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Section16.2Integral Domains and Fields

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 \(p\).

Proof