Principal ideal domain

Written by Patrick Stevens, et al. last updated

In ring theory, an integral domain is a principal ideal domain (or PID) if every ideal can be generated by a single element. That is, for every ideal there is an element such that ; equivalently, every element of is a multiple of .

Since ideals are kernels of ring homomorphisms (proof), this is saying that a PID has the special property that every ring homomorphism from acts "nearly non-trivially", in that the collection of things it sends to the identity is just "one particular element, and everything that is forced by that, but nothing else".

Examples

  • Every Euclidean domain is a PID. (Proof.)
  • Therefore is a PID, because it is a Euclidean domain. (Its Euclidean function is "take the modulus".)
  • Every field is a PID because every ideal is either the singleton (i.e. generated by ) or else is the entire ring (i.e. generated by ).
  • The ring of polynomials over a field is a PID, because it is a Euclidean domain. (Its Euclidean function is "take the degree of the polynomial".)
  • The ring of Gaussian integers, , is a PID because it is a Euclidean domain. (Proof; its Euclidean function is "take the norm".)
  • The ring (of integer-coefficient polynomials) is not a PID, because the ideal is not principal. This is an example of a unique_factorisation_domain which is not a PID.
    proof of this
  • The ring is not a PID, because it is not an integral domain. (Indeed, in this ring.)

There are examples of PIDs which are not Euclidean domains, but they are mostly uninteresting. One such ring is . (Proof.)

Properties

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