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 $(d)$.
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We have that $R$ is a U.F.D. iff $\forall r\in R\setminus \{0\}$, $r\notin U(R)$ the following hold:
- $r=a_1 \cdots a_k$ with $a_i$ irreducible
- If $r=a_1 \cdots a_k=b_1 \cdots b_t$ with $a_i, b_i$ irreducible then $k=t$ and $a_i=b_iu_i$ with $u_i\in U(R), \forall u=1, \dots , k$.
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A principal ideal is generated by a single element, say $i$, and so that it contains the ideal $(d)$, $i$ must divide $d$, right?
To show that $d$ has finitely many divisors do we use the definition of a U.F.D. ?
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EDIT:
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 $0\leq j_i\leq k_i$ ?
From that we get that the set of the divisors is finite, right?