Integral Domains

Introduction

An integral domain is a commutative ring with $1 \neq 0$ in which a product of two nonzero elements is nonzero. The hypothesis is one line long, and almost the whole multiplicative theory of the category rests on it: it gives cancellation, it makes divisibility a partial order on associate classes, it forces the characteristic to be $0$ or a prime, and it makes a nonzero polynomial of degree $n$ have at most $n$ roots. The rung that it occupies is the fourth of the commutative chain, above Reduced Rings and the Nilradical, of which a domain is the special case whose nilradical vanishes and whose zero ideal is prime.

Throughout, $R$ is a commutative ring with identity $1 \neq 0$, the corpus default, and $K$ is a field wherever a field is needed; each place where a field rather than a general ring is required is marked. Divisibility is the heart of the article, and the three rungs above it — GCD Domains, Bézout Domains and Unique Factorisation Domains, below this article in this category — are each a statement about how well divisibility in a domain behaves. The two constructions that a domain carries but does not develop here are the fraction field, deferred to Localization and the Fraction Field, below this article in this category, and the greatest common divisor, deferred to GCD Domains.

Definition and Elementary Properties

Cancellation

Definition. An integral domain is a commutative ring $R$ with $1 \neq 0$ such that

$$ a b = 0 \quad \Longrightarrow \quad a = 0 \ \text{or}\ b = 0 . $$

Equivalently, $R$ has no zero divisors in the sense of Rings, §Zero Divisors, the element $0$ being the only element that annihilates a nonzero element. A field is the special case in which every nonzero element is a unit; the field axioms belong to Fields, later in this category.

Proposition (cancellation). Let $R$ be an integral domain, $a, b, c \in R$ and $a \neq 0$. If $ab = ac$ then $b = c$.

Proof. From $a(b-c) = 0$ and $a \neq 0$ the definition gives $b - c = 0$.

Proposition. Every integral domain is a reduced ring in the sense of Reduced Rings and the Nilradical, above this article in this category, and every domain other than the zero ring is connected, its only idempotents being $0$ and $1$.

Proof. A nonzero nilpotent $a$ with $a^n = 0$ and $n$ minimal satisfies $a \cdot a^{n-1} = 0$ with $a^{n-1} \neq 0$, so $a$ is a zero divisor, which a domain does not have. If $e$ is idempotent and $e \neq 0$ then $e(1-e) = 0$, so $1 - e = 0$ and $e = 1$.

Remark. The condition "$1 \neq 0$" is not redundant: the zero ring satisfies $ab = 0 \Rightarrow a = 0$ or $b = 0$ vacuously, and is excluded in order that the prime subring of a domain be well defined.

Subrings, Products and Polynomial Rings

Proposition. Let $R$ be an integral domain.

(a) Every subring of $R$ containing $1$ is an integral domain.

(b) If $S$ is a nonzero commutative ring with $1 \neq 0$, then $R \times S$ is not an integral domain.

(c) The polynomial ring $R[x]$ is an integral domain, and $\deg(fg) = \deg f + \deg g$ for nonzero $f, g \in R[x]$.

(d) $(R[x])^{\times} = R^{\times}$.

Proof. (a) A subring with the same $1$ inherits the absence of zero divisors. (b) $(1, 0)(0, 1) = (0,0)$ with both factors nonzero. (c) If $f$ and $g$ are nonzero with leading coefficients $a$ and $b$, then the term of degree $\deg f + \deg g$ in $fg$ is $ab x^{\deg f + \deg g}$, and $ab \neq 0$ in the domain $R$; no higher-degree term occurs, so the coefficient is nonzero and $fg \neq 0$. (d) If $fg = 1$ then $\deg f + \deg g = \deg 1 = 0$, so $f$ and $g$ are nonzero constants, and a constant is a unit of $R[x]$ exactly when it is a unit of $R$.

