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Section16.1Rings

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

$a + b = b + a$ for $a, b \in R\text{.}$
$(a + b) + c = a + ( b + c)$ for $a, b, c \in R\text{.}$
There is an element $0$ in $R$ such that $a + 0 = a$ for all $a \in R\text{.}$
For every element $a \in R\text{,}$ there exists an element $-a$ in $R$ such that $a + (-a) = 0\text{.}$
$(ab) c = a ( b c)$ for $a, b, c \in R\text{.}$
For $a, b, c \in R\text{,}$
\begin{align*} a( b + c)&= ab +ac\\ (a + b)c & = ac + bc. \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 or . A ring $R$ for which $ab = ba$ for all $a, b$ in $R$ is called a . A commutative ring $R$ with identity is called an if, for every $a, b \in R$ such that $ab = 0\text{,}$ either $a = 0$ or $b = 0\text{.}$ A is a ring $R\text{,}$ with an identity, in which every nonzero element in $R$ is a ; 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 . The relationship among rings, integral domains, division rings, and fields is shown in Figure 16.1.

Figure16.1Types of rings

Example16.2

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 $b=0\text{.}$ However, ${\mathbb Z}$ is not a field. There is no integer that is the multiplicative inverse of 2, since $1/2$ is not an integer. The only integers with multiplicative inverses are 1 and $-1\text{.}$

Example16.3

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

Example16.4

We can define the product of two elements $a$ and $b$ in ${\mathbb Z}_n$ by $ab \pmod{n}\text{.}$ 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 ring $R$ is called a 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 ${\mathbb Z}_{12}\text{.}$

Example16.5

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

Example16.6

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

Example16.7

For an example of a noncommutative division ring, let

\begin{equation*} 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}, \end{equation*}

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}. \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{equation*} \begin{pmatrix} \alpha & \beta \\ -\overline{\beta} & \overline{\alpha } \end{pmatrix}, \end{equation*}

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, ${\mathbf i}\text{,}$ ${\mathbf j}\text{,}$ 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

\begin{equation*} (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}, \end{equation*}

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. \end{align*}

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 .

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

\begin{equation*} ( 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. \end{equation*}

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

\begin{equation*} ( 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. \end{equation*}

Proposition16.8

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

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

Proof

To prove (1), observe that

\begin{equation*} a0 = a(0+0)= a0+ a0; \end{equation*}

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 $S$ of a ring $R$ is a subset $S$ of $R$ such that $S$ is also a ring under the inherited operations from $R\text{.}$

Example16.9

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

\begin{equation*} {\mathbb Z} \subset {\mathbb Q} \subset {\mathbb R} \subset {\mathbb C}. \end{equation*}

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.)

Proposition16.10

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.

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

Example16.11

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.,

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

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

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

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

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

is in $T\text{.}$