§3.1 Introduction to Rings

Definition of Rings

We begin the process of abstracting the common features of familiar systems with this definition:

A ring is a nonempty set \(R\) equipped with two operations* (usually written as addition and multiplication) that satisfy the following axioms. For all \(a, b, c \in R\):

1. If \(a \in R\) and \(b \in R\), then \(a + b \in R\). [Closure for addition]
2. \(a + (b + c) = (a + b) + c\). [Associative addition]
3. \(a + b = b + a\). [Commutative addition]
4. There is an element \(0_R\) in \(R\) such that \(a + 0_R = 0_R + a = a\) for every \(a \in R\). [Additive identity or zero element]
5. For each \(a \in R\), the equation \(a + x = 0_R\) has a solution in \(R\). [Additive inverse]
6. If \(a \in R\) and \(b \in R\), then \(ab \in R\). [Closure for multiplication]
7. \(a(bc) = (ab)c\). [Associative for multiplication]
8. \(a(b + c) = ab + ac\) and \((a + b)c = ac + bc\). [Distributive laws]

These axioms are the bare minimum needed for a system to resemble \(\mathbb{Z}\) and \(\mathbb{Z}_n\). But \(\mathbb{Z}\) and \(\mathbb{Z}_n\) have several additional properties that are worth special mention:

Commutative Rings

A commutative ring is a ring \(R\) that satisfies this axiom:

9. \(ab = ba\) for all \(a, b \in R\). [Commutative multiplication]

Ring with Identity

A ring with identity is a ring \(R\) that contains an element \(1_R\) satisfying this axiom:

10. \(a1_R = 1_Ra = a\) for all \(a \in R\). [Multiplicative identity]

Using terms defined later in the course, we simply say that a ring is an abelian group under addition.

Example 3.1.1: Commutative Rings with Identity

With the usual addition and multiplication:

\(\mathbb{Z} \ (\text{the integers}) \quad \text{and} \quad \mathbb{R} \ (\text{the real numbers})\)

are commutative rings with identity.

Example 3.1.2: Commutative Rings with Identity in \(\mathbb{Z}_m\)

The set \(\mathbb{Z}_m\) with the usual addition and multiplication of classes is a commutative ring with identity by Theorem 2.7.

Example 3.1.3: Even Integers as a Commutative Ring without Identity

Let \(E\) be the set of even integers with the usual addition and multiplication. Since the sum or product of two even integers is also even, the closure axioms (1 and 6) hold. Since 0 is an even integer, \(E\) has an additive identity element (Axiom 4). If \(a\) is even, then the solution of \(a + x = 0\) (namely \(-a\)) is also even, and so Axiom 5 holds.

The remaining axioms (2, 3, 7, 8, and 9) hold for all integers and, therefore, are true whenever \(a\), \(b\), and \(c\) are even. Consequently, \(E\) is a commutative ring. However, \(E\) does not have an identity because no even integer \(e\) has the property that \(ae = a = ea\) for every even integer \(a\).

Example 3.1.4: Odd Integers as a Non-Ring

The set of odd integers with the usual addition and multiplication is not a ring. Among other things, Axiom 1 fails: The sum of two odd integers is not odd.

Expanding Beyond \(\mathbb{Z}\) and \(\mathbb{Z}_m\)

Although the definition of ring was constructed with \(\mathbb{Z}\) and \(\mathbb{Z}_m\) as models, there are many rings that aren’t at all like these models. In these rings, the elements may not be numbers or classes of numbers, and their operations may have nothing to do with “ordinary” addition and multiplication.

Example 3.1.5: Ring with Addition and Multiplication Defined by Tables

The set \(T = \{r, s, t, z\}\) equipped with the addition and multiplication defined by the following tables is a ring:

You may take our word for it that associativity and distributivity hold (Axioms 2, 7, and 8). The remaining axioms can be easily verified from the operation tables above. In particular, they show that \(T\) is closed under both addition and multiplication (Axioms 1 and 6) and that addition is commutative (Axiom 3).

The element \(z\) is the additive identity—the element denoted \(0_R\) in Axiom 4. It behaves in the same way the number 0 does in \(\mathbb{Z}\) (that’s why the notation \(0_R\) is used in the axiom), but \(z\) is not the integer 0—in fact, it’s not any kind of number. Nevertheless, we shall call \(z\) the “zero element” of the ring \(T\).

