$D$ be a UFD having infinitely many maximal ideals , then does $D$ have infinitely many irreducible elements which are pairwise non-associate?

Question

Let $D$ be a UFD having infinitely many maximal ideals. Then is it true that $D$ has infinitely many irreducible elements which are pairwise non-associate ?

I can see that the infinite collection of maximal ideals would give infinitely many irreducibles but I can't figure out whether that would produce infinitely many non-associated elements ... Please help . Thanks in advance

Answer 1

The answer is Yes. In fact, if $D$ is a UFD that has finitely many irreducible elements up to associates, then $D$ has finitely many prime ideals.

To see this, choose a prime ideal $P$ of $D$ and some element $a\in P$. Let $a = q_1\cdots q_k$ be a factorization into irreducibles. Since $q_1\cdots q_k\in P$, we get that $q_i\in P$ for some $i$, so $(a)\subseteq (q_i)\subseteq P$. Taking the union of these inclusions over all choices of $a\in P$ yields

$P \subseteq \bigcup_{q\in P, q \;\textrm{irred}} (q)\subseteq P$,

which asserts that $P$ is the union of all principal ideals $(q)$ where $q\in P$ and $q$ is irreducible.

Now, if there are only finitely many irreducible elements up to associates, then there are only finitely many ideals of the form $(q)$ where $q$ is irreducible, hence there are finitely many ideals of the form $\bigcup_{q\in P, q \;\textrm{irred}} (q)$. This shows that there are only finitely many prime ideals. (Then there are only finitely many maximal ideals, too.)