the center of a ring is a subring.

4. Show that 4 divides 2 in Z 6 ; 3 divides 7 in Z 8 ; and 9 divides 21 in Z 15 .

5. Give an example of ring elements a and b with the properties that a b = 0 but b a 6= 0 .

6. Suppose that R is a ring such that x 3 = x for all x ∈ R . Prove that 6x = 0 for all x ∈ R .

7. Is the set T =

{[0 a

b 0

]: a , b ∈ R

}a commutative ring ?

C

R

Z [ i ]Q [√

2 ]

Q

Z

5Z 2Z 3Z 7Z

10Z

11Z 13Z

4Z 6Z 9Z

8Z 12Z 18Z 14Z20Z


A partial lattice of subrings of C

♣ Ideals. A left ideal I of a ring R is a subring of R which absorbs all the elements of the ring

from the left side; i.e.,

∀ x ∈ R, ∀ a ∈ I, x a ∈ I

A right ideal J of a ring R is a subring of R which absorbs all the elements of the ring from the

right side; i.e.,

∀ x ∈ R, ∀ a ∈ J , a x ∈ J

An Ideal is both a right ideal and a left ideal.

Note that the trivial subring {θ} is an ideal, since ∀ x ∈ R , x θ = θ x = θ . This ideal is called the

trivial ideal. Clearly the ring R is an ideal of itself. A non-trivial ideal which is not the ring itself

is said to be proper. For n ∈ N∗, nZ is an ideal of Z .

The subring Z of the rational numbers is not an ideal in Q because Z fails to have the absorption

property. For instance,1

2∈ Q and 3 ∈ Z , but their product,

3

2is not in Z .

Theorem 7. A nonempty subset I of a ring R is an ideal if and only if it has these properties :

(i) If a, , b ∈ I , then a− b ∈ I ;

(ii) if r ∈ R and a ∈ I , then r a ∈ I and a r ∈ I .

Proof. Every ideal certainly has these two properties.

Conversely, suppose I has properties (i) and (ii). Then I absorbs products by (ii), so we need

only verify that I is a subring. Property (i) states that I is closed under subtraction. Since I is

a subset of R , the product of any two elements of I must be in I by (ii). In other words, I is

closed under multiplication. Therefore, I is a subring of R

Lemma 6. Let I be an ideal (left, right, or two-sided) of a ring R with unity e. If e ∈ I , then

I = R .

Proof. We only show for the left ideal. If e ∈ I , then ∀x ∈ R , we have x = x e ∈ I ; hence

R ⊂ I . Thus I = R.

Theorem 8. A division ring R has no proper ideal.

Proof. Suppose I is a proper ideal, then there is at least a non-zero x ∈ I such that x ∈ R . Since

R is a division ring, x−1 exists and e = x−1x ∈ I . Then by the previous lemma, I = R .

Definitions 3. A proper ideal I is a prime ideal, if a b ∈ I , then either a ∈ I or b ∈ I . For

example, the ideal pZ of Z , where p is a prime number, is a prime ideal.

A proper ideal I is a primary ideal, if whenever a b ∈ I , then at least one of a an bn is in I , for

some natural number n . Every prime ideal is primary, but not conversely. For example, in the ring

of integers Z , the set { pn | p prime and n ∈ N } is a primary ideal.

A proper ideal I is called semi-prime if, whenever J n ⊂ I for an ideal J of R and some positive

integer, then J ⊂ I .

For a commutative ring R , a proper ideal I is a semi-prime ideal if I satisfies either of the

following equivalent conditions :

(i) If x k ∈ I for some positive integer k and element x ∈ R , then x ∈ I .

(ii) If y ∈ R but not in I , all positive integer powers of y are not in I .


For example, in the ring of integers, the semi-prime ideals are the zero ideal, along with those ideals

of the form nZ where n is a square-free integer. So, 6Z is a semi-prime ideal of Z , but 12Z is

not.

An ideal I is maximal, if it is proper and there is no proper ideal J which is a strictly greater set

than I . The ideal pZ of Z , where p is a prime number, is a maximal ideal.

A non-zero ideal is called minimal, if it contains no other non-zero ideal.

The radical of an ideal I in a commutative ring R , denoted by Rad (I ) or√I , is defined as

√I = { r ∈ R | r n ∈ I for some positive integer n } .

For example the radical of 12Z is 6Z . The radical of an ideal is always at least as big as the

original ideal.

A left principal ideal I of R is a subset of R , generated by a single element a , of the form

R a = { r a | r ∈ R} .

A right principal ideal J is a subset of R , generated by a single element a , of the form

aR = { a r | r ∈ R} .

A two-sided principal ideal is a subset of R of the form

R aR = { r 1 a s 1 + r 2 a s 2 + · · ·+ rn a sn | r 1 , s 1 , r 2 , s 2 , . . . , rn , sn ∈ R} .

If R is a commutative ring, then the above three notions are all the same. In that case, it is common

to write the ideal generated by the element a , as 〈 a 〉 .

A ring in which every ideal is principal is called a principal ring. A ring like Z , whose ideals are

all principal is a principal ring.

A principal ideal domain (PID) is an integral domain that is principal. The distinction is that a

principal ring may have zero divisors whereas a principal ideal domain cannot. The ring of Gaussian

integers Z [ i ] is a PID .

In all unital rings, maximal ideals are prime. In principal ideal domains a near converse holds:

every nonzero prime ideal is maximal.

♥ Examples.

1. Although Z is a subring of the field Q , but it is not an ideal of Q, because a division ring or a

field has no proper ideal.

2. In the ring C(R ) of all continuous functions f from R to R , the set of all f with f(a) = 0 for

some real number a , is an ideal of C(R ) .

3. Let

IL =

{[a 0

b 0

]: a, b ∈ R

}and JL =

{[0 c

0 d

]: c, d ∈ R

}.

