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

Since ideals are kernels of 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 is a PID. (Proof.)

Therefore $Z$ is a PID, because it is a 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 ring $F [X]$ of polynomials over a field $F$ is a PID, because it is a Euclidean domain. (Its Euclidean function is "take the degree of the polynomial".)

The ring of Gaussian integers, $Z [i]$ , is a PID because it is a Euclidean domain. (Proof; its Euclidean function is "take the norm".)

The ring $Z [X]$ (of integer-coefficient polynomials) is not a PID, because the ideal $⟨ 2, X ⟩$ is not principal. This is an example of a unique_factorisation_domain which is not a PID.
proof of this

The ring $Z_{6}$ 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 $Z [\frac{1}{2} (1 + \sqrt{- 19})]$ . (Proof.)

Properties

Every PID is a unique_factorisation_domain. (Proof; this fact is not trivial.) The converse is false; see the case $Z [X]$ above.

In a PID, "prime" and "irreducible" coincide. (Proof.) This fact also characterises the maximal ideals of PIDs.

Every PID is trivially Noetherian: every ideal is not just finitely generated, but generated by a single element.