Principal Ideal Domains

Introduction

A principal ideal domain is an integral domain in which every ideal is generated by a single element. The definition is the strongest of the three divisibility rungs above it in this category, and it is the rung at which ideal theory and divisibility theory coincide exactly: an ideal, an arbitrary set of divisibility conditions, is the same thing as the set of multiples of one element, and the containment order of the ideals of $R$ is the divisibility order of $R$ reversed. Everything in the ring is then read off from elements, and the arithmetic that Unique Factorisation Domains, above this article, develops by factorisation is available for free.

The article establishes the three facts that make the rung usable: that the principal ideal domains are exactly the Bézout domains that are Noetherian, that a principal ideal domain is a unique factorisation domain, and that in a principal ideal domain every nonzero prime ideal is maximal, so that the quotient by a nonzero prime ideal is a field. The rung is strict: $\mathbb{Z}[x]$ is a unique factorisation domain that is not principal, and it is worked at the end. Throughout, $R$ is an integral domain; ideals, quotient rings and the correspondence theorem are Rings, above, and divisibility, irreducibles and primes are Integral Domains and Unique Factorisation Domains, above this article.

The Definition and Its Equivalents

Definition. An integral domain $R$ is a principal ideal domain (PID) if every ideal of $R$ is principal: for every ideal $I \subseteq R$ there is an $a \in R$ with $I = (a)$.

Proposition. For an integral domain $R$ the following are equivalent.

(a) $R$ is a principal ideal domain.

(b) $R$ is a Bézout domain and is Noetherian.

(c) Every ideal of $R$ is finitely generated and every finitely generated ideal of $R$ is principal.

Proof. (a) $\Rightarrow$ (b): every ideal is generated by one element, hence finitely generated, so $R$ is Noetherian; and a finitely generated ideal is an ideal, hence principal, so $R$ is Bézout. (b) $\Rightarrow$ (c): a Bézout domain has every finitely generated ideal principal, and a Noetherian ring has every ideal finitely generated. (c) $\Rightarrow$ (a): every ideal is finitely generated, hence principal.

The notions of a Bézout domain and of a Noetherian ring, and the theorem above, are those of Bézout Domains and Noetherian and Artinian Rings, the first above and the second below this article in this category.

Corollary. $\mathbb{Z}$ and, for a field $F$ (the field axioms are those of Fields, later in this category), the polynomial ring $F[x]$ are principal ideal domains.

Proof. $\mathbb{Z}$ is a Bézout domain by Bézout Domains, above, and it is Noetherian since every ideal of $\mathbb{Z}$ is generated by the least positive element it contains. For $F[x]$ the division algorithm with remainder, which belongs to Polynomial Rings and Rational Functions, below this article in this category, shows that every ideal is generated by a polynomial of least degree in it; and $F[x]$ is Euclidean, hence Bézout, by Euclidean Domains, below this article in this category.

Corollary. In a principal ideal domain every nonzero non-unit is a product of finitely many irreducibles, and every pair of elements has a gcd, which is a Bézout combination $ax + by$.

Proof. A principal ideal domain is Bézout, so the gcd statement is that of Bézout Domains, above. For the first statement, from Unique Factorisation Domains, above, it suffices to rule out an infinite strictly increasing chain of principal ideals; if $a$ were a nonzero non-unit without a factorisation into irreducibles, then $a$ would fail to be irreducible, so $a = a_1 b_1$ with both factors non-units and at least one of them, say $a_1$, again without a factorisation; then $(a) \subsetneq (a_1) \subsetneq (a_2) \subsetneq \cdots$ would be an infinite strictly increasing chain of ideals, impossible in a Noetherian ring.

The Ideal Structure of a Principal Ideal Domain

Prime and Maximal Ideals

Theorem. Let $R$ be a principal ideal domain and let $p \in R$ be nonzero and a non-unit. Then the following are equivalent.

(a) $p$ is irreducible.

(b) $(p)$ is a maximal ideal of $R$.

(c) $(p)$ is a prime ideal of $R$.