Then both IL and IL are principal left ideals of the ring M2 (R ) , generated by the matrices

A =

[1 0

0 0

]and B =

[0 0

0 1

]respectively.


IL =

{[a 0

b 0

]: a, b ∈ R

}= M2 (R)A =

{[a c

b d

][1 0

0 0

]: a, b, c, d ∈ R

};

JL =

{[0 c

0 d

]: c, d ∈ R

}= M2 (R)B =

{[a c

b d

][0 0

0 1

]: a, b, c, d ∈ R

};

These two ideals are the only proper left ideals of M2 (R ) . The only proper right ideals of M2 (R )

are:

KR =

{[a c

0 0

]: a, b ∈ R

}= AM2 (R) and HR =

{[0 0

b d

]: c, d ∈ R

}= BM2 (R ) .

An interesting fact about these ideals, is that all four of them are at the same time, minimal and

maximal.

It is not difficult to show that any non-zero matrix in any of these ideals generates the ideal con-

taining that matrix.

4. Consider the ring R=M 2 (R )=

{[a b

c d

]: a , b , c , d ∈ R

}. Then define the following subrings :

S1 =

{[a 0

0 0

]: a ∈ R

}, S2 =

{[a 0

c 0

]: a , c ∈ R

}, S3 =

{[a b

0 0

]: a , b ∈ R

},

S4 =

{[a 0

0 d

]: a , d ∈ R

}, S5 =

{[a b

0 d

]: a , b , d ∈ R

}, S6 =

{[a 0

c d

]: a , c , d ∈ R

}.

We have

S1 < S2 < S6 < R , S1 < S3 < S5 < R , S1 < S4 < S5 < R , and S1 < S4 < S6 < R .

a. S2 is a maximal and minimal left ideal of R and a prime two-sided maximal and minimal ideal

of S6 .

b. S3 is a maximal and minimal right ideal of R and a prime two-sided maximal and minimal ideal

of S5 .

c. S1 is a principal, prime, maximal and minimal left ideal of S3 , a principal, prime, maximal and

a minimal right ideal of S2 , and a principal, prime, maximal and minimal two-sided ideal of S4 .

d. Notice that S2 , S3 , S5 and S6 absorb S4 ; S2 · S1 = S2 and S1 · S3 = S3 .

5. Although the ring M 2(R ) has left and right ideals, but it is a simple ring. This means that it

doesn’t have any proper two-sided ideal.

Suppose I is an ideal of M 2(R ) , then it contains a non-zero matrix A =

[a b

c d

]. At least one of

the entries of A is not zero. Let E1 =

[1 0

0 0

], E2 =

[0 1

0 0

], E3 =

[0 0

1 0

]and E4 =

[0 0

0 1

].

1. If a 6= 0 , then1

aE1AE1 = E1 ∈ I and

1

aE3AE2 = E4 ∈ I . Thus E1 + E2 =

[1 0

0 1

]∈ I .


2. If b 6= 0 , then1

bE1AE3 = E1 ∈ I and

1

bE1AE4 = E4 ∈ I . Thus E1 + E2 =

[1 0

0 1

]∈ I .

3. If c 6= 0 , then1

cE2AE1 = E1 ∈ I and

1

cE4AE2 = E4∈ I . Thus E1 + E2 =

[1 0

0 1

]∈ I .

4. If d 6= 0 , then1

dE4AE4 = E4∈ I and

1

dE2AE3 = E4∈ I . Thus E1 + E2 =

[1 0

0 1

]∈ I .

Therefore I = R . Hence M 2(R ) is simple.

6. The set of all polynomials with real coefficients which are divisible for example by the polynomial

x 2 + 1 is an ideal in the ring of all polynomials R [X ] .

7. The ring of all polynomials with integer coefficients Z [X ] is not principal because the ideal

generated by 2 and X is an example of an ideal that cannot be generated by a single polynomial.

Solution. Suppose 〈 2 , X 〉 = 〈 f (X ) 〉 . Then f(X) = 2 g(X) + X h(X) . For X = 0 , we have

f (0) = 2 g (0) = 2n , for some integer n . Since the polynomials 2 and X are in 〈 f (X ) 〉 ; we have

2 = f (X ) p (X ) and X = f (X ) q (X ) , for some p (X ) and q (X ) in Z [X ] . Thus f (X ) is a

constant polynomial equal to f (0) = 2n , We have X = f (X ) q (X ) = 2n q (X ) .

Let X = 1 , then 1 = 2n q ( 1 ) ∈ 2Z . This is a contradiction with the fact that 1 is odd

Definitions 4. A ring R is a local ring if it has any one of the following equivalent properties:

• R has a unique maximal left ideal.

• R has a unique maximal right ideal.

• 1 6= 0 and the sum of any two non-units in R is a non-unit.

• 1 6= 0 and if x is any element of R , then x or x− 1 is a unit.

Examples.

1. All fields (and skew fields) are local rings, since { 0 } is the only maximal ideal in these rings.

2. A nonzero ring in which every element is either a unit or nilpotent is a local ring.

♣ Prime and Irreducible Elements.

Definitions 5. Recall that a unit in a ring R is an element with a multiplicative inverse. The

elements a and b are associates if a = u b for some unit u.

Let a be a nonzero non-unit; a is said to be irreducible if it cannot be represented as a product of

non-units. In other words, if a = b c , then either b or c must be a unit.

Again let a be a nonzero non-unit; a is said to be prime if whenever a divides a product of terms,

it must divide one of the factors. In other words, if a divides b c , then a divides b or a divides c

( a divides b means that b = a r for some r ∈ R ).

It follows from the definition that if p is any nonzero element of R , then p is prime if and only if

〈 p 〉 is a prime ideal. The units of Z are 1 and −1 , and the irreducible and the prime elements

coincide. But these properties are not the same in an arbitrary integral domain.

Definitions 6. Quadratic integers are the solutions of equations of the form

x2 +B x+ C = 0


with B and C integers. They are thus algebraic integers of the degree two. Common examples of

quadratic integers are the square roots of integers, such as√

2 , and the complex number i =√−1 .

