## Section16.2Integral Domains and Fields

Let us briefly recall some definitions. If $$R$$ is a commutative 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)\text{.} \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}\text{.} \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\text{.} \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.