# Principal ideal domain

https://arbital.com/p/principal_ideal_domain

by Patrick Stevens Aug 3 2016 updated Aug 4 2016

A principal ideal domain is a kind of ring, in which all ideals have a certain nice form.

[summary: A principal ideal domain is an Integral domain in which every [ideal_ring_theory ideal] has a single generator.]

In ring theory, an Integral domain is a principal ideal domain (or PID) if every [ideal_ring_theory ideal] can be generated by a single element. That is, for every ideal $I$ there is an element $i \in I$ such that $\langle i \rangle = I$; equivalently, every element of $I$ is a multiple of $i$.

Since ideals are kernels of [ring_homomorphism ring homomorphisms] (proof), this is saying that a PID $R$ has the special property that every ring homomorphism from $R$ acts "nearly non-trivially", in that the collection of things it sends to the identity is just "one particular element, and everything that is forced by that, but nothing else".

# Examples

• Every [euclidean_domain Euclidean domain] is a PID. (Proof.)
• Therefore $\mathbb{Z}$ is a PID, because it is a [euclidean_domain Euclidean domain]. (Its Euclidean function is "take the modulus".)
• Every field is a PID because every ideal is either the singleton $\{ 0 \}$ (i.e. generated by $0$) or else is the entire ring (i.e. generated by $1$).
• The [polynomial_ring ring $F[X]$ of polynomials] over a field $F$ is a PID, because it is a Euclidean domain. (Its Euclidean function is "take the [polynomial_degree degree] of the polynomial".)
• The ring of [gaussian_integer Gaussian integers], $\mathbb{Z}[i]$, is a PID because it is a Euclidean domain. ([gaussian_integers_is_pid Proof]; its Euclidean function is "take the [norm_complex_number norm]".)
• The ring $\mathbb{Z}[X]$ (of integer-coefficient polynomials) is not a PID, because the ideal $\langle 2, X \rangle$ is not principal. This is an example of a Unique factorisation domain which is not a PID. [todo: proof of this]
• The ring $\mathbb{Z}_6$ is not a PID, because it is not an integral domain. (Indeed, $3 \times 2 = 0$ in this ring.)

There are examples of PIDs which are not Euclidean domains, but they are mostly uninteresting. One such ring is $\mathbb{Z}[\frac{1}{2} (1+\sqrt{-19})]$. (Proof.)

# Properties

• Every PID is a Unique factorisation domain. ([principal_ideal_domain_has_unique_factorisation Proof]; this fact is not trivial.) The converse is false; see the case $\mathbb{Z}[X]$ above.
• In a PID, "prime" and "irreducible" coincide. (Proof.) This fact also characterises the [maximal_ideal maximal ideals] of PIDs.
• Every PID is trivially [noetherian_ring Noetherian]: every ideal is not just finitely generated, but generated by a single element.