The norm of such a quadratic integer is

N ( a+ b√D ) = a2 + b2D .

A nonzero non-unit element in an integral domain is said to be irreducible if it is not a product of

two non-units.

Every quadratic integer may be written as a+ ω b , where a and b are integers, ω is defined by:

ω =

√D if D ≡ 2 , 3 (mod 4)

1 +√D

2if D ≡ 1 (mod 4)

(D must be a square-free integer – not being divisible by a square number.)

Eisenstein integers, also known as Eulerian integers, denoted by Z [ω ] are complex numbers of the

form z = a+ b ω , where a and b are integers and

ω =1

2(−1 + i

√3) = e2πi/3

is a primitive (non-real) cube root of unity.

Note that each Eisenstein integer z = a+ b ω is a root of the monic polynomial

z2 − (2a− b)z + (a2 − ab+ b2) .

In particular, ω satisfies the equation

ω2 + ω + 1 = 0 .

Thus the Eisenstein integers form a commutative ring of algebraic integers. The product of two

Eisenstein integers a+ bω and c+ dω is given explicitly by

( a+ bω ) · ( c+ dω ) = ( a c− b d ) + ( b c+ a d− b d )ω .

The norm of an Eisenstein integer is given by

N ( a+ b ω ) = | a+ b ω |2 = a2 − a b+ b2 .

Definitions 7. A Euclidean domain (E , + , · , d ) is an integral domain (E , + , · ) which can be

endowed with at least one Euclidean function. d : E 7→ N , such that for all a , b ∈ E with b 6= 0 ,

there exist q , r ∈ E , satisfying

a = q b+ r , where either r = 0 or d ( r ) < d ( b ) and ∀ a , b ∈ E , d ( a ) ≤ d ( a b ) .

In general, a Euclidean domain will admit many different Euclidean functions.

Examples of Euclidean domains include:

Z , the ring of integers. Define d (n ) = |n | , the absolute value of n .

Any field. Define d (x ) = 1 for all nonzero x .


Z [√−2 ]. Define d ( a+ b i ) as N ( a+ b i ) = a 2 + 2 b 2 .

Z [ i ], the ring of Gaussian integers. Define d ( a+ b i ) as N ( a+ b i ) = a 2 + b 2 .

Z [ω ], the ring of Eisenstein integers. Define d ( a+ b i ) as N ( a+ b i ) = a 2 − a b+ b 2 .

K [X ],, the ring of polynomials over a field K . For each nonzero polynomial p(X ) , define d ( p )

to be the degree of p(X ) .

Theorem 9. If a is prime, then a is irreducible, but not conversely.

Proof. We use the standard notation r | s to indicate that r divides s . Suppose that a is prime,

and that a = b c . Then certainly a | b c , so by definition of prime, a | b or a | c , say a | b . If

b = a d then b = b c d , so c d = 1 and therefore c is a unit. (Note that b cannot be 0 , for if so,

a = b c = 0 , which is not possible since a is prime.) Similarly, if a | c with c = a d then c = b c d ,

so b d = 1 and b is a unit. Therefore a is irreducible.

To give an example of an irreducible element that is not prime, consider

R = Z[√− 3

]={a+ b

√−3 : a , b ∈ Z

}.

In R , 2 is irreducible but not prime. To see this, first suppose that we have a factorization of the

form

2 =(a+ b

√−3

) (c+ d

√−3

);

take complex conjugates of 2 to get

2 =(a− b

√−3

) (c− d

√−3

).

Now multiply these two equations to obtain

4 =(a 2 + 3 b 2

) (c2 + 3 d 2

).

Each factor on the right must be a divisor of 4 , and there is no way that a 2 + 3 b 2 can be 2 . Thus

one of the factors must be 4 and the other must be 1 . If, say, a 2 + 3 b 2 = 1 , then a = ±1 and

b = 0 . Thus in the original factorization of 2 , one of the factors must be a unit, so 2 is irreducible.

Finally, 2 divides 4 which is equal to(1 +√−3

) (1−√−3

).

If 2 were prime, it would divide one of the factors, which is a contradiction since none of these

factor is an integer.

The distinction between irreducible and prime elements disappears in the presence of unique factor-

ization.

Definitions 8. A unique factorization domain (UFD) is an integral domain R satisfying the fol-

lowing properties:

(UF1) : Every nonzero element in a ∈ R can be expressed as a = u p1 p2 . . . pn , where u is a unit

and the pi s are irreducible.

(UF2) : If a has another factorization, say a = v q1 q2 . . . qm , where v is a unit and the qi s are

irreducible, then n = m and, after reordering if necessary, pi and qi are associates for each i .

Property UF1 asserts the existence of a factorization into irreducibles, and UF2 asserts uniqueness.


Theorem 10. In a unique factorization domain, a is irreducible if and only if a is prime.

Proof. By the previous theorem, prime implies irreducible, so assume that a is irreducible, and let

a divide b c . Then we have a d = b c for some d ∈ R . We factor d , b and c into irreducibles to

obtain

a u d1 . . . dr = ( v b1 . . . bs ) (w c1 . . . ct ) ,

where u , v and w are units and the di , bj , and ck are irreducible. By uniqueness of factorization,

a, which is irreducible, must be an associate of some bj or ck . Thus a divides b or c .

Exercise. Prove that Z[√− 5

]is not a UFD.

♣ Cosets. For a two-sided ideal I of a ring R and an element r of R , define the coset

r + I = { r + j : j ∈ I } .

Theorem 11. Let I be an ideal in a ring R . Then the relation of congruence modulo I is

(1) Reflexive : a ≡ a (mod I) for every a ∈ R .

(2) Symmetric : If a ≡ b (mod I) , then b ≡ a (mod I) .

(3) Transitive : If a ≡ b (mod I) and b ≡ c (mod I) , then a ≡ c (mod I) .

