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Notes on Ring Theory
S. F. Ellermeyer
Department of Mathematics
Kennesaw State University
January 15, 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.
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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
 · ( + ) =  ·  +  · 
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and
( + ) ·  =  ·  +  · .
Let us look at a few examples of rings.
Example 1 The trivial ring is the one—element set  = {0} with addition
operation 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 numbers, , 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 1
0 1
1 2
2 3
3 0
2
2
3
0
1
·
0
1
2
3
3
3
0
1
2
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 and
operations defined by
∙
¸ ∙
¸ ∙
¸
1 1
2 2
1 + 2 1 + 2
+
=
1 1
2 2
1 + 2 1 + 2
and
is a ring.
∙
1 1
1 1
¸∙
2 2
2 2
¸
=
∙
1 2 + 1 2 1 2 + 1 2
1 2 + 1 2 1 2 + 1 2
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  + (−). The
following 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
which gives
·0+0=0
 · 0 = 0.
The proof that 0 ·  = 0 is similar.
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2. Let  and  ∈ . Then
·(−)+· =  (− + ) = ·0 = 0 (by Part 1 which has just been proved).
Since  · (−) +  ·  = 0 and since the additive inverse of  ·  is
unique (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 multiplication operation have an identity element in  or that (if the multiplication
operation does have an identity element) that each element in  have a multiplicative 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 multiplicative 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 multiplicative identity element of  and that 2 is a multiplicative identity element
of . Then
1 2 = 2 (since 1 is a multiplicative identity)
and
1 2 = 1 (since 2 is a multiplicative identity).
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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—trivial
ring with unity and suppose that 0 is invertible. Then there exists  ∈ 
such that 0 ·  = 1 but Proposition 6 then gives 0 = 1 which contradicts what
was stated above. Therefore 0 is not invertible.
A commutative ring with unity in which every non—zero element is invertible 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  = ) .
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All fields have the cancellation property.  also has the cancellation property
but 2 () 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. As
examples note that 2 is a divisor of 0 in the ring 6 because 2 6= 0 and 3 6= 0
but 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= 0
but  = 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= 0
and there exists  ∈  with  6= 0 such that  = 0 (or  = 0 but we will
assume 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.
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