16.3: Rings

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A nonempty set $$R$$ is a ring if it has two closed binary operations, addition and multiplication, satisfying the following conditions.

1. $$a + b = b + a$$ for $$a, b \in R\text{.}$$
2. $$(a + b) + c = a + ( b + c)$$ for $$a, b, c \in R\text{.}$$
3. There is an element $$0$$ in $$R$$ such that $$a + 0 = a$$ for all $$a \in R\text{.}$$
4. For every element $$a \in R\text{,}$$ there exists an element $$-a$$ in $$R$$ such that $$a + (-a) = 0\text{.}$$
5. $$(ab) c = a ( b c)$$ for $$a, b, c \in R\text{.}$$
6. For $$a, b, c \in R\text{,}$$

\begin{align*} a( b + c)&= ab +ac\\ (a + b)c & = ac + bc\text{.} \end{align*}

This last condition, the distributive axiom, relates the binary operations of addition and multiplication. Notice that the first four axioms simply require that a ring be an abelian group under addition, so we could also have defined a ring to be an abelian group $$(R, +)$$ together with a second binary operation satisfying the fifth and sixth conditions given above.

If there is an element $$1 \in R$$ such that $$1 \neq 0$$ and $$1a = a1 = a$$ for each element $$a \in R\text{,}$$ we say that $$R$$ is a ring with unity or identity. A ring $$R$$ for which $$ab = ba$$ for all $$a, b$$ in $$R$$ is called a commutative ring. A commutative ring $$R$$ with identity is called an integral domain if, for every $$a, b \in R$$ such that $$ab = 0\text{,}$$ either $$a = 0$$ or $$b = 0\text{.}$$ A division ring is a ring $$R\text{,}$$ with an identity, in which every nonzero element in $$R$$ is a unit; that is, for each $$a \in R$$ with $$a \neq 0\text{,}$$ there exists a unique element $$a^{-1}$$ such that $$a^{-1} a = a a^{-1} = 1\text{.}$$ A commutative division ring is called a field. The relationship among rings, integral domains, division rings, and fields is shown in Figure $$16.1$$.

$$Figure \text { } 16.1.$$ Types of rings

Example $$16.2$$

As we have mentioned previously, the integers form a ring. In fact, $${\mathbb Z}$$ is an integral domain. Certainly if $$a b = 0$$ for two integers $$a$$ and $$b\text{,}$$ either $$a=0$$ or $$b=0\text{.}$$

Solution

However, $${\mathbb Z}$$ is not a field. There is no integer that is the multiplicative inverse of $$2\text{,}$$ since $$1/2$$ is not an integer. The only integers with multiplicative inverses are $$1$$ and $$-1\text{.}$$

Example $$16.3$$

Under the ordinary operations of addition and multiplication, all of the familiar number systems are rings:

Solution

the rationals, $${\mathbb Q}\text{;}$$ the real numbers, $${\mathbb R}\text{;}$$ and the complex numbers, $${\mathbb C}\text{.}$$ Each of these rings is a field.

Example $$16.4$$

We can define the product of two elements $$a$$ and $$b$$ in $${\mathbb Z}_n$$ by $$ab \pmod{n}\text{.}$$

Solution

For instance, in $${\mathbb Z}_{12}\text{,}$$ $$5 \cdot 7 \equiv 11 \pmod{12}\text{.}$$ This product makes the abelian group $${\mathbb Z}_n$$ into a ring. Certainly $${\mathbb Z}_n$$ is a commutative ring; however, it may fail to be an integral domain. If we consider $$3 \cdot 4 \equiv 0 \pmod{12}$$ in $${\mathbb Z}_{12}\text{,}$$ it is easy to see that a product of two nonzero elements in the ring can be equal to zero.