The general theory of polynomial rings — degree, the division algorithm, roots in the applications sense and the arithmetic of $R[x]$ as a ring — is Polynomial Rings and Rational Functions, below this article in this category, and only the two facts above and the root theorem of this article are used before it is reached.

Example. $\mathbb{Z}$ is an integral domain, since a product of two nonzero integers has nonzero absolute value. For a field $K$ the polynomial ring $K[x]$ is an integral domain by (c), and its units are the nonzero constants. The rings $\mathbb{Z}[i]$, $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{-5}]$ are integral domains, each being a subring of $\mathbb{C}$; the last is a domain that is not a unique factorisation domain, as the rung below records.

Remark. $\mathbb{Z}/n\mathbb{Z}$ is an integral domain only when $n$ is prime, in which case it is the field $\mathbb{F}_p$; for composite $n$ it has zero divisors. So being a domain is a genuine restriction on a quotient, and a prime ideal in the sense of Commutative Rings, above this article, is exactly one whose quotient is a domain.

Divisibility

The Divisibility Preorder

Definition. For $a, b \in R$, write $a \mid b$, and say that $a$ divides $b$, if $b = ac$ for some $c \in R$. An element $a$ is a divisor of $b$ in that case, and $b$ is a multiple of $a$.

Proposition. Let $R$ be an integral domain and $a, b, c \in R$, with $u \in R^{\times}$ a unit.

(a) $a \mid a$, and $1 \mid a$, and $a \mid 0$.

(b) If $a \mid b$ and $b \mid c$ then $a \mid c$.

(c) $a \mid b$ if and only if the principal ideal satisfies $(b) \subseteq (a)$.

(d) $u \mid a$ and $a \mid u a$; if $a \neq 0$ then $a \mid u a$ has $u$ as its only cofactor in the sense that $ua = av$ forces $v = u$.

(e) If $a \mid b$ and $a \mid c$ then $a \mid (bx + cy)$ for all $x, y \in R$.

Proof. (a) $a = a \cdot 1$, $a = 1 \cdot a$ and $0 = a \cdot 0$. (b) If $b = ax$ and $c = by$ then $c = a(xy)$. (c) $b = ac$ says exactly that $b \in (a)$, that is, $(b) \subseteq (a)$. (d) Clear from the definition; cancellation gives $ua = av \Rightarrow u = v$ when $a \neq 0$. (e) $bx + cy = a(x'x + y'y)$ if $b = ax'$ and $c = ay'$.

Thus divisibility is a reflexive and transitive relation, a preorder; by (c) it is the inclusion order of principal ideals read backwards.

Associates

Definition. Two elements $a, b \in R$ are associates, written $a \sim b$, if $a = ub$ for some unit $u \in R^{\times}$.

Proposition (associates). Let $R$ be an integral domain and $a, b \in R$. Then $a \sim b$ if and only if $a \mid b$ and $b \mid a$.

Proof. If $a = ub$ with $u$ a unit then $b = u^{-1}a$, so each divides the other. Conversely suppose $b = ax$ and $a = by$. If $a = 0$ then $b = a x = 0$, and $a = 1 \cdot b$. If $a \neq 0$, then $a = axy$ and cancellation gives $1 = xy$, so $x$ is a unit and $a \sim b$.

This is the point at which the domain hypothesis is used: in a general commutative ring the conclusion fails, and what breaks it is that the two cofactors need not be units. In $R = k[s,t]/(s^2, st^2)$ the element $s$ divides $s + st$, because $s + st = s(1+t)$, and $s + st$ divides $s$, because $s = (s+st)(1-t)$; but the units of $R$ are the classes $c + s(b + et)$ with $c \neq 0$, and these multiply $s$ to $cs$, so $s+st$ is not an associate of $s$.

Corollary. Association is an equivalence relation, and divisibility induces a partial order on the associate classes of an integral domain: $[a] \leq [b]$ exactly when $a \mid b$. The class of $0$ is the largest element, and the class of the units is the smallest.

