Let us briefly recall some definitions. If $R$ is a 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.
Example16.12
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
\n\\begin{equation*}\n1 = \\alpha \\beta \\overline{\\alpha} \\overline{\\beta} = (a^2 + b^2 )(c^2 + d^2).\n\\end{equation*}\n
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.
Example16.13
The set of matrices
\n\\begin{equation*}\nF =\n\\left\\{\n\\begin{pmatrix}\n1 & 0 \\\\\n0 & 1\n\\end{pmatrix},\n\\begin{pmatrix}\n1 & 1 \\\\\n1 & 0\n\\end{pmatrix},\n\\begin{pmatrix}\n0 & 1 \\\\\n1 & 1\n\\end{pmatrix},\n\\begin{pmatrix}\n0 & 0 \\\\\n0 & 0\n\\end{pmatrix}\n\\right\\}\n\\end{equation*}\n
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
\n\\begin{equation*}\n\\frac{a}{a^2 - 2 b^2} +\\frac{- b}{ a^2 - 2 b^2} \\sqrt{2}.\n\\end{equation*}\n
We have the following alternative characterization of integral domains.
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 $ab = ac\\text{,}$ 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.
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 $D\\text{.}$ 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 $d_1, d_2 \\in D^*\\text{,}$
\n\\begin{equation*}\nad_1 = \\lambda_a(d_1) = \\lambda_a(d_2) = ad_2\n\\end{equation*}\n
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. We will denote the characteristic of $R$ by $\\chr R\\text{.}$
Example16.17
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 $p\\text{.}$
Lemma16.18
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 $r \\in R\\text{,}$
\n\\begin{equation*}\nnr = n(1r) = (n 1) r = 0r = 0.\n\\end{equation*}\n
On the other hand, if no positive $n$ exists such that $n1 = 0\\text{,}$ then the characteristic of $R$ is zero.
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 $n \\neq 0\\text{.}$ 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.