3

"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?

hardmath
  • 37,015
KingDingeling
  • 1,642
  • See my answer in the dupe, which give $5$ such characterizations. – Bill Dubuque Jan 09 '19 at 19:31
  • 1
    @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 09 '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 09 '19 at 20:04
  • 1
    @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 09 '19 at 20:49
  • @BillDubuque: Thanks for pinning down the argument and for searching out a precise duplicate. – hardmath Jan 09 '19 at 20:56
  • Thank you very much @BillDubuque. – KingDingeling Jan 10 '19 at 08:23

2 Answers2

3

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?

Wojowu
  • 26,600
  • 1
    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 09 '19 at 21:04
1

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.

KingDingeling
  • 1,642
  • 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 09 '19 at 19:42