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.
1Rings
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 ( + ) = + . · · · Letuslookatafewexamplesofrings.
Example 1 The trivial ring is the one—element set = 0 with addition operation defined by 0+0 = 0 and multiplication operation de{fi}ned 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 1 2 3 · 0 0 1 2 3 0 0 0 0 0 1 1 2 3 0 1 0 1 2 3 2 2 3 0 1 2 0 2 0 2 3 3 0 1 2 3 0 3 2 1 is a ring. (In general, for any given integer 1,theset = 0 1 2 1 with addition and multiplication operations de≥fined “modulo ”isaring.){ − }
Example 4 The set, 2 (),ofall2 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 + + 1 1 2 2 = 1 2 1 2 1 2 1 2 1 1 2 2 12 + 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,forany 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=0and 0 =0for 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,wecansubtract 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=0 · which gives 0=0. · The proof that 0 =0is similar. · 3 2. Let and .Then ∈ ( )+ = ( + )= 0=0(by Part 1 which has just been proved). · − · − · Since ( )+ =0and 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 ( ).Inotherwords ( )= ( ). · · − − · · − 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 andisalsoreferredtoastheunity 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. Returningtotheexamplesofringsthatweregivenaboveweseethat
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 10 = . 01 ∙ ¸ () is a commutative ring with unity. The unity is the function • 1 () defined by 1 ()=1for 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,∈,thentheinverseof must be unique (which 1 the reader can prove) and is usually denoted by − . 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 =0.Nowsupposethat is a non—trivial ring with unity and suppose that 6 0 is invertible. Then there exists such that 0 =1but Proposition 6 then gives 0=1which contradicts what∈ was 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.Examplesoffields are , , and 5.(More generally, is a field if and only if is a prime number.) , 2 () and () are not fields. Aring,,issaidtohavethecancellation property if for all , and we have ∈ ( = or = )= ( =0or = ) . ⇒ 5 All fields have the cancellation property. also has the cancellation property but 2 () and () do not. Neither does when isnotaprimenumber. 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 =0and there exists∈ such that =0and =0.As 6 ∈ 6 examples note that 2 is a divisor of 0 in the ring 6 because 2 =0and 3 =0 6 6 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 =0and =0 6 6 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 Aring,, 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 =0 and there∈ exists with =0such that =0(or =0but we6 will assume without loss∈ of generality6 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 =0or =0. That is, either =0or = .Wehavethusshownthat has the cancellation− property. We have definedanintegraldomaintobeacommutativeringwithunity 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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