16.4: 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 \(\beta = c + di\text{,}\)

Solution

then

\[ 1 = \alpha \beta \overline{\alpha} \overline{\beta} = (a^2 + b^2 )(c^2 + d^2)\text{.} \nonumber \]

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

\[ F = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \right\} \nonumber \]

Solution

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

Solution

\[ \frac{a}{a^2 - 2 b^2} +\frac{- b}{ a^2 - 2 b^2} \sqrt{2}\text{.} \nonumber \]

We have the following alternative characterization of integral domains.

Proposition \(16.15\). Cancellation Law

Let \(D\) be a commutative ring with identity. Then \(D\) is an integral domain if and only if for all nonzero elements \(a \in D\) with \(ab = ac\text{,}\) we have \(b=c\text{.}\)

Proof

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.

Theorem \(16.16\)

Every finite integral domain is a field.

Proof

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

\[ ad_1 = \lambda_a(d_1) = \lambda_a(d_2) = ad_2 \nonumber \]

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 \(p\text{.}\)

Solution

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 \(p\text{.}\)

Lemma \(16.18\)

Let \(R\) be a ring with identity. If \(1\) has order \(n\text{,}\) then the characteristic of \(R\) is \(n\text{.}\)

Proof

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

\[ nr = n(1r) = (n 1) r = 0r = 0\text{.} \nonumber \]

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

Theorem \(16.19\)

The characteristic of an integral domain is either prime or zero.

Proof

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.