Proof. (a) $\Rightarrow$ (b): suppose $(p) \subseteq I \subseteq R$ with $I = (a)$, which is possible by the definition of the rung. Then $p = ab$; since $p$ is irreducible, $a$ or $b$ is a unit, so $I = (a) = R$ or $I = (p)$. (b) $\Rightarrow$ (c) is the proposition of Commutative Rings, above, that a maximal ideal is prime. (c) $\Rightarrow$ (a): let $p = ab$. Since $ab = p \in (p)$ and $(p)$ is prime, $a \in (p)$ or $b \in (p)$. If $a = pc$ then $p = pcb$, and cancelling $p$ gives $cb = 1$, so $b$ is a unit; if $b = pc$ then $ac = 1$ and $a$ is a unit. So $p$ is irreducible.

Corollary. In a principal ideal domain every nonzero prime ideal is maximal. Consequently a principal ideal domain that is not a field has Krull dimension one, in the sense that every strictly increasing chain of prime ideals has length at most one. Krull dimension is Integral Extensions and Krull Dimension, below this article in this category.

Proof. Every nonzero prime ideal is $(p)$ with $p \neq 0$, and then $p$ is irreducible by (c) $\Rightarrow$ (a), so the ideal is maximal by (a) $\Rightarrow$ (b). A field has no nonzero prime ideal, and in a non-field the zero ideal is prime and every nonzero prime is maximal.

Corollary. $R$ is a field if and only if its only ideals are $(0)$ and $R$; in particular a principal ideal domain is a field if and only if it has no irreducible element.

Proof. The condition on ideals says that every nonzero $a$ generates $R$, that is, every nonzero element is a unit, which is the field property. A field has no nonzero non-unit and hence no irreducible element. Conversely, if the principal ideal domain $R$ is not a field, choose a nonzero non-unit $a$; by the corollary above $a$ is a product of irreducibles, so an irreducible exists.

Example. The maximal ideals of $\mathbb{Z}$ are the ideals $(p)$ for $p$ prime, and the maximal ideals of $F[x]$ are the $(f)$ for $f$ an irreducible polynomial; in both rings these are exactly the nonzero prime ideals.

Quotients by Prime and Maximal Ideals

Theorem. Let $R$ be a principal ideal domain and let $p \in R$ be nonzero.

(a) $R/(p)$ is an integral domain if and only if $p$ is prime, and in that case $R/(p)$ is a field.

(b) If $p = q_1^{e_1} \cdots q_r^{e_r}$ with the $q_i$ pairwise non-associate irreducibles and $e_i \geq 1$, then the ideals of $R/(p)$ correspond to the divisors of $p$, and

$$ R/(p) \cong R/(q_1^{e_1}) \times \cdots \times R/(q_r^{e_r}), $$

a product of rings in each of which the ideals form a chain of $e_i + 1$ elements, namely the ideals generated by the images of $q_i^k$ for $k = 0, 1, \ldots, e_i$, the chain of inclusions having length $e_i$.

Proof. (a) The quotient by an ideal is a domain exactly when the ideal is prime and a field exactly when the ideal is maximal, by the quotient characterisations of Commutative Rings, above; a nonzero prime ideal of a principal ideal domain is maximal by the corollary above. (b) The ideals of a quotient correspond to the ideals of $R$ containing $(p)$ by the correspondence theorem of Rings, above; an ideal is $(d)$ with $(p) \subseteq (d)$, that is, $d \mid p$, and the divisors of $p$ are the products $q_1^{f_1}\cdots q_r^{f_r}$ with $0 \leq f_i \leq e_i$. The product decomposition is the Chinese remainder theorem of Commutative Rings, above, applied to the pairwise coprime ideals $(q_i^{e_i})$.

Example. For $n = p_1^{e_1} \cdots p_r^{e_r}$ the factorisation of $\mathbb{Z}/n\mathbb{Z}$ as a product of the rings $\mathbb{Z}/p_i^{e_i}\mathbb{Z}$ is the case $R = \mathbb{Z}$; the ring $\mathbb{Z}/p^n\mathbb{Z}$ has its ideals in a chain, with the nilradical $(p)/(p^n)$ of Reduced Rings and the Nilradical, above, and is therefore not reduced for $n \geq 2$.