Proof. Reflexivity, symmetry and transitivity follow from $1 \in R^{\times}$ and the closure of $R^{\times}$ under inverses and products; antisymmetry on classes is the proposition above.

Irreducible and Prime Elements

Immediately above a domain in the chain one asks which elements cannot be factored further and which control divisibility of products.

Definition. Let $R$ be an integral domain. A nonzero non-unit $a \in R$ is

(a) irreducible if every factorisation $a = bc$ has $b$ or $c$ a unit;

(b) prime if $a \mid bc$ always implies $a \mid b$ or $a \mid c$.

Proposition. Let $R$ be an integral domain.

(a) A prime element is irreducible.

(b) The ideal $(p)$ generated by a prime element $p$ is a prime ideal, and conversely if $(p)$ is prime and $p \neq 0$ then $p$ is prime.

Proof. (a) Let $p$ be prime and $p = bc$. Since $p \mid bc = p$, primality gives $p \mid b$ or $p \mid c$. If $p \mid b$, say $b = pd$, then $p = pdc$ and cancellation of $p \neq 0$ gives $dc = 1$, so $c$ is a unit; the other case is symmetric. (b) $ab \in (p)$ means $p \mid ab$, which for prime $p$ means $p \mid a$ or $p \mid b$, that is $a \in (p)$ or $b \in (p)$; conversely $(p)$ prime says $p \mid ab$ implies $p \mid a$ or $p \mid b$.

Remark. The converse of (a) fails, and its failure is the arithmetic obstruction that the rung Unique Factorisation Domains, below this article in this category, is defined to exclude: in $\mathbb{Z}[\sqrt{-5}]$ the element $2$ is irreducible but not prime. What is missing in a general domain is not the factorisation theory but the existence of greatest common divisors, which GCD Domains, below this article in this category, adds.

Characteristic

The Characteristic of a Domain

Definition. The characteristic $\operatorname{char} R$ of a unital ring $R$ is the least positive integer $n$ with $n \cdot 1 = 0$, if such an $n$ exists, and $0$ otherwise. Equivalently, $\operatorname{char} R$ is the order of $1$ in the additive group $(R,+)$ when that order is finite, the two conventions agreeing under the first one, and it generates the kernel of the unique unital homomorphism

$$ \chi : \mathbb{Z} \to R, \qquad \chi(n) = n \cdot 1 . $$

Theorem. Let $R$ be an integral domain. Then $\operatorname{char} R$ is either $0$ or a prime number. In particular $\operatorname{char} R \neq 1$, and no domain has composite characteristic.

Proof. Suppose $n = \operatorname{char} R$ is finite and $n = ab$ with $a, b > 1$. Then $(a \cdot 1)(b \cdot 1) = (ab) \cdot 1 = n \cdot 1 = 0$, and since $R$ is a domain one factor vanishes, so $a \cdot 1 = 0$ or $b \cdot 1 = 0$, contradicting the minimality of $n$. Hence $n$ has no proper factorisation.

Corollary. $\mathbb{Z}/6\mathbb{Z}$ is not an integral domain, since it has characteristic $6$; a domain of characteristic $0$ contains a copy of $\mathbb{Z}$ and no finite subring, and a domain of characteristic $p$ contains a copy of $\mathbb{F}_p$. The quotient $\mathbb{Z}/p\mathbb{Z}$ is the field $\mathbb{F}_p$, the field axioms being those of Fields, later in this category.

Proposition (freshman's dream). Let $R$ be a commutative ring of prime characteristic $p$. Then for all $x, y \in R$ and all $n \geq 1$,

$$ (x + y)^{p^n} = x^{p^n} + y^{p^n}, \qquad (xy)^{p^n} = x^{p^n} y^{p^n} . $$

Proof. For $n = 1$, expand $(x+y)^p$ by the binomial theorem of Commutative Rings. Each binomial coefficient $\binom{p}{k}$ with $0 < k < p$ is divisible by $p$ and so vanishes in $R$, leaving $x^p + y^p$; the second identity is the commutativity of $R$. The general case follows by iterating the case $n = 1$.

