Notes on Ring Theory

S. F. Ellermeyer

Department of Mathematics

Kennesaw State University

February 10, 2016

Abstract

These notes contain an outline of essential definitions and theorems

from Ring Theory but only contain a minimal number of examples.

Many examples are presented in class and thus it is important to

come to class. Also of utmost importance is that the student put

alot of effort into doing the assigned homework exercises.The material

in these notes is outlined in the order of Chapters 17—33 of Charles

Pinter’s “A Book of Abstract Algebra”, which is the textbook we are

using in the course.

1 Rings

A ring is a non—empty set, , along with two operations called addition

(usually denoted by the symbol +) and multiplication (usually denoted by

the symbol · or by no symbol) such that

1. is an abelian group under addition. (The additive identity element

of is usually denoted by 0.)

2. The multiplication operation is associative.

3. Multiplication is distributive over addition, meaning that for all

and in we have

· (+ ) = · + ·

1

and

(+ ) · = · + · .

Let us look at a few examples of rings.

Example 1 The trivial ring is the one—element set = {0} with additionoperation defined by 0+0 = 0 and multiplication operation defined by 0·0 = 0.

Example 2 The set of integers, , with the standard operations of addition

and multiplication is a ring. The set of real numbers, , with the standard

operations of addition and multiplication is a ring. The set of rational num-

bers, , with the standard operations of addition and multiplication is a ring.

The set of complex numbers, , with the standard operations of addition and

multiplication is a ring.

Example 3 The set 4 = {0 1 2 3} with operations defined by

+ 0 1 2 3

0 0 1 2 3

1 1 2 3 0

2 2 3 0 1

3 3 0 1 2

· 0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

is a ring. (In general, for any given integer ≥ 1, the set = {0 1 2 − 1}with addition and multiplication operations defined “modulo ” is a ring.)

Example 4 The set, 2 (), of all 2 × 2 matrices with real entries andoperations defined by∙

1 11 1

¸+

∙2 22 2

¸=

∙1 + 2 1 + 21 + 2 1 + 2

¸and ∙

1 11 1

¸ ∙2 22 2

¸=

∙12 + 12 12 + 1212 + 12 12 + 12

¸is a ring.

2

Example 5 Let () denote the set of all functions from into with

operations defined as follows: For and ∈ (), + is the function in

() defined by

( + ) () = () + () for all ∈

and · is the function in () defined by

( · ) () = () () for all ∈ .

() with these operations is a ring.

The additive identity element of a ring, , is usually denoted by the

symbol 0 and the additive inverse of any element ∈ is usually denoted

by −. Also, for any and ∈ we interpret − to mean + (−). Thefollowing Proposition gives some basic algebraic properties of multiplication

in rings.

Proposition 6 If is a ring with additive identity element 0 then

1. · 0 = 0 and 0 · = 0 for all ∈ .

2. · (−) = − ( · ) and (−) · = − ( · ) for all and ∈ .

3. (−) · (−) = for all and ∈ .

Proof. Suppose that is a ring with additive identity element 0.

1. Let ∈ . Then

· 0 = · (0 + 0) = · 0 + · 0.Since · 0 + · 0 = · 0, we can subtract · 0 from both sides of this

equation (which really means adding − · 0 to both sides) to obtain · 0 + · 0− · 0 = · 0− · 0

which gives

· 0 + 0 = 0which gives

· 0 = 0.The proof that 0 · = 0 is similar.

3

2. Let and ∈ . Then

·(−)+· = (−+ ) = ·0 = 0 (by Part 1 which has just been proved).

Since · (−) + · = 0 and since the additive inverse of · isunique (because is a group under addition), then it must be the case

that the additive inverse of · is in fact · (−). In other words− ( · ) = · (−).The proof that (−) · = − ( · ) is similar.

3. Let and ∈ . Then, by using Part 2 of this Proposition twice, we

obtain

(−) · (−) = − ((−) · ) = − (− ( · )) = · .

Note that our definition of the term “ring” above does not require that

the multiplication operation of a ring be commutative or that the multipli-

cation operation have an identity element in or that (if the multiplication

operation does have an identity element) that each element in have a mul-

tiplicative inverse. If is a ring for which the multiplication operation is

commutative (meaning that · = · for all and ∈ ), then is

said to be a commutative ring. If has a multiplicative identity element

(meaning an element, ∈ , such that · = · = for all ∈ ), then

is said to be a ring with unity. In any ring with unity the multiplica-

tive identity must be unique (as is proved in the following proposition). The

multiplicative identity element, if it exists, is usually denoted by the symbol

1 and is also referred to as the unity of .

Proposition 7 If is a ring with a multiplicative identity element (a ring

with unity), then the multiplicative identity element (the unity) is unique.

Proof. Suppose that is a ring with unity and suppose that 1 is a multi-

plicative identity element of and that 2 is a multiplicative identity element

of . Then

12 = 2 (since 1 is a multiplicative identity)

and

12 = 1 (since 2 is a multiplicative identity).

4

We conclude that 1 = 2 and hence that the multiplicative identity is

unique.

Returning to the examples of rings that were given above we see that

• The trivial ring is a commutative ring with unity.• , , and are all commutative rings with unity.

• 2 () is not a commutative ring but it is a ring with unity. The unity

is

=

∙1 0

0 1

¸.