Proof. (1) a− a = 0 ∈ I ; hence, a ≡ a (mod I) .

(2) a ≡ b (mod I) means that a − b = s for some s ∈ I . Therefore, b − a = − ( a − b ) = − s .

Since I is an ideal, the negative of an element of I is also in I , and so b−a ≡ s (mod I) . Hence,

b ≡ a (mod I) .

(3) If a ≡ b (mod I) and b ≡ c (mod I) , then by the definition of congruence, there are elements s

and t in I such that a−b = s ∈ I and b−c = t ∈ I . Therefore, a−c = (a−b)+(b−c) = s+t ∈ I .

Since the ideal I is closed under addition, s+ t ∈ I ; hence, a ≡ c (mod I) .

Theorem 12. Let I be an ideal in a ring R . If a ≡ b (mod I) and c ≡ d (mod I) , then

(1) a+ c ≡ b+ d (mod I) .

(2) a c ≡ b d (mod I) .

Proof. By the definition of congruence, a−b = s ∈ I and c−d = t ∈ I , for some elements s and t

in I . Therefore, ( a + c )− ( b + d ) = ( a − b ) + ( c − d ) = s + t ∈ I . Hence, a + c = b + d ∈ I .

(2) a c − b d = a c − b c + b c − b d = ( a − b ) c + b ( c − d ) = s c + b t . Since the ideal Iabsorbs products on both left and right, s c ∈ I and b t ∈ I . Hence, a c − b d ≡ s c + b t (mod I) .

Therefore, a c − b d ≡ 0 (mod I) . Hence a c ≡ b d (mod I) .

Theorem 13. Let I be an ideal in a ring R and let a , c ∈ R . Then a ≡ c (mod I) if and only

if a + I = c + I .

Proof. First, assume that a ≡ c (mod I) . To prove that a + I = c + I , we first show that

a + I ⊂ c + I . To do this, let b ∈ a + I . Then by definition b ≡ a (mod I) . Since a ≡ c

(mod I) , we have b ≡ c (mod I) by transitivity. Therefore, b ∈ c + I and a + I ⊂ c + I .

Reversing the roles of a and c in this argument and using the fact that c ≡ a by symmetry, show

that c + I ⊂ a + I . Therefore, a + I = c + I .

Conversely, assume that a + I = c + I . Since a ≡ a (mod I) by reflexivity, we have a ∈ a + Iand, hence, a ∈ c + I . By the definition of c + I , we see that a ≡ c (mod I) .


♣ Quotient Rings . If I is a two-sided ideal of the ring R , we would like (by analogy with the

group case) to define the quotient ring R/I . Since I is, in particular, a normal subgroup of the

additive group of R , we can define the quotient group R/I . Its element are the additive cosets

r + I , where r ∈ R . The product will be defined as ( r + I)(s + I) = r s + I . A quotient ring is

also known as a factor ring.

Consider the ring R[X ] of polynomials in the variable x with coefficients R and let p(x ) be a

polynomial in R[X ]. According to the division algorithm for polynomials, for a polynomial f(x ) ,

there exist polynomials q(x ) and r(x ) , where 0 ≤ deg( r(x) ) < deg( f(x) ), such that f(x) =

p(x) q(x) + r(x). The ideal I , generated by p(x )

I =< p(x) >= {a p(x) : a ∈ R}

produces a quotient ring R [X ] /I of R [X ] , with [p(x)] = [0] and [f(x)] = [r(x)].

Using the ideal 8Z of the ring Z , we construct the quotient ring Z /8Z , denoted Z8. The set

S1 = {[0], [4]} and S2 = {[0], [2], [4], [6]} are the only proper subrings which are also the only

proper ideals of the ring.

♣ Homomorphisms. Given two rings R1 and R2; a ring homomorphism from R1 to R2; is a

mapping ϕ from R1 to R2, such that

∀x, y ∈ R1 , ϕ(x + y) = ϕ(x) + ϕ(y) and ϕ(x y) = ϕ(x)ϕ(y) .

From this property, one may deduce that ϕ(θ1) = θ2 and ϕ(−x) = −ϕ(x ) .

Suppose both R1 and R2 have identities. Then ∀x ∈ R ,

ϕ(x) = ϕ(x e1) = ϕ(x)ϕ(e1).

This implies that ϕ(e1) = e2.

Suppose x−1 exists, then from

ϕ(e1) = ϕ(xx−1) = ϕ(x)ϕ(x−1) = e2 ,

we conclude that ϕ(x−1) = ϕ(x)−1 .

The set of all ring homomorphisms from R1 to R2 is denoted by Hom(R1 , R2 ) .

If ϕ ,ψ ∈ Hom(R1 , R2 ) , then ∀u ∈ R1 ,

(a) (ϕ + ψ)(u) = ϕ(u) + ψ(u)⇒ ϕ + ψ ∈ Hom(R1 , R2) ;

(b) (ϕψ)(u) = ϕ(u)ψ(u)⇒ ϕψ ∈ Hom(R1 , R2) .

The above identities imply that Hom(R1 , R2 ) is a ring. If R2 is commutative, then Hom(R1 , R2 )

becomes a commutative ring.

We define the kernel of ϕ, denoted by Ker(ϕ ) , to be the set of elements in R1 which are mapped

to the zero θ2 in R2.

Ker(ϕ) = {x ∈ R1 : ϕ(x) = θ2} ;

and the image of ϕ, denoted by Im(ϕ ) to be

Im(ϕ) = {ϕ(x) : x ∈ R1} .


Theorem 14. If ϕ ∈ Hom(R1 , R2 ) , then Ker(ϕ ) is an ideal of R1.

Proof. Let x, y ∈ Ker(ϕ ) , then

ϕ(x− y) = ϕ(x)− ϕ(y) = θ2 − θ2 = θ2 and ϕ(x y) = ϕ(x)ϕ(y) = θ2 θ2 = θ2 .

