§3.2 Basic Properties of Rings

Arithmetic in Rings

When you do arithmetic in \(\mathbb{Z}\), you often use far more than the axioms for an integral domain. For instance, subtraction appears regularly, as do cancellation and the various rules for multiplying negative numbers. We begin by showing that many of these same properties hold in every ring.

Subtraction is not mentioned in the axioms for a ring, and we cannot just assume that such an operation exists in an arbitrary ring. If we want to define a subtraction in a ring, we must do so in terms of addition, multiplication, and the ring axioms. The first step is:

Theorem 3.3: Existence of a Unique Solution

For any element \(a\) in a ring \(R\), the equation \(a + x = 0_R\) has a unique solution.

Proof of Theorem 3.3

We know that \(a + x = 0_R\) has at least one solution \(u\) by Axiom 5. If \(v\) is also a solution, then \(a + u = 0_R\) and \(a + v = 0_R\), so that:

\(v = 0_R + v = (a + u) + v = (u + a) + v = u + (a + v) = u + 0_R = u\).

Therefore, \(u\) is the only solution. \(\blacksquare\)

Defining Negatives and Subtraction in Rings

We can now define negatives and subtraction in any ring by copying what happens in familiar rings such as \(\mathbb{Z}\). Let \(R\) be a ring and \(a \in R\). By Theorem 3.3, the equation \(a + x = 0_R\) has a unique solution. Using notation adapted from \(\mathbb{Z}\), we denote this unique solution by the symbol "\(-a\)."

\(-a\) is the unique element of \(R\) such that:

\(a + (-a) = 0_R = (-a) + a\).

In familiar rings, this definition coincides with the well-known concept of the negative of an element. More importantly, it provides a meaning for "negative" in any ring.

Example 3.2.1: Solutions in \(\mathbb{Z}_9\) and \(\mathbb{Z}_6\)

In the ring \(\mathbb{Z}_9\), the solution of the equation \(2 + x = 0\) is \(x = 7\), and so in this ring \(-2 = 7\). Similarly, \(9 = 5\) in \(\mathbb{Z}_6\) because \(5\) is the solution of \(9 + x = 0\).

Subtraction in a ring is now defined by the rule:

\(b - a \text{ means } b + (-a)\).

In \(\mathbb{Z}\) and other familiar rings, this is just ordinary subtraction. In other rings, we have a new operation.

Example 3.2.2: Subtraction in \(\mathbb{Z}_6\)

In \(\mathbb{Z}_6\), we have \(1 - 2 = 1 + (-2) = 1 + 4 = 5\).

Theorem 3.4: Cancellation Property in Rings

If \(a + b = a + c\) in a ring \(R\), then \(b = c\).

Proof of Theorem 3.4

Adding \(-a\) to both sides of \(a + b = a + c\) and then using associativity and negatives shows that:

\(-a + (a + b) = -a + (a + c)\)

\((-a + a) + b = (-a + a) + c\)

\(0_R + b = 0_R + c\)

\(b = c\). \(\blacksquare\)

Theorem 3.5: Properties of Zero and Negatives in Rings

For any elements \(a\) and \(b\) of a ring \(R\), the following properties hold:

\(a \cdot 0_R = 0_R = 0_R \cdot a\). In particular, \(0_R \cdot 0_R = 0_R\).
\(a \cdot (-b) = -(a \cdot b)\) and \((-a) \cdot b = -(a \cdot b)\).
\(-(-a) = a\).
\(-(a + b) = (-a) + (-b)\).
\(- (a - b) = -a + b\).
\(-(a \cdot b) = (-a) \cdot b\).
If \(R\) has an identity, then \((-1_R) \cdot a = -a\).

Proof of Theorem 3.5

(1) Since \(0_R + 0_R = 0_R\), the distributive law shows that:

\(a \cdot 0_R = a \cdot (0_R + 0_R) = a \cdot 0_R + a \cdot 0_R\).

Applying Theorem 3.4 to the first and last parts of this equation shows that \(a \cdot 0_R = 0_R\). The proof that \(0_R \cdot a = 0_R\) is similar.

