Unique factorization conjecture?

Question

Let $A_p$ be an Integral domain.

Conjecture :

If every $a$ in $A_p$ that equals $b \space c$ for irreducible elements $b,c$ in $A_p$ , has Unique factorization then the Integral domain $A_p$ is a Unique factorization domain ( UFD ).

Is this true ?

How to show that ?

Answer 1

No, because a domain without irreducible elements vacuously satisfies this condition, but there are domains without irreducible elements which are not fields. Namely, the integral closure of $\Bbb Z$ in $\Bbb C$.

Answer 2

Suppose that the domain is atomic, i.e. all nonzero nonunits factor into atoms (irreducibles). Such domains satisfying your condition are called $2$-factorial. They are called square-factorial if $p^2\! = qq'\,\Rightarrow\, q,q'$ are associate, for atoms $p,q,q'$. Generally $2$-factorial $\,\Rightarrow\,$ square factorial. Further square factorial $\Rightarrow$ factorial (UFD) for rings of integers of an algebraic number field, so your conjecture is true for number rings. But it is not true generally.

For further discussion see D. D. Anderson et al., Criteria for unique factorizatio in integral domains, Jnl. Pure & Applied Algebra, 127 (1998) 205-218.