By Theorem 1 , Ker(ϕ ) is subring of R1. Now, chose any s ∈ Ker (ϕ ) and any x ∈ R1; then

ϕ(x s) = ϕ(x)ϕ(s) = ϕ(x) θ2 = θ2 and ϕ(s x) = ϕ(s)ϕ(x) = θ2 ϕ(x) = θ2 .

This proves that Ker (ϕ ) is a two-sided ideal of R1, .

Theorem 15. Let ϕ : R1 7→ R2 be a homomorphism of rings. Then Ker (ϕ) is a prime ideal if

R2 is an integral domain.

Proof. Suppose a b ∈ Ker (ϕ) . Then 0 = ϕ (a b) = ϕ (a)ϕ (b) . Since R2 is an integral domain,

then ϕ (a) = θ2 or ϕ (b) = θ2 . This implies that a ∈ Ker (ϕ ) or b ∈ Ker (ϕ ) . Thus Ker (ϕ) is

a prime ideal of R1 .

Theorem 16. Let ϕ ∈ Hom(R1 , R2 ) . Then

1. Im(ϕ ) = { y ∈ R2 | ∃x ∈ R1, ϕ (x) = y } is a subring of R2 but not necessary an ideal.

2. Let J be an ideal of R2 and I = ϕ−1(J ) ={x ∈ R1 | ∃ y ∈ J , ϕ−1 (y) = x

}be the inverse

image of J . Then I is an ideal of R1.

Proof. 1. Let y1 , y2 ∈ Im(ϕ ) , then there exist x1 , x2 ∈ R1 such that ϕ (x1) = y1 and ϕ (x2) =

y2 . Hence

y1 ± y2 = ϕ (x1)± ϕ (x2) = ϕ (x1 ± x2) and y1 y2 = ϕ (x1)ϕ (x2) = ϕ (x1 x2) .

Thus Im(ϕ ) is a subring of R2 .

If ϕ is not surjective, then there exists a z ∈ R2, which is not in Im(ϕ ) . Then for any y ∈ Im(ϕ ) ,

z y is not the image of an element of R1 Thus Im(ϕ ) is not an ideal of R2 .

2. Let x1 , x2 ∈ I , then there exist y1 , y2 ∈ J such that ϕ (x1) = y1 and ϕ (x2) = y2 . We have

y1 ± y2 = ϕ (x1)± ϕ (x2) = ϕ (x1 ± x2) and y1 y2 = ϕ (x1)ϕ (x2) = ϕ (x1 x2) .

Hence x1 ± x2 ∈ I and x1 x2 ∈ I . Thus I = ϕ−1 (J ) is a subring of RI .

Let z ∈ R1 and x ∈ I . Then ϕ (z x) = ϕ (z)ϕ (x) . Since J is an ideal of R2 , we have

ϕ (z)ϕ (x) ∈ J . Thus z x ∈ ϕ−1 (J ) . Hence I is an ideal in R1 .

Definitions 9.

1. A homomorphism from R into itself is called an endomorphism.

2. A bijective homomorphism is called an isomorphism.

3. An isomorphism from R into itself is called an automorphism.

4. A injective homomorphism is called an monomorphism.

5. A surjective homomorphism is called an epimorphism.

The isomorphism is denoted by the symbol ∼=.

The mapping R :−→ R / I, where a −→ [ a ] is a homomorphism, and is called the natural homo-

morphism.

Theorem 17. The epimorphism ϕ : R1 :−→ R2 is an isomorphism, if Ker (ϕ ) = { θ1 } .


Proof. Suppose for some x, y ∈ R1, ϕ(x) = ϕ(y ) , then

θ2 = ϕ(x)− ϕ(y) = ϕ(x− y) = ϕ(θ1) .

If Ker(ϕ) = {θ} , then x − y = θ1; hence x = y, so ϕ is injective. Thus the epimorphism ϕ is an

isomorphism.

Theorem 18. Consider the ring R[X ] of polynomials in x with real coefficients, and the ideal

I =< x 2 + 1 >={a (x 2 + 1) : a ∈ R

}.

Then R [X ] / I ∼= C

Proof. By defining the quotient ring R [X ] / I , we ‘force’ x 2 + 1 to be 0 , i.e. x 2 = −1 , which

is the defining property of the imaginary number i =√−1 . Consider the homomorphism:

ϕ : R [X ] / I −→ C with ϕ( a + b x ) = a + b i .

This is clearly an isomorphism from R [X ] /I to C .

Fundamental Theorems of Homomorphisms. Let ϕ : R 7→ S be a homomorphism of rings

with I = Ker (ϕ) . Then there exists a unique monomorphism ψ : R / I 7→ S such that the diagram

R S

R / I

ϕ

δ ψ

commutes, i.e., ϕ = ψ ◦ δ , where δ is the natural map from R to R / I , i.e., δ(x) = x+ I .

Moreover, ψ is an isomorphism if and only if ϕ is an epimorphism.

Proof. Let ψ : R / I 7→ S be defined by ψ (x+ I) = ϕ (x), for all x ∈ R .

Step 1. ψ is well-defined.

Let x + I = y + I for some x , y ∈ R . This means that x − y ∈ I , i.e., x − y ∈ Ker (ϕ) . Thus

ϕ (x− y) = 0 , i.e., ϕ (x) = ϕ(y) , hence ψ (x+ I) = ψ (y + I) . as required.

Step 2. ψ is injective.

Let ψ (x+ I) = ψ (y + I) . Then

ϕ (x) = ψ (x+ I) = ψ (y + I) = ϕ (y) =⇒ ϕ (x)− ϕ (y) = ϕ (x− y) = 0 .

Thus x− y ∈ Ker (ϕ) = I . Hence x+ I = y + I, as required.

Step 3. ψ is a homomorphism.

