MHB Show Finite Principal Ideals Contain $(d)$ in U.F.D.

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Hey! :o

Let $R$ be a U.F.D. and $0\neq d\in R$.
I want to show that there are finitely many different principal ideals that contain the ideal $(d)$.

A principal ideal is generated by a single element, say $i$, and so that it contains the ideal $(d)$, $i$ must divide $d$, right? (Wondering)
We have that $d=a_1^{k_1}\cdots a_r^{k_r}$ with $a_i$ irreducible.
Since $R$ is a U.F.D. the irreducible elements are prime. Does it follow from that that the divisors of $d$ are of the form $a_1^{j_1}\cdots a_r^{j_r}$ with $0\leq j_i\leq k_i$ ? (Wondering)
From that we get that the set of the divisors is finite, right? (Wondering)
 
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Yes.
 
Deveno said:
Yes.

Nice... Thank you! (Yes)
 

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