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Recently, I learnt this nice theorem of Kaplansky:

Let $R$ be an integral domain. Then $R$ is a UFD iff every nonzero prime ideal contains a nonzero prime principal ideal.

Using this theorem, it follows easily that all PIDs are UFDs.

Now let $R$ be an integral domain such that every ideal $I$ in $R$ is of the form $(a,b)$ for some $a,b\in R$. Is there a necessary and sufficient condition about when the ring $R$ is a UFD ? In other words, is there a good way to characterize which rings like $R$ are UFDs ?

Amr
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  • IIRC such a ring must be Noetherian and of Krull dimension 1. Note that rings such as Dedekind domains satisfy your condition above but many of them are not PIDs (and hence not UFDs). –  Jan 20 '13 at 18:09
  • Why does not PID imply not UFD – Amr Jan 20 '13 at 18:11
  • If an integral domain satisifies your condition and is a GCD domain, then its a PID. Any UFD is a GCD domain. – JSchlather Jan 20 '13 at 18:11
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    @Amr A Dedekind domain is a PID iff it is a UFD. –  Jan 20 '13 at 18:12
  • @BenjaLim I see. Are all rings like $R$ (as described above) a Dedekind domain ? – Amr Jan 20 '13 at 18:14
  • @JacobSchlather Isee. This means that my question is not good. – Amr Jan 20 '13 at 18:15
  • @Amr No I don' think so; but can't come up with an example right now. –  Jan 20 '13 at 18:15
  • @BenjaLim Why can't $R$ have krull dimension $2$? – JSchlather Jan 20 '13 at 18:15
  • @JacobSchlather IIRC this is a result of I.S. Cohen. I can look for the paper if you want. –  Jan 20 '13 at 18:16
  • @BenjaLim I see. I think this mathoverflow question is very relevant to the discussion. – JSchlather Jan 20 '13 at 18:17
  • @JacobSchlather OK let me ask a similar question, Let $a+bi\in\mathbb{Z}[i]$ is there a way to characterize which $\mathbb{Z}[a+bi]$ are PIDs ? – Amr Jan 20 '13 at 18:20
  • @Amr Even the ring $\Bbb{Z}[a+ bi]$ is the ring of integers of some quadratic number field $\Bbb{Q}(\sqrt{d})$ it is not so easy to say which are PIDs. When $d < 0$ those that are PIDs are completely classified, given by the Heegner numbers. For $d > 0$ I believe the problem is still open. –  Jan 20 '13 at 18:28
  • @BenjaLim This is great to know. Finally let $R$ be a Notherian ring. Is there a necessary and sufficent condition that tells when R is a UFD ? (I want to see if there is a theorem like the theorem that I gave at the beginnning). – Amr Jan 20 '13 at 18:31
  • @BenjaLim I don't know if you quite get that. Since something like $\mathbb Z[8i]$ isn't integrally closed. – JSchlather Jan 20 '13 at 18:32
  • @BenjaLim Ah the last question that I asked is not good again. – Amr Jan 20 '13 at 19:43
  • @JacobSchlather Sorry I meant to type "even *if* the ring..." –  Jan 21 '13 at 06:10
  • @JacobSchlather You said:"If an integral domain satisifies your condition and is a GCD domain, then its a PID....." Why is this true ? – Amr Jan 21 '13 at 23:10
  • @Amr, Actually you need a Bezout domain, my mistake. – JSchlather Jan 22 '13 at 00:55

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This answer is intended to make some order between so many good comments above.

If $R$ is a commutative ring with the property that there exists $k\geq 1$ such that every ideal of $R$ is $k$-generated, then $\dim R\le 1$. (Cohen)

Proof. One can assume that $R$ is local with the maximal ideal $\mathfrak m$. One knows that the Hilbert-Samuel polynomial of $R$ with respect to $\mathfrak m$ satisfies: $P(n)=l_R(R/\mathfrak m^{n+1})$ for $n$ large enough. Since we have $P(n)=\sum_{i=0}^nl_R(\mathfrak m^i/\mathfrak m^{i+1})$, it follows that $P(n)\le (n+1)k$ for $n$ large enough. This shows that $\deg P\le 1$, so $\dim R\le 1$.

In particular, if $R$ is an integral domain with the property above, then $\dim R=1$. As we can learn from this topic, in this case $R$ is an UFD if and only if $R$ is a PID.

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