• () is a commutative ring with unity. The unity is the function

1 ∈ () defined by 1 () = 1 for all ∈ .

If is a ring with unity, then an element ∈ is said to be invertible

if there exists an element ∈ such that = = 1. If is an invertible

element of a ring with unity, , then the inverse of must be unique (which

the reader can prove) and is usually denoted by −1.In a non—trivial ring with unity, the additive identity element, 0, is never

invertible. The reasoning is as follows: If is a ring with unity and 1 = 0,

then every element ∈ must satisfy

= · 1 = · 0 = 0

by Proposition 6. Thus every element of must be equal to 0 and hence

is the trivial ring. Stated in contrapositive form this means that if is a

non—trivial ring with unity then 1 6= 0. Now suppose that is a non—trivialring with unity and suppose that 0 is invertible. Then there exists ∈

such that 0 · = 1 but Proposition 6 then gives 0 = 1 which contradicts whatwas stated above. Therefore 0 is not invertible.

A commutative ring with unity in which every non—zero element is in-

vertible is called a field. Examples of fields are , , and 5. (More

generally, is a field if and only if is a prime number.) , 2 () and

() are not fields.

A ring, , is said to have the cancellation property if for all , and

∈ we have

( = or = ) =⇒ ( = 0 or = ) .

5

All fields have the cancellation property. also has the cancellation property

but2 () and () do not. Neither does when is not a prime number.

An integral domain is a commutative ring with unity which has the

cancellation property. Thus all fields are integral domains and is also an

integral domain (though it is not a field).

If is a ring then an element ∈ is called a divisor of zero (or a

zero divisor) if 6= 0 and there exists ∈ such that 6= 0 and = 0. Asexamples note that 2 is a divisor of 0 in the ring 6 because 2 6= 0 and 3 6= 0but 2 · 3 = 0 (which also means that 3 is a divisor of 0 in 6). In 2 () we

can find many examples of matrices and such that 6= 0 and 6= 0but = 0 (the zero matrix). Thus 2 () contains many divisors of zero.

In (), the function : → defined by () = 2 is a divisor of zero.

Why?

Theorem 8 A ring, , has the cancellation property if and only if has no

divisors of zero.

Proof. Suppose that has the cancellation property. We will use proof by

contradiction to show that has no divisors of zero.

Let ∈ and suppose that is a divisor of 0. Then there exists 6= 0and there exists ∈ with 6= 0 such that = 0 (or = 0 but we willassume without loss of generality that = 0). Since = 0, then = 0

by Proposition 6. The cancellation property then gives us that either = 0

or = 0. However this is a contradiction to what was deduced above. We

conclude that no element of can be a divisor of 0.

Next suppose that has not divisors of 0. We will show via a direct

proof that must have the cancellation property.

Let , and ∈ and suppose that = (if we suppose =

, the remainder of the proof is handled similarly). Since = , then

+ (− ()) = 0 and by Proposition 6 we obtain + (−) = 0. The

distributive property then gives

(− ) = 0.

Since has no zero divisors then either = 0 or − = 0. That is, either

= 0 or = . We have thus shown that has the cancellation property.

We have defined an integral domain to be a commutative ring with unity

which has the cancellation property. According to the above theorem, it is

equivalent to define an integral domain to be a commutative ring with unity

which has no divisors of zero.

6

2 Subrings and Ideals

Suppose that is a ring. A subring of is a subset, , of which is itself

a ring under the operations of .

Given that is a ring and that ⊆ , what must be checked in order

to see whether is a subring of ?

1. We must check to see if is closed under addition ( ∈ and ∈

implies that + ∈ ).

2. We don’t need to check that is associative under addition because

the associativity of addition is inherited from .

3. We must check that 0 ∈ .

4. We must check that is closed under negation ( ∈ implies that

− ∈ ).

5. We don’t need to check that addition is commutative in because

commutativity of addition is inherited from .

6. We must check that is closed under multiplication ( ∈ and ∈

implies that ∈ and ∈ ).

7. We don’t need to check that multiplication is associative in because

associativity of multiplication is inherited from .

8. We don’t need to check that multiplication is distributive over addition

in because this property is inherited from .

In summary, if is a ring and ⊆ , then is a subring of if and

only if

1. is closed under addition.

2. 0 ∈ .

3. is closed under negation.

4. is closed under multiplication.

7

The following Proposition gives two shorter ways to verify that a non—

empty subset of a ring, , is a subring of .

Proposition 9 Suppose that is a ring and that ⊆ with 6= ∅. Then:

1. is a subring of if and only if is closed under addition, negation

and multiplication.

2. is a subring of if and only if is closed under subtraction and

multiplication.

Proof. Suppose that is a ring and that ⊆ with 6= ∅. We will firstprove Statement 1:

If is a subring of , then must be closed under addition, negation

and multiplication (by the definition of ring).

Suppose that is closed under addition negation and multiplication.

We just need to verify that this implies that 0 ∈ and we will be done.

Since 6= ∅ then there exists and element ∈ . Since is closed under

negation, the we also have that − ∈ . Since is closed under addition

then 0 = + (−) ∈ . Thus is a subring of .

Now we prove Statement 2:

Suppose that is a subring of . Then is closed under multiplication

(by the definition of ring) so we just need to verify that is closed under

subtraction. Let and ∈ . Then, since is closed under negation

(because it is a ring), then − ∈ . Since is also closed under addition

(because it is a ring) then + (−) ∈ which means that − ∈ .

