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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?

Mary Star
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    You're correct that $i$ must divide $d$. So it's enough to show that $d$ has finitely many divisors (up to multiplication by a unit). – carmichael561 Apr 13 '16 at 23:09
  • I thought about it again.... so that $(i)$ contains $(d)$, should $d$ divides $i$ ? I got stuck right now @carmichael561 – Mary Star Apr 13 '16 at 23:15
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    Try proving it... – David Wheeler Apr 13 '16 at 23:32
  • We have that $a\mid b\Leftrightarrow b\in (a)$, Does it follow from that that when $(x)\subseteq (y)$ then $x\in (y)$, and so $y\mid x$ ? Therefore, $i$ must divide $d$, right? @DavidWheeler – Mary Star Apr 14 '16 at 00:13
  • To show that $d$ has finitely many divisors do we use the definition of a U.F.D. ? – Mary Star Apr 14 '16 at 00:14
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    Yes, suppose that $d=\pi_1^{k_1}\cdots \pi_r^{k_r}$ with the $\pi_i$ being irreducible, and show that the divisors of $d$ (up to units) have the form $\pi_1^{j_1}\cdots \pi_r^{j_r}$ with $0\leq j_r\leq k_r$. It might be helpful to recall that irreducible elements are prime in a UFD. – carmichael561 Apr 14 '16 at 00:38
  • The linked duplicate was in the related list on the right, and I believe it was probably in the list you saw as you titled the question. Please pay attention to those. – rschwieb Apr 14 '16 at 00:53
  • We have that $d=\pi_1^{k_1}\cdots \pi_r^{k_r}$ with $\pi_i$ being 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 $\pi_1^{j_1}\cdots \pi_r^{j_r}$ with $0\leq j_i\leq k_i$ ? From that we get that the set of the divisors is finite, right? @carmichael561 – Mary Star Apr 14 '16 at 14:30

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