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?
Best Answer
(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.)