Therefore is closed under subtraction.

Suppose that is closed under subtraction and multiplication. Since

6= ∅, then there exists some element ∈ . Since is closed under

subtraction then − ∈ which implies that 0 ∈ . Next we show that

closed under negation. Suppose ∈ . Then since 0 ∈ and is

closed under subtraction we obtain that 0 − ∈ meaning that − ∈ .

Thus is closed under negation. Finally we show that is closed under

addition. Let and ∈ . Since is closed under negation then − ∈ .

Since is closed under subtraction then − (−) ∈ . This means that

− (−) = + (− (−)) = + ∈ . Thus is closed under addition. We

have thus shown that is a subring of .

Next we discuss certain special subrings that are called ideals. An ideal,

, of a ring, , is a subring of that absorbs products in . When we

8

say that absorbs products we mean that for any ∈ and ∈ we have

∈ and ∈ . (Note that if is a commutative ring to begin with then

only one of the conditions ∈ and ∈ needs to be checked because

= for all and in a commutative ring.) The following Proposition

gives necessary and sufficient conditions for a non—empty subset of a ring to

be an ideal.

Proposition 10 Suppose that is a ring and that ⊆ with 6= ∅.Then is an ideal in if and only if is closed under subtraction and

absorbs products in .

Proof. Suppose that is a ring and that ⊆ with 6= ∅.If is an ideal in then is a subring of and hence is closed under

subtraction (because all rings are closed under subtraction). Also ab-

sorbs products in (by definition of ideal). This proves one direction of the

implication.

Now suppose that is closed under subtraction and that absorbs

products in . Since 6= ∅ and is closed under subtraction then we

will know that is a subring of if we can prove that is closed under

multiplication (by Statement 2 of the preceding Proposition). Let ∈ and

∈ . Then ∈ (because ⊆ ) and since absorbs products in

we conclude that ∈ and ∈ . This shows that is closed under

multiplication and hence that is a subring of . We are already assuming

that absorbs products in and thus is an ideal in . This concludes

the proof.

If is a commutative ring with unity, then one way to construct an ideal

in is to choose any element ∈ and form the set

hi = { | ∈ } .To see that hi as defined above is an ideal in we will employ Proposition

10. First observe that hi 6= ∅ because 1 = ∈ hi. Next let us verify thathi is closed under subtraction and that hi absorbs products in . Let

and ∈ hi. Then there exist and ∈ such that = and = .

This gives

− = − = (− ) ∈ hiand thus hi is closed under subtraction. Now let ∈ and ∈ hi. Thenthere exists ∈ such that = and we obtain

= () = () ∈ hi

9

and thus hi absorbs products in . Therefore hi is an ideal in . For any

∈ , the ideal hi is called the principal ideal in generated by .

We remark that earlier in the Pinter textbook (and in our Modern Alge-

bra I course) the notation hi was used to denote the cyclic subgroup of acertain group generated by the element . We are now using this notation

differently. However it does turn out as we will see that sometimes hi is boththe principal ideal generated by and the cyclic subgroup of the additive

abelian group of generated by .

2.1 Examples of Subrings and Ideals

Example 11 In the ring of integers, , we see that the set of even integers

= {2 | ∈ } = { −4−2 0 2 4 }is an ideal in . This is easily seen using Proposition 10. Clearly 6= ∅.Also the difference of any two even integers is an even integer so is closed

under subtraction and the product of any integer and an even integer is an

even integer so absorbs products in . Thus is an ideal in . In fact

is the principal ideal generated by the element 2 ∈ . That is = h2i.Notice that is also generated by −2. Also notice that is in fact the cyclic

subgroup of the additive abelian group of generated by the element 2 (or by

−2).

Example 12 In the ring of integers, , we see that the set of odd integers

= {2+ 1 | ∈ } = { −3−1 1 3 }is an not a subring of . One reason for this is that is not closed under

addition.

Example 13 Recall that the set of complex numbers, , with standard ad-

dition and multiplication is a commutative ring with unity (in fact a field).

The set of Gaussian integers, , is the set of all complex numbers whose real

and imaginary parts are integers. That is

= { ∈ | = + where ∈ and ∈ } .Clearly 6= ∅ (for example 1 ∈ ) and it is also easy to verify that is

closed under subtraction and closed under multiplication. Thus is a subring

10

of . However does not absorb products in . For example 12 ∈ and

1 + ∈ but1

2(1 + ) =

1

2+1

2 ∈ .

We conclude that is not an ideal in .

Example 14 If is any commutative ring with unity, then is an ideal in

. In fact is the principal ideal generated by 1 because

= { | ∈ } = { · 1 | ∈ } = h1i On the other extreme {0} is also a principal ideal in because

{0} = { · 0 | ∈ } = h0i .

3 Ring Homomorphisms

Suppose that and are rings. A ring homomorphism (or just homo-

morphism) from into is a function : → such that for all 1 and

2 ∈ we have

(1 + 2) = (1) + (2)

and

(12) = (1) (2) .

It is an easy exercise to verify that if : → is a homomorphism then

(0) = 0 and (−) = − () for all ∈ .