In order to verify Axiom 5, you must show that each of the equations:

\(r + x = z \quad s + x = z \quad t + x = z\)

has a solution in \(T\). This is easily seen to be the case from the addition table; for example, \(x = r\) is the solution of \(r + x = z\) because \(r + r = z\).

Finally, note that \(T\) is not a commutative ring; for instance, \(rs = r\) and \(sr = z\), so that \(rs \neq sr\).

Example 3.1.6: Matrix Ring \(M(\mathbb{R})\)

Let \(M(\mathbb{R})\) be the set of all \(2 \times 2\) matrices over the real numbers, that is, \(M(\mathbb{R})\) consists of all arrays:

\(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\)

where \(a, b, c, d\) are real numbers. Two matrices are equal provided that the entries in corresponding positions are equal; that is:

\(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} r & s \\ t & u \end{pmatrix}\)

if and only if \(a = r, b = s, c = t, d = u\).

For example:

\(\begin{pmatrix} -4 & 0 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 2 + 2 & 0 \\ 1 - 4 & 0 + 1 \end{pmatrix}\)

but

\(\begin{pmatrix} 1 & 3 \\ 5 & 2 \end{pmatrix} \neq \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\)

Addition of matrices is defined by:

\(\begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix} = \begin{pmatrix} a + a' & b + b' \\ c + c' & d + d' \end{pmatrix}\)

For example:

\(\begin{pmatrix} 3 & -2 \\ 5 & 1 \end{pmatrix} + \begin{pmatrix} 4 & 7 \\ 6 & 0 \end{pmatrix} = \begin{pmatrix} 3 + 4 & -2 + 7 \\ 5 + 6 & 1 + 0 \end{pmatrix} = \begin{pmatrix} 7 & 5 \\ 11 & 1 \end{pmatrix}\)

Multiplication of matrices is defined by:

\(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} w & x \\ y & z \end{pmatrix} = \begin{pmatrix} aw + by & ax + bz \\ cw + dy & cx + dz \end{pmatrix}\)

For example:

\(\begin{pmatrix} 2 & 3 \\ 0 & -4 \end{pmatrix} \begin{pmatrix} 1 & -5 \\ 6 & 7 \end{pmatrix} = \begin{pmatrix} 2 \cdot 1 + 3 \cdot 6 & 2 \cdot (-5) + 3 \cdot 7 \\ 0 \cdot 1 + (-4) \cdot 6 & 0 \cdot (-5) + (-4) \cdot 7 \end{pmatrix} = \begin{pmatrix} 20 & 11 \\ -24 & -28 \end{pmatrix}\)

Reversing the order of the factors in matrix multiplication may produce a different answer, as is the case here:

\(\begin{pmatrix} 1 & -5 \\ 6 & 7 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 0 & -4 \end{pmatrix} = \begin{pmatrix} 2 & 23 \\ 12 & -10 \end{pmatrix}\)

So this multiplication is not commutative. With a bit of work, you can verify that \(M(\mathbb{R})\) is a ring with identity. The zero element is the zero matrix:

\(\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

which is denoted \(0\), and

\(X = \begin{pmatrix} -a & -b \\ -c & -d \end{pmatrix}\)

is a solution of

\(\begin{pmatrix} a & b \\ c & d \end{pmatrix} + X = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

We claim that the multiplicative identity element (Axiom 10) is the matrix \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).

To prove this claim, we first multiply a typical matrix in \(M(\mathbb{R})\) on the right by \(I\):

\(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a \cdot 1 + b \cdot 0 & a \cdot 0 + b \cdot 1 \\ c \cdot 1 + d \cdot 0 & c \cdot 0 + d \cdot 1 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)

Since multiplication is not commutative here, we also need to check left multiplication by \(I\) as well:

\(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 \cdot a + 0 \cdot c & 1 \cdot b + 0 \cdot d \\ 0 \cdot a + 1 \cdot c & 0 \cdot b + 1 \cdot d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)

This proves that \(I\) satisfies Axiom 10.\* Consequently, \(I\) is called the identity matrix. Note that the product of nonzero elements of \(M(\mathbb{R})\) may be the zero element; for example:

\(\begin{pmatrix} 4 & 6 \\ -3 & 2 \end{pmatrix} \begin{pmatrix} -3 & -9 \\ 2 & 6 \end{pmatrix} = \begin{pmatrix} 4 \cdot (-3) + 6 \cdot 2 & 4 \cdot (-9) + 6 \cdot 6 \\ (-3) \cdot (-3) + 2 \cdot 2 & (-3) \cdot (-9) + 2 \cdot 6 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