(2) By definition, \(-ab\) is the unique solution of the equation:

\(ab + x = 0_R\),

and so any other solution of this equation must be equal to \(-ab\). But \(x = a(-b)\) is a solution because, by the distributive law and (1):

\(ab + a(-b) = a[b + (-b)] = a[0_R] = 0_R\).

Therefore, \(a(-b) = -ab\). The other part is proved similarly.

(3) By definition, \(-(-a)\) is the unique solution of \((-a) + x = 0_R\). But \(a\) is a solution of this equation since \((-a) + a = 0_R\). Hence, \(-(-a) = a\) by uniqueness.

(4) By definition, \(-(a + b)\) is the unique solution of \((a + b) + x = 0_R\), but \((-a) + (-b)\) is also a solution because addition is commutative, so that:

\((a + b) + [(-a) + (-b)] = a + (-a) + b + (-b) = 0_R + 0_R = 0_R\).

Therefore, \(-(a + b) = (-a) + (-b)\) by uniqueness.

(5) By the definition of subtraction and using (4) and (3):

\(-(a - b) = -(a + (-b)) = (-a) + (-( -b)) = (-a) + b\).

(6) \( -(a \cdot b) = -(a \cdot (-b)) \)

[By the second equation in (2), with \(-b\) in place of \(b\)]

= \(-( - (ab)) \quad \text{[By the first equation in (2)]}\)

= \(ab \quad \text{[By (3), with \(ab\) in place of \(a\)]}\)

(7) By (2):

\((-1_R) \cdot a = -(1_R \cdot a) = -(a) = -a\). \(\blacksquare\)

Exponentiation and Repeated Addition in Rings

When doing ordinary arithmetic, exponent notation is a definite convenience, as is its additive analogue (for instance, \(a + a + a = 3a\)). We now carry these concepts over to arbitrary rings. If \(R\) is a ring, \(a \in R\), and \(n\) is a positive integer, then we define:

\(a^n = a \cdot a \cdot a \cdot \cdots \cdot a \quad (n \text{ factors})\).

It is easy to verify that for any \(a \in R\) and positive integers \(m\) and \(n\):

\(a^m \cdot a^n = a^{m+n} \quad \text{and} \quad (a^m)^n = a^{mn}\).

If \(R\) has an identity and \(a \neq 0_R\), then we define \(0a\) to be the element \(1_R\). In this case, the exponent rules are valid for all \(m, n \geq 0\), and we define:

\(na = a + a + a + \cdots + a \quad (n \text{ summands})\),

\(-na = (-a) + (-a) + (-a) + \cdots + (-a) \quad (n \text{ summands})\).

Finally, we define \(0a = 0_R\). In familiar rings, this is nothing new, but in other rings, it gives meaning to the "product" of an integer \(n\) and a ring element \(a\).

Example 3.2.3: Expanding Exponentiation in Rings

Let \(R\) be a ring and \(a, b \in R\). Then:

\((a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b)\)

= \(aa + ab + ba + bb = a^2 + ab + ba + b^2\).

Be careful here. If \(ab \neq ba\), then you can't combine the middle terms. If \(R\) is a commutative ring, however, then \(ab = ba\) and we have the familiar pattern:

\((a + b)^2 = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2\).

For a calculation of \((a + b)^n\) in a commutative ring, with \(n > 2\), see the Binomial Theorem in Appendix E.

It's worth noting that subtraction provides a faster method than Theorem 3.4 for showing that a subset of a ring is actually a subring.

Theorem 3.6: Subring Conditions

Let \(S\) be a nonempty subset of a ring \(R\) such that:

\(S\) is closed under subtraction (if \(a, b \in S\), then \(a - b \in S\));
\(S\) is closed under multiplication (if \(a, b \in S\), then \(ab \in S\)).

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

Proof of Theorem 3.6

We show that \(S\) satisfies conditions (i)–(iv) of Theorem 3.2 and hence is a subring. The conditions will be proved in this order: (ii), (iii), (iv), and (i).

(ii) Hypothesis (2) here is identical with condition (ii) of Theorem 3.2. Hence, \(S\) satisfies condition (ii).

(iii) Since \(S\) is nonempty, there is some element \(c\) with \(c \in S\). Applying (1) (with \(a = c\) and \(b = c\)), we see that \(c - c = 0_R\) is in \(S\). Therefore, \(S\) satisfies condition (iii) of Theorem 3.2.

