: any ,
field
: in one variable with coefficients in a field. (The converse is also true, i.e. if is a PID then is a field.) Furthermore, a ring of formal power series in one variable over a field is a PID since every ideal is of the form ,
![{\displaystyle K[x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a9e6c2ac2830d6a9abe078b47450777c41d69a9)
![{\displaystyle A[x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7d5d3b3ae4cfb46d3ef305e5d7dcb130f10e468)
rings of polynomials
![{\displaystyle \mathbb {Z} [i]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ffa94e9e2e6d9e5e5373d5fafb954b902743fde)
![{\displaystyle \mathbb {Z} [\omega ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ae955a9a0d0f342fc73aaafe28af604d23267f7)
In a principal ideal domain, any two elements a,b have a greatest common divisor, which may be obtained as a generator of the ideal (a, b).
All Euclidean domains are principal ideal domains, but the converse is not true.
An example of a principal ideal domain that is not a Euclidean domain is the ring ,[6][7] this was proved by Theodore Motzkin and was the first case known.[8] In this domain no q and r exist, with 0 ≤ |r| < 4, so that , despite and having a greatest common divisor of 2.
Every principal ideal domain is a unique factorization domain (UFD).[9][10][11][12] The converse does not hold since for any UFD K, the ring K[X, Y] of polynomials in 2 variables is a UFD but is not a PID. (To prove this look at the ideal generated by It is not the whole ring since it contains no polynomials of degree 0, but it cannot be generated by any one single element.)
The previous three statements give the definition of a Dedekind domain, and hence every principal ideal domain is a Dedekind domain.
Let A be an integral domain. Then the following are equivalent.
Any Euclidean norm is a Dedekind-Hasse norm; thus, (5) shows that a Euclidean domain is a PID. (4) compares to:
An integral domain is a Bézout domain if and only if any two elements in it have a gcd that is a linear combination of the two. A Bézout domain is thus a GCD domain, and (4) gives yet another proof that a PID is a UFD.
![{\displaystyle \mathbb {Z} \left[{\frac {1+{\sqrt {-19}}}{2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb17dc804d1adeb142f5d624f5a59ef5d419c61a)
