A commutative group. The set of integers. The set of rational numbers. The set of real numbers. The set residue classes mod n. The set of reduced residue classes mod n.

### The Concept of a Ring

Definition 1: A ring is a nonempty set R on which there are defined two binary operations + and ·, which satisfy the following axioms:

R1. With respect to +, R is an abelian group.

R2. · is associative:a · (b · c) = (a · b) · c for all a,b,c R.

R3. The following distributive laws are satisfied:

for all a, b, c R.

The operation + is called addition and the operation · is called multiplication.

In a ring R, we will denote the identity with respect to addition by 0, and we will denote the inverse of a with respect to + by -a. We will call the identity with respect to addition the zero element of R or just zero

Note that although we have assumed that R contains a an identity with respect to addition, we have made no such assumption with regards to multiplication. An identity element i with respect to multiplication is an element of R such that

i · a = a · i = 1

for all a R. A given ring may or may not contain such an element. We will give examples of both phenomena below. However, if i and i' are two identity elements with respect to multiplication, then we must have

i = i · i' = i'.

Therefore, any two identities of R are equal and R contains at most one identity with respect to multiplication.

Definition 2: Let R be a ring. If R contains an identity with respect to multiplication, then we say that R is a ring with identity. The unique identity with respect to multiplication will be denoted 1R (or just 1 if R is clear from context) and will be call the identity of R. (Note that some texts require that all rings have an identity. We will adopt that convention later.)

We will see below that the operation of multiplication need not satisfy the commutative law. Therefore, let us make the following definition.

Definition 3: Let R be a ring. We say that R is a commutative ring if

a · b = b · a

for all a, b R.

Let us now get some feel for the variety of possible rings by studying a number of examples.

Example 1: Let R = Z, with respect to the usual operations of addition and multiplication. Then as mentioned at the outset, all the ring axioms are verified. The ring Z is commutative and has identity 1.

Example 2: Let R = Q, the set of all rational numbers, and let addition and multiplication, respectively, be addition and multiplication of fractions. With respect to these operations, Q becomes a ring. Then ring Q is commutative and has identity 1.

Example 3: Let R = R, the set of all real numbers. Let addition and multiplication be addition and multiplication of real numbers. Then R becomes a commutative ring with identity 1.

Example 4: Let R = {2n | n Z} = {..., -4, -2, 0, 2, 4, ...}, and let the operations of addition and multiplication be the same as in Z. Since the sum, difference, and product of even integers is even, the addition and multiplication in Z define binary operations on R, with respect to which R becomes a ring. R is clearly commutative, but has no identity. For if 2n R is an identity, then (2n)·2 = 2, which leads to a contradiction, since the left hand side is divisible by 4, but the right hand side is not. The ring R is sometimes demoted 2Z. It is possible to generalize this example as follows: Let r be an integer, and let

rZ = {rn | n Z}.

Then, with respect to the usual operations of addition and multiplication of integers, rZ is a commutative ring having no identity unless r = ±1.

Example 5: Let R = {0}, and let us define addition and multiplication by

0 + 0 = 0,
0 · 0 = 0.

Then the axioms R1-R3 are satisfied in a trivial way. Thus, R is a ring called the trivial ring. Note that R is a commutative ring with identity 0. Thus, in this example, 1R = 0R.

Example 6: Let R = Zn for some positive integer n, and let addition and multiplication be defined as addition and multiplication of residue classes. It was established earlier that R1-R3 are satisfied, so that R becomes a ring. Note that R is a finite ring with n elements. Moreover, R is commutative with 1. The most peculiar property exhibited by some of the rings Zn is that it often happens that the product of two nonzero elements of Zn equals 0, which is the zero element of Zn. For example, consider 2, 3 Z6. Then 2 0, 3 0, but nevertheless 2 · 3 = 6 = 0. This situation should be viewed in contrast with the situation which occurs in examples 1-5. In each of these examples, the product of two nonzero ring elements is nonzero.

Example 7: Let R denote the set of functions f :RR. If f,g R, let us define f + g and the product f · g by

(f + g)(x) = f(x) + g(x)   (x R),
(f · g)(x) = f(x) · g(x).