(iv) If \(a\) is any element of \(S\), then by (1), \(0_R - a = -a\) is also in \(S\). Since \(-a\) is the solution of \(a + x = 0_R\), condition (iv) of Theorem 3.2 is satisfied.

(i) If \(a, b \in S\), then \(-b\) is in \(S\) by the proof of (iv). By (1), \(a - (-b) = a + b\) is in \(S\). So \(S\) satisfies condition (i) of Theorem 3.2.

Therefore, \(S\) is a subring of \(R\) by Theorem 3.2. \(\blacksquare\)

Units and Zero Divisors

Units and zero divisors in \(\mathbb{Z}_n\) were introduced in Section 2.3. We now carry these concepts over to arbitrary rings.

An element \(a\) in a ring \(R\) with identity is called a unit if there exists \(u \in R\) such that \(au = 1_R = ua\). In this case, the element \(u\) is called the (multiplicative) inverse of \(a\) and is denoted \(a^{-1}\).

Example 3.2.4: Units in \(\mathbb{Z}\)

The only units in \(\mathbb{Z}\) are \(1\) and \(-1\).

Example 3.2.5: Units in \(\mathbb{Z}_{15}\)

By Theorem 2.10, the units in \(\mathbb{Z}_{15}\) are \(1, 2, 4, 7, 8, 11, 13\), and \(14\). For instance, \(2 \cdot 8 = 1\), so \(2^{-1} = 8\) and \(8^{-1} = 2\).

Example 3.2.6: Units in Fields

Every nonzero element of the field \(\mathbb{R}\) is a unit: If \(a \neq 0\), then \(a \cdot \frac{1}{a} = 1\). The same is true for every field \(F\). By definition, \(F\) satisfies Axiom 12: If \(a \neq 0_F\), then the equation \(ax = 1_F\) has a solution in \(F\). Hence:

Every nonzero element of a field is a unit.

Example 3.2.7: Invertible Matrices

A matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) in \(M(\mathbb{R})\) such that \(ad - bc \neq 0\) is a unit because, as you can easily verify:

\(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \frac{d}{ad - bc} & \frac{-b}{ad - bc} \\ \frac{-c}{ad - bc} & \frac{a}{ad - bc} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

and

\(\begin{pmatrix} \frac{d}{ad - bc} & \frac{-b}{ad - bc} \\ \frac{-c}{ad - bc} & \frac{a}{ad - bc} \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).

In particular, each of these matrices is a unit:

\(A = \begin{pmatrix} 3 & 2 \\ 5 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 4 & 3 \\ -2 & 5 \end{pmatrix}, \quad C = \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & \frac{1}{6} \end{pmatrix}\).

Units in a matrix ring are called invertible matrices.

Example 3.2.8: Invertible Matrices in a Field

Let \(F\) be a field and \(M(F)\) the ring of \(2 \times 2\) matrices with entries in \(F\). If:

\(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M(F)\)

and \(ad - bc \neq 0_F\), then \(ad - bc\) is a unit in \(F\) by Example 3.2.6. The computations in Example 3.2.7, with \(\frac{1}{ad - bc}\) replaced by \((ad - bc)^{-1}\), show that \(A\) is an invertible matrix (unit in \(M(F)\)) with inverse:

\(\begin{pmatrix} d(ad - bc)^{-1} & -b(ad - bc)^{-1} \\ -c(ad - bc)^{-1} & a(ad - bc)^{-1} \end{pmatrix}\).

Zero Divisors

An element \(a\) in a ring \(R\) is a zero divisor provided that:

\(a \neq 0_R\);
There exists a nonzero element \(c \in R\) such that \(ac = 0_R\) or \(ca = 0_R\).

Note that in requirement (2), the element \(c\) is not unique: Many elements in the ring may satisfy the equation \(ax = 0_R\) or the equation \(xa = 0_R\) (see Exercise 3.2.6). Furthermore, ...

Example 3.2.9: Zero Divisors in \(\mathbb{Z}_6\) and \(\mathbb{Z}_{12}\)

Both 2 and 3 are zero divisors in \(\mathbb{Z}_6\) because \(2 \cdot 3 = 0\). Similarly, 4 and 9 are zero divisors in \(\mathbb{Z}_{12}\) because \(4 \cdot 9 = 0\). For a zero divisor \(A\) in a matrix ring, it is possible to find a matrix \(C\) such that \(AC = 0\) and \(CA = 0\).