The Prime Subring and the Prime Field

The image of $\chi$ is the smallest subring of $R$ containing $1$.

Definition. The prime subring of a unital ring $R$ is the image of $\chi : \mathbb{Z} \to R$, the subring generated by $1$.

Proposition. Let $R$ be an integral domain, with prime subring $P$.

(a) If $\operatorname{char} R = 0$ then $\chi$ is injective, so $P \cong \mathbb{Z}$.

(b) If $\operatorname{char} R = p$ then $P \cong \mathbb{F}_p$, a field, and $P$ is the prime field of $R$.

(c) The characteristic of a subring of $R$ that contains $1$ equals $\operatorname{char} R$, and its prime subring is $P$.

Proof. (a) If $\chi(m) = \chi(n)$ with $m > n$ then $(m-n) \cdot 1 = 0$ with $m - n > 0$, contradicting $\operatorname{char} R = 0$. (b) The kernel of $\chi$ is $p\mathbb{Z}$ and $\mathbb{Z}/p\mathbb{Z}$ is a field. (c) The element $1$ and the subring it generates are the same in $R$ and in a subring containing $1$.

Proposition (embedding of the prime field). Let $R$ be an integral domain. Then $R$ contains a smallest subfield, namely $\mathbb{Q}$ if $\operatorname{char} R = 0$ and $\mathbb{F}_p$ if $\operatorname{char} R = p$.

Proof. In prime characteristic the prime subring is the field $\mathbb{F}_p$ by the proposition above. In characteristic $0$ the prime subring is $\mathbb{Z}$, which is not a field; it is contained in every subfield of $R$, and the smallest subfield containing it is its fraction field.

The fraction field of a domain is constructed in Localization and the Fraction Field, below this article in this category: every integral domain $R$ embeds in a field $\operatorname{Frac}(R)$, its fraction field, characterised by the property that every injective homomorphism from $R$ into a field extends uniquely to $\operatorname{Frac}(R)$. In characteristic $0$ the fraction field of $\mathbb{Z}$ is $\mathbb{Q}$; in prime characteristic the fraction field of $\mathbb{F}_p$ is $\mathbb{F}_p$ itself.

Roots of Polynomials

The Root Theorem

The domain hypothesis converts a statement about divisibility of polynomials into the finite bound on roots that the applications below this article rely on.

Definition. Let $f \in R[x]$ and $a \in R$. Then $a$ is a root of $f$ if $f(a) = 0$, where $f(a) = \sum_k c_k a^k$ for $f = \sum_k c_k x^k$.

Lemma. Let $R$ be a commutative ring, $f \in R[x]$ and $a \in R$ with $f(a) = 0$. Then $f = (x - a)g$ for some $g \in R[x]$, and $\deg g = \deg f - 1$ when $f \neq 0$.

Proof. For every $k \geq 1$, $x^k - a^k = (x - a)(x^{k-1} + x^{k-2}a + \cdots + a^{k-1})$; the identity is verified by expanding the right-hand side, and it uses only distributivity. Hence, with $f = \sum_{k=0}^{n} c_k x^k$,

$$ f(x) = f(x) - f(a) = \sum_{k=1}^{n} c_k (x^k - a^k) = (x-a) \sum_{k=1}^{n} c_k (x^{k-1} + x^{k-2}a + \cdots + a^{k-1}), $$

which exhibits $g$ of the stated degree when $c_n \neq 0$.

Theorem (root theorem). Let $R$ be an integral domain and let $f \in R[x]$ have degree $< m$. If $f$ has $m$ distinct roots $a_1, \ldots, a_m$ in $R$, then $f = 0$.

