Irreducible element not implies prime

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 $\mathbb {Z} [{\sqrt {-5}}]$. In this ring, we have:

$2\cdot 3=(1+{\sqrt {-5}})(1-{\sqrt {-5}})$.

Thus, $2|(1+{\sqrt {-5}})(1-{\sqrt {-5}})$, but $2$ does not divide either factor, so $2$ is not prime.

On the other hand, $2$ is irreducible, as can be verified using the algebraic norm. If $(a+b{\sqrt {-5}})|2$, then $N(a+b{\sqrt {-5}})|N(2)$, yielding:

$a^{2}+5b^{2}|4$.

But the only possibilities for this are $a=\pm 2,b=0$ or $a=\pm 1,b=0$.

Example of a ring of integer-valued polynomials

Further information: ring of integer-valued polynomials over rational integers

Let $R$ be the ring of integer-valued polynomials over rational integers: this is the ring of those polynomials in $\mathbb {Q} [x]$ that send integers to integers. Then, any binomial polynomial, i.e., any polynomial of the form:

${\binom {x}{r}}={\frac {x(x-1)\dots (x-r+1)}{r!}}$

is irreducible but not prime in $R$.

For full proof, refer: Binomial polynomial is irreducible but not prime in ring of integer-valued polynomials over rational integers