For all x , y ∈ R , we have :

• ψ ( (x+ I) + (y + I) ) = ψ (x+ y + I) = ϕ (x+ y) = ϕ (x) + ϕ (y) = ψ (x+ I) + ψ (x+ I) and

• ψ ( (x+ I) (y + I) ) = ψ (x y + I) = ϕ (x y) = ϕ (x)ϕ (y) = ψ (x+ I)ψ (x+ I), as required.

Step 4. ψ is an isomorphism if and only if ϕ is an epimorphism.

Note that natural map δ is an epimorphism, ψ is a monomorphism, and ϕ = ψ ◦ δ .

If ψ is an isomorphism, then ϕ must be an epimorphism. Similarly ϕ an epimorphism implies that

ψ is an epimorphism, hence an isomorphism.


Endomorphism Ring of Abelian Group. Let (G , + ) be an abelian group and let End (G )

be the set of all group homomorphisms of G into itself, i.e.,

End (G ) = {ϕ : G 7→ G | ϕ (x+ y) = ϕ (x) + ϕ (y) , ∀x , y ∈ G } .

Then End (G , + , ◦ ) , where

ϕ± ψ (x) = ϕ (x)± ψ (x) and ϕ ◦ ψ (x) = ϕ (ψ (x) )

is a non-commutative ring . The identity of End (G ) is I(x) = x and units of End (G ) are precisely

all automorphisms of G . Thus the set of all automorphisms of G , noted Aut (G ) along with the

zero homomorphism, form a subfield of End (G ) .

Endomorphism Ring of Vector Space. Let K be a field and V be a vector space over K and

let EndK (V ) be the set of all K-linear endomorphisms of V , i.e., ∀x , y ∈ V and ∀λ ∈ V ,

ϕ (x± y) = ϕ (x)± ϕ (y) and ϕ (λx) = λϕ (x) .

Since V is an abelian group , the ring EndK (V ) is a subring of the group-endomorphism , End (V , + , ◦ ) ,

where

ϕ± ψ (x) = ϕ (x)± ψ (x) and ϕ ◦ ψ (x) = ϕ (ψ (x) ) .

♣ Direct Product of Rings. One way of building a new ring from two old rings, say R1 and R2

is to take the set R1 ×R2 of ordered pairs of elements, the first from R1 and the second from R2,

using the operation of R1 for the first component and the operation of R2 for the second component.

This operation is analogue of the Cartesian product of sets.

The direct product, also known as the external direct product is an operation that takes n ≥ 2 rings

R1 , R2, R3, . . . , Rn and constructs a new ring, denoted

R = R1 ×R2 ×R3 × · · · × Rn .

The binary operations on R are defined component-wise:

(x1 , x2 , x3 , . . . , xn) + (y1 , y2 , y3 , . . . , yn) = (x1 + y1 , x2 + y2 , x3 + y3 , · · · , yn + yn),

(x1 , x2 , x3 , . . . , xn) . (y1 , y2 , y3 , . . . , yn) = (x1 . y1 , x2 . y2 , x3 . y3 , . . . , yn . yn).

This operation makes the set R a ring, where θ = (θ1 , θ2, θ3, . . . , θn ) .

Let δk = (x1 , x2 , x3 , . . . , xn) ∈ R , where xi = 1 and ∀j 6= i, xj = 0. Then if i 6= j , δi δj = θ .

Hence R is not an integral domain.

Remark 3. The subring S = Z× {0} of the ring R = Z× Z does not contain the unity ( 1 , 1 ) ;

but it has its own unity ( 1 , 0 ) .

♣ Embedding of Rings. Sometimes it is worthwhile to study the properties of a ring by consid-

ering it as a subring of some ring with more ring properties than itself. A ring without identity lacks

important arithmetic properties. For example, in the ring 2Z of even integers, we cannot say that

2 divides 2 because 1 /∈ 2Z . Now 2Z is a subring of Z and 1 ∈ Z . In Z , it is true that 2 divides

2 . The main aim of this section is to embed a ring into a suitable ring with additional properties.

Definitions 10. A ring R is said to be embedded in a ring S if there exists a monomorphism of

R into S .


From the above definition, it follows that a ring R can be embedded in a ring S if there exists a

subring T of S such that R ' T .

Theorem 19. Any ring R can be embedded in a ring S = R × Z with unity such that R is an

ideal of S . If R is commutative, then S is commutative.

Proof. Define addition and multiplication in S = R× Z as follows:

( a , m ) + ( b , n ) = ( a+ b , m+ n ) and ( a ,m ) · ( b , n ) = ( a b+ na+mb , mn ) ,

for all a , b ∈ R and m, n ∈ Z . Then S forms a ring under these definitions of addition and

multiplication. Note that ( 0 , 0 ) is the additive identity and that ( 0 , 1 ) is the multiplicative

identity of S = R× Z .

Consider the subset R×{ 0 } of S . Clearly (0, 0) ∈ R× 0 ; also for all ( a , 0 ) , ( b , 0 ) ∈ R×{ 0 } ,

( a , 0 )− ( b , 0 ) = ( a− b , 0 ) ∈ R× { 0 } and ( a , 0 ) · ( b , 0 ) = ( a b , 0 ) ∈ R× { 0 }

Thus, R× { 0 } is a subring of S . Now for all ( a , 0 ) ∈ R× { 0 } and ( c , n ) ∈ S , we have

( a , 0 ) · ( c , n ) = ( a c+ na , 0 ) ∈ R× { 0 } and ( c , n ) · ( a , 0 ) = ( c a+ na , 0 ) ∈ R× { 0 }

This proves that R× { 0 } is an ideal of S .

Now define f : R 7→ R× { 0 } by f ( a ) = ( a , 0 ) for all a ∈ R . Then f is an isomorphism of Ronto R× { 0 } , so R ' R× { 0 } . Therefore, R can be embedded in S .

By identifying a ∈ R with ( a , 0 ) ∈ R× { 0 } , we can regard R to be an ideal of S .

