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Section 16.2 Integral 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.

Example 16.12.

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.

Example 16.13.

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.

Example 16.14.

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 \(d_1, d_2 \in D^*\text{,}\)

\begin{equation*} ad_1 = \lambda_a(d_1) = \lambda_a(d_2) = ad_2 \end{equation*}

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 \(\chr R\text{.}\)

Example 16.17.

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.