[Math] Does there exist a UFD having only finitely many irreducibles

Question

Does there exist a UFD (which is not a field) having only finitely many irreducible elements?

Definition of a UFD is: $R$ is an integral domain ($R$ is a commutative ring having unity and no zero-divisors) such that each $a\in R \setminus (R^* \cup {0})$ factors into a product of irreducible elements.

I've tried to look for one, but without luck so far.

Answer 1

If you localize $\mathbb Z$ at the multiplicative set $S=\mathbb Z-\bigcup_{i=1}^n p_i\mathbb Z$, where $p_i$ are distinct primes, then you get a UFD having exactly $n$ prime elements.

Edit. (To answer some questions raised in the comments.)

There are no UFDs (excepting fields) having only finitely many distinct primes.

Say $q_1,\dots,q_m$ are all primes. Set $a=q_1\cdots q_m$ and consider $a+1,a^2+1,\dots$. Since all these elements are invertible there exist $i\ne j$ such that $a^i=a^j$, a contradiction.

As a side note, a UFD having finitely many primes is necessarily a PID.