Unique factorization domain and principal ideals

abstract-algebra ring-theory ideals unique-factorization-domains

If R was a unique factorization domain, can we deduce that for a nonzero element d in R, d has a finite number of divisors?

I need this in solving this question "If R is a unique factorization domain then there are only finite number of distinct principal ideals that contain the ideal (d)."

My idea to solve it: if a divides d then (d) is contained in (a), so if we have finite divisors then we have finite principal ideals. Am I wrong?


(This is a CW answer to get this question out of the unanswered queue.)

Yes, the ideas are basically right.

The divisors of $d$ (up to associates) determine the principal ideals containing $(d)$ in the way you describe. The fact that $d$ has a unique set of irreducibles that it factors into limits the number of containing principal ideals to finitely many (corresponding to divisors determined by those irreducibles.)

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