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

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

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.)