Principal ideal domain

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:

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.

Metaproperties
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.

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.

Module theory
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.