Example. For $n = 12 = 2^2 \cdot 3$ the divisors of $12$ are $1, 2, 3, 4, 6, 12$, so $\mathbb{Z}/12\mathbb{Z}$ has exactly the six ideals generated by their images, and $\mathbb{Z}/12\mathbb{Z} \cong \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The ideals of $\mathbb{Z}/4\mathbb{Z}$ form the chain $(0) \subsetneq (\bar 2) \subsetneq \mathbb{Z}/4\mathbb{Z}$ of $3 = e + 1$ elements, and $\mathbb{Z}/3\mathbb{Z}$ is a field, with the two ideals $(0) \subsetneq \mathbb{Z}/3\mathbb{Z}$; the lattice of ideals of $\mathbb{Z}/12\mathbb{Z}$ is the product of these two chains and has $3 \cdot 2 = 6$ elements, matching the number of divisors of $12$.

Ideals and Divisibility

Proposition. Let $R$ be a principal ideal domain and let $a, b \in R$ be nonzero. Then

(a) $(a) \subseteq (b)$ if and only if $b \mid a$;

(b) $(a) = (b)$ if and only if $a \sim b$;

(c) $(a) \cap (b) = (\operatorname{lcm}(a,b))$ and $(a) + (b) = (\gcd(a,b))$, both principal.

Proof. For (a), $(a) \subseteq (b)$ means $a = bc$ for some $c \in R$, that is, $b \mid a$; (b) is (a) applied in both directions, since $a \sim b$ means that each divides the other. For (c), the identities are the corollary of Bézout Domains, above, which holds in every Bézout domain and therefore in every principal ideal domain.

Corollary. The map $a \mapsto (a)$ is an order-reversing bijection from the associate classes of nonzero elements of $R$ onto the nonzero ideals of $R$; the nonzero ideals form a lattice under $+$ and $\cap$, with $(a) + (b)$ the join and $(a) \cap (b)$ the meet.

Corollary (unique factorisation of ideals). Let $a \in R$ be a nonzero non-unit with factorisation $a = u q_1^{e_1} \cdots q_r^{e_r}$ into pairwise non-associate irreducibles. Then

$$ (a) = (q_1)^{e_1} (q_2)^{e_2} \cdots (q_r)^{e_r} $$

as a product of nonzero prime ideals, and this expression is unique up to the order of the factors. The nonzero prime ideals of $R$ are exactly the maximal ideals.

Proof. The product identity is the multiplicativity $(x)(y) = (xy)$ of generation by products, of Commutative Rings, above, applied repeatedly. Uniqueness follows from the uniqueness of the factorisation of $a$ in the unique factorisation domain $R$ together with $(x) = (y)$ if and only if $x \sim y$, proved above. The final statement is the theorem on prime and maximal ideals of this article.

Corollary. Let $a \in R$ be a nonzero non-unit with $a = u q_1^{e_1} \cdots q_r^{e_r}$ as above. Then the maximal ideals of $R$ containing $a$ are exactly $(q_1), \ldots, (q_r)$, and there are finitely many of them.

Proof. An ideal contains $a$ exactly when it contains $(a)$, and a maximal ideal is $(q)$ with $q$ irreducible; $(a) \subseteq (q)$ means $q \mid a$, and by the uniqueness of the factorisation of $a$ the irreducible divisors of $a$ are the associates of $q_1, \ldots, q_r$.

A Principal Ideal Domain Is a Unique Factorisation Domain

Theorem. Every principal ideal domain is a unique factorisation domain.

Proof. By the criterion of Unique Factorisation Domains, above, it suffices to show that $R$ satisfies the ascending chain condition on principal ideals and that every irreducible element of $R$ is prime. The chain condition holds because $R$ is Noetherian, by the proposition above. If $p$ is irreducible then $(p)$ is maximal by the theorem above, hence prime by Commutative Rings, above, so $p$ is prime by the definition of a prime element.