Example 3.2.10: Zero Divisors in Matrices

Let \(F\) be a field. A nonzero matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) in \(M(F)\) such that \(ad - bc = 0_F\) is a zero divisor because, as you can easily verify:

\(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 0_F & 0_F \\ 0_F & 0_F \end{pmatrix}\)

and

\(\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 0_F & 0_F \\ 0_F & 0_F \end{pmatrix}\).

In particular, each of these matrices is a zero divisor in the given ring:

\(A = \begin{pmatrix} 3 & 2 \\ 9 & 6 \end{pmatrix} \in M(\mathbb{R}), \quad B = \begin{pmatrix} 4/3 & -8 \\ -2 & 12 \end{pmatrix} \in M(\mathbb{Q}), \quad C = \begin{pmatrix} 4 & 2 \\ 1 & 5 \end{pmatrix} \in M(\mathbb{Z}_6)\).

Example 3.2.11: Zero Divisors in an Integral Domain

Every integral domain \(R\) satisfies Axiom 11: If \(ab = 0_R\), then \(a = 0_R\) or \(b = 0_R\). In other words, the product of two nonzero elements cannot be 0. Therefore:

An integral domain contains no zero divisors.

Theorem 3.7: Cancelation in an Integral Domain

Cancelation is valid in any integral domain \(R\): If \(a \neq 0_R\) and \(ab = ac\) in \(R\), then \(b = c\).

Cancelation may fail in rings that are not integral domains. In \(\mathbb{Z}_{12}\), for instance, \(2 \cdot 4 = 2 \cdot 10\), but \(4 \neq 10\).

Proof of Theorem 3.7

If \(ab = ac\), then \(ab - ac = 0_R\), so that \(a(b - c) = 0_R\). Since \(a \neq 0_R\), we must have \(b - c = 0_R\) (if not, then \(a\) is a zero divisor, contradicting Axiom 11). Therefore, \(b = c\). \(\blacksquare\)

Theorem 3.8: Every Field is an Integral Domain

Every field \(F\) is an integral domain.

Proof of Theorem 3.8

Since a field is a commutative ring with identity by definition, we need only show that \(F\) satisfies Axiom 11: If \(ab = 0_F\), then \(a = 0_F\) or \(b = 0_F\). So suppose that \(ab = 0_F\). If \(b = 0_F\), there is nothing to prove. If \(b \neq 0_F\), then \(b\) is a unit (Example 3.2.6). Consequently, by the definition of unit and part (1) of Theorem 3.5:

\(a = a \cdot 1_F = ab \cdot b^{-1} = 0_F \cdot b^{-1} = 0_F\).

So in every case, \(a = 0_F\) or \(b = 0_F\). Hence, Axiom 11 holds and \(F\) is an integral domain. \(\blacksquare\)

The converse of Theorem 3.8 is false in general (\(\mathbb{Z}\) is an integral domain that is not a field), but true in the finite case.

Theorem 3.9: Every Finite Integral Domain is a Field

Every finite integral domain \(R\) is a field.

Proof of Theorem 3.9

Since \(R\) is a commutative ring with identity, we need only show that for each \(a \neq 0_R\), the equation \(a x = 1_R\) has a solution. Let \(a_1, a_2, \dots, a_n\) be the distinct elements of \(R\) and suppose \(a x = 0_R\). To show that \(a x = 1_R\) has a solution, consider the products \(a a_1, a a_2, a a_3, \dots, a a_n\). If \(a \neq a_j\), then we must have \(a a_i \neq a a_j\) (because \(a a_i = a a_j\) would imply that \(a_i = a_j\) by cancelation). Therefore, \(a a_1, a a_2, \dots, a a_n\) are distinct multiples of \(a\) in \(R\). However, \(R\) has exactly \(n\) elements altogether, and so these must be all the elements of \(R\) in some order. In particular, for some \(j\), \(a a_j = 1_R\). Therefore, the equation \(a x = 1_R\) has a solution and \(R\) is a field. \(\blacksquare\)