If : → is a homomorphism then range() is said to be a homomor-

phic image of . If the homomorphism is one—to—one then we say that

is isomorphic to range(). The kernel of a homomorphism : → ,

denoted by ker (), is the set of all elements in that are mapped by to

the zero element of . That is

ker () = { ∈ | () = 0} .We are going to prove a fundamental homomorphism theorem for rings

that is analogous to the fundamental homomorphism theorem for groups

(given in Chapter 16 of Pinter). We will need the following lemma.

Lemma 15 Suppose that and are rings and that : → is a

homomorphism. Then

11

1. range() is a subring of .

2. ker () is an ideal in .

Proof. Suppose that and are rings and that : → is a homomor-

phism. We will first show that range() is a subring of . It is clear that

range() 6= ∅. Also for any 1 and 2 ∈ range() there exist 1 and 2 ∈

such that 1 = (1) and 2 = (2). Thus

1 − 2 = (1)− (2)

= (1) + (− (2))= (1) + (−2)= (1 − 2)

and thus 1 − 2 ∈ range(). This shows that range() is closed under

subtraction. Also

12 = (1) (2) = (12)

and thus 12 ∈ range() which shows that range() is closed under multi-plication. Therefore range() is a subring of .

Now we will show that ker () is an ideal in. Clearly ker () 6= ∅ because0 ∈ ker (). Also for any 1 and 2 ∈ ker () we have

(1 − 2) = (1)− (2) = 0− 0 = 0so 1 − 2 ∈ ker () which shows that ker () is closed under subtraction.Now suppose that 1 ∈ and 2 ∈ ker (). Then

(12) = (1) (2) = (1) 0 = 0

which means that 12 ∈ ker () and hence ker () absorbs products in .

Therefore ker () is an ideal of .

Let us look at an example of a ring homomorphism.

Example 16 Let = {} with operations defined by + =

+ =

+ = + =

=

=

= = .

12

with these operations is called the “parity ring”. Now define the function

: → to be

() =

½ if is even

if is odd.

To check that is a homomorphism we consider three cases:

If 1 and 2 are both even then

(1 + 2) = = + = (1) + (2)

and

(12) = = = (1) (2) .

If 1 and 2 are both odd then

(1 + 2) = = + = (1) + (2)

and

(12) = = = (1) (2) .

If 1 is even and 2 is odd then

(1 + 2) = = + = (1) + (2)

and

(12) = = = (1) (2) .

Since is onto we see that is a homomorphic image of . Note that

is the additive identity element (the zero) of and thus

ker () = { ∈ | () = } = {all even integers} .

4 Quotient Rings

Suppose that is an abelian group with the operation of addition (denoted

by +) and suppose that is a subgroup of . Then is a normal subgroup

of because is abelian and all subgroups of abelian groups are normal.

This means that left cosets of are equal to their corresponding right cosets

(+ = + for all ∈ ). Recall that

+ = {+ | ∈ } .

13

Lemma 17 If is an abelian group with operation + and is a subgroup

of then for any elements and ∈ we have

1. ∈ + if and only if + = + .

2. + = + if and only if − ∈ .

3. + = if and only if ∈ .

Proof. 1) Suppose that ∈ + . Then there exists ∈ such that

= +. We want to show that this implies that + = + . To do this,

we choose an arbitrary element ∈ + . Then we know that there exists

∈ such that = +. This gives = + = ++ and since + ∈

(because is a subgroup of and hence closed under addition) we see that

∈ + . This shows that + ⊆ + . The proof that + ⊆ + is

similar. Thus + = + .

Now suppose that + = + . Since = +0 ∈ + then ∈ + .

This completes the proof of the first statement.

2) Suppose that + = + . Then ∈ + by part 1 of this

lemma. Hence there exists ∈ such that = + and we conclude that

− = ∈ . To prove the converse suppose that − ∈ . Then there

exists ∈ such that − = . We want to show that this implies that

+ = + . Choosing an arbitrary element ∈ + , we know that there

exists ∈ such that = + . This implies that = + = + +

and since + ∈ then ∈ + . This shows that + ⊆ + . The

proof that + ⊆ + is similar and we conclude that + = + .

3) Suppose that + = . Since = 0 + then + = 0 + which

implies that ∈ 0 + = by part 1 of this lemma. Conversely, suppose

that ∈ . Then ∈ 0 + and thus + = 0 + = by part 1.

The collection of sets = {+ | ∈ } is a partition of . Bydefining the operation

(+ ) + (+ ) = (+ ) +

on we obtain an additive abelian group called the quotient group of

by . We remark that it is necessary to verify that the above—defined addition

operation is well—defined. That is, we must verify that if + = + and

+ = + , then (+ ) + (+ ) = (+ ) + (+ ). The verification

of this is left as an exercise to the reader.

14

Now we will see that if is a ring and is an ideal in , then we can

also make into a ring, called the quotient ring of by , by defining

the multiplication operation by

(+ ) (+ ) = + .

To see that this multiplication operation is well defined, suppose that +

= + and + = + . We need to show that (+ ) (+ ) =

(+ ) (+ ). By Lemma 17 we know that − ∈ and − ∈ . Since

absorbs products in and ∈ and − ∈ then (− ) ∈ which

implies that − ∈ . Likewise we obtain that − ∈ . Since

is closed under addition then (− ) + (− ) ∈ which implies that

− ∈ . By Lemma 17 we thus obtain that + = + or equivalently

+ = + . This shows that (+ ) (+ ) = (+ ) (+ ).