Corollary. The chain of implications

$$ \text{Euclidean} \Longrightarrow \text{principal} \Longrightarrow \text{unique factorisation} \Longrightarrow \text{integral domain} $$

holds, and the second implication is strict: $\mathbb{Z}[x]$ is a unique factorisation domain that is not a principal ideal domain.

Proof. The first implication is Euclidean Domains, below this article in this category, the middle is the theorem, and the last is the definition of an integral domain. The strictness is the section below.

Corollary. In a principal ideal domain the gcd of $a$ and $b$ is computed from their factorisations as in Unique Factorisation Domains, above, and is also a Bézout combination $ax + by$ by Bézout Domains, above. Consequently $\gcd(a,b) \sim 1$ if and only if $1 = ax + by$ for some $x, y$.

Remark. The rung is the meeting point of the two incomparable strengthenings of the GCD domains, of GCD Domains and Bézout Domains, above: a principal ideal domain is both a unique factorisation domain and a Bézout domain, and it is the largest class that is both, in the sense that a Bézout domain that is a unique factorisation domain is principal by the theorem of Bézout Domains, above.

The Strictness: $\mathbb{Z}[x]$ and Other Examples

$\mathbb{Z}[x]$ is a UFD and Not Principal

Theorem. $\mathbb{Z}[x]$ is a unique factorisation domain that is not a principal ideal domain.

Proof. $\mathbb{Z}[x]$ is a unique factorisation domain by Gauss's lemma, in Unique Factorisation Domains, above, applied to the unique factorisation domain $\mathbb{Z}$. The ideal $(2, x)$ is not principal. Suppose $(2,x) = (d)$; then $d \mid 2$ and $d \mid x$. From $d \mid x$ and $\deg x = 1$ it follows that $d$ is either a nonzero constant or an associate of $x$, and an associate of $x$ has degree $1$ and cannot divide the constant polynomial $2$; so $d$ is a constant, and $d \mid x$ means that $d$ divides the coefficient $1$ of $x$, hence $d$ is a unit and $(2,x) = \mathbb{Z}[x]$. But $(2,x)$ is proper: the evaluation homomorphism $\mathbb{Z}[x] \to \mathbb{Z}/2\mathbb{Z}$, $f \mapsto f(0) \bmod 2$, vanishes on $2$ and on $x$ and is surjective, since it fixes the residue of $1$. Hence $(2,x)$ is not principal and $\mathbb{Z}[x]$ is not a principal ideal domain.

Corollary. For a commutative ring $R$ the polynomial ring $R[x]$ is a principal ideal domain if and only if $R$ is a field.

Proof. If $R$ is a field then $R[x]$ is Euclidean, hence principal, by Euclidean Domains, below this article in this category. Conversely, suppose $R[x]$ is a principal ideal domain and let $a \in R$ be a nonzero non-unit; then the ideal $(a, x)$ is proper, since evaluation at $0$ maps it into the proper ideal $(a)$ of $R$, the element $a$ being a non-unit. If $(a,x) = (d)$ then $d \mid a$ and $d \mid x$, and the degree argument above shows that $d$ is either a nonzero constant or an associate of $x$; the second case is impossible because a nonzero constant is not divisible by an associate of $x$. So $d$ is a constant dividing the coefficient $1$ of $x$, hence a unit, and $(a,x) = R[x]$, a contradiction with properness. Hence $R$ has no nonzero non-unit, every nonzero element of $R$ is a unit, and $R$ is a field.

The Examples

Ring	PID	UFD	Bézout	Euclidean
$\mathbb{Z}$	yes	yes	yes	yes
$F[x]$, $F$ a field	yes	yes	yes	yes
$\mathbb{Z}[i]$	yes	yes	yes	yes
$\mathbb{Z}[(1+\sqrt{-19})/2]$	yes	yes	yes	no
$\mathbb{Z}[\sqrt{-5}]$	no	no	no	no
$\mathbb{Z}[x]$	no	yes	no	no
$F[x,y]$	no	yes	no	no

The first four rows are principal ideal domains; the fourth is the standard principal ideal domain that is not Euclidean, whose unique factorisation follows from the theorem of this article. The last three rows are the failures, in increasing order of severity: $\mathbb{Z}[x]$ and $F[x,y]$ are unique factorisation domains that are not principal, and $\mathbb{Z}[\sqrt{-5}]$ fails at the level of the gcd itself. The rows for $\mathbb{Z}[i]$ and $\mathbb{Z}[(1+\sqrt{-19})/2]$ are established in Euclidean Domains, below this article in this category, and the row for $\mathbb{Z}[\sqrt{-5}]$ in Unique Factorisation Domains, above.