A nonzero element $$a$$ in a commutative ring $$R$$ is called a zero divisor if there is a nonzero element $$b$$ in $$R$$ such that $$ab = 0\text{.}$$ In the previous example, $$3$$ and $$4$$ are zero divisors in $${\mathbb Z}_{12}\text{.}$$

Example $$16.5$$

In calculus the continuous real-valued functions on an interval $$[a,b]$$ form a commutative ring. We add or multiply two functions by adding or multiplying the values of the functions. If $$f(x) = x^2$$ and $$g(x) = \cos x\text{,}$$

Solution

then $$(f+g)(x) = f(x) + g(x) = x^2 + \cos x$$ and $$(fg)(x) = f(x) g(x) = x^2 \cos x\text{.}$$

Example $$16.6$$

The $$2 \times 2$$ matrices with entries in $${\mathbb R}$$ form a ring under the usual operations of matrix addition and multiplication.

Solution

This ring is noncommutative, since it is usually the case that $$AB \neq BA\text{.}$$ Also, notice that we can have $$AB = 0$$ when neither $$A$$ nor $$B$$ is zero.

Example $$16.7$$

For an example of a noncommutative division ring, let

$1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad {\mathbf i} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \quad {\mathbf j} = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, \quad {\mathbf k} = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}\text{,} \nonumber$

where $$i^2 = -1\text{.}$$ These elements satisfy the following relations:

\begin{align*} {\mathbf i}^2 = {\mathbf j}^2 & = {\mathbf k}^2 = -1\\ {\mathbf i} {\mathbf j} & = {\mathbf k}\\ {\mathbf j} {\mathbf k} & = {\mathbf i}\\ {\mathbf k} {\mathbf i} & = {\mathbf j}\\ {\mathbf j} {\mathbf i} & = - {\mathbf k}\\ {\mathbf k} {\mathbf j} & = - {\mathbf i}\\ {\mathbf i} {\mathbf k} & = - {\mathbf j}\text{.} \end{align*}

Let $${\mathbb H}$$ consist of elements of the form $$a + b {\mathbf i} + c {\mathbf j} +d {\mathbf k}\text{,}$$ where $$a, b , c, d$$ are real numbers. Equivalently, $${\mathbb H}$$ can be considered to be the set of all $$2 \times 2$$ matrices of the form

$\begin{pmatrix} \alpha & \beta \\ -\overline{\beta} & \overline{\alpha } \end{pmatrix}\text{,} \nonumber$

where $$\alpha = a + di$$ and $$\beta = b + ci$$ are complex numbers. We can define addition and multiplication on $${\mathbb H}$$ either by the usual matrix operations or in terms of the generators $$1\text{,}$$ $${\mathbf i}\text{,}$$ $${\mathbf j}\text{,}$$ and $${\mathbf k}\text{:}$$

\begin{gather*} (a_1 + b_1 {\mathbf i} + c_1 {\mathbf j} +d_1 {\mathbf k} ) + ( a_2 + b_2 {\mathbf i} + c_2 {\mathbf j} +d_2 {\mathbf k} )\\ = (a_1 + a_2) + ( b_1 + b_2) {\mathbf i} + ( c_1 + c_2) \mathbf j + (d_1 + d_2) \mathbf k \end{gather*}

and

$(a_1 + b_1 {\mathbf i} + c_1 {\mathbf j} +d_1 {\mathbf k} ) ( a_2 + b_2 {\mathbf i} + c_2 {\mathbf j} +d_2 {\mathbf k} ) = \alpha + \beta {\mathbf i} + \gamma {\mathbf j} + \delta {\mathbf k}\text{,} \nonumber$

where

\begin{align*} \alpha & = a_1 a_2 - b_1 b_2 - c_1 c_2 -d_1 d_2\\ \beta & = a_1 b_2 + a_2 b_1 + c_1 d_2 - d_1 c_2\\ \gamma & = a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2\\ \delta & = a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2\text{.} \end{align*}

Solution

Though multiplication looks complicated, it is actually a straightforward computation if we remember that we just add and multiply elements in $${\mathbb H}$$ like polynomials and keep in mind the relationships between the generators $${\mathbf i}\text{,}$$ $${\mathbf j}\text{,}$$ and $${\mathbf k}\text{.}$$ The ring $${\mathbb H}$$ is called the ring of quaternions.

