We know the following theorem about UFD.
Theorem.
Let $R$ be a factorization domain. Then, $R$ is a UFD iff every irreducible element is prime.
The forward part is okay, i.e., if $R$ is a UFD, then every irreducible is prime.
Now, my question is if we modify the statement a bit, then does the converse part remain true?
Statement.
Let $R$ be an integral domain. Then, $R$ is a UFD only if every irreducible element is prime.
The above statement is no true. We are not giving the guarantee that factorization exists, but we are asking can still every irreducible be prime? I am asking this because we need to use the existence of factorization in the converse part of the theorem I mentioned.
I had searched the site Non-UFD integral domain such that prime is equivalent to irreducible?
and found counterexamples there but nobody mentions how such examples can be constructed.
So,I am looking for a general method to construct such rings.
