Principal ideal domain

This article defines a property of integral domains, viz., a property that, given any integral domain, is either true or false for that.
The corrresponding general property for commutative unital rings is: principal ideal ring
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VIEW RELATED: Commutative unital ring property implications | Commutative unital ring property non-implications |Commutative unital ring metaproperty satisfactions | Commutative unital ring metaproperty dissatisfactions | Commutative unital ring property satisfactions | Commutative unital ring property dissatisfactions

Definition

A commutative unital ring is termed a principal ideal domain (also known as PID) if it is an integral domain and a principal ideal ring. Explicitly, it must satisfy the following equivalent conditions:

No.	Shorthand	A commutative unital ring is termed a principal ideal domain if ...	A commutative unital ring $R$ is termed a principal ideal domain if ...
1	every ideal principal	it is an integral domain and every ideal in it is principal	$R$ is an integral domain (i.e., $ab = 0$ implies $a = 0$ or $b = 0$) and for every ideal $I \subseteq R$, there exists $x \in I$ such that $I = (x)$, i.e., $I = Rx$.
2	every prime ideal principal	it is an integral domain and every prime ideal in it is principal	$R$ is an integral domain (i.e., $ab = 0$ implies $a = 0$ or $b = 0$) and for every prime ideal $P \subseteq R$, there exists $x \in P$ such that $P = (x)$, i.e., $P = Rx$.
3	Dedekind-Hasse norm	it admits a Dedekind-Hasse norm	there is a Dedekind-Hasse norm on $R$.

Equivalence of definitions

• For the equivalence of definitions (1) and (2), refer Principal ideal ring iff every prime ideal is principal.
• For (1) implies (3), refer principal ideal domain admits multiplicative Dedekind-Hasse norm.
• For (3) implies (1), refer Dedekind-Hasse norm implies principal ideal ring.

Examples

Since Euclidean implies PID, all the typical examples of Eulidean domains are also examples of principal ideal domains. There are some PIDs that are not Euclidean domains -- in particular, the domain $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is not a Euclidean domain because it does not have a universal side divisor.

Relation with other properties

Expression as a conjunction of other properties

First component of conjunction	Second component of conjunction	Proof
principal ideal ring	integral domain	by definition
Noetherian domain	Bezout domain	Noetherian and Bezout iff principal ideal
unique factorization domain	Bezout domain	Unique factorization and Bezout iff principal ideal
unique factorization domain	Dedekind domain	(see next)
unique factorization domain	one-dimensional domain	Unique factorization and one-dimensional iff principal ideal

Stronger properties

Property	Meaning	Proof of implication	Proof of strictness (reverse implication failure)	Intermediate notions
Euclidean domain	has a Euclidean norm	Euclidean implies PID	PID not implies Euclidean	click here
Polynomial ring over a field	$k[x]$ for a field $k$	(via Euclidean domain)	(via Euclidean domain)	click here

Weaker properties

Property	Meaning	Proof of implication	Proof of strictness (reverse implication failure)	Intermediate notions
Dedekind domain		PID implies Dedekind	Dedekind not implies PID	click here
Bezout domain	integral domain where every finitely generated ideal is principal	PID implies Bezout	Bezout not implies PID	click here
Bezout ring				click here
Noetherian domain	integral domain that's also a Noetherian ring	PID implies Noetherian	Noetherian not implies principal ideal	click here
Noetherian ring				click here
unique factorization domain	integral domain with unique factorization	PID implies UFD	UFD not implies PID	click here
normal domain	integral domain that is integrally closed in its field of fractions	(via UFD)	(via UFD)	click here
Noetherian normal domain	Noetherian ring and normal domain			click here
One-dimensional domain	integral domain in which every nonzero prime ideal is maximal	PID implies one-dimensional	One-dimensional domain not implies PID	click here
One-dimensional Noetherian domain	one-dimensional and Noetherian			click here
elementary divisor domain				click here
gcd domain				click here
principal ideal ring	every ideal is principal, not necessarily a domain			click here
catenary ring	any two saturated chains of primes with the same starting and ending point have the same length			click here

Metaproperties

Metaproperty name	Satisfied?	Proof	Statement with symbols
polynomial-closed property of commutative unital rings	No	follows from Polynomial ring over a ring is a PID iff the ring is a field	It is possible to have a principal ideal domain $R$ such that the polynomial ring $R[x]$ is not a principal ideal domain. In fact, $R[x]$ is a principal ideal domain if and only if $R$ is a field, so for instance $R = \mathbb{Z}$ or $R = \mathbb{R}[x]$ give counterexamples.

Closure under taking localizations

This property of integral domains is closed under taking localizations: the localization at a multiplicatively closed subset of a commutative unital ring with this property, also has this property. In particular, the localization at a prime ideal, and the localization at a maximal ideal, have the property.
View other localization-closed properties of integral domains | View other localization-closed properties of commutative unital rings

If we localize a principal ideal domain at any multiplicatively closed subset that does not contain zero, and in particular if we localize relative to a prime ideal, we continue to get a principal ideal domain. The reason is, roughly, that any ideal in the localization is generated by the element in the contraction that generates it.

Closure under quotients by prime ideals

This property of integral domains is prime-quotient-closed: the quotient of any integral domain satisfying this property by a prime ideal also satisfies the property. Note that we need the ideal to be prime for the quotient to also be an integral domain.
View other prime-quotient-closed properties of integral domains

The quotient of a principal ideal domain by any prime ideal is again a principal ideal domain. In fact, the quotient is either equal to the original domain (in case the prime ideal is zero) or is a field (in case the prime ideal is maximal. This is because in a principal ideal domain, every nonzero prime ideal is maximal.

However, it is true in slightly greater generality that the quotient of a principal ideal ring by any ideal is again a principal ideal ring. For full proof, refer: Principal ideal ring is quotient-closed

Module theory

Further information: structure theory of modules over PIDs

Any finitely generated module $M$ over a PID $R$ can be expressed as follows:

$M = R/(d_1) \oplus R/(d_2) \oplus R/(d_n)$

where $d_1|d_2|\ldots|d_n$. Some of the $d_n$ could be zero.

The $d_i$ are unique upto units; the principal ideals they generate are unique.

There is another equivalent formulation:

$M = R/(p_1^{k_1}) \oplus R/(p_2^{k_2}) \oplus R/(p_r^{k_r})$

Where all the $p_i$ are prime.

Thus, a finitely generated module over a PID is projective if and only if it is free.