Next we show that the commutativity of R implies that of S . Let ( a , m ) , ( b , n ) ∈ S and Rbe commutative. Then

( a , m ) · ( b , n ) = ( a b+ na+mb , mn ) = ( b a+mb+ na, nm ) = ( b , n ) · ( a , m ) .

Hence, S is commutative.

♣ Rings of Polynomials. Let R commutative ring with unity or F be a field, and let X be an

arbitrary symbol. Every expression of the form

p (X ) = a0 + a1X + a2X2 + · · ·+ anX

n

is called a polynomial in X with coefficients in R or F . The sets of all polynomials in X with

coefficients in R are denoted by R [X ] and F [X ] , respectively

Although in elementary algebra, descending order is preferred, but for our purpose ascending order

is more convenient.

The expressions akXk , for k = 0 , 1 , 2 , . . . , n are called the terms of the polynomial and ak is

said to be the coefficient of Xk . By the degree of the polynomial p (X ) we mean the greatest n

such that the coefficient of Xk is not zero. In other words, if the polynomial has degree n , then

an 6= 0 , but am = 0 for all m > 0 . The degree of the polynomial p (X ) is denoted by deg p (X ) .

The coefficient an is called the leading coefficient. If the leading coefficient of a polynomial is one,

then the polynomial is call monic.

The polynomial p (X ) = a0 is called a ıconstant polynomial of degree zero. The polynomial whose

all coefficients are zero is called the zero polynomial. It is the only polynomial whose degree is


undefined (because it has no nonzero coefficient).

The familiar sigma notation for sum is useful for polynomial, Thus

p (X ) = a0 + a1X + a2X2 + · · ·+ anX

n =n∑

k=0

akXk .

Equality of Polynomials. Given two polynomials p (X ) = a0 + a1X + a2X2 + · · · + amX

m

and q (X ) = b0 + b1X + b2X2 + · · ·+ bnX

n in R [X ] , we say that p (X ) = q(X ) if m = n and

ak = bk , for all k ∈ Z .

Addition and Multiplication of Polynomials : Let p (X ) and q (X ) be as above. (We may

assume m ≤ n .) Define

p (X ) + q (X ) = ( a0 + a1X + a2X2 + · · ·+ amX

m ) + ( b0 + b1X + b2X2 + · · ·+ bnX

n )

= ( a0 + b0 ) + ( a1X + b1X ) + ( a2X2 + b2X

2 ) + · · ·+ ( amXm + bmX

m )

bm+1Xm+1 + · · ·+ anX

n , and

p (X ) q (X ) = ( a0 + a1X + a2X2 + · · ·+ amX

m ) ( b0 + b1X + b2X2 + · · ·+ bnX

n )

= ( a0 b0 ) + ( a0 b1 + a1 b0 )X + ( a0 b2 + a1 b1 + a2 b0 )X2 + · · ·+ · · ·+ (a0 bk + a1 bk−1 + · · ·+ ak b0 ) + · · ·+ ( am bn )Xm+n .

Definitions 11. Elements of R are called constants. Polynomials of degree zero are precisely the

non-zero constants. Polynomials of degree 1 , 2 , 3 , 4 , 5 , . . . are called linear, quadratic, cubic,

biquadraitc, quintic, . . . etc.

Theorem 20. Let R be a commutative ring with unity, then R [X ] is a commutative ring with

unity.

The proof of the theorem is not difficult but tedious.

Theorem 21. If R is an integral domain, then R [X ] is also an integral domain.

Proof. If p (X ) and q (X ) are nonzero polynomials, then we must show that the polynomial

p (X ) q (X ) is not zero.

Let an and bn be the leading coefficients of p (X ) and q (X ) respectively. Since R is an integral

domain, an bn 6= 0 . It follows that p (X ) q (X ) has a nonzero coefficient. Therefore is not the zero

polynomial.

Remark 4. It would be tempting to believe that if F is a field, then F [X ] is also a field. However

this is not true for one simple reason, the multiplicative inverse of a polynomial is usually not a

polynomial (except for constant polynomials).

Division Algorithm. If f and g are polynomials in R [X ] , with g monic, there are unique

polynomials q and r in R [X ] such that f = q g + r and deg r < deg g . If R is a field, g can

be any nonzero polynomial.

Proof. The above procedure, which works in any ring R , shows that q and r exist.

If f = q g + r = q1 g + r1 where r and r1 are of degree less than deg g , then g ( q − q1 ) = r1 − r .

But if q − q1 = 0 , then since g is monic, the degree of the left side is at least deg g , while the

degree of the right side is less than deg g , a contradiction. Therefore q = q1 , and consequently

r = r1 .


Example. In Z [X ] , we can divide 2X3 + 10X2 + 16X + 10 by X2 + 3X + 5 :

2X3 + 10X2 + 16X + 10 = 2X (X2 + 3X + 5 ) + ( 4X2 + 6X + 10 ) .

The remainder 4X2 + 6X + 10 does not have degree less than 2 , so we divide it by X2 + 3X + 5 :

4X2 + 6X + 10 = 4 (X2 + 3X + 5 )− 6X − 10 .

Combining the two calculations, we have

2X3 + 10X2 + 16X + 10 = ( 2X + 4 ) (X2 + 3X + 5 ) + (− 6X − 10 ) .

Remainder Theorem. If f ∈ R [X ] and a ∈ R , then for some unique polynomial q (X ) in

R [X ] we have f (X ) = q (X ) (X − a ) + f ( a ) ; hence f(a) = 0 if and only if X − a divides

f (X ) .

Proof. By the division algorithm, we may write f (X ) = q (X ) (X − a ) + r (X ) where the degree

of r is less than 1,i.e., r is a constant. Apply the evaluation homomorphism X 7→ a to show that

r = f ( a ) .

Remark 5. If R is an integral domain then so is the polynomial ring R [X ] because the leading

coefficient of product of non-zero polynomials in R [X ] is equal to the product of their leading

coefficients, hence is nonzero.

