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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 \(a, b \in R\).

  2. \((a + b) + c = a + ( b + c)\) for \(a, b, c \in R\).

  3. There is an element \(0\) in \(R\) such that \(a + 0 = a\) for all \(a \in R\).

  4. For every element \(a \in R\), there exists an element \(-a\) in \(R\) such that \(a + (-a) = 0\).

  5. \((ab) c = a ( b c)\) for \(a, b, c \in R\).

  6. For \(a, b, c \in R\), \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 \(a \in R\), 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 \(ab = 0\), either \(a = 0\) or \(b = 0\). A division ring is a ring \(R\), with an identity, in which every nonzero element in \(R\) is a unit; that is, for each \(a \in R\) with \(a \neq 0\), there exists a unique element \(a^{-1}\) such that \(a^{-1} a = a a^{-1} = 1\). A commutative division ring is called a field. The relationship among rings, integral domains, division rings, and fields is shown in Figure 16.1.

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Figure16.1Types of rings

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 \(b\), either \(a=0\) or \(b=0\). 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 \(-1\).


Under the ordinary operations of addition and multiplication, all of the familiar number systems are rings: the rationals, \({\mathbb Q}\); the real numbers, \({\mathbb R}\); and the complex numbers, \({\mathbb C}\). Each of these rings is a field.


We can define the product of two elements \(a\) and \(b\) in \({\mathbb Z}_n\) by \(ab \pmod{n}\). For instance, in \({\mathbb Z}_{12}\), \(5 \cdot 7 \equiv 11 \pmod{12}\). 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 \({\mathbb Z}_{12}\), 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 zero divisor if there is a nonzero element \(b\) in \(R\) such that \(ab = 0\). In the previous example, 3 and 4 are zero divisors in \({\mathbb Z}_{12}\).


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


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 \(AB \neq BA\). Also, notice that we can have \(AB = 0\) when neither \(A\) nor \(B\) is zero.


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 \(i^2 = -1\). 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 \(a + b {\mathbf i} + c {\mathbf j} +d {\mathbf k}\), 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, \({\mathbf i}\), \({\mathbf j}\), and \({\mathbf k}\): \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 \({\mathbf i}\), \({\mathbf j}\), and \({\mathbf k}\). 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 \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 \(a\), \(b\), \(c\), and \(d\) are all zero. So if \(a + b {\mathbf i} + c {\mathbf j} +d {\mathbf k} \neq 0\), \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*}

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 \(R\).


The ring \(n {\mathbb Z}\) is a subring of \({\mathbb Z}\). 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.)


Let \(R ={\mathbb M}_2( {\mathbb R} )\) be the ring of \(2 \times 2\) matrices with entries in \({\mathbb R}\). If \(T\) is the set of upper triangular matrices in \(R\); 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 \(R\). 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 \(T\), then clearly \(A-B\) is also in \(T\). Also, \begin{equation*}AB = \begin{pmatrix} a a' & ab' + bc' \\ 0 & cc' \end{pmatrix}\end{equation*} is in \(T\).