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Why is it true that if a unique factorization domain has an irreducible then it has infinitely many irreducibles? I am guessing that it has something to do with them being primes?

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Euclid's proof of the infinitude of the primes works in any UFD. – Fredrik Meyer Mar 4 '12 at 1:54
@Fredrik It is not true that Euclid's proof generalizes to any UFD since there are UFDs with only finitely many primes, e.g. the subring of $\mathbb Q$ obtained by adjoining inverses of all but finitely many primes. Euclid's proof breaks down because $\rm\:1 + p_1 p_2\:\cdots\:p_n\:$ might be a unit, so need not yield a new prime. But one can avoid this problem in a ring with fewer units than elements, see my proof here. – Bill Dubuque Mar 4 '12 at 8:10
@Jakucha Please see my comment above. – Bill Dubuque Mar 4 '12 at 8:12
up vote 9 down vote accepted

The assertion is not quite true as it stands, nor are the comments quite correct. That is, the Euclid-style argument easily fails: for example, the ring R obtained from the integers by allowing arbitrary odd denominators (=localization of $\mathbb Z$ at 2) is a PID (the only ideals are generated by powers of 2), and is a UFD, therefore, but has only one irreducible/prime element (modulo units), namely, 2.

How does Euclid's argument fail here? For example, 2n+1 is a unit in this ring, so has no non-unit divisors. It is "big", but a unit, nevertheless.

Edit: and/or perhaps I misunderstood the intent, namely, as commented, the multiples 2/odd are distinct, although associate. Bill Dubuque's discussion of rings with fewer elements (see above link) may address the question's intent?

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Right; but it has infinitely many irreducible elements, since $2/m$ is irreducible for every odd $m$... – Arturo Magidin Mar 4 '12 at 6:54
@Paul Yes, Euclid's argument may fail if there are too many units. However, it does generalize as follows: An infinite ring R has infinitely many max ideals if it has fewer units than elements, see my proof here. – Bill Dubuque Mar 4 '12 at 8:15

Note that if $x$ is irreducible, then so is $ux$ for any unit $u$.

Let $D$ be a UFD, and let $x$ be an irreducible in $D$.

If a domain is finite, then it is a field, so the fact that $D$ contains an irreducible means that $D$ is infinite.

  1. If $D$ has infinitely many units, then the elements $ux$ with $u$ a unit form an infinite set of pairwise distinct irreducibles. (Note that since $x$ is irreducible, it is nonzero, so $ux=vx$ implies $u=v$).

  2. If $D$ has only finitely many units, then we can proceed along the lines of Euclid's proof: let $x_1,\ldots,x_n$ be any finite list of irreducibles. Then the collection of all elements of the form $1+(x_1\cdots x_n)k$ with $k$ in $D$ is infinite. They can't all be units (since there are only finitely many units), so at least one is a nonunit, hence divisible by some irreducible. But it cannot be divisible by any of the $x_i$ (same argument as Euclid's), so there must be some irreducible not on our original list.

I'm pretty sure I've seen Bill Dubuque give the proof above... ah, yes, as he's mentioned before, if the cardinality of the set of units is smaller than the cardinality of the ring, then the argument above goes through.

Note that, in a sense, case 1 is a bit of a cheat, since even though we found "infinitely many irreducibles", we did not guarantee infinitely many pairwise non-associated irreducibles. But as Paul Garrett's example shows, we cannot hope to achieve that in general; in fact, for every finite $n\gt 0$ there are UFDs that have exactly $n$ associate classes of irreducibles: e.g., localize $\mathbb{Z}$ away from $n$ primes.

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