Theorem 22. If R is an integral domain, then a nonzero polynomial f ∈ R [X ] of degree n has

at most n roots in R [X ] , counting multiplicity.

Proof. If f ( a1 ) = 0 , then by the Remainder Theorem, possibly applied several times, we have

f (X ) = q1 (X ) (X − a1 )n1 , where q1 ( a1 ) = 0 and the degree of q1 is n− n1 .

If a2 is another root of f , then 0 = f ( a2 ) = q1 ( a2 ) ( a2 − a1 )n1 . But a1 = a2 and R is an

integral domain, so q1(a2) must be 0 , i.e. a2 is a root of q1 (X) .

Repeating the argument, we have q1 (X ) = q2 (X ) (X − a2 )n2 , where q2 ( a2 ) = 0 and

deg q2 = n − n1 − n2 . After n applications of the Remainder Theorem, the quotient becomes

constant, and we have f (X ) = c (X− a1 )n1 . . . (X− ak )nk where c ∈ R and n1+ · · · +nk = n .

Since R is an integral domain, the only possible roots of f are a1, . . . , ak.

Note that a polynomial of degree n in a non-integral domain may have more roots. For example

in R = Z8 , which is not an integral domain; the polynomial f (X ) = X3 has four roots, namely

0 , 2 , 4 , and 6.

Definitions 12. An irreducible polynomial is a non-constant polynomial that cannot be factored into

the product of two non-constant polynomials. The property of irreducibility depends on the field or

ring to which the coefficients are considered to belong. For example, the polynomial p (X ) = X2−2

is irreducible in Z [X ] and Q [X ] , but it factors as (X −√

2 ) (X +√

2 ) in R [X ] .

The polynomial q (X ) = X2 +1 is irreducible in R [X ] . but it factors as (X−√−1 ) (X+

√−1 )

in C [X ] . Actually in C [X ] , any polynomial of degree greater than or equal to two is reducible.

This fact is known as the fundamental theorem of algebra in the case of the complex numbers. It

follows that every non-constant polynomial p (X ) = a0 + a1X + a2X2 + · · · + anX

n can be

factored as

p (X ) = an (X − z1 ) (X − z1 ) · · · (X − zn ) ,


where z1 , z2 , . . . , zn are the zeros of p (X ) (not necessarily distinct).

Definitions 13. let p (X ) = a0 + a1X + a2X2 + · · · + anX

n ∈ Z [X ] . The content of p (X ) is

defined to be c ( p ) = g c d ( a0, , a2 , . . . , an ) .

Definitions 14. A polynomial p (X ) in R [X ] , where R is a commutative ring is primitive if the

only elements of R that divide all coefficients of p (X ) at once are the invertible elements of R .

In the case where R is the ring Z of the integers, this is equivalent to the condition that c ( p ) = 1 .

Lemma 7. (Gauss Lemma) Consider the ring Z [X ] . Then

(1) The product of two primitive polynomials is also primitive.

(2) For any f (x ) and g (x ) , c ( f g ) = c ( f ) c ( g ) .

Proof. (1) Suppose there is a prime p which divides all the coefficients of c ( f ) c ( g ) . Let Zp [X ]

be the set of all polynomials with coefficients in Zp . By reducing the coefficients f (x ) and g (x )

modulo p we obtain f1 (x ) and g1 (x ) in Zp [X ] .

Since Zp [X ] is an integral domain, either f1 (x ) = 0 or g1 (x ) = 0 . This implies that p divides

all the coefficients of either f (x ) or g (x ) , which contradicts the fact that f (x ) and g (x ) .

(2) By the definition, f (x ) = c ( f ) f1 (x ) and g (x ) = c ( g ) g1 (x ) , where f1 (x ) and g1 (x ) are

primitive. So f (x ) g (x ) = c ( f ) c ( g ) f1 (x ) g1 (x ) . By the first part, we know that f1 (x ) g1 (x )

is primitive. Therefore c ( f g ) = c ( f ) c ( g ) .

♣ Ring of Power Series. Given a ring R , let R [ [X ] ] be the formal power series with coefficients

in R , i.e.,

R [ [X ] ] ={a0 + a1X + a2X

2 + · · · | ak ∈ R}.

Equality of Power Series : Let f (X ) = a0+a1X+a2X2+· · · and g (X ) = b0+b1X+b2X

2+· · ·be in R [ [X ] ] . Then f (X ) = g(X ) if and only if ak = bk , for all k ∈ Z .

Addition and Multiplication of Power Series : Let f (X ) and g (X ) be as above. Then we

define

f (X ) + g (X ) = ( a0 + a1X + a2X2 + · · · ) + ( b0 + b1X + b2X

2 + · · · )= ( a0 + b0 ) + ( a1X + b1X ) + ( a2X

2 + b2X2 ) + · · · and

f (X ) g (X ) = ( a0 + a1X + a2X2 + · · · ) ( b0 + b1X + b2X

2 + · · · )= ( a0 b0 ) + ( a0 b1 + a1 b0 )X + ( a0 b2 + a1 b1 + a2 b0 )X2 + · · ·

Under these operations R [ [X ] ] is a ring, called the ring of (formal) power series over R in the

variable X .

Definitions 15. The term a0 of f (X ) = a0 + a1X + a2X2 + · · · is called the constant term

of the power series. If a power series is of the form f (X ) = anXn + · · · + Xn+1 + · · · with

an 6= 0 but ak = 0 for all k = 0 , 1 , . . . , n− 1 , then n is called the order of the power series and

is denoted by ord ( f(X ) ) .

♣ Laurent Polynomials. Given a ring R , a formal expression of the form

L (X ) = a−nX−n + a−n+1X

−n+1 + · · · + a−1X−1 + a0 + a1X + · · · + amX

m ,

where ak ∈ R and m, n ∈ Z , is called a Laurent polynomial in the variable X with coefficients in

R . The set of all Laurent polynomials is denoted by R [X , X−1 ] or R [X ±1 ].

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