To show that the quaternions are a division ring, we must be able to find an inverse for each nonzero element. Notice that

$( a + b {\mathbf i} + c {\mathbf j} + d {\mathbf k} )( a - b {\mathbf i} - c {\mathbf j} - d {\mathbf k} ) = a^2 + b^2 + c^2 + d^2\text{.} \nonumber$

This element can be zero only if $$a\text{,}$$ $$b\text{,}$$ $$c\text{,}$$ and $$d$$ are all zero. So if $$a + b {\mathbf i} + c {\mathbf j} +d {\mathbf k} \neq 0\text{,}$$

$(a + b {\mathbf i} + c {\mathbf j} + d {\mathbf k})\left( \frac{a - b {\mathbf i} - c {\mathbf j} - d {\mathbf k} }{a^2 + b^2 + c^2 + d^2} \right) = 1\text{.} \nonumber$

Proposition $$16.8$$

Let $$R$$ be a ring with $$a, b \in R\text{.}$$ Then

1. $$a0 = 0a = 0\text{;}$$
2. $$a(-b) = (-a)b = -ab\text{;}$$
3. $$(-a)(-b) =ab\text{.}$$
Proof

To prove (1), observe that

$a0 = a(0+0)= a0+ a0; \nonumber$

hence, $$a0=0\text{.}$$ Similarly, $$0a = 0\text{.}$$ For (2), we have $$ab + a(-b) = a(b-b) = a0 = 0\text{;}$$ consequently, $$-ab = a(-b)\text{.}$$ Similarly, $$-ab = (-a)b\text{.}$$ Part (3) follows directly from (2) since $$(-a)(-b) = -(a(- b)) = -(-ab) = ab\text{.}$$

Just as we have subgroups of groups, we have an analogous class of substructures for rings. A subring $$S$$ of a ring $$R$$ is a subset $$S$$ of $$R$$ such that $$S$$ is also a ring under the inherited operations from $$R\text{.}$$

Example $$16.9$$

The ring $$n {\mathbb Z}$$ is a subring of $${\mathbb Z}\text{.}$$

Solution

Notice that even though the original ring may have an identity, we do not require that its subring have an identity. We have the following chain of subrings:

${\mathbb Z} \subset {\mathbb Q} \subset {\mathbb R} \subset {\mathbb C}\text{.} \nonumber$

The following proposition gives us some easy criteria for determining whether or not a subset of a ring is indeed a subring. (We will leave the proof of this proposition as an exercise.)

Proposition $$16.10$$

Let $$R$$ be a ring and $$S$$ a subset of $$R\text{.}$$ Then $$S$$ is a subring of $$R$$ if and only if the following conditions are satisfied.

1. $$S \neq \emptyset\text{.}$$
2. $$rs \in S$$ for all $$r, s \in S\text{.}$$
3. $$r-s \in S$$ for all $$r, s \in S\text{.}$$

Example $$16.11$$

Let $$R ={\mathbb M}_2( {\mathbb R} )$$ be the ring of $$2 \times 2$$ matrices with entries in $${\mathbb R}\text{.}$$ If $$T$$ is the set of upper triangular matrices in $$R\text{;}$$ i.e.,

$T = \left\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} : a, b, c \in {\mathbb R} \right\}\text{,} \nonumber$

Solution

then $$T$$ is a subring of $$R\text{.}$$ If

$A = \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} a' & b' \\ 0 & c' \end{pmatrix} \nonumber$

are in $$T\text{,}$$ then clearly $$A-B$$ is also in $$T\text{.}$$ Also,

$AB = \begin{pmatrix} a a' & ab' + bc' \\ 0 & cc' \end{pmatrix} \nonumber$

is in $$T\text{.}$$

This page titled 16.3: Rings is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Thomas W. Judson (Abstract Algebra: Theory and Applications) via source content that was edited to the style and standards of the LibreTexts platform.