Example 3.1.7: Matrix Rings with Entries in a Commutative Ring

If \(R\) is a commutative ring with identity, then \(M(R)\) denotes the set of all \(2 \times 2\) matrices with entries in \(R\). With addition and multiplication defined as in Example 6, \(M(\mathbb{R})\) is a noncommutative ring with identity, as you can readily verify. For instance, \(M(\mathbb{Z})\) is the ring of \(2 \times 2\) matrices with integer entries, \(M(\mathbb{Q})\) the ring of \(2 \times 2\) matrices with rational number entries, and \(M(\mathbb{Z}_n)\) the ring of \(2 \times 2\) matrices with entries from \(\mathbb{Z}_n\).

Example 3.1.8: Ring of Functions from \(\mathbb{R}\) to \(\mathbb{R}\)

Let \(T\) be the set of all functions from \(\mathbb{R}\) to \(\mathbb{R}\), where \(\mathbb{R}\) is the set of real numbers. As in calculus, \(f + g\) and \(fg\) are the functions defined by:

\((f + g)(x) = f(x) + g(x) \quad \text{and} \quad (fg)(x) = f(x)g(x)\).

You can readily verify that \(T\) is a commutative ring with identity. The zero element is the function \(h\) given by \(h(x) = 0\) for all \(x \in \mathbb{R}\). The identity element is the function \(e\) given by \(e(x) = 1\) for all \(x \in \mathbb{R}\). Once again, the product of nonzero elements of \(T\) may turn out to be the zero element; see Exercise 3.1.36.

Integral Domains

We have seen that some rings do not have the property that the product of two nonzero elements is always nonzero. But some of the rings that do have this property, such as \(\mathbb{Z}\), occur frequently enough to merit a title.

An integral domain is a commutative ring \(R\) with identity \(1_R \neq 0_R\) that satisfies this axiom:

11. Whenever \(a, b \in R\) and \(ab = 0_R\), then \(a = 0_R\) or \(b = 0_R\).

\* Checking a possible identity element under both right and left multiplication is essential. There are rings in which an element acts like an identity when you multiply on the right, but not when you multiply on the left. See Exercise 3.1.11.

The condition \(1_R \neq 0_R\) is needed to exclude the zero ring (that is, the single-element ring \(\{0_R\}\)) from the class of integral domains. Note that Axiom 11 is logically equivalent to its contrapositive:

Whenever \(a \neq 0_R\) and \(b \neq 0_R\), then \(ab \neq 0_R\).

Example 3.1.9: Integers and Integral Domains

The ring \(\mathbb{Z}\) of integers is an integral domain. If \(p\) is prime, then \(\mathbb{Z}_p\) is an integral domain by Theorem 2.8. On the other hand, \(\mathbb{Z}_6\) is not an integral domain because \(4 \cdot 3 = 0\), even though \(4 \neq 0\) and \(3 \neq 0\).

You should be familiar with the set \(\mathbb{Q}\) of rational numbers, which consists of all fractions \(\frac{a}{b}\) with \(a, b \in \mathbb{Z}\) and \(b \neq 0\). Equality of fractions, addition, and multiplication are given by the usual rules:

\(\frac{a}{b} = \frac{r}{s} \quad \text{if and only if} \quad as = br\)

\(\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} \quad \text{and} \quad \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}\)

It is easy to verify that \(\mathbb{Q}\) is an integral domain. But \(\mathbb{Q}\) has an additional property that does not hold in \(\mathbb{Z}\): Every equation of the form \(ax = 1\) (with \(a \neq 0\)) has a solution in \(\mathbb{Q}\). Therefore, \(\mathbb{Q}\) is an example of the next definition.

Fields

A field is a commutative ring \(R\) with identity \(1_R \neq 0_R\) that satisfies this axiom:

12. For each \(a \neq 0_R\) in \(R\), the equation \(ax = 1_R\) has a solution in \(R\).

Once again, the condition \(1_R \neq 0_R\) is needed to exclude the zero ring. Note that Axiom 11 is not mentioned explicitly in the definition of a field. However, Axiom 11 does hold in fields, as we shall see in Theorem 3.8 below.

