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In abstract algebra, a principal ideal domain, or PID is an integral domain in which every ideal is principal, i.e., can be generated by a single element.## Examples

Examples include:## Modules

The key result here is the structure theorem for finitely generated modules over a principal ideal domain. This yields that if R is a principal ideal domain, and M is a finitely
generated R-module, then a minimal generating set for M has properties somewhat akin to those of a basis for a finite-dimensional vector space over a field. This is something of an over-simplification, since there can be nonzero elements r, m of R and M respectively
such that r.m = 0, unlike the case of a vector-space over a field, and this necessitates a more complicated statement.## Properties

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).## References

Principal ideal domains are thus mathematical objects which behave somewhat like the integers, with respect to divisibility: any element of a PID has a unique decomposition into prime elements (so an analogue of the fundamental theorem of arithmetic holds); any two elements of a PID have a greatest common divisor.

A principal ideal domain is a specific type of integral domain, and can be characterized by the following (not necessarily exhaustive) chain of class inclusions:

- integral domains ⊃ unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields

- K: any field,
- Z: the ring of integers,
- K[x]'': rings of polynomials in one variable with coefficients in a field.
- Z[i]: the ring of Gaussian integers
- Z[ω] (where ω is a cube root of 1): the Eisenstein integers

Examples of integral domains that are not PIDs:

- Z[x]: the ring of all polynomials with integer coefficients.

It is not principal because the ideal generated by 2 and X is an example of an ideal that cannot be generated by a single polynomial.

- K[x,y]: The ideal (x,y) is not principal.

If M is a free module over a principal ideal domain R, then every submodule of M is again free. This does not hold for modules over arbitrary rings, as the example $(2,X)\; subseteq\; Bbb\{Z\}[X]$ of modules over $Bbb\{Z\}[X]$ shows.

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 $Bbb\{Z\}left[frac\{1+sqrt\{-19\}\}\{2\}right]$ .

Every principal ideal domain is a unique factorization domain (UFD). The converse does not hold since for any field K, K[X,Y] is a UFD but is not a PID (to prove this look at the ideal generated by $leftlangle\; X,Y\; rightrangle.$ It is not the whole ring since it contains no polynomials of degree 0, but it cannot be generated by any one single element).

- Every principal ideal domain is Noetherian.
- In all rings, maximal ideals are prime. In principal ideal domains a near converse holds: every nonzero prime ideal is maximal.
- All principal ideal domains are integrally closed.

The previous three statements give the definition of a Dedekind domain, and hence every principal ideal domain is a Dedekind domain.

So that PID ⊆ Dedekind∩UFD . However there is another theorem which states that any unique factorisation domain that is a Dedekind domain is also a principal ideal domain. Thus we get the reverse inclusion Dedekind∩UFD ⊆ PID, and then this shows equality and hence, Dedekind∩UFD = PID. (Note that condition (3) above is redundant in this equality, since all UFDs are integrally closed.)

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Last updated on Tuesday September 30, 2008 at 19:00:34 PDT (GMT -0700)

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This article is licensed under the GNU Free Documentation License.

Last updated on Tuesday September 30, 2008 at 19:00:34 PDT (GMT -0700)

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