Proof. Induction on $m$. For $m = 0$ the hypothesis on the degree says $f = 0$. For $m \geq 1$, the lemma applied to the root $a_1$ gives $f = (x - a_1)g$ with $\deg g < m - 1$. For $i \geq 2$,

$$ 0 = f(a_i) = (a_i - a_1) g(a_i), $$

and $a_i - a_1 \neq 0$ since the roots are distinct, so $g(a_i) = 0$ because $R$ is a domain. Thus $g$ has the $m - 1$ distinct roots $a_2, \ldots, a_m$ and degree $< m - 1$, and the induction hypothesis gives $g = 0$, hence $f = 0$.

Consequences

Corollary. Let $R$ be an integral domain and $f \in R[x]$ nonzero of degree $n$. Then $f$ has at most $n$ roots in $R$.

Proof. If $f$ had $n+1$ distinct roots, the root theorem with $m = n+1$ would give $f = 0$, since $\deg f = n < n+1$.

Corollary. Let $R$ be an integral domain and let $f, g \in R[x]$ have degree $\leq n$. If $f(a) = g(a)$ for $n+1$ distinct elements $a \in R$, then $f = g$.

Proof. Apply the previous corollary to $f - g$, of degree $\leq n$.

Example. Over $\mathbb{Z}/4\mathbb{Z}$, which is not a domain, the polynomial $2x$ of degree $1$ has the two roots $0$ and $2$; over $\mathbb{Z}/8\mathbb{Z}$, the polynomial $x^2 - 1$ has the four roots $1, 3, 5, 7$, since each of these is its own inverse modulo $8$. So the number of roots can exceed the degree when the ring has zero divisors, and the domain hypothesis in the root theorem and in both corollaries cannot be dropped. Over a domain the bound is sharp: $x^2 - 1$ has exactly the two roots $\pm 1$ in $\mathbb{Z}$, and $x^2 + 1$ has none.

Corollary. A finite integral domain is a field.

Proof. Let $R$ be a finite domain and $a \in R$ nonzero. The powers $a, a^2, a^3, \ldots$ cannot all be distinct in the finite set $R$, so $a^i = a^j$ for some $i > j \geq 1$. Then $a^j (a^{i-j} - 1) = 0$ with $a^j \neq 0$, so $a^{i-j} = 1$ and $a \cdot a^{i-j-1} = 1$ exhibits $a$ as a unit.

The Shape of the Rung

Examples

Example ($\mathbb{Z}$). The integers form an integral domain of characteristic $0$, whose units are $\pm 1$ and whose irreducible elements are the primes. Divisibility in $\mathbb{Z}$ is governed by the absolute value of the divisor, and the prime factorisation of an integer is the model case of the rungs above this article in this category.

Example ($K[x]$, $K$ a field). The polynomial ring over a field is an integral domain of characteristic $\operatorname{char} K$, with units the nonzero constants, by the proposition above. Its irreducibles are the irreducible polynomials, and the root theorem gives the standard test: a polynomial of degree $2$ or $3$ over a field is irreducible exactly when it has no root in the field.

Example (quadratic rings). For a squarefree integer $d$, $\mathbb{Z}[\sqrt{d}]$ is an integral domain, being a subring of $\mathbb{C}$. Its multiplicative structure is governed by the norm

$$ N(a + b\sqrt{d}) = a^2 - d b^2 = (a + b\sqrt{d})(a - b\sqrt{d}), $$

which is multiplicative, so that $N(xy) = N(x)N(y)$, and an element is a unit exactly when its norm is a unit of $\mathbb{Z}$, that is, when $N(x) = \pm 1$; when $d < 0$ the norm is non-negative and the condition is $N(x) = 1$. For $d = -1$ and $d = -5$ the norm reads $N(a+bi) = a^2 + b^2$ and $N(a + b\sqrt{-5}) = a^2 + 5b^2$. This $N$ is the norm on a quadratic ring; it is not the Euclidean degree function of Euclidean Domains, below this article in this category, although the letter is the same.