For example, if f(x) = x, g(x) = 2x2, we have (f + g)(x) = x + 2x2, and (f · g)(x) = 2x3. Note that we have defined the sum and product of functions by specifying the values of the sum and products for all values x R. Thus, by the sum and product functions in R also belong to R, and we have so defined two binary operations on R. With respect to these operations, R is a ring. We have already verified earlier that with respect to +, R is an abelian group. Thus it suffices to check axioms R2 and R3, which is fairly simply and we will leave to the reader. It is easy to see that R is a commutative ring with identity. The identity element is the function I defined by I(x) = 1 for all x R, Note that in example 6, it can happen that the product of two nonzero elements of R is the zero element of R. For the zero element of R is the function 0 defined by 0(x) = 0 for all x R. Let f, g R be defined by

f(x) = 0 (x 0),   f(0) = 1,
g(x) = 1 (x 0),   g(0) = 0.

Then f 0, g 0, but f · g = 0.

Example 8: Let R be the set of all functions f :RR of the form

f(x) = a0 + a1x + a2x2 + ... + anxn   (a0,...,an R).

Such a function is called a polynomial function. Let addition and multiplication of functions be defined as in Example 7. Then the sum (and product) of two polynomial functions is again a polynomial function, so we have two binary operations defined on R. With respect to these operations, R is a commutative ring with identity.

Example 9: Let R denote the set of all 2 2 matrices with real entries. A typical element of R looks like

where a, b,c, d, R. Let us define addition and multiplication of matrices by

With respect to these operations, R becomes a ring with identity. The zero element and the identity of R are given by

Note that

Therefore, R is not a commutative ring. We will henceforth denote the ring of 2 2 matrices with real entries by M2(R). Note that the last computation also shows that the product of two nonzero matrices may be zero.

Example 10: Let d be an integer, and let R consist of the set of all complex numbers of the form a + b, where a, b Z. Let addition and multiplication in R be the usual addition and multiplication for complex numbers, respectively. Since

(a + b) + (a' + b') = (a + a') + (b + b'),
(a + b)(a' + b') = (aa' + bb'd) + (ab' +ba'),

we see that addition and multiplication define binary operations on R. The fact that R forms a ring is now trivial to verify from the properties of the complex numbers. The ring R will be denoted Z[].

Example 11: Let R be the set of all complex numbers of the form a + b, where d is given and a, b Q. Then the ring R is a ring which contains Z[]. We will denote the ring of this example by Q().

Example 12: Let R and S be any two rings, and let R S denote the Cartesian product of R and S as sets). Let us define addition and multiplication in R S by

(r,s) + (r',s') = (r + r', s + s'),
(r,s)(r',s') = (rr', ss').

Then it is easy to verify that R S is a ring, called the Cartesian product of R and S. Sometimes, the Cartesian product is called the direct sum of R and S and is denoted by R S.

Now that we have surveyed the wide possibilities, which the ring axioms create, let us establish some elementary properties of rings. The following proposition will start us along that line.

Proposition 4: Let R be a ring. a, b R. Then

(1) a · 0 = 0 · a = 0.

(2) a · (-b) = (-a) · b = -(a · b).

(3) (-a) · (-b) = a · b.

Note that these rules of arithmetic in an arbitrary ring are just generalizations of well-known facts from the algebra of the integers, Z.

Proof:

(1) Since 0 is the identity with respect to addition, 0 + 0 = 0. Therefore, by the distributive law,

a · 0 = a · (0 + 0) = a · 0 + a · 0.

Therefore, by adding -(a · 0) to both sides of this last equation, we get a · 0 = 0. Similarly 0 · a = 0.

(2) Let y = -(a · b). Then a · b + y = 0. Therefore,

(-a) · b = (-a) · b + 0

= (-a) · b + [a · b + y]

= (-a + a) · b + y   (by associative and distributive laws)

= 0 · b + y   (since -a + a = 0)

= 0 + y   (by (1))

= y.

Therefore, (-a) · b = -(a · b).

(3) By part (2),

(-a) · (-b) = -a · (-b)

= -(-a · b)

= a · b   (since R is a group with respect to addition)

Let us now examine some of the phenomena which were exhibited by our examples. In Examples 6 and 7 we found that there exist nonzero ring elements a and b such that a · b = 0. Note that if a or b is zero, then a · b = 0 by Part (1) above. However, it is somewhat surprising that nonzero elements can have a zero product. Therefore, we are prompted to make the following definition.

Definition 5: Let R be a ring, a R, a 0. We say that a is a right zero divisor (respectively, left zero divisor) if there exists b R, b 0, such that a · b = 0 (respectively, b · a = 0). An element of R is called a zero divisor of R if it is either a right or left zero divisor.