In summary, if is a ring and is an ideal in then the set =

{+ | ∈ } with operations defined by(+ ) + (+ ) = (+ ) +

(+ ) (+ ) = +

is a ring. (It is left an exercise to verify that satisfies all of the require-

ments of the definition of a ring.)

Example 18 Let be the ring of integers and consider the principal ideal

h4i = { −8−4 0 4 8 } .The distinct cosets of h4i in are

0 = 0 + h4i = { −8−4 0 4 8 }1 = 1 + h4i = { −7−3 1 5 9 }2 = 2 + h4i = { −6−2 2 6 10 }3 = 3 + h4i = { −5−1 3 7 11 }

and thus h4i = ©0 1 2 3ª. Note that although this quotient group has onlyfour distinct elements, each element can be named in many different ways.

For example 2 = −6 and 3 = 15. Nonetheless we know that coset additionand multiplication are well—defined. For example

2 + 3 = 5 = 1 and −6 + 15 = 9 = 1and

2 · 3 = 6 = 2 and −6 · 15 = −90 = 2.

15

Example 19 If is any ring then we know that = {0} is an ideal in .

The corresponding quotient ring is

{0} = {+ | ∈ } = {{} | ∈ } .

Likewise we know that is an ideal in and we see that the corresponding

quotient ring is

= {+ | ∈ } = {} .

These are, respectively, the largest and smallest quotient rings that can be

formed from a ring. The ring {0} is isomorphic to itself (the actual

difference between and {0} is subtle — the elements of {0} are sin-gleton subsets of ). The ring contains only one element which is the

set itself. In fact is the trivial ring.

The following theorem precisely identifies all of the possible homomorphic

images of a given ring.

Theorem 20 (The Fundamental Theorem of Ring Homomorphisms)

If is a ring and is an ideal in , then is a homomorphic image of .

Conversely, if is a homomorphic image of then there exists an ideal, ,

in such that is isomorphic to . Specifically, if is a homomorphism

from onto then is isomorphic to ker ().

Proof. Suppose that is a ring and that is an ideal in . Define : → by () = + . (Each element of is mapped by to the coset of

to which it belongs.) It is clear that maps onto . We will now

show that is also a homomorphism. To to this we take any and ∈

and observe that

(+ ) = (+ ) + = (+ ) + (+ ) = () + ()

and

() = + = (+ ) (+ ) = () () .

Since is a homomorphism of onto then is a homomorphic image

of .

Now suppose that is a homomorphic image of . Then there exists a

homomorphism, , which maps onto . Recalling that ker () is an ideal

16

in (by Lemma 15) we will now show that is isomorphic to ker ().

To do this we define the function : ker ()→ by

(+ ker ()) = () .

To see that is well—defined observe that if + ker () = + ker () then

− ∈ ker () and thus (− ) = 0. Since is a homomorphism then

()− () = 0 which means that () = () and we obtain

(+ ker ()) = () = () = (+ ker ()) .

This shows that is well—defined.

To see that is a homomorphism observe that for any + ker () and

+ ker () ∈ ker () we have

((+ ker ()) + (+ ker ())) = ((+ ) + ker ())

= (+ )

= () + ()

= (+ ker ()) + (+ ker ())

and

((+ ker ()) (+ ker ())) = (+ ker ())

= ()

= () ()

= (+ ker ()) (+ ker ()) .

In order to complete the proof we need to show that is one—to—one and

that maps ker () onto . If + ker () and + ker () ∈ ker ()

and (+ ker ()) = (+ ker ()) then () = () which implies that

(− ) = 0 and hence − ∈ ker (). This implies that + ker () = + ker (). We have thus shown that is one—to—one. Now let ∈

and note that since maps onto then there exists ∈ such that

= (). We then see that = () = (+ ker ()). This shows that

maps ker () onto .

An illustration of Theorem 20 is given in the following two examples.

17

Example 21 Let us identify all homomorphic images of 6. Since the only

ideals in 6 are

h0i = 0h1i = 6

h2i = {0 2 4}h3i = {0 3} ,

then the only homomorphic images of 6 (up to isomorphism) are

6

{0} = {{} | ∈ 6} which is isomorphic to 6

6

6= {6} which is the trivial ring

6

{0 2 4} = {{0 2 4} {1 3 5}} which is isomorphic to 26

{0 3} = {{0 3} {1 4} {2 5}} which is isomorphic to 3.

The specific homomorphisms that give the above four results are : 6 →6 {0} defined by

() = {} , : 6 → 66 defined by

() = {6} ,the function : 6 → 6 {0 2 4} defined by

(0) = (2) = (4) = {0 2 4} (1) = (3) = (5) = {1 3 5}

and : 6 → 6 {0 3} defined by (0) = (3) = {0 3} (1) = (4) = {1 4} (2) = (5) = {2 5} .

Example 22 Consider the function : → 2 defined by

() =

½0 if is even

1 if is odd.

18

This is a homomorphism of onto 2 which means that 2 is a homomorphic

image of . Also the kernel of is

ker () = {all even integers}

and we see that 2 is isomorphic to ker ().