Example 3.1.10: Real Numbers as a Field

The set \(\mathbb{R}\) of real numbers, with the usual addition and multiplication, is a field. If \(p\) is a prime, then \(\mathbb{Z}_p\) is a field by Theorem 2.8.

Example 3.1.11: Complex Numbers as a Field

The set \(\mathbb{C}\) of complex numbers consists of all numbers of the form \(a + bi\), where \(a, b \in \mathbb{R}\) and \(i^2 = -1\). Equality in \(\mathbb{C}\) is defined by:

\(a + bi = r + si \quad \text{if and only if} \quad a = r \text{ and } b = s\).

The set \(\mathbb{C}\) is a field with addition and multiplication given by:

\((a + bi) + (c + di) = (a + c) + (b + d)i\)

\((a + bi)(c + di) = (ac - bd) + (ad + bc)i\)

The field \(\mathbb{R}\) of real numbers is contained in \(\mathbb{C}\) because \(\mathbb{R}\) consists of all complex numbers of the form \(a + 0i\). If \(a + bi \neq 0\) in \(\mathbb{C}\), then the solution of the equation \((a + bi)x = 1\) is \(x = c + di\), where:

\(c = \frac{a}{a^2 + b^2} \in \mathbb{R} \quad \text{and} \quad d = \frac{-b}{a^2 + b^2} \in \mathbb{R} \quad (\text{verify!})\).

Example 3.1.12: Matrices in \(K\)

Let \(K\) be the set of all \(2 \times 2\) matrices of the form:

\(\begin{pmatrix} a & b \\ -b & a \end{pmatrix}\)

where \(a\) and \(b\) are real numbers. We claim that \(K\) is a field. For any two matrices in \(K\):

\(\begin{pmatrix} a & b \\ -b & a \end{pmatrix} + \begin{pmatrix} c & d \\ -d & c \end{pmatrix} = \begin{pmatrix} a + c & b + d \\ -(b + d) & a + c \end{pmatrix}\)

In each case, the matrix on the right is in \(K\) because the entries along the main diagonal (upper left to lower right) are the same, and the entries on the opposite diagonal (upper right to lower left) are negatives of each other. Therefore, \(K\) is closed under addition and multiplication. \(K\) is commutative because:

\(\begin{pmatrix} c & d \\ -d & c \end{pmatrix} \begin{pmatrix} a & b \\ -b & a \end{pmatrix} = \begin{pmatrix} ac - bd & ad + bc \\ -(ad + bc) & ac - bd \end{pmatrix} = \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \begin{pmatrix} c & d \\ -d & c \end{pmatrix}\)

Clearly, the zero matrix and the identity matrix \(I\) are in \(K\). If:

\(A = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}\)

is not the zero matrix, then verify that the solution of \(AX = I\) is:

\(X = \begin{pmatrix} \frac{a}{a^2 + b^2} & \frac{-b}{a^2 + b^2} \\ \frac{b}{a^2 + b^2} & \frac{a}{a^2 + b^2} \end{pmatrix} \in K\), where \(d = a^2 + b^2\).

Whenever the rings in the preceding examples are mentioned, you may assume that addition and multiplication are the operations defined above unless there is some specific statement to the contrary. You should be aware, however, that a given set (such as \(\mathbb{Z}\)) may be made into a ring in many different ways by defining different addition and multiplication operations on it. See Exercises 3.1.17 and 22–26 for examples.

Now that we know a variety of different kinds of rings, we can use them to produce new rings in the following way.

Example 3.1.13: Cartesian Product of Rings

Let \(T\) be the Cartesian product \(\mathbb{Z}_6 \times \mathbb{Z}\), as defined in Appendix B. Define addition in \(T\) by the rule:

\((a, z) + (a', z') = (a + a', z + z')\).

The plus sign is being used in three ways here: In the first coordinate on the right-hand side of the equal sign, \(+\) denotes addition in \(\mathbb{Z}_6\); in the second coordinate, \(+\) denotes addition in \(\mathbb{Z}\); the \(+\) on the left of the equal sign is the addition in \(T\) that is being defined. Since \(\mathbb{Z}_6\) is a ring and \(a, a' \in \mathbb{Z}_6\), the first coordinate on the right, \(a + a'\), is in \(\mathbb{Z}_6\). Similarly, \(z + z' \in \(\mathbb{Z}\). Therefore, addition in \(T\) is closed. Multiplication is defined similarly:

\((a, z)(a', z') = (aa', zz')\).