Remark. For a nonzero $a \in R$ the quotient $R/(a)$ has finitely many ideals: they correspond to the divisors of $a$, and there are finitely many of those by the unique factorisation of $a$ in $R$. For $a = q^n$ a power of an irreducible, the count is $n + 1$, the ideals forming the chain of the theorem above.

Remark. The rung is not automatic for the rings of number theory. The ring of integers of a number field of class number greater than one is a Dedekind domain that is neither principal nor a unique factorisation domain, and the ideal class group measures the failure; that theory is Dedekind Domains and Ideal Class Groups, below this article in this category. The example $\mathbb{Z}[\sqrt{-5}]$ of the table is the quadratic case of class number two, of which the non-principal ideal $(2, 1+\sqrt{-5})$ is the witness.

Remark. The correspondence between ideals and elements that defines the rung also makes every quotient ring $R/(a)$ of a principal ideal domain a principal ideal ring, in the sense that all of its ideals are principal, each generated by the image of a generator: the ideals of $R/(a)$ are the ideals $(d)/(a)$ for $d \mid a$, as computed above. Such a quotient need not be a domain, so it is not itself a principal ideal domain. Nothing here uses the polynomial-ring theory beyond the division algorithm, which belongs to Polynomial Rings and Rational Functions, below this article in this category.

Summary

A principal ideal domain is an integral domain in which every ideal is principal; equivalently it is a Bézout domain that is Noetherian, equivalently a domain in which every ideal is finitely generated and every finitely generated ideal is principal. In such a ring every nonzero prime ideal is maximal, the maximal ideals are exactly the ideals $(p)$ with $p$ irreducible, and the quotient by a nonzero prime ideal is a field. Every principal ideal domain is a unique factorisation domain: the ascending chain condition on principal ideals holds because the ring is Noetherian, and every irreducible is prime because its principal ideal is maximal. Conversely a Bézout domain that is a unique factorisation domain is principal, so the rung is the intersection of the two strengthenings of the GCD domains.

The rung is strict in both directions. $\mathbb{Z}$ and $F[x]$ are principal ideal domains; $\mathbb{Z}[(1+\sqrt{-19})/2]$ is a principal ideal domain that is not Euclidean, and $\mathbb{Z}[x]$ is a unique factorisation domain that is not principal because $(2,x)$ has no single generator. The chain Euclidean $\Rightarrow$ principal $\Rightarrow$ unique factorisation $\Rightarrow$ domain is therefore strict at every step.

Summary of Notation

Symbol	Meaning
$R$	Integral domain, and a principal ideal domain where stated
$R^{\times}$	Group of units
$(a)$, $(a,b)$	Principal ideal generated by $a$; ideal generated by $a$ and $b$
PID	Principal ideal domain
UFD	Unique factorisation domain
$(p)$, $\mathrm{P}$, $\mathrm{M}$	The ideal generated by $p$; a prime ideal; a maximal ideal
$R/I$	Quotient ring of Rings, above
irreducible, prime	As in Integral Domains and Unique Factorisation Domains, above
$F$	A field
$F[x]$	Polynomial ring over $F$, of Polynomial Rings and Rational Functions, below
$\mathbb{Z}[i]$, $\mathbb{Z}[\sqrt{-5}]$, $\mathbb{Z}[(1+\sqrt{-19})/2]$	Quadratic rings used as examples; the last is the ring of integers of $\mathbb{Q}(\sqrt{-19})$
$\operatorname{Frac}(R)$	Fraction field, of Localization and the Fraction Field, below