4.1 Maximal Ideals

An ideal, , in a ring is said to be a proper ideal in if ⊂ ( is a

proper subset of ). A maximal ideal in is defined to be a proper ideal,

, in which is not properly contained in any other proper ideal of . In

other words is a maximal ideal in if

1. is a proper ideal in and

2. There does not exist any ideal, , in such that ⊂ ⊂ .

Theorem 23 Suppose that is a commutative ring with unity and that

is a proper ideal in . Then is a field if and only if is a maximal ideal

in .

Before giving the proof of this theorem, let us consider a couple of exam-

ples.

Example 24 We know that

h4i = { −8−4 0 4 8 }

is a proper ideal in . However it is not a maximal ideal because h2i is alsoa proper ideal in and h4i ⊂ h2i. Consequently the quotient group h4i isnot a field. (In particular we know that the element 2 ∈ 4 is not invertible.)

On the other hand the ideal h2i is maximal in and we find that h2iis a field. In general we find that hi is a field if and only if the integer is prime. This is because the ideal hi is maximal in if and only if is

prime.

Here is the proof of the theorem: Suppose that is a commutative ring

with unity and that is a maximal ideal in . We want to prove that is

a field. This means that we must show that every non—zero element of is

19

invertible. To this end suppose that ∈ and that 6= . (Recall that

is the zero element of .) We would like to find an element ∈

such that = 1+ (since that 1 + is the unity element of ). Since

6= then = + for some ∈ with ∈ . Now define

= { − | ∈ and ∈ } .We see that ⊂ . This is because for any ∈ , we have − ∈ which

means that = 0− (−) ∈ and hence ⊆ . However = · 1− 0 ∈

but ∈ so ⊂ . Also it can be seen that is closed under subtraction

and that absorbs products in . Thus is an ideal in . Since is

assumed to be a maximal ideal in and ⊂ then it must be the case

that = . Thus 1 ∈ which means that there must exist some ∈

and some ∈ such that − = 1. We claim that the element = +

is the inverse of = + . To see this note that

= (+ ) ( + )

= +

= (1 + ) +

= (1 + ) + ( + )

= (1 + ) + (0 + )

= 1 + .

This shows that is invertible and we have thus shown that is a field.

To prove the converse suppose that is a commutative ring with unity

and is a proper ideal in and is a field. We want to show that is a

maximal ideal in . To do this let us suppose that is an ideal in with

⊂ and let us show that it must be then be the case that = . Here

is why: Since ⊂ then there exists an element ∈ such that ∈

which means that + 6= and hence + is invertible (because is a

field). Thus there exists an element ∈ such that ( + ) (+ ) = 1 +

which means that + = 1+ which means that 1− ∈ which means

that 1 − ∈ (since ⊂ ). Since 1 − ∈ then 1 + = +.

However since ∈ and ∈ and absorbs products in then ∈

which means that + = and we now see that 1 + = and hence

that 1 ∈ . Since 1 ∈ and absorbs products in then = 1 ∈ for

all ∈ which means that = . We have now proved that must be

maximal in .

20

5 Integral Domains

For any ring with unity, , we define the characteristic of to be the

smallest positive integer such that

1 + 1 + · · ·+ 1| {z } times

= 0

if such an exists and define the characteristic of to be 0 if no such

exists.

Example 25 The set of integers, , with the usual addition and multipli-

cation has zero element 0 and unity element 1. Since there is no positive

integer such that

1 + 1 + · · ·+ 1| {z } times

= 0

then has characteristic 0. It can likewise be seen that the rings , and

all have characteristic 0.

Example 26 The trivial ring, = {0} with operations 0+0 = 0 and 0·0 = 0is a ring in which 1 = 0. Thus has characteristic 1.

Example 27 The ring 5 has characteristic 5 because in 5 we have

1 6= 01 + 1 6= 0

1 + 1 + 1 6= 01 + 1 + 1 + 1 6= 0

1 + 1 + 1 + 1 + 1 = 0.

In general, the ring has characteristic .

Before proceeding it will be convenient to introduce some useful notation.

If is any ring, ∈ and is a positive integer we define

· = + + · · ·+ | {z } times

.

In addition we define

0 · = 0

21

and for any positive integer we define

(−) · = − ( · ) .

With these definitions we have the following useful algebraic rules: For any

ring , element ∈ and any integers and ,

+ = (+ )

() = () .

Recall that if is any abelian group with operation of addition the addi-

tive order of an element ∈ is defined to be the smallest positive integer

such that · = 0. The order of is denoted by ord(). If no such positiveinteger exists then we say that ord() = ∞. In any group, , we haveord(0) = 1 because 1 · 0 = 0.

Theorem 28 If is an integral domain then all non—zero elements of

have the same additive order.

Proof. Suppose that is an integral domain and let ∈ with 6= 0. Forany positive integer we have

· = + + · · ·+ | {z } times

= 1 + 1 + · · ·+ 1| {z } times

=

Ã1 + 1 + · · ·+ 1| {z }

times

!= ( · 1) .

Thus for any non—zero ∈ and any positive integer we have · =( · 1) . If · = 0 then ( · 1) = 0 and since is an integral domain

(and hence has no divisors of zero) and 6= 0 then it must be the case that · 1 = 0. Conversely, if · 1 = 0 then · = 0. Hence for any non—zero

∈ and any positive integer we have · = 0 if and only if · 1 = 0.This means that the additive order of must equal the additive order of 1

(make sure you see why this is so).