Where the Rung Sits

The integral domain is the point at which cancellation and a well-behaved divisibility theory are available, and it is the weakest rung of the commutative chain at which that is so: a reduced ring such as $\mathbb{Z} \times \mathbb{Z}$ has cancellation only for elements that are not zero divisors, and its divisibility is not a partial order on associate classes.

Remark (the rungs above). Every field is an integral domain, and the converse fails, witness $\mathbb{Z}$; the division rings that generalise fields by dropping commutativity are Division Rings, later in this category, and a non-commutative division ring is not an integral domain, since the terminology reserves the word for the commutative case.

Remark (what the rung does not yet supply). Two elements of a domain need not have a greatest common divisor, and an irreducible element need not be prime. The first failure is repaired by GCD Domains, below this article in this category; the second, and with it unique factorisation, by the chain Unique Factorisation Domains, Principal Ideal Domains, Euclidean Domains, each below this article in this category. The standard witness in both cases is $\mathbb{Z}[\sqrt{-5}]$, which is an integral domain, which is not a unique factorisation domain because

$$ 6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}) $$

are two factorisations into irreducibles differing neither by order nor by units; the verification is carried out in Unique Factorisation Domains, below this article in this category.

Summary

An integral domain is a commutative ring with $1 \neq 0$ and no zero divisors, equivalently a commutative ring in which $ab = 0$ forces $a = 0$ or $b = 0$. Cancellation holds for nonzero factors, subrings and polynomial rings over a domain remain domains, and a product of two nonzero rings is never a domain. Divisibility is a preorder, equivalent to reverse inclusion of principal ideals, and it becomes a partial order on associate classes because in a domain $a \mid b$ and $b \mid a$ force $a \sim b$; this is the point at which domains are separated from general commutative rings.

A nonzero non-unit is irreducible when it admits no nontrivial factorisation and prime when it divides a product only through its factors; every prime is irreducible, and the converse can fail. The characteristic of a domain is $0$ or a prime, and the prime subring is $\mathbb{Z}$ in characteristic $0$ and the field $\mathbb{F}_p$ in characteristic $p$; every domain contains exactly one smallest field, $\mathbb{Q}$ or $\mathbb{F}_p$. Over a domain a polynomial of degree $< m$ with $m$ distinct roots vanishes identically, so a nonzero polynomial of degree $n$ has at most $n$ roots, and a finite domain is a field. The greatest common divisor, the fraction field and the factorisation rungs are the content of articles below this one in the category, each of which is a statement about how well divisibility in a domain behaves.

Summary of Notation

Symbol	Meaning
$R$	Integral domain (commutative ring with $1 \neq 0$ and no zero divisors)
$K$, $F$	Fields, where the field axioms are needed
$R^{\times}$	Group of units
$a \mid b$	$a$ divides $b$, that is, $b = ac$ for some $c \in R$
$a \sim b$	Associates, $a = ub$ with $u \in R^{\times}$
$(a)$, $(a, b)$	Principal ideal generated by $a$; ideal generated by $a$ and $b$
irreducible, prime	Nonzero non-unit admitting no nontrivial factorisation; dividing products only through factors
$R[x]$	Polynomial ring in one indeterminate
$f(a)$	Evaluation of $f$ at $a$; $a$ a root when $f(a) = 0$
$\operatorname{char} R$	Characteristic of $R$
$n \cdot 1$	Sum of $n$ copies of $1$
$\chi : \mathbb{Z} \to R$	The unique unital homomorphism, $n \mapsto n \cdot 1$
$\mathbb{F}_p$	The field $\mathbb{Z}/p\mathbb{Z}$ of $p$ elements
$\operatorname{Frac}(R)$	Fraction field of $R$
$N(a + b\sqrt{d}) = a^2 - d b^2$	Norm on a quadratic ring $\mathbb{Z}[\sqrt{d}]$