
## Section16.2Integral Domains and Fields

Let us briefly recall some definitions. If $R$ is a ring and $r$ is a nonzero element in $R\text{,}$ then $r$ is said to be a zero divisor if there is some nonzero element $s \in R$ such that $rs = 0\text{.}$ A commutative ring with identity is said to be an integral domain if it has no zero divisors. If an element $a$ in a ring $R$ with identity has a multiplicative inverse, we say that $a$ is a unit. If every nonzero element in a ring $R$ is a unit, then $R$ is called a division ring. A commutative division ring is called a field.

###### Example16.12

If $i^2 = -1\text{,}$ then the set ${\mathbb Z}[ i ] = \{ m + ni : m, n \in {\mathbb Z} \}$ forms a ring known as the Gaussian integers. It is easily seen that the Gaussian integers are a subring of the complex numbers since they are closed under addition and multiplication. Let $\alpha = a + bi$ be a unit in ${\mathbb Z}[ i ]\text{.}$ Then $\overline{\alpha} = a - bi$ is also a unit since if $\alpha \beta = 1\text{,}$ then $\overline{\alpha} \overline{\beta} = 1\text{.}$ If $\beta = c + di\text{,}$ then

\begin{equation*} 1 = \alpha \beta \overline{\alpha} \overline{\beta} = (a^2 + b^2 )(c^2 + d^2). \end{equation*}

Therefore, $a^2 + b^2$ must either be $1$ or $-1\text{;}$ or, equivalently, $a + bi = \pm 1$ or $a + bi = \pm i\text{.}$ Therefore, units of this ring are $\pm 1$ and $\pm i\text{;}$ hence, the Gaussian integers are not a field. We will leave it as an exercise to prove that the Gaussian integers are an integral domain.

###### Example16.13

The set of matrices

\begin{equation*} F = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \right\} \end{equation*}

with entries in ${\mathbb Z}_2$ forms a field.

###### Example16.14

The set ${\mathbb Q}( \sqrt{2}\, ) = \{ a + b \sqrt{2} : a, b \in {\mathbb Q} \}$ is a field. The inverse of an element $a + b \sqrt{2}$ in ${\mathbb Q}( \sqrt{2}\, )$ is

\begin{equation*} \frac{a}{a^2 - 2 b^2} +\frac{- b}{ a^2 - 2 b^2} \sqrt{2}. \end{equation*}

We have the following alternative characterization of integral domains.

Let $D$ be an integral domain. Then $D$ has no zero divisors. Let $ab = ac$ with $a \neq 0\text{.}$ Then $a(b - c) =0\text{.}$ Hence, $b - c = 0$ and $b = c\text{.}$

Conversely, let us suppose that cancellation is possible in $D\text{.}$ That is, suppose that $ab = ac$ implies $b=c\text{.}$ Let $ab = 0\text{.}$ If $a \neq 0\text{,}$ then $ab = a 0$ or $b=0\text{.}$ Therefore, $a$ cannot be a zero divisor.

The following surprising theorem is due to Wedderburn.

Let $D$ be a finite integral domain and $D^\ast$ be the set of nonzero elements of $D\text{.}$ We must show that every element in $D^*$ has an inverse. For each $a \in D^\ast$ we can define a map $\lambda_a : D^\ast \rightarrow D^\ast$ by $\lambda_a(d) = ad\text{.}$ This map makes sense, because if $a \neq 0$ and $d \neq 0\text{,}$ then $ad \neq 0\text{.}$ The map $\lambda_a$ is one-to-one, since for $d_1, d_2 \in D^*\text{,}$

implies $d_1 = d_2$ by left cancellation. Since $D^\ast$ is a finite set, the map $\lambda_a$ must also be onto; hence, for some $d \in D^\ast\text{,}$ $\lambda_a(d) = ad = 1\text{.}$ Therefore, $a$ has a left inverse. Since $D$ is commutative, $d$ must also be a right inverse for $a\text{.}$ Consequently, $D$ is a field.

For any nonnegative integer $n$ and any element $r$ in a ring $R$ we write $r + \cdots + r$ ($n$ times) as $nr\text{.}$ We define the characteristic of a ring $R$ to be the least positive integer $n$ such that $nr = 0$ for all $r \in R\text{.}$ If no such integer exists, then the characteristic of $R$ is defined to be $0\text{.}$ We will denote the characteristic of $R$ by $\chr R\text{.}$

###### Example16.17

For every prime $p\text{,}$ ${\mathbb Z}_p$ is a field of characteristic $p\text{.}$ By Proposition 3.4, every nonzero element in ${\mathbb Z}_p$ has an inverse; hence, ${\mathbb Z}_p$ is a field. If $a$ is any nonzero element in the field, then $pa =0\text{,}$ since the order of any nonzero element in the abelian group ${\mathbb Z}_p$ is $p\text{.}$

If $1$ has order $n\text{,}$ then $n$ is the least positive integer such that $n 1 = 0\text{.}$ Thus, for all $r \in R\text{,}$

\begin{equation*} nr = n(1r) = (n 1) r = 0r = 0. \end{equation*}

On the other hand, if no positive $n$ exists such that $n1 = 0\text{,}$ then the characteristic of $R$ is zero.

Let $D$ be an integral domain and suppose that the characteristic of $D$ is $n$ with $n \neq 0\text{.}$ If $n$ is not prime, then $n = ab\text{,}$ where $1 \lt a \lt n$ and $1 \lt b \lt n\text{.}$ By Lemma 16.18, we need only consider the case $n 1 = 0\text{.}$ Since $0 = n 1 = (ab)1 = (a1)(b1)$ and there are no zero divisors in $D\text{,}$ either $a1 =0$ or $b1=0\text{.}$ Hence, the characteristic of $D$ must be less than $n\text{,}$ which is a contradiction. Therefore, $n$ must be prime.