22

Example 29 To illustrate the above theorem note that all non—zero elements

in 5 have additive order 5: For example

3 6= 03 + 3 = 1 6= 0

3 + 3 + 3 = 4 6= 03 + 3 + 3 + 3 = 2 6= 0

3 + 3 + 3 + 3 + 3 = 0

so ord(3) = 5. (The reader can verify that ord() = 5 for all non—zero

∈ 5.) This fact is true because 5 is an integral domain. Note that the

same is not true in rings that are not integral domains. For example in 6we have ord(2) = 3 and ord(3) = 2.

Two more theorems of interest are the following:

Theorem 30 If is a non—trivial integral domain with non—zero character-

istic then the characteristic is a prime number.

Proof. Suppose that is a non—trivial integral domain with a non—zero

characteristic. Since is non—trivial then 1 6= 0 in . Also, since the

characteristic of is non—zero then the characteristic must be some integer

≥ 2. (The characteristic cannot equal 1 because 1 6= 0). Suppose, for thesake of obtaining a contradiction, that is not a prime number. Then there

exist positive integers and with 1 and 1 such that

= . Since · 1 = 0 then () · 1 = 0. However, note that () · 1 =( · 1) ( · 1) (make sure you understand why this is so perhaps by writing · 1 and · 1 out the long way) and hence we have ( · 1) ( · 1) = 0. Since has no divisors of zero then either · 1 = 0 or · 1 = 0. However this is acontradiction because is the characteristic of and both of the integers

and are positive and less than . This completes the proof.

Theorem 31 Every finite integral domain is a field.

Proof. Suppose that is a finite integral domain. If is the trivial integral

domain then it is a field. Furthermore, if = {0 1} is the two—elementintegral domain (2) then is a field. Having taken care of these two cases,

we now suppose that contains at least three elements. One of the elements

23

must be 0 and one must be 1 6= 0 and there is at least one other element thatis not equal to 0 or to 1.

Since is finite then we can list the elements of as

= {0 1 1 2 } .

Thus we are assuming that 6= for any and with 6= and that

6= 1 for any and that 6= 0 for any and that 1 6= 0. This means that contains exactly + 2 members. In order to show that is a field we

must show that all non—zero elements of are invertible. It is clear that 1

is invertible because 1 · 1 = 1 and hence 1−1 = 1. Thus let us choose one ofthe elements with 6= 0 and 6= 1 and show that is invertible. To dothis we multiply every element in the above listing of by to obtain the

new listing

= {0 1 2 } .We claim that = . To see this note that clearly ⊆ because is

closed under multiplication. Also, no two elements in are equal to each

other because = implies that = by the cancellation property

which implies that = because each element of is listed only once in the

listing of given above. By similar reasoning it can be seen that 6= 0and 6= for any = 1 2 and we also know that 6= 0. Thus notwo elements of are equal to each other. Since no two elements of are

equal to each other then contains exactly + 2 elements. However, since

⊆ and also contains exactly +2 elements then = . To complete

the proof we observe that since = then 1 ∈ . We know that 6= 1and that 0 6= 1 so it must be the case that = 1 for some = 1 2 and this shows that is invertible. The proof is complete.

6 The Integers

The integers are the most familiar set of numbers to us beginning from our

days in grade school. Although we did not think of them this way in grade

school, we now know that they are an integral domain. Integral domains are

defined by their algebraic properties alone and we are very familiar with the

algebraic properties of the integers (how to add and subtract and multiply

with them and many other observed algebraic properties such as the can-

cellation property). We also now know that there are many other integral

24

domains besides the integers. The examples that first come to mind are the

rings where is a prime number. An obvious difference between and

is that has infinitely many members and has only finitely many

members. Another property that has is that its members can be ordered

in a way that is compatible with many pleasant algebraic properties. When

we order in the usual way

−4 −3 −2 −1 0 1 2 3 4

we observe properties such as, for example, the fact that if , and ∈

with then + + . On the other hand, integral domains such as

5 =©0 1 2 3 4

ªcan’t be ordered in any way that goes along nicely with

the algebra. For example suppose that we were to just try what might seem

to be the obvious ordering

0 ≤ 1 2 3 4.

After thinking about this it immediately does not seem to make much sense

because the elements of 5 are sets and for example

0 = { −10−5 0 5 10 }

and

1 = { −9−4 1 6 11 }so there seems to be no good reason to say that 0 1 because each member

of 0 is greater than many of the members of 1. We will now examine the

properties of the integral domain that makes it different from any other

integral domain.

To begin, we define an ordered integral domain to be an integral

domain, , that also has an order relation, denoted by , such that

1. For any and ∈ , exactly one of the following is true: either

or = or .

2. For any , and ∈ , if and then .

3. For any , and ∈ , if then + + .

4. For any , and ∈ , if and 0 then .

25

Given that is an ordered integral domain with order relation , we

will also use the symbols and ≤ and ≥ in the usual way. For example ≤ means that either or = . Some basic properties of the order

relation are given in the following Proposition. (There are many other basic

properties that you are asked to prove in the homework exercises.)

