Principal ideal domain


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


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