For example, \((3, 5) + (4, 9) = (3 + 4, 5 + 9) = (1, 14)\) and \((3, 5)(4, 9) = (3 \cdot 4, 5 \cdot 9) = (0, 45)\). You can readily verify that \(T\) is a commutative ring with identity. The zero element is \((0, 0)\), and the multiplicative identity is \((1, 1)\). What was done here can be done for any two rings.

Theorem 3.1: Cartesian Product of Rings

Let \(R\) and \(S\) be rings. Define addition and multiplication on the Cartesian product \(R \times S\) by:

\((r, s) + (r', s') = (r + r', s + s') \quad \text{and} \quad (r, s)(r', s') = (rr', ss')\).

Then \(R \times S\) is a ring. If \(R\) and \(S\) are both commutative, then so is \(R \times S\). If both \(R\) and \(S\) have an identity, then so does \(R \times S\).

Proof: See Exercise 3.1.33.

Subrings

If \(R\) is a ring and \(S\) is a subset of \(R\), then \(S\) may or may not itself be a ring under the operations in \(R\). In the ring \(\mathbb{Z}\) of integers, for example, the subset \(E\) of even integers is a ring, but the subset \(O\) of odd integers is not, as we saw in Examples 3 and 4. When a subset \(S\) of a ring \(R\) is itself a ring under the addition and multiplication in \(R\), then we say that \(S\) is a subring of \(R\).

Example 3.1.14: Subfields and Subrings

\(\mathbb{Z}\) is a subring of the ring \(\mathbb{Q}\) of rational numbers, and \(\mathbb{Q}\) is a subring of the field \(\mathbb{R}\) of all real numbers. Since \(\mathbb{Q}\) is itself a field, we say that \(\mathbb{Q}\) is a subfield of \(\mathbb{R}\). Similarly, \(\mathbb{R}\) is a subfield of the field \(\mathbb{C}\) of complex numbers.

Example 3.1.15: Rings of Matrices as Subrings

The matrix rings \(M(\mathbb{Z})\) and \(M(\mathbb{Q})\) in Example 3.1.7 are subrings of \(M(\mathbb{R})\).

Example 3.1.16: Subring of a Ring of Functions

The ring \(K\) in Example 3.1.12 is a subring of the ring of \(M(\mathbb{R})\).

Example 3.1.17: Subring of Continuous Functions

Let \( T \) be the ring of all functions from \(\mathbb{R}\) to \(\mathbb{R}\) in Example 3.1.8. Then the subset \( S \) consisting of all continuous functions from \(\mathbb{R}\) to \(\mathbb{R}\) is a subring of \( T \). To prove this, you need one fact proved in calculus: the sum and product of continuous functions are also continuous. So \( S \) is closed under addition and multiplication (Axioms 1 and 6). You can readily verify the other axioms.

Proving that a subset \( S \) of a ring \( R \) is actually a subring is easier than proving directly that \( S \) is a ring. For instance, since \( a + b = b + a \) for all elements of \( R \), this fact is also true when \( a, b \) happen to be in the subset \( S \). Thus, Axiom 3 (commutative addition) automatically holds in any subset \( S \) of a ring. In fact, to prove that a subset of a ring is actually a subring, you need only verify a few of the axioms for a ring, as the next theorem shows.

Theorem 3.2: Subring Test

Suppose that \( R \) is a ring and that \( S \) is a subset of \( R \) such that:

1. \( S \) is closed under addition (if \( a, b \in S \), then \( a + b \in S \));
2. \( S \) is closed under multiplication (if \( a, b \in S \), then \( ab \in S \));
3. \( 0_R \in S \);
4. If \( a \in S \), then the solution of the equation \( a + x = 0_R \) is in \( S \).

Then \( S \) is a subring of \( R \).

Note: Condition (iv) is important. To verify it, you need not show that the equation \( a + x = 0_R \) has a solution—we already know that it does because \( R \) is a ring. You need only show that this solution is an element of \( S \) (which implies that Axiom 5 holds for \( S \)).

Proof of Theorem 3.2

As noted before the theorem, Axioms 2, 3, 7, and 8 hold for all elements of \( R \), and so they necessarily hold for the elements of the subset \( S \). Axioms 1, 6, 4, and 5 hold by (i)–(iv).

Example 3.1.18: Subring of Integers