Irreducible element not implies prime
Contents
Statement
An irreducible element in an integral domain need not be a prime element.
Related facts
An integral domain in which every irreducible is prime is an integral domain where irreducible elements are all prime. Such integral domains are very common. In fact:
- Bezout implies every irreducible is prime: In a Bezout domain, i.e., an integral domain where every finitely generated ideal is principal, every irreducible element is prime.
- Unique factorization implies every irreducible is prime: In a unique factorization domain, every irreducible element is prime.
Proof
Example of a quadratic integer ring
Consider the ring . In this ring, we have:
.
Thus, , but
does not divide either factor, so
is not prime.
On the other hand, is irreducible, as can be verified using the algebraic norm. If
, then
, yielding:
.
But the only possibilities for this are or
.
Note that this ring is a Dedekind domain.
Example of a ring of integer-valued polynomials
Further information: ring of integer-valued polynomials over rational integers
Let be the ring of integer-valued polynomials over rational integers: this is the ring of those polynomials in
that send integers to integers. Then, any binomial polynomial, i.e., any polynomial of the form:
is irreducible but not prime in .
For full proof, refer: Every binomial polynomial is irreducible but not prime in the ring of integer-valued polynomials over rational integers
Unlike the previous example, the ring is not a Noetherian ring.