# When is a Unique Factorization Domain a Principal Ideal domain [duplicate]

"Let $$R$$ be a Unique Factorization Domain and $$(a,b)=(c)$$ for $$a,b,c \in R$$. Show that $$R$$ is a Principal Ideal domain."

To be honest I found this very hard, here is my naive try:

Lets assume $$R$$ is a Unique Factorization Domain and $$(a,b)=(c)$$ $$\Rightarrow c$$ is gct of $$a$$ and $$b$$ then $$\Rightarrow (b)\subseteq (c)$$ and $$(a)\subseteq (c)$$ . Since this is true for every pair of Elements of $$R$$ we can construct a series $$(a_1) \subseteq (a_2)\subseteq (a_3) ... \subseteq(a_n)$$ in $$R$$. Because $$R$$ is a Unique Factorization Domain there is a $$N$$ with $$a_N \in R$$ and $$(a_i)=(a_N)$$ for $$i \geq N$$.

Let $$I$$ be an Ideal but not a Principal Ideal, so $$I=(d,e,f,...z)$$ with $$d,e,f,...z \in R$$. Then $$I= r_1 \cdot d + r_2 \cdot e + r_3 \cdot f ... r_j \cdot z$$ for all $$r_i \in R$$. But because $$d,e,f,...z \in R$$ and $$(d),(e),(f),...,(z) \in (a_N)$$, so $$d,e,f,...z \in (a_N)$$ so there are $$\alpha, \beta, ..., \omega$$ so that $$d=\alpha \cdot a_N$$, $$e=\beta \cdot a_N$$,..., $$z=\omega \cdot a_N$$. This means:

\begin{align*} I &= r_1 \cdot d + r_2 \cdot e + r_3 \cdot f \ldots r_j \cdot z \\ &= r_1 \cdot \alpha \cdot a_N + r_2 \cdot \beta \cdot a_N + \ldots r_j \cdot \omega \cdot a_N \\ &= a_N(r_1 \cdot \alpha + r_2 \cdot \beta \cdot + \ldots r_j \cdot \omega) \in (a_N) \end{align*}

I know that I am wrong somewhere, but I don't really know how to show it. Maybe there is a way to show that $$R$$ is an Euclidean Ring?

• See my answer in the dupe, which give $5$ such characterizations. – Bill Dubuque Jan 9 '19 at 19:31
• @BillDubuque: Since you've used your Golden Hammer to unilaterally close this as a duplicate, I feel you have a responsibility to give more explanation of how the earlier Question addresses the problem here. Do you claim that the condition that finitely generated ideals in a UFD makes it both one-dimensional and Noetherian? Even if you direct attention to your Answer there, rather than to the Question itself, some exposition of the problem here would be helpful. – hardmath Jan 9 '19 at 19:46
• @hardmath The sought inference is $(5)\Rightarrow (6)$ in my answer there. I'd rather not duplicate that Theorem yet again here, so I linked to the closely related question. There's really no good alternative solution for things like this that doesn't cause immense duplication (which is very bad). – Bill Dubuque Jan 9 '19 at 20:04
• @hardmath Here is a precise dupe target. But I can't change the dupe link above by myself. If someone with a gold badge wants to do that then ping me and we can coordinate to do so. – Bill Dubuque Jan 9 '19 at 20:49
• @BillDubuque: Thanks for pinning down the argument and for searching out a precise duplicate. – hardmath Jan 9 '19 at 20:56

Here are some hints hints: every finitely generated ideal is principal (why?), so it's enough to show there is no ideal which is not finitely generated. If $$I$$ is such an ideal, take elements $$a_1,a_2,\dots\in I$$ such that $$(a_1)\subsetneq(a_1,a_2)\subsetneq(a_1,a_2,a_3)\subsetneq\dots$$. Each of those ideals is generated by some element, say $$(a_1,\dots,a_n)=(b_n)$$. What relations are there between $$b_n$$? Writing them in terms of their unique factorizations, can you see why we reach a contradiction?

• Another way to view this is that ideals are generated by any "shortest" element, i.e. any element having fewest number of prime factors (= gcd of all elements). This is closely connected to the Euclidean-like characterization of PIDs given by the Dedekind-Hasse test - see here. – Bill Dubuque Jan 9 '19 at 21:04

Thank you very much for taking the time and helping:

1.) "Every finitely generated ideal is principal":

This follows with induction since $$(a,b)=(c)$$, then $$(a,b,d)=(c,d)=(e)$$... Since $$(a,b,d) = a\cdot R + b\cdot R + d\cdot R$$ and $$a\cdot R +b\cdot R = c\cdot R$$.

2.) "What relations are between $$b_n$$?"

$$(b_i) \subsetneq (b_{i+1}) \Leftrightarrow (b_{i+1})$$ is non-trivial divisor of $$(b_i)$$

3.) "...can you see why we reach a contradiction":

Unfortunately not. This last step is still unclear for me.

• In short for 3.: by writing out the prime factorization of $b_1$, you can deduce that up to multiplication by units, it has only finitely many factors. – Wojowu Jan 9 '19 at 19:42