If R is a commutative ring, then the distinction between right and left zero divisors disappears. An element of a commutative ring R is a right zero divisor if and only if it is a left zero divisor.

Definition 6: A commutative ring R is said to be an integral domain if R contains no zero divisors. In other words, R is an integral domain if the product of any two nonzero elements of R is nonzero. (For convenience in stating various results, we will always assume that an integral domain always contains an identity.)

For example Z6 is not an integral domain because 2 is a zero divisor. Similarly, Example 7 shows the ring of functions f :RR is not an integral domain. On the other hand Z is an integral domain. For if a and b are nonzero integers, then a · b 0. Moreover if p is prime, then Zp is an integral domain.

Proposition 7: Suppose that R is an integral domain, a,b,c R, a 0. Suppose that a · b = a · c. Then b = c.

Proof: By Proposition 4 and the distributive law, we have

0 = a · b - a · c = a · b + a · (-c)
= a · (b - c).

However, since a 0 and R is an integral domain, b - c = 0, and thus b = c.

If R is a ring with identity 1, it makes sense to ask whether an element a of R has an inverse with respect to multiplication. That is, does there exist b R such that

a · b = b · a = 1?

If such an element b exists, it is unique. We have already worked through this sort of argument several times, and therefore we will leave the verification that b is unique as an exercise for the reader. If a has an inverse with respect to multiplication, then the unique inverse will be denoted a-1.

Definition 8: Let R be a ring with identity 1, and let a R. If a has an inverse with respect to multiplication, then we say that a is a unit of R. (Some texts will also say that a is invertible.)

Example 13: Let R be any ring with identity. Then 1 and -1 are units. In fact, since 1 · 1 = 1, (-1) · (-1) = 1 · 1 = 1, we see that

1-1 = 1, (-1)-1 = -1.

Example 14: Let R = Z. Then the only units of R are 1 and -1. For if a R is a unit, there exists b R such that a · b = 1. Then a = b = 1 or a = b = -1

Example 15: Let R = Q. Then 0 is not a unit of R since 0 · b = 0 1 for all b R. However, if a/b Q - {0}, then a/b is a unit of R. In fact, since

a/b · b/a =1,

we see that (a/b)-1 = b/a.

Example 16: Let R = the trivial ring. Then, since 1 = 0 in this ring, 0 is a unit.

Example 17: Let R be any ring with identity which is not the trivial ring. In this ring 1 0. For if R is not the trivial ring, there exists a nonzero element in R. Pick one such element a. Then, if 1 = 0, we would have

1 · a = 0 · a = 0.

However, 1 · a = a 0 Thus we reach a contradiction, so that 1 0. Let us now show that 0 is not a unit of R. Indeed, for all b R, we have 0 · b = 0 1. Thus in any ring that is not the trivial ring, 0 is not a unit.

Example 18: Let R = {a + b | a,b Z} be the ring of Example 10. The units of R are not quite so easy to write down as the following special cases show:

d = -1: 1 + 0, -1 + 0, 0 + , 0 - are units with respective inverses 1 + 0, -1 + 0, 0 - , 0 + .

d = 2: 1 + 0, -1 + 0, 1 + , 1 - , 3 + 2, 3 - 2, -3 + 2, -3 - 2 are unit with respect to inverses 1 + 0, -1 + 0, -1 + , -1 - , 3 - 2, 3 + 2, -3 - 2, -3 + 2 [All these units are powers of ±(1 + )n for various values of n.]

It is worthwhile to record the observation made in example 17 as a proposition.

Proposition 19: Let R be a ring with identity which is not the trivial ring. Then 1R 0R.

For any ring with identity, let us denote by UR the set of all units of R. From our above examples, we know that 1, -1 UR and that if R is not the trivial ring, then 0 UR.

Proposition 20: Let R be a ring with identity, and let a, b UR.

(1) a-1 UR.

(2) a · b UR and (a · b)-1 = b-1 · a-1.

(3) UR is a group with respect to multiplication.

Proof:

(1) Since a UR, a has an inverse a-1 and a·a-1 = a-1·a = 1. But these last equations imply that a-1 is a unit of R and that its inverse is a. Thus, we have proven part (1).

