- Principal ideal domain
In

abstract algebra , a**principal ideal domain**, or**PID**is anintegral 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 thefundamental theorem of arithmetic holds); any two elements of a PID have agreatest 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 domain s⊃unique factorization domain s⊃principal ideal domain s⊃Euclidean domain s**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 ofGaussian integers

***Z**[ω] (where ω is a cube root of 1): theEisenstein 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 ideal s are prime. In principal ideal domains a near converse holds: every nonzero prime ideal is maximal.

#All principal ideal domains areintegrally 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**

*Wikimedia Foundation.
2010.*

### Look at other dictionaries:

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