Proposition 32 Suppose that is a non—trivial (meaning that 1 6= 0) or-dered integral domain and suppose that and ∈ . Then

1. If 0 then − 0 and if 0 then − 0.2. If 0 and 0 then 0.

3. If 0 and 0 then 0.

4. If 0 and 0 then 0.

5. If 6= 0 then 2 0.

6. 1 0

7. If then − −.It is easy to see that the system of integers with its usual ordering is

an ordered integral domain. (We have been used to working with these

properties for a long time.) Another property that the integers have is what

is called the well—ordering property. This property says that if we take any

non—empty subset, , of the set of positive integers, +, then must have

a smallest member. More formally, if ⊆ + and 6= ∅ then there existsan element ∈ such that ≤ for all ∈ . (Another way to say

this is for all ∈ such that 6= .) The well—ordering property

of the integers may seem very obvious to us and hardly worth mentioning.

However it is the property that makes the integers be different from other

integral domains. Hence let us make a fewmore definitions before proceeding:

If is an ordered integral domain and is an element of such that 0

then we will say that is positive. (Likewise if 0 then we will say

that is negative.) The set of all positive elements of will be denoted

by +. We will say that has the well—ordering property or that is

an integral system if every non—empty subset of + has a least member.

This means, specifically, that if ⊆ + and 6= ∅ then there exists ∈

such that ≤ for all ∈ .

26

Lemma 33 If is a non—trivial ordered integral domain with the well—

ordering property then there does not exist any element ∈ such that

0 1.

Proof. Suppose that is a non—trivial ordered integral domain with the

well—ordering property and suppose there does exist an element ∈ such

that 0 1. Then define = { ∈ | 0 1}. Clearly ⊆ +

because every member of is positive and also we see that 6= ∅ because ∈ . By the well—ordering property there exists ∈ such that ≤ for

all ∈ . Clearly 0 1. Since 6= 0 then 2 0 and since 1 then

1 which means that 2 . We now have that 0 2 1. This is

a contradiction because we see that 2 ∈ and 2 but was assumed to

be the least member of . Our conclusion is that there is no element ∈

such that 0 1.

Theorem 34 If is a non—trivial ordered integral domain with the well—

ordering property then every element of is an integer multiple of 1 (the

unity element of ) and the elements of are ordered as follows:

−3 · 1 −2 · 1 −1 · 1 0 · 1 1 · 1 2 · 1 3 · 1 .

Proof. To begin the proof we first observe that every integer multiple of 1 is

a member of and that these members are ordered as shown above. To see

this note that because is closed under addition we have

1 ∈

2 · 1 = 1 + 1 ∈

3 · 1 = 1 + 1 + 1 ∈

etc.

and no two of these elements are equal to each other. For example we can’t

have 2 · 1 = 3 · 1 because this means that 1 + 1 = 1 + 1 + 1 from which we

would obtain 0 = 1 (which is not possible since is non—trivial). In fact we

have by definition of the order relation and the fact that 0 1 that

(1 + 1) + 0 (1 + 1) + 1

and hence 2 · 1 3 · 1. Although perhaps this is not a rigorous argument,we can use similar reasoning to conclude that · 1 · 1 for any positive

27

integers and with . By property 7 of Lemma 32 this gives us that

− ( · 1) − ( · 1) or equivalently − · 1 − · 1 for all positive integers and with which establishes the order of the multiples of 1 shown

in the statement of this theorem.

The only thing left to prove is that contains no elements other than

the integer multiples of 1. Suppose that does contain such an element

which we will call . Without loss of generality we can assume that 0

(because if 0 we could take the negative of and just call that ). We

now define = { ∈ + | is not an integer multiple of 1}. Then ⊆ +

and 6= ∅ because ∈ . By the well—ordering property, there exists an

element ∈ such that ≤ for all ∈ . Since Lemma 34 tells us that

there is no element of that lies strictly between 0 and 1 and since 6= 1

(since is not an integer multiple of 1) then 1. We obtain from this that

+ (−1) 1 + (−1) which means that − 1 0. Furthermore − 1 cannotbe an integer multiple of 1 because if it were then

= (− 1) + 1

would also be an integer multiple of 1 which we have said it is not. We now

see that − 1 ∈ but also, since −1 0 we have that − 1 . This is a

contradiction because was assumed to be the least member in . The proof

of the theorem is now complete.

The theorem that has just been proved tells us that the ring of integers

is essentially the only ordered integral domain that has the well—ordering

property. By “essentially” we mean that any other ordered integral domain,

, that has the well—ordering property is isomorphic to . If is any ordered

integral domain (with unity element 1) with the well—ordering property then

an isomorphism from to is the function : → defined by () = ·1.The Principal of Mathematical Induction comes from the following theo-

rem which relies on the well—ordering property of the integers.

Theorem 35 Suppose that ⊆ + and that

1. 1 ∈

2. For any ∈ +, ∈ implies that + 1 ∈ .

Then = +.

28

Proof. Suppose that 6= +. Then there is some number ∈ + such

that ∈ . Let = { ∈ + | ∈ }. Then ⊆ + and 6= ∅ (because ∈ ). By the well—ordering of the integers there exists some element ∈

such that ≤ for all ∈ . Since 1 ∈ then 1 ∈ so 6= 1 which meansthat 1. However this implies that −1 ∈ for if −1 ∈ then −1 ∈

which would mean that = (− 1) + 1 ∈ by property 2 that defines .

Since − 1 ∈ and − 1 then we have obtained a contradiction of the

fact that is the least member of and the proof is complete.

29