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.

  • $\begingroup$ Nice question. None spring to mind immediately. (You forgot the word "uniquely" in your definition btw). $\endgroup$ – Tobias Kildetoft Sep 10 '14 at 17:50

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.

  • $\begingroup$ I'm not sure this is precisely true. For example, if you localize $\mathbb{Z}$ away from $(3)$, the ring you get contains infinitely many distinct units, and each of them times $3$ is a distinct irreducible element. $\endgroup$ – exitmouse Sep 10 '14 at 18:02
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    $\begingroup$ @exitmouse As far as I know, elements which are associates are considered as not being distinct from arithmetical point of view. $\endgroup$ – user26857 Sep 10 '14 at 18:05
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    $\begingroup$ On the other hand, if the ring has finitely many units and finitely many irreducible elements and all elements factor, then it is finite and thus a field, so no ring with this stricter version of distinctness exists. $\endgroup$ – Tobias Kildetoft Sep 10 '14 at 18:07
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    $\begingroup$ There are two questions here, but now we have the answer to both. $\endgroup$ – exitmouse Sep 10 '14 at 18:09
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    $\begingroup$ For those not familiar with localization, one can also describe the ring in the answer as consisting of all rationals with numerator not divisible by a fixed finite set of primes. $\endgroup$ – Tobias Kildetoft Sep 10 '14 at 18:10

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