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*.

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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.

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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.

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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.

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Proposition16.15Cancellation 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 \(ab = ac\), we have \(b=c\).

Let \(D\) be an integral domain. Then \(D\) has no zero divisors. Let \(ab = ac\) with \(a \neq 0\). Then \(a(b - c) =0\). Hence, \(b - c = 0\) and \(b = c\).

Conversely, let us suppose that cancellation is possible in \(D\). That is, suppose that \(ab = ac\) implies \(b=c\). Let \(ab = 0\). If \(a \neq 0\), then \(ab = a 0\) or \(b=0\). Therefore, \(a\) cannot be a zero divisor.

The following surprising theorem is due to Wedderburn.

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Theorem16.16

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 \(D\). 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 \(\lambda_a(d) = ad\). This map makes sense, because if \(a \neq 0\) and \(d \neq 0\), then \(ad \neq 0\). The map \(\lambda_a\) is one-to-one, since for \(d_1, d_2 \in D^*\),
\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 \(d \in D^\ast\), \(\lambda_a(d) = ad = 1\). Therefore, \(a\) has a left inverse. Since \(D\) is commutative, \(d\) must also be a right inverse for \(a\). 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 \(nr\). 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 \(\chr R\).

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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 \(p\).

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Lemma16.18

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

If 1 has order \(n\), then \(n\) is the least positive integer such that \(n 1 = 0\). Thus, for all \(r \in R\),
\begin{equation*}nr = n(1r) = (n 1) r = 0r = 0.\end{equation*}
On the other hand, if no positive \(n\) exists such that \(n1 = 0\), then the characteristic of \(R\) is zero.

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Theorem16.19

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 \(n \neq 0\). If \(n\) is not prime, then \(n = ab\), where \(1 \lt a \lt n\) and \(1 \lt b \lt n\). By Lemma 16.18, we need only consider the case \(n 1 = 0\). Since \(0 = n 1 = (ab)1 = (a1)(b1)\) and there are no zero divisors in \(D\), either \(a1 =0\) or \(b1=0\). Hence, the characteristic of \(D\) must be less than \(n\), which is a contradiction. Therefore, \(n\) must be prime.