(2) Since a · a-1 = a-1 · a = 1, b · b-1 = b-1 · b = 1, we see that

(a · b) · (b-1· a-1) = a · 1 · a-1 = 1,
(b-1 · a-1) · (a · b) = b-1 · 1 · b = 1.

These last two equations show how a · b is a unit of R and that its inverse is b-1·a-1. This proves part (2)

(3) By part (2), UR is closed with respect to multiplication. Moreover, we observed above that 1 UR, so that UR has an identity with respect to multiplication. Multiplication in UR is associative since multiplication in R is associative. Finally, by part (1), if a UR, then a has an inverse with respect to multiplication in UR, namely a-1. Thus, UR satisfies the group axioms.

The group UR is called the group of units of R.

In order to introduce the last of special classes of rings which we will examine in this section, let us confine our discussion to a ring R with identity which is not the trivial ring. Let us denote the subset R - {0} of R by Rx. We showed earlier that 0 is not a unit of R. Therefore UR Rx. Note that it is not always true that UR = Rx. For example, we saw that if R = Z, then UR = {1, -1}. However, we also found that if R = Q, then UR = Q - {0} = Rx. Thus for some rings we have UR = Rx. Such rings play an exceeding important role in algebra, for reasons we will discuss below. Before we give this class of rings a name, let us observe that UR = Rx if and only if every nonzero element of R has an inverse with respect to multiplication. Thus let us make the following definition.

Definition 21: A field is a nontrivial commutative ring with identity in which every nonzero element has an inverse with respect to multiplication.

Note that a field must contain at least two elements, since it is a nontrivial ring. Therefore, in particular, in a field, we have 1 0. Note also that we have assumed that a field is commutative. Some authors define a field without this condition. And, indeed, there are many systems which satisfy the axioms for a field, with the exception of commutativity. Such systems are called skew fields, division algebras, or division rings, all of these terms being used interchangeably. In general we will not be discussing any of these object for the remainder of this document. We will assume the commutativity of a field unless otherwise stated.

Example 23: R, the set of real numbers, is a field.

Example 24: Let p be a prime and let Zp denote the ring of residue classes modulo p. Recall that Zp = {0,1, ..., p-1}. Since p is a prime, we have (i,p) = 1 for i = 1,2,...,p-1. Therefore, each of the residue classes 1,2,...,p-1 is reduced. Therefore, since 0 is clearly not reduced, we have Zpx = {1, 2, ..., p-1}. That is every nonzero element of Zp is a reduced residue class. We showed in the previous section on congruences, that every reduced residue class has an inverse with respect to multiplication in Zp. Therefore, Zp is a field. In fact Zp is a field with p elements. A field with a finite number of elements is called a finite field. Therefore, we can summarize this example by saying that Zp is a finite field.

Proposition 25: Let F be a nontrivial commutative ring with identity. Then F is a field if and only if Fx is a group with respect to multiplication.

Proof: . Assume that F is a field. Then by definition of a field Fx = UF. However, we have shown that UF is a group. Therefore, Fx is a group.

. Assume that Fx is a group. Then every element of Fx has an inverse with respect to multiplication. Therefore, F is a field.

Corollary 26: Let F be a field. Then F is an integral domain.

Proof: If F is a field, then FX is a group. Fx is closed with respect to multiplication. Therefore, F is an integral domain.

The extreme importance of fields in mathematics is due to the fact that in a field it is possible to combine elements using all the operations which one meets in elementary arithmetic. That is, given any two elements a and b of a field, it is possible to form their sum (a + b), difference (a - b), product (a · b), and, if b 0, their quotient (a · b-1.

If R is a ring with identity, then it is convenient to introduce the language of exponents in R. Let a R, n is a nonnegative integer. Then we define an inductively by

a0 = 1,
an+1 = a · an.

Then the following laws of exponents hold:

(1)
anam = an+m,
(2)
(an)m = anm.

Moreover, if R is commutative, then

(3)
(ab)n = anbn.

Another useful definition is:

Definition 27:The characteristic of a ring R (written Char R) is the smallest positive integer such that n1 = 0,where n1 is an abbreviation for 1 + 1 + ... + 1 (n times). If n1 is never 0,we say that R has characteristic 0. Note that the characteristic can never be 1,since 1R 0. If R is an integral domain and Char R 0, then Char R must be a prime number. For if Char R = n = rs where r and s are positive integers greater than 1,then (r1)(s1) = n1 = 0, so either r1 or s1 is 0,contradicting the minimality of n.