# Section16.2Integral Domains and Fields¶ permalink

Let us briefly recall some definitions. If $R$ is a ring and $r$ is a nonzero element in $R$, then $r$ is said to be a zero divisor if there is some nonzero element $s \in R$ such that $rs = 0$. 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 $i^2 = -1$, 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 ${\mathbb Z}[ i ]$. Then $\overline{\alpha} = a - bi$ is also a unit since if $\alpha \beta = 1$, then $\overline{\alpha} \overline{\beta} = 1$. If $\beta = c + di$, 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 $-1$; or, equivalently, $a + bi = \pm 1$ or $a+ bi = \pm i$. Therefore, units of this ring are $\pm 1$ and $\pm i$; 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.

The following surprising theorem is due to Wedderburn.

##### Proof

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

##### Example16.17

For every prime $p$, ${\mathbb Z}_p$ is a field of characteristic $p$. 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 $pa =0$, since the order of any nonzero element in the abelian group ${\mathbb Z}_p$ is $p$.