Principal ideal domain

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.

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

Examples

Examples include:
* "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.

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 finitelygenerated "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" respectivelysuch that "r"."m" = "0", unlike the case of a vector-space over a field, and this necessitates a more complicated statement.

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.

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)".

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 $Bbb\{Z\}left\; [frac\{1+sqrt\{-19\{2\}\; ight]$ [Wilson, Jack C. "A Principal Ring that is Not a Euclidean Ring." Math. Mag 46 (Jan 1973) 34-38 [http://links.jstor.org/sici?sici=0025-570X(197301)46%3A1%3C34%3AAPIRTI%3E2.0.CO%3B2-U] ] [George Bergman, "A principal ideal domain that is not Euclidean - developed as a series of exercises" [http://math.berkeley.edu/~gbergman/grad.hndts/nonEucPID.ps PostScript file] ] .

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 $leftlangle\; X,Y\; ight\; angle.$ 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 areintegrally closed.)

References