# 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.