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 I there is an element i∈I such that ⟨i⟩=I; equivalently, every element of I is a multiple of i.
Since ideals are kernels of ring homomorphisms (proof), this is saying that a PID R has the special property that every ring homomorphism from R 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 Z 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 {0} (i.e. generated by 0) or else is the entire ring (i.e. generated by 1).
- The ring F[X] of polynomials over a field F is a PID, because it is a Euclidean domain. (Its Euclidean function is "take the degree of the polynomial".)
- The ring of Gaussian integers, Z[i], is a PID because it is a Euclidean domain. (Proof; its Euclidean function is "take the norm".)
- The ring Z[X] (of integer-coefficient polynomials) is not a PID, because the ideal ⟨2,X⟩ is not principal. This is an example of a unique_factorisation_domain which is not a PID.
proof of this
- The ring Z6 is not a PID, because it is not an integral domain. (Indeed, 3×2=0 in this ring.)
There are examples of PIDs which are not Euclidean domains, but they are mostly uninteresting.
One such ring is Z[12